Questions about the Lorentz Transformation (LT)

[Alan Grayson]

Using the LT, we have the following transformations of Length, Time, and Mass, that is,

x --->x',  t ---> t',  m ---> m', where the primed quantities are the transformed values in the primed frame, given their values in the unprimed frame. The question is this; which of the quantities in the primed frame are actually measured in the primed frame, and which are appearances in the primed frame as seen by unprimed frame? How is this consistent, if it is, with the fact that when doing EM measurements, E' and B' are the actual measurements of the fields in the primed frame, given that E and B are measured in the unprimed frame, but the same cannot be said of some, or all of the measurements of Length, Time, and Mass? TY, AG

[Brent Meeker]

On 1/13/2025 9:02 PM, Alan Grayson wrote:

Using the LT, we have the following transformations of Length, Time, and Mass, that is,

x --->x',  t ---> t',  m ---> m', where the primed quantities are the transformed values in the primed frame, given their values in the unprimed frame. The question is this; which of the quantities in the primed frame are actually measured in the primed frame, and which are appearances in the primed frame as seen by unprimed frame?

All of them.  That's why it's relativity theory.  x and t are measurements in one frame and x' and t' are measurements in another frame moving relative to the unprimed frame.  And note the use of "measurements"  not "as seen".  The two are different when you consider things moving at a significant fraction of the speed of light.

How is this consistent, if it is, with the fact that when doing EM measurements, E' and B' are the actual measurements of the fields in the primed frame, given that E and B are measured in the unprimed frame, but the same cannot be said of some, or all of the measurements of Length, Time, and Mass? TY,

The same applies except the electromagnetic field is a tensor, so it transforms by tensor like a sequence of LTs.  How is that not consistent?

Brent

AG

[Alan Grayson]

On Monday, January 13, 2025 at 10:21:28 PM UTC-7 Brent Meeker wrote:

On 1/13/2025 9:02 PM, Alan Grayson wrote:

Using the LT, we have the following transformations of Length, Time, and Mass, that is,

x --->x',  t ---> t',  m ---> m', where the primed quantities are the transformed values in the primed frame, given their values in the unprimed frame. The question is this; which of the quantities in the primed frame are actually measured in the primed frame, and which are appearances in the primed frame as seen by unprimed frame?

All of them.  That's why it's relativity theory.  x and t are measurements in one frame and x' and t' are measurements in another frame moving relative to the unprimed frame.  And note the use of "measurements"  not "as seen".  The two are different when you consider things moving at a significant fraction of the speed of light.

But length in primed frame is contracted from the pov of unprimed frame, but in primed frame it isn't measured as contracted, so it APPEARS contracted from the pov of unprimed frame, but NOT measured in primed frame. Same with time and mass? AG 

[Brent Meeker]

On 1/13/2025 10:04 PM, Alan Grayson wrote:

On Monday, January 13, 2025 at 10:21:28 PM UTC-7 Brent Meeker wrote:

On 1/13/2025 9:02 PM, Alan Grayson wrote:

Using the LT, we have the following transformations of Length, Time, and Mass, that is,

x --->x',  t ---> t',  m ---> m', where the primed quantities are the transformed values in the primed frame, given their values in the unprimed frame. The question is this; which of the quantities in the primed frame are actually measured in the primed frame, and which are appearances in the primed frame as seen by unprimed frame?

All of them.  That's why it's relativity theory.  x and t are measurements in one frame and x' and t' are measurements in another frame moving relative to the unprimed frame.  And note the use of "measurements"  not "as seen".  The two are different when you consider things moving at a significant fraction of the speed of light.

But length in primed frame is contracted from the pov of unprimed frame, but in primed frame it isn't measured as contracted, so it APPEARS contracted from the pov of unprimed frame

No, it will appear rotated (c.f. Terrell rotation).  It will measure contracted (using light and clocks, as with radar).

Brent

but NOT measured in primed frame. Same with time and mass? AG 

How is this consistent, if it is, with the fact that when doing EM measurements, E' and B' are the actual measurements of the fields in the primed frame, given that E and B are measured in the unprimed frame, but the same cannot be said of some, or all of the measurements of Length, Time, and Mass? TY,

The same applies except the electromagnetic field is a tensor, so it transforms by tensor like a sequence of LTs.  How is that not consistent?

Brent

AG

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[Alan Grayson]

On Monday, January 13, 2025 at 11:20:05 PM UTC-7 Brent Meeker wrote:

On 1/13/2025 10:04 PM, Alan Grayson wrote:

On Monday, January 13, 2025 at 10:21:28 PM UTC-7 Brent Meeker wrote:

On 1/13/2025 9:02 PM, Alan Grayson wrote:

Using the LT, we have the following transformations of Length, Time, and Mass, that is,

x --->x',  t ---> t',  m ---> m', where the primed quantities are the transformed values in the primed frame, given their values in the unprimed frame. The question is this; which of the quantities in the primed frame are actually measured in the primed frame, and which are appearances in the primed frame as seen by unprimed frame?

All of them.  That's why it's relativity theory.  x and t are measurements in one frame and x' and t' are measurements in another frame moving relative to the unprimed frame.  And note the use of "measurements"  not "as seen".  The two are different when you consider things moving at a significant fraction of the speed of light.

But length in primed frame is contracted from the pov of unprimed frame, but in primed frame it isn't measured as contracted, so it APPEARS contracted from the pov of unprimed frame

No, it will appear rotated (c.f. Terrell rotation).  It will measure contracted (using light and clocks, as with radar).

Brent

Terrell rotation over my head, but length contraction allegedly measured in primed frame contradicts the discussion of the paradox, where car and garage lengths aren't contracted when viewing each other.  IOW, when car measures garage's length, garage's length, from pov of garage, isn't contracted but only appears contracted from pov of car frame. Here the garage is in the primed frame but isn't actually contracted. So the LT seems to deal in appearances, not what's actually measured in the primed or transformed frame. AG

[Brent Meeker]

On 1/13/2025 10:39 PM, Alan Grayson wrote:

On Monday, January 13, 2025 at 11:20:05 PM UTC-7 Brent Meeker wrote:

On 1/13/2025 10:04 PM, Alan Grayson wrote:

On Monday, January 13, 2025 at 10:21:28 PM UTC-7 Brent Meeker wrote:

On 1/13/2025 9:02 PM, Alan Grayson wrote:

Using the LT, we have the following transformations of Length, Time, and Mass, that is,

x --->x',  t ---> t',  m ---> m', where the primed quantities are the transformed values in the primed frame, given their values in the unprimed frame. The question is this; which of the quantities in the primed frame are actually measured in the primed frame, and which are appearances in the primed frame as seen by unprimed frame?

All of them.  That's why it's relativity theory.  x and t are measurements in one frame and x' and t' are measurements in another frame moving relative to the unprimed frame.  And note the use of "measurements"  not "as seen".  The two are different when you consider things moving at a significant fraction of the speed of light.

But length in primed frame is contracted from the pov of unprimed frame, but in primed frame it isn't measured as contracted, so it APPEARS contracted from the pov of unprimed frame

No, it will appear rotated (c.f. Terrell rotation).  It will measure contracted (using light and clocks, as with radar).

Brent

Terrell rotation over my head,

It's probably within your capability to Google it.

but length contraction allegedly measured in primed frame contradicts the discussion of the paradox, where car and garage lengths aren't contracted when viewing each other. 

That doesn't even parse.

Here the garage is in the primed frame but isn't actually contracted.

No object is ever contracted in it's own frame, but you haven't said which is the primed frame, thus introducing ambiguity.

So the LT seems to deal in appearances, not what's actually measured in the primed or transformed frame. AG

I already told you that LT transforms what it measured NOT what appears.

Brent

but NOT measured in primed frame. Same with time and mass? AG 

How is this consistent, if it is, with the fact that when doing EM measurements, E' and B' are the actual measurements of the fields in the primed frame, given that E and B are measured in the unprimed frame, but the same cannot be said of some, or all of the measurements of Length, Time, and Mass? TY,

The same applies except the electromagnetic field is a tensor, so it transforms by tensor like a sequence of LTs.  How is that not consistent?

Brent

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[Alan Grayson]

On Monday, January 13, 2025 at 11:54:56 PM UTC-7 Brent Meeker wrote:

On 1/13/2025 10:39 PM, Alan Grayson wrote:

On Monday, January 13, 2025 at 11:20:05 PM UTC-7 Brent Meeker wrote:

On 1/13/2025 10:04 PM, Alan Grayson wrote:

On Monday, January 13, 2025 at 10:21:28 PM UTC-7 Brent Meeker wrote:

On 1/13/2025 9:02 PM, Alan Grayson wrote:

Using the LT, we have the following transformations of Length, Time, and Mass, that is,

x --->x',  t ---> t',  m ---> m', where the primed quantities are the transformed values in the primed frame, given their values in the unprimed frame. The question is this; which of the quantities in the primed frame are actually measured in the primed frame, and which are appearances in the primed frame as seen by unprimed frame?

All of them.  That's why it's relativity theory.  x and t are measurements in one frame and x' and t' are measurements in another frame moving relative to the unprimed frame.  And note the use of "measurements"  not "as seen".  The two are different when you consider things moving at a significant fraction of the speed of light.

But length in primed frame is contracted from the pov of unprimed frame, but in primed frame it isn't measured as contracted, so it APPEARS contracted from the pov of unprimed frame

No, it will appear rotated (c.f. Terrell rotation).  It will measure contracted (using light and clocks, as with radar).

Brent

Terrell rotation over my head,

It's probably within your capability to Google it.

but length contraction allegedly measured in primed frame contradicts the discussion of the paradox, where car and garage lengths aren't contracted when viewing each other. 

That doesn't even parse.

Here the garage is in the primed frame but isn't actually contracted.

No object is ever contracted in it's own frame, but you haven't said which is the primed frame, thus introducing ambiguity.

So the LT seems to deal in appearances, not what's actually measured in the primed or transformed frame. AG

I already told you that LT transforms what it measured NOT what appears.

Brent

Yes, the LT transforms what is measured, and what is transformed is called the primed frame, and as you wrote, no object is ever contracted in its own frame. So the LT doesn't give us what's measured in the primed frame, only how it appears from the pov of unprimed frame. Correct? AG

[Alan Grayson]

On Tuesday, January 14, 2025 at 12:02:11 AM UTC-7 Alan Grayson wrote:

On Monday, January 13, 2025 at 11:54:56 PM UTC-7 Brent Meeker wrote:

On 1/13/2025 10:39 PM, Alan Grayson wrote:

On Monday, January 13, 2025 at 11:20:05 PM UTC-7 Brent Meeker wrote:

On 1/13/2025 10:04 PM, Alan Grayson wrote:

On Monday, January 13, 2025 at 10:21:28 PM UTC-7 Brent Meeker wrote:

On 1/13/2025 9:02 PM, Alan Grayson wrote:

Using the LT, we have the following transformations of Length, Time, and Mass, that is,

x --->x',  t ---> t',  m ---> m', where the primed quantities are the transformed values in the primed frame, given their values in the unprimed frame. The question is this; which of the quantities in the primed frame are actually measured in the primed frame, and which are appearances in the primed frame as seen by unprimed frame?

All of them.  That's why it's relativity theory.  x and t are measurements in one frame and x' and t' are measurements in another frame moving relative to the unprimed frame.  And note the use of "measurements"  not "as seen".  The two are different when you consider things moving at a significant fraction of the speed of light.

But length in primed frame is contracted from the pov of unprimed frame, but in primed frame it isn't measured as contracted, so it APPEARS contracted from the pov of unprimed frame

No, it will appear rotated (c.f. Terrell rotation).  It will measure contracted (using light and clocks, as with radar).

Brent

Terrell rotation over my head,

It's probably within your capability to Google it.

but length contraction allegedly measured in primed frame contradicts the discussion of the paradox, where car and garage lengths aren't contracted when viewing each other. 

That doesn't even parse.

Here the garage is in the primed frame but isn't actually contracted.

No object is ever contracted in it's own frame, but you haven't said which is the primed frame, thus introducing ambiguity.

So the LT seems to deal in appearances, not what's actually measured in the primed or transformed frame. AG

I already told you that LT transforms what it measured NOT what appears.

Brent

Yes, the LT transforms what is measured, and what is transformed is called the primed frame, and as you wrote, no object is ever contracted in its own frame. So the LT doesn't give us what's measured in the primed frame, only how it appears from the pov of unprimed frame. Correct? AG

I mean the contracted length is what the LT gives us from the pov of the unprimed frame, what the unprimed frame concludes is measured in the primed frame, x --> x', but what's actually measured in that frame is the un-contracted or rest length. From this I conclude that the LT gives us what appears from the pov of the unprimed frame, not what's actually measured measured in the prime or transformed frame. Is this not correct? AG 

[Alan Grayson]

As I see it, observer measuring x thinks the LT tells him that observer in primed frame will measure x', but that's not what observer in primed frame measures. He measures his length unchanged. AG 

[Alan Grayson]

On Monday, January 13, 2025 at 11:54:56 PM UTC-7 Brent Meeker wrote:

On 1/13/2025 10:39 PM, Alan Grayson wrote:

On Monday, January 13, 2025 at 11:20:05 PM UTC-7 Brent Meeker wrote:

On 1/13/2025 10:04 PM, Alan Grayson wrote:

On Monday, January 13, 2025 at 10:21:28 PM UTC-7 Brent Meeker wrote:

On 1/13/2025 9:02 PM, Alan Grayson wrote:

Using the LT, we have the following transformations of Length, Time, and Mass, that is,

x --->x',  t ---> t',  m ---> m', where the primed quantities are the transformed values in the primed frame, given their values in the unprimed frame. The question is this; which of the quantities in the primed frame are actually measured in the primed frame, and which are appearances in the primed frame as seen by unprimed frame?

All of them.  That's why it's relativity theory.  x and t are measurements in one frame and x' and t' are measurements in another frame moving relative to the unprimed frame.  And note the use of "measurements"  not "as seen".  The two are different when you consider things moving at a significant fraction of the speed of light.

But length in primed frame is contracted from the pov of unprimed frame, but in primed frame it isn't measured as contracted, so it APPEARS contracted from the pov of unprimed frame

No, it will appear rotated (c.f. Terrell rotation).  It will measure contracted (using light and clocks, as with radar).

Brent

Terrell rotation over my head,

It's probably within your capability to Google it.

but length contraction allegedly measured in primed frame contradicts the discussion of the paradox, where car and garage lengths aren't contracted when viewing each other. 

That doesn't even parse.

Here the garage is in the primed frame but isn't actually contracted.

No object is ever contracted in it's own frame, but you haven't said which is the primed frame, thus introducing ambiguity.

So the LT seems to deal in appearances, not what's actually measured in the primed or transformed frame. AG

I already told you that LT transforms what it measured NOT what appears.

Brent

I'm referring to the primed frame in the LT formula x --> x'. The LT gives us the length contracted from the pov of the moving frame, of the primed frame, but the primed frame never measures its length contracted. If this is correct, isn't it reasonable and accurate to say the LT give us appearances of what the moving frame measures, but not what is actually measured in the stationary or primed frame? For example, on a near light speed trip to Andromeda, the distance is hugely contracted from the pov of the traveler, what the traveler measures, but from the pov of the stationary observer, the distance remainS 2.5 MLY. AG  

[Jesse Mazer]

On Tue, Jan 14, 2025 at 12:02 AM Alan Grayson <agrays...@gmail.com> wrote:

Using the LT, we have the following transformations of Length, Time, and Mass, that is,

x --->x',  t ---> t',  m ---> m'

The length contraction equation is not part of the Lorentz transformation equations, the x --> x' equation in the LT is just about the position coordinate assigned to a *single* event in each frame. The length contraction equation can be derived from the LT but only by considering worldlines of the front and back of an object, and looking at *pairs* of events (one on each of the two worldlines) which are simultaneous in each frame--length in a given frame is just defined as the difference in position coordinate between the front and back of an object at a single time-coordinate in that frame, so it requires looking at a pair of events that are simultaneous in that frame. The result is that for any inertial object, it has its maximum length L in the frame where the object is at rest (the object's own 'rest frame'), and a shorter length L*sqrt(1 - v^2/c^2) in a different frame where the object has nonzero velocity v.

The t ---> t' equation is likewise not the same as the time dilation equation, it's just about the time coordinate assigned to a single event in each frame, although it has a simpler relation to time dilation since you can consider an event on the worldline that passes through the origin where both t and t' are equal to 0, and then the time coordinates t and t' assigned to some other event E on this worldline tell you the time elapsed in each frame between the origin and E. And the LT don't include any mass transformation equation.

Jesse

 

[Alan Grayson]

You're right of course. TY. I see the LT as giving appearances because, say for length contraction, the reduced length is not measured in the primed frame, but that is the length measurement from the pov of the unprimed or stationary frame. About mass, since the measured mass grows exponentially to infinity as v --> c, isn't this derivable from the LT, but in which frame? AG 

 

[Jesse Mazer]

In relativity one does not normally designate any particular frame to be the "stationary frame", since all concepts of motion and rest are defined in purely relative way; if one has two objects A and B in relative motion, one could talk about the frame where A is stationary (A's 'rest frame') or the frame where B is stationary (B's rest frame), but that's all. I'm not sure what you mean by "the reduced length is not measured in the primed frame"--which object's length are you talking about? If A's rest frame is the unprimed frame and B's rest frame is the primed frame, then the length of object A in the primed frame is reduced relative to its length in its own rest frame, i.e. the unprimed frame.

 

About mass, since the measured mass grows exponentially to infinity as v --> c, isn't this derivable from the LT, but in which frame? AG 

The notion of a variable relativistic mass is just an alternate way of talking about relativistic momentum, often modern textbooks talk solely about the latter and the only mass concept they use is the rest mass. For example the page at https://courses.lumenlearning.com/suny-physics/chapter/28-5-relativistic-momentum has a box titled "Misconception alert: relativistic mass and momentum" which says the following (note that they are using u to denote velocity):

"The relativistically correct definition of momentum as p = γmu is sometimes taken to imply that mass varies with velocity: m_var = γm, particularly in older textbooks. However, note that m is the mass of the object as measured by a person at rest relative to the object. Thus, m is defined to be the rest mass, which could be measured at rest, perhaps using gravity. When a mass is moving relative to an observer, the only way that its mass can be determined is through collisions or other means in which momentum is involved. Since the mass of a moving object cannot be determined independently of momentum, the only meaningful mass is rest mass. Thus, when we use the term mass, assume it to be identical to rest mass."

I'd say there's nothing strictly incorrect about defining a variable relativistic mass, it's just a cosmetically different formalism, but it may be that part of the reason it was mostly abandoned is because for people learning relativity it can lead to misconceptions that there is more to the concept than just a difference in how momentum is calculated, whereas in fact there is no application of relativistic mass that does not involve relativistic momentum. Momentum is needed for situations like collisions or particle creation/annihilation where there's a change in which objects have which individual momenta, but total momentum must be conserved. It's also used in the more general form of the relation of energy to rest mass m and relativistic momentum p, given by the equation E^2 = (mc^2)^2 + (pc)^2, which reduces to the more well-known E=mc^2 in the special case where p=0. 

By the way, since relativistic momentum is given by p=mv/sqrt(1 - v^2/c^2), you can substitute this into the above equation to get E^2 = (m^2)(c^4) + (m^2)(v^2)(c^2)/(1 - v^2/c^2), and then if you take the first term on the right hand side, (m^2)(c^4), and multiply it by (1 - v^2/c^2)/(1 - v^2/c^2) and gather terms, you get E^2 = [(m^2)(c^4) - (m^2)(v^2)(c^2) + (m^2)(v^2)(c^2)]/(1 - v^2/c^2), and two terms cancel each other out so this simplifies to E^2 = (m^2)(c^4)/(1 - v^2/c^2), and then if you take the square root of both sides you get E = γmc^2. So the original equation for energy as a function fo rest mass m and relativistic momentum p can be rewritten as E=Mc^2 where M is the relativistic mass defined as M = γm, again showing that relativistic mass is only useful for rewriting equations involving relativistic momentum.

Jesse

 

 

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[Alan Grayson]

On Thursday, January 16, 2025 at 2:39:55 PM UTC-7 Jesse Mazer wrote:

On Thu, Jan 16, 2025 at 2:43 PM Alan Grayson <agrays...@gmail.com> wrote:

On Thursday, January 16, 2025 at 11:36:48 AM UTC-7 Jesse Mazer wrote:

On Tue, Jan 14, 2025 at 12:02 AM Alan Grayson <agrays...@gmail.com> wrote:

Using the LT, we have the following transformations of Length, Time, and Mass, that is,

x --->x',  t ---> t',  m ---> m'

The length contraction equation is not part of the Lorentz transformation equations, the x --> x' equation in the LT is just about the position coordinate assigned to a *single* event in each frame. The length contraction equation can be derived from the LT but only by considering worldlines of the front and back of an object, and looking at *pairs* of events (one on each of the two worldlines) which are simultaneous in each frame--length in a given frame is just defined as the difference in position coordinate between the front and back of an object at a single time-coordinate in that frame, so it requires looking at a pair of events that are simultaneous in that frame. The result is that for any inertial object, it has its maximum length L in the frame where the object is at rest (the object's own 'rest frame'), and a shorter length L*sqrt(1 - v^2/c^2) in a different frame where the object has nonzero velocity v.

The t ---> t' equation is likewise not the same as the time dilation equation, it's just about the time coordinate assigned to a single event in each frame, although it has a simpler relation to time dilation since you can consider an event on the worldline that passes through the origin where both t and t' are equal to 0, and then the time coordinates t and t' assigned to some other event E on this worldline tell you the time elapsed in each frame between the origin and E. And the LT don't include any mass transformation equation.

Jesse

You're right of course. TY. I see the LT as giving appearances because, say for length contraction, the reduced length is not measured in the primed frame, but that is the length measurement from the pov of the unprimed or stationary frame.

In relativity one does not normally designate any particular frame to be the "stationary frame", since all concepts of motion and rest are defined in purely relative way; if one has two objects A and B in relative motion, one could talk about the frame where A is stationary (A's 'rest frame') or the frame where B is stationary (B's rest frame), but that's all. I'm not sure what you mean by "the reduced length is not measured in the primed frame"--which object's length are you talking about? If A's rest frame is the unprimed frame and B's rest frame is the primed frame, then the length of object A in the primed frame is reduced relative to its length in its own rest frame, i.e. the unprimed frame.

Let's consider a concrete example of a traveler moving at near light speed to Andromeda. From the traveler's frame, the distance to Andromeda is hugely reduced from its length of 2.5 MLY from the pov of a non-traveling observer. This seems to imply that the reduced length is only measured from the pov of the traveler, but not from the pov of the non-traveler, because of which I describe the measurement from the pov of the traveler as APPARENT. Do you agree that the traveler's measurement is apparent because the non-traveler measures the distance to Andromeda as unchanged? TY, AG  

[Jesse Mazer]

On Thu, Jan 16, 2025 at 7:33 PM Alan Grayson <agrays...@gmail.com> wrote:

On Thursday, January 16, 2025 at 2:39:55 PM UTC-7 Jesse Mazer wrote:

On Thu, Jan 16, 2025 at 2:43 PM Alan Grayson <agrays...@gmail.com> wrote:

On Thursday, January 16, 2025 at 11:36:48 AM UTC-7 Jesse Mazer wrote:

On Tue, Jan 14, 2025 at 12:02 AM Alan Grayson <agrays...@gmail.com> wrote:

Using the LT, we have the following transformations of Length, Time, and Mass, that is,

x --->x',  t ---> t',  m ---> m'

The length contraction equation is not part of the Lorentz transformation equations, the x --> x' equation in the LT is just about the position coordinate assigned to a *single* event in each frame. The length contraction equation can be derived from the LT but only by considering worldlines of the front and back of an object, and looking at *pairs* of events (one on each of the two worldlines) which are simultaneous in each frame--length in a given frame is just defined as the difference in position coordinate between the front and back of an object at a single time-coordinate in that frame, so it requires looking at a pair of events that are simultaneous in that frame. The result is that for any inertial object, it has its maximum length L in the frame where the object is at rest (the object's own 'rest frame'), and a shorter length L*sqrt(1 - v^2/c^2) in a different frame where the object has nonzero velocity v.

The t ---> t' equation is likewise not the same as the time dilation equation, it's just about the time coordinate assigned to a single event in each frame, although it has a simpler relation to time dilation since you can consider an event on the worldline that passes through the origin where both t and t' are equal to 0, and then the time coordinates t and t' assigned to some other event E on this worldline tell you the time elapsed in each frame between the origin and E. And the LT don't include any mass transformation equation.

Jesse

You're right of course. TY. I see the LT as giving appearances because, say for length contraction, the reduced length is not measured in the primed frame, but that is the length measurement from the pov of the unprimed or stationary frame.

In relativity one does not normally designate any particular frame to be the "stationary frame", since all concepts of motion and rest are defined in purely relative way; if one has two objects A and B in relative motion, one could talk about the frame where A is stationary (A's 'rest frame') or the frame where B is stationary (B's rest frame), but that's all. I'm not sure what you mean by "the reduced length is not measured in the primed frame"--which object's length are you talking about? If A's rest frame is the unprimed frame and B's rest frame is the primed frame, then the length of object A in the primed frame is reduced relative to its length in its own rest frame, i.e. the unprimed frame.

Let's consider a concrete example of a traveler moving at near light speed to Andromeda. From the traveler's frame, the distance to Andromeda is hugely reduced from its length of 2.5 MLY from the pov of a non-traveling observer. This seems to imply that the reduced length is only measured from the pov of the traveler, but not from the pov of the non-traveler, because of which I describe the measurement from the pov of the traveler as APPARENT. Do you agree that the traveler's measurement is apparent because the non-traveler measures the distance to Andromeda as unchanged? TY, AG  

I don't know what you mean by "apparent", but there is no asymmetry in the way Lorentz contraction works in each frame--if we assume there is a frame A where Milky Way and Andromeda are both at rest (ignoring the fact that in reality they have some motion relative to one another), and another frame B where the rocket ship of the traveler is at rest, then in frame B the Milky Way/Andromeda distance is shortened relative to the distance in their rest frame, and the rocket has its maximum length; in frame A the the rocket's length is shortened relative to its length in its rest frame, and the Milky Way/Andromeda distance has its maximum value. The only asymmetry here is in the choice of the two things to measure the length of (the distance between the Milky Way and Andromeda in their rest frame is obviously huge compared to the rest length of a rocket moving between them), the symmetry might be easier to see if we consider two rockets traveling towards each other (their noses facing each other), and each wants to know the distance it must traverse to get from the nose of the other rocket to its tail. Then for example if each rocket is 10 meters long in its rest frame, and the two rockets have a relative velocity of 0.8c, each will measure only a 6 meter distance between the nose and tail of the other rocket, and the time they each measure to cross that distance is just (6 meters)/(0.8c).

Jesse

 

 

About mass, since the measured mass grows exponentially to infinity as v --> c, isn't this derivable from the LT, but in which frame? AG 

The notion of a variable relativistic mass is just an alternate way of talking about relativistic momentum, often modern textbooks talk solely about the latter and the only mass concept they use is the rest mass. For example the page at https://courses.lumenlearning.com/suny-physics/chapter/28-5-relativistic-momentum has a box titled "Misconception alert: relativistic mass and momentum" which says the following (note that they are using u to denote velocity):

"The relativistically correct definition of momentum as p = γmu is sometimes taken to imply that mass varies with velocity: m_var = γm, particularly in older textbooks. However, note that m is the mass of the object as measured by a person at rest relative to the object. Thus, m is defined to be the rest mass, which could be measured at rest, perhaps using gravity. When a mass is moving relative to an observer, the only way that its mass can be determined is through collisions or other means in which momentum is involved. Since the mass of a moving object cannot be determined independently of momentum, the only meaningful mass is rest mass. Thus, when we use the term mass, assume it to be identical to rest mass."

I'd say there's nothing strictly incorrect about defining a variable relativistic mass, it's just a cosmetically different formalism, but it may be that part of the reason it was mostly abandoned is because for people learning relativity it can lead to misconceptions that there is more to the concept than just a difference in how momentum is calculated, whereas in fact there is no application of relativistic mass that does not involve relativistic momentum. Momentum is needed for situations like collisions or particle creation/annihilation where there's a change in which objects have which individual momenta, but total momentum must be conserved. It's also used in the more general form of the relation of energy to rest mass m and relativistic momentum p, given by the equation E^2 = (mc^2)^2 + (pc)^2, which reduces to the more well-known E=mc^2 in the special case where p=0. 

By the way, since relativistic momentum is given by p=mv/sqrt(1 - v^2/c^2), you can substitute this into the above equation to get E^2 = (m^2)(c^4) + (m^2)(v^2)(c^2)/(1 - v^2/c^2), and then if you take the first term on the right hand side, (m^2)(c^4), and multiply it by (1 - v^2/c^2)/(1 - v^2/c^2) and gather terms, you get E^2 = [(m^2)(c^4) - (m^2)(v^2)(c^2) + (m^2)(v^2)(c^2)]/(1 - v^2/c^2), and two terms cancel each other out so this simplifies to E^2 = (m^2)(c^4)/(1 - v^2/c^2), and then if you take the square root of both sides you get E = γmc^2. So the original equation for energy as a function fo rest mass m and relativistic momentum p can be rewritten as E=Mc^2 where M is the relativistic mass defined as M = γm, again showing that relativistic mass is only useful for rewriting equations involving relativistic momentum.

Jesse

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[Brent Meeker]

On 1/16/2025 10:07 AM, Alan Grayson wrote:

On Monday, January 13, 2025 at 11:54:56 PM UTC-7 Brent Meeker wrote:

On 1/13/2025 10:39 PM, Alan Grayson wrote:

On Monday, January 13, 2025 at 11:20:05 PM UTC-7 Brent Meeker wrote:

On 1/13/2025 10:04 PM, Alan Grayson wrote:

On Monday, January 13, 2025 at 10:21:28 PM UTC-7 Brent Meeker wrote:

On 1/13/2025 9:02 PM, Alan Grayson wrote:

Using the LT, we have the following transformations of Length, Time, and Mass, that is,

x --->x',  t ---> t',  m ---> m', where the primed quantities are the transformed values in the primed frame, given their values in the unprimed frame. The question is this; which of the quantities in the primed frame are actually measured in the primed frame, and which are appearances in the primed frame as seen by unprimed frame?

All of them.  That's why it's relativity theory.  x and t are measurements in one frame and x' and t' are measurements in another frame moving relative to the unprimed frame.  And note the use of "measurements"  not "as seen".  The two are different when you consider things moving at a significant fraction of the speed of light.

But length in primed frame is contracted from the pov of unprimed frame, but in primed frame it isn't measured as contracted, so it APPEARS contracted from the pov of unprimed frame

No, it will appear rotated (c.f. Terrell rotation).  It will measure contracted (using light and clocks, as with radar).

Brent

Terrell rotation over my head,

It's probably within your capability to Google it.

but length contraction allegedly measured in primed frame contradicts the discussion of the paradox, where car and garage lengths aren't contracted when viewing each other. 

That doesn't even parse.

Here the garage is in the primed frame but isn't actually contracted.

No object is ever contracted in it's own frame, but you haven't said which is the primed frame, thus introducing ambiguity.

So the LT seems to deal in appearances, not what's actually measured in the primed or transformed frame. AG

I already told you that LT transforms what it measured NOT what appears.

Brent

I'm referring to the primed frame in the LT formula x --> x'. The LT gives us the length contracted from the pov of the moving frame, of the primed frame, but the primed frame never measures its length contracted. If this is correct, isn't it reasonable and accurate to say the LT give us appearances of what the moving frame measures, but not what is actually measured in the stationary or primed frame?

Roughly speaking, yes.  So long as you mean "appearance" broadly to include what you measure, not just what you would see.'

For example, on a near light speed trip to Andromeda, the distance is hugely contracted from the pov of the traveler, what the traveler measures, but from the pov of the stationary observer, the distance remainS 2.5 MLY. AG 

Right.

Brent

but NOT measured in primed frame. Same with time and mass? AG 

How is this consistent, if it is, with the fact that when doing EM measurements, E' and B' are the actual measurements of the fields in the primed frame, given that E and B are measured in the unprimed frame, but the same cannot be said of some, or all of the measurements of Length, Time, and Mass? TY,

The same applies except the electromagnetic field is a tensor, so it transforms by tensor like a sequence of LTs.  How is that not consistent?

Brent

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[Alan Grayson]

"Houston, we have a problem!" Now let's consider time dilation using SR in the Twin Paradox. Imagine the traveling twin moving in a circle and returning to Earth, and imagine the circle contains a polygon consisting of straight paths, which will later be infinitely partitioned, whose limit will be that circle. As measured by the stationary twin, the traveling twin's clock is dilated along each segment, so when the twins are juxtaposed, the traveling twin's elapsed time is LESS than clock readings for the stationary twin. If this is correct, it demostrates that what the stationary twin measures, is actually what the traveling twin's clock reads. IOW, what happens to time dilation in this case is OPPOSITE to what happens to the frames for the trip to Andromeda! Do you understand what I am alleging -- that length contraction acts in an opposite manner compared to time dilation, when I would expect them to behave similarly? AG 

[John Clark]

On Thu, Jan 16, 2025 at 7:33 PM Alan Grayson <agrays...@gmail.com> wrote:

> I describe the measurement from the pov of the traveler as APPARENT. Do you agree that the traveler's measurement is apparent because the non-traveler measures the distance to Andromeda as unchanged?  

You can call it anything you like provided you play fair and use the same word to describe the measurement of both observers; there is one and only one true value for the speed of light but there is NOT one and only one true value for the distance between the Earth and Andromeda. 

I could be wrong but I get the impression you believe the opposite of the word "apparent" is "real", but it is not. According to my Mac "apparent" means obvious or clear while its opposite means unclear or obscure. 

John K Clark    See what's on my new list at  Extropolis

air

 

 

[Alan Grayson]

The distance to Andromeda the traveler measures using the LT, is not the measured value in the target frame. IOW, in this case, the LT does not produce what's actually measured in the target frame of the transformation. This behavior negates the claim that the LT produces what is actually measured in the target frame. So I referred to the resultant measurement predicted by the LT as apparent. You can give it another description. I am not wedded to that word. AG 

a

 

[Alan Grayson]

BTW, apparent can also mean illusory, or unreal, and applicable in this case since the LT does NOT transform to any measurement in the transformed frame. AG 

[Alan Grayson]

On Thursday, January 16, 2025 at 5:52:52 PM UTC-7 Jesse Mazer wrote:

On Thu, Jan 16, 2025 at 7:33 PM Alan Grayson <agrays...@gmail.com> wrote:

On Thursday, January 16, 2025 at 2:39:55 PM UTC-7 Jesse Mazer wrote:

On Thu, Jan 16, 2025 at 2:43 PM Alan Grayson <agrays...@gmail.com> wrote:

On Thursday, January 16, 2025 at 11:36:48 AM UTC-7 Jesse Mazer wrote:

On Tue, Jan 14, 2025 at 12:02 AM Alan Grayson <agrays...@gmail.com> wrote:

Using the LT, we have the following transformations of Length, Time, and Mass, that is,

x --->x',  t ---> t',  m ---> m'

The length contraction equation is not part of the Lorentz transformation equations, the x --> x' equation in the LT is just about the position coordinate assigned to a *single* event in each frame. The length contraction equation can be derived from the LT but only by considering worldlines of the front and back of an object, and looking at *pairs* of events (one on each of the two worldlines) which are simultaneous in each frame--length in a given frame is just defined as the difference in position coordinate between the front and back of an object at a single time-coordinate in that frame, so it requires looking at a pair of events that are simultaneous in that frame. The result is that for any inertial object, it has its maximum length L in the frame where the object is at rest (the object's own 'rest frame'), and a shorter length L*sqrt(1 - v^2/c^2) in a different frame where the object has nonzero velocity v.

The t ---> t' equation is likewise not the same as the time dilation equation, it's just about the time coordinate assigned to a single event in each frame, although it has a simpler relation to time dilation since you can consider an event on the worldline that passes through the origin where both t and t' are equal to 0, and then the time coordinates t and t' assigned to some other event E on this worldline tell you the time elapsed in each frame between the origin and E. And the LT don't include any mass transformation equation.

Jesse

You're right of course. TY. I see the LT as giving appearances because, say for length contraction, the reduced length is not measured in the primed frame, but that is the length measurement from the pov of the unprimed or stationary frame.

In relativity one does not normally designate any particular frame to be the "stationary frame", since all concepts of motion and rest are defined in purely relative way; if one has two objects A and B in relative motion, one could talk about the frame where A is stationary (A's 'rest frame') or the frame where B is stationary (B's rest frame), but that's all. I'm not sure what you mean by "the reduced length is not measured in the primed frame"--which object's length are you talking about? If A's rest frame is the unprimed frame and B's rest frame is the primed frame, then the length of object A in the primed frame is reduced relative to its length in its own rest frame, i.e. the unprimed frame.

Let's consider a concrete example of a traveler moving at near light speed to Andromeda. From the traveler's frame, the distance to Andromeda is hugely reduced from its length of 2.5 MLY from the pov of a non-traveling observer. This seems to imply that the reduced length is only measured from the pov of the traveler, but not from the pov of the non-traveler, because of which I describe the measurement from the pov of the traveler as APPARENT. Do you agree that the traveler's measurement is apparent because the non-traveler measures the distance to Andromeda as unchanged? TY, AG  

I don't know what you mean by "apparent", but there is no asymmetry in the way Lorentz contraction works in each frame--if we assume there is a frame A where Milky Way and Andromeda are both at rest (ignoring the fact that in reality they have some motion relative to one another), and another frame B where the rocket ship of the traveler is at rest, then in frame B the Milky Way/Andromeda distance is shortened relative to the distance in their rest frame, and the rocket has its maximum length; in frame A the the rocket's length is shortened relative to its length in its rest frame, and the Milky Way/Andromeda distance has its maximum value. The only asymmetry here is in the choice of the two things to measure the length of (the distance between the Milky Way and Andromeda in their rest frame is obviously huge compared to the rest length of a rocket moving between them), the symmetry might be easier to see if we consider two rockets traveling towards each other (their noses facing each other), and each wants to know the distance it must traverse to get from the nose of the other rocket to its tail. Then for example if each rocket is 10 meters long in its rest frame, and the two rockets have a relative velocity of 0.8c, each will measure only a 6 meter distance between the nose and tail of the other rocket, and the time they each measure to cross that distance is just (6 meters)/(0.8c).

Jesse 

By apparent I just mean that the measurement the LT gives in this case, is not what is actually measured in the target frame. Moreover, this is differnt from the situation in the Twin Paradox as discussed in another recent post on this thread. AG 

[John Clark]

On Fri, Jan 17, 2025 at 7:08 AM Alan Grayson <agrays...@gmail.com> wrote:

> The distance to Andromeda the traveler measures using the LT, is not the measured value in the target frame.

That's because distance is relative, not absolute, and the traveler is moving relative to both the Earth and Andromeda, but the Earth and Andromeda happen not to be moving relative to each other ,... well they are but for the purposes of this thought experiment we can pretend that they are not.  

 > This behavior negates the claim that the LT produces what is actually measured in the target frame

If the Earth astronomer understands Lorentz Transformations and he knows how fast the traveler is moving relative to himself then he can predict what number the traveler will obtain when he measures the distance to Andromeda. 

 

I suggest you take a look at this video, the guy has the ability to explain complicated things as simply as possible. 

Why the twin paradox is NOT about acceleration

John K Clark    See what's on my new list at  Extropolis

naa

[Alan Grayson]

On Friday, January 17, 2025 at 6:37:46 AM UTC-7 John Clark wrote:

On Fri, Jan 17, 2025 at 7:08 AM Alan Grayson <agrays...@gmail.com> wrote:

> The distance to Andromeda the traveler measures using the LT, is not the measured value in the target frame.

That's because distance is relative, not absolute, and the traveler is moving relative to both the Earth and Andromeda, but the Earth and Andromeda happen not to be moving relative to each other ,... well they are but for the purposes of this thought experiment we can pretend that they are not.  

 > This behavior negates the claim that the LT produces what is actually measured in the target frame

If the Earth astronomer understands Lorentz Transformations and he knows how fast the traveler is moving relative to himself then he can predict what number the traveler will obtain when he measures the distance to Andromeda. 

What this to do with anything I posted? AG 

 

I suggest you take a look at this video, the guy has the ability to explain complicated things as simply as possible. 

Why the twin paradox is NOT about acceleration

Did say anything about acceleration? Obviously, in your anxiousness to respond, you omitted to actually reading what I wrote. I used SR purposely so I could compare the result to that obtained in the Andromeda case. It's obvious that in the Andromeda case, what the LT predicts for measurements in the target frame isn't what is measured in that frame, so what you wrote above has nothing to do with nothing. BTW, one can use acceleration in the Twin Paradox. It can be used to show the frames are not symmetric, where the assumption of symmetry is the cause of the paradox. AG 

[John Clark]

On Fri, Jan 17, 2025 at 9:02 AM Alan Grayson <agrays...@gmail.com> wrote:

> BTW, one can use acceleration in the Twin Paradox. It can be used to show the frames are not symmetric, where the assumption of symmetry is the cause of the paradox. AG

Look at the video.  

 

Why the twin paradox is NOT about acceleration

John K Clark    See what's on my new list at  Extropolis

3zp

[Jesse Mazer]

On Fri, Jan 17, 2025 at 7:51 AM Alan Grayson <agrays...@gmail.com> wrote:

On Thursday, January 16, 2025 at 5:52:52 PM UTC-7 Jesse Mazer wrote:

On Thu, Jan 16, 2025 at 7:33 PM Alan Grayson <agrays...@gmail.com> wrote:

On Thursday, January 16, 2025 at 2:39:55 PM UTC-7 Jesse Mazer wrote:

On Thu, Jan 16, 2025 at 2:43 PM Alan Grayson <agrays...@gmail.com> wrote:

On Thursday, January 16, 2025 at 11:36:48 AM UTC-7 Jesse Mazer wrote:

On Tue, Jan 14, 2025 at 12:02 AM Alan Grayson <agrays...@gmail.com> wrote:

Using the LT, we have the following transformations of Length, Time, and Mass, that is,

x --->x',  t ---> t',  m ---> m'

The length contraction equation is not part of the Lorentz transformation equations, the x --> x' equation in the LT is just about the position coordinate assigned to a *single* event in each frame. The length contraction equation can be derived from the LT but only by considering worldlines of the front and back of an object, and looking at *pairs* of events (one on each of the two worldlines) which are simultaneous in each frame--length in a given frame is just defined as the difference in position coordinate between the front and back of an object at a single time-coordinate in that frame, so it requires looking at a pair of events that are simultaneous in that frame. The result is that for any inertial object, it has its maximum length L in the frame where the object is at rest (the object's own 'rest frame'), and a shorter length L*sqrt(1 - v^2/c^2) in a different frame where the object has nonzero velocity v.

The t ---> t' equation is likewise not the same as the time dilation equation, it's just about the time coordinate assigned to a single event in each frame, although it has a simpler relation to time dilation since you can consider an event on the worldline that passes through the origin where both t and t' are equal to 0, and then the time coordinates t and t' assigned to some other event E on this worldline tell you the time elapsed in each frame between the origin and E. And the LT don't include any mass transformation equation.

Jesse

You're right of course. TY. I see the LT as giving appearances because, say for length contraction, the reduced length is not measured in the primed frame, but that is the length measurement from the pov of the unprimed or stationary frame.

In relativity one does not normally designate any particular frame to be the "stationary frame", since all concepts of motion and rest are defined in purely relative way; if one has two objects A and B in relative motion, one could talk about the frame where A is stationary (A's 'rest frame') or the frame where B is stationary (B's rest frame), but that's all. I'm not sure what you mean by "the reduced length is not measured in the primed frame"--which object's length are you talking about? If A's rest frame is the unprimed frame and B's rest frame is the primed frame, then the length of object A in the primed frame is reduced relative to its length in its own rest frame, i.e. the unprimed frame.

Let's consider a concrete example of a traveler moving at near light speed to Andromeda. From the traveler's frame, the distance to Andromeda is hugely reduced from its length of 2.5 MLY from the pov of a non-traveling observer. This seems to imply that the reduced length is only measured from the pov of the traveler, but not from the pov of the non-traveler, because of which I describe the measurement from the pov of the traveler as APPARENT. Do you agree that the traveler's measurement is apparent because the non-traveler measures the distance to Andromeda as unchanged? TY, AG  

I don't know what you mean by "apparent", but there is no asymmetry in the way Lorentz contraction works in each frame--if we assume there is a frame A where Milky Way and Andromeda are both at rest (ignoring the fact that in reality they have some motion relative to one another), and another frame B where the rocket ship of the traveler is at rest, then in frame B the Milky Way/Andromeda distance is shortened relative to the distance in their rest frame, and the rocket has its maximum length; in frame A the the rocket's length is shortened relative to its length in its rest frame, and the Milky Way/Andromeda distance has its maximum value. The only asymmetry here is in the choice of the two things to measure the length of (the distance between the Milky Way and Andromeda in their rest frame is obviously huge compared to the rest length of a rocket moving between them), the symmetry might be easier to see if we consider two rockets traveling towards each other (their noses facing each other), and each wants to know the distance it must traverse to get from the nose of the other rocket to its tail. Then for example if each rocket is 10 meters long in its rest frame, and the two rockets have a relative velocity of 0.8c, each will measure only a 6 meter distance between the nose and tail of the other rocket, and the time they each measure to cross that distance is just (6 meters)/(0.8c).

Jesse 

By apparent I just mean that the measurement the LT gives in this case, is not what is actually measured in the target frame. Moreover, this is differnt from the situation in the Twin Paradox as discussed in another recent post on this thread. AG 

What do you mean by target frame? If the unprimed frame is the frame where Milky Way/Andromeda are at rest and the primed frame is the frame where the rocket is at rest, are you saying the primed frame does not actually measure a shorter distance from Milky Way to Andromeda if we use the LT starting from the coordinates of everything in the unprimed frame? Or are you arguing something different? Are you using primed or unprimed as the "target frame"?

Jesse

 

 

 

About mass, since the measured mass grows exponentially to infinity as v --> c, isn't this derivable from the LT, but in which frame? AG 

The notion of a variable relativistic mass is just an alternate way of talking about relativistic momentum, often modern textbooks talk solely about the latter and the only mass concept they use is the rest mass. For example the page at https://courses.lumenlearning.com/suny-physics/chapter/28-5-relativistic-momentum has a box titled "Misconception alert: relativistic mass and momentum" which says the following (note that they are using u to denote velocity):

"The relativistically correct definition of momentum as p = γmu is sometimes taken to imply that mass varies with velocity: m_var = γm, particularly in older textbooks. However, note that m is the mass of the object as measured by a person at rest relative to the object. Thus, m is defined to be the rest mass, which could be measured at rest, perhaps using gravity. When a mass is moving relative to an observer, the only way that its mass can be determined is through collisions or other means in which momentum is involved. Since the mass of a moving object cannot be determined independently of momentum, the only meaningful mass is rest mass. Thus, when we use the term mass, assume it to be identical to rest mass."

I'd say there's nothing strictly incorrect about defining a variable relativistic mass, it's just a cosmetically different formalism, but it may be that part of the reason it was mostly abandoned is because for people learning relativity it can lead to misconceptions that there is more to the concept than just a difference in how momentum is calculated, whereas in fact there is no application of relativistic mass that does not involve relativistic momentum. Momentum is needed for situations like collisions or particle creation/annihilation where there's a change in which objects have which individual momenta, but total momentum must be conserved. It's also used in the more general form of the relation of energy to rest mass m and relativistic momentum p, given by the equation E^2 = (mc^2)^2 + (pc)^2, which reduces to the more well-known E=mc^2 in the special case where p=0. 

By the way, since relativistic momentum is given by p=mv/sqrt(1 - v^2/c^2), you can substitute this into the above equation to get E^2 = (m^2)(c^4) + (m^2)(v^2)(c^2)/(1 - v^2/c^2), and then if you take the first term on the right hand side, (m^2)(c^4), and multiply it by (1 - v^2/c^2)/(1 - v^2/c^2) and gather terms, you get E^2 = [(m^2)(c^4) - (m^2)(v^2)(c^2) + (m^2)(v^2)(c^2)]/(1 - v^2/c^2), and two terms cancel each other out so this simplifies to E^2 = (m^2)(c^4)/(1 - v^2/c^2), and then if you take the square root of both sides you get E = γmc^2. So the original equation for energy as a function fo rest mass m and relativistic momentum p can be rewritten as E=Mc^2 where M is the relativistic mass defined as M = γm, again showing that relativistic mass is only useful for rewriting equations involving relativistic momentum.

Jesse

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[Alan Grayson]

On Friday, January 17, 2025 at 7:29:19 AM UTC-7 Jesse Mazer wrote:

On Fri, Jan 17, 2025 at 7:51 AM Alan Grayson <agrays...@gmail.com> wrote:

On Thursday, January 16, 2025 at 5:52:52 PM UTC-7 Jesse Mazer wrote:

On Thu, Jan 16, 2025 at 7:33 PM Alan Grayson <agrays...@gmail.com> wrote:

On Thursday, January 16, 2025 at 2:39:55 PM UTC-7 Jesse Mazer wrote:

On Thu, Jan 16, 2025 at 2:43 PM Alan Grayson <agrays...@gmail.com> wrote:

On Thursday, January 16, 2025 at 11:36:48 AM UTC-7 Jesse Mazer wrote:

On Tue, Jan 14, 2025 at 12:02 AM Alan Grayson <agrays...@gmail.com> wrote:

Using the LT, we have the following transformations of Length, Time, and Mass, that is,

x --->x',  t ---> t',  m ---> m'

The length contraction equation is not part of the Lorentz transformation equations, the x --> x' equation in the LT is just about the position coordinate assigned to a *single* event in each frame. The length contraction equation can be derived from the LT but only by considering worldlines of the front and back of an object, and looking at *pairs* of events (one on each of the two worldlines) which are simultaneous in each frame--length in a given frame is just defined as the difference in position coordinate between the front and back of an object at a single time-coordinate in that frame, so it requires looking at a pair of events that are simultaneous in that frame. The result is that for any inertial object, it has its maximum length L in the frame where the object is at rest (the object's own 'rest frame'), and a shorter length L*sqrt(1 - v^2/c^2) in a different frame where the object has nonzero velocity v.

The t ---> t' equation is likewise not the same as the time dilation equation, it's just about the time coordinate assigned to a single event in each frame, although it has a simpler relation to time dilation since you can consider an event on the worldline that passes through the origin where both t and t' are equal to 0, and then the time coordinates t and t' assigned to some other event E on this worldline tell you the time elapsed in each frame between the origin and E. And the LT don't include any mass transformation equation.

Jesse

You're right of course. TY. I see the LT as giving appearances because, say for length contraction, the reduced length is not measured in the primed frame, but that is the length measurement from the pov of the unprimed or stationary frame.

In relativity one does not normally designate any particular frame to be the "stationary frame", since all concepts of motion and rest are defined in purely relative way; if one has two objects A and B in relative motion, one could talk about the frame where A is stationary (A's 'rest frame') or the frame where B is stationary (B's rest frame), but that's all. I'm not sure what you mean by "the reduced length is not measured in the primed frame"--which object's length are you talking about? If A's rest frame is the unprimed frame and B's rest frame is the primed frame, then the length of object A in the primed frame is reduced relative to its length in its own rest frame, i.e. the unprimed frame.

Let's consider a concrete example of a traveler moving at near light speed to Andromeda. From the traveler's frame, the distance to Andromeda is hugely reduced from its length of 2.5 MLY from the pov of a non-traveling observer. This seems to imply that the reduced length is only measured from the pov of the traveler, but not from the pov of the non-traveler, because of which I describe the measurement from the pov of the traveler as APPARENT. Do you agree that the traveler's measurement is apparent because the non-traveler measures the distance to Andromeda as unchanged? TY, AG  

I don't know what you mean by "apparent", but there is no asymmetry in the way Lorentz contraction works in each frame--if we assume there is a frame A where Milky Way and Andromeda are both at rest (ignoring the fact that in reality they have some motion relative to one another), and another frame B where the rocket ship of the traveler is at rest, then in frame B the Milky Way/Andromeda distance is shortened relative to the distance in their rest frame, and the rocket has its maximum length; in frame A the the rocket's length is shortened relative to its length in its rest frame, and the Milky Way/Andromeda distance has its maximum value. The only asymmetry here is in the choice of the two things to measure the length of (the distance between the Milky Way and Andromeda in their rest frame is obviously huge compared to the rest length of a rocket moving between them), the symmetry might be easier to see if we consider two rockets traveling towards each other (their noses facing each other), and each wants to know the distance it must traverse to get from the nose of the other rocket to its tail. Then for example if each rocket is 10 meters long in its rest frame, and the two rockets have a relative velocity of 0.8c, each will measure only a 6 meter distance between the nose and tail of the other rocket, and the time they each measure to cross that distance is just (6 meters)/(0.8c).

Jesse 

By apparent I just mean that the measurement the LT gives in this case, is not what is actually measured in the target frame. Moreover, this is differnt from the situation in the Twin Paradox as discussed in another recent post on this thread. AG 

What do you mean by target frame? If the unprimed frame is the frame where Milky Way/Andromeda are at rest and the primed frame is the frame where the rocket is at rest, are you saying the primed frame does not actually measure a shorter distance from Milky Way to Andromeda if we use the LT starting from the coordinates of everything in the unprimed frame? Or are you arguing something different? Are you using primed or unprimed as the "target frame"?

Jesse

The target frame is the primed frame, the result of the LT. The unprimed frame is the traveler's frame moving at some speed toward Andromeda. It's often claimed that the result of applying the LT will yield the actual measurement in the primed frame, but this isn't the case in this example. AG 

[Jesse Mazer]

On Fri, Jan 17, 2025 at 9:38 AM Alan Grayson <agrays...@gmail.com> wrote:

On Friday, January 17, 2025 at 7:29:19 AM UTC-7 Jesse Mazer wrote:

On Fri, Jan 17, 2025 at 7:51 AM Alan Grayson <agrays...@gmail.com> wrote:

On Thursday, January 16, 2025 at 5:52:52 PM UTC-7 Jesse Mazer wrote:

On Thu, Jan 16, 2025 at 7:33 PM Alan Grayson <agrays...@gmail.com> wrote:

On Thursday, January 16, 2025 at 2:39:55 PM UTC-7 Jesse Mazer wrote:

On Thu, Jan 16, 2025 at 2:43 PM Alan Grayson <agrays...@gmail.com> wrote:

On Thursday, January 16, 2025 at 11:36:48 AM UTC-7 Jesse Mazer wrote:

On Tue, Jan 14, 2025 at 12:02 AM Alan Grayson <agrays...@gmail.com> wrote:

Using the LT, we have the following transformations of Length, Time, and Mass, that is,

x --->x',  t ---> t',  m ---> m'

The length contraction equation is not part of the Lorentz transformation equations, the x --> x' equation in the LT is just about the position coordinate assigned to a *single* event in each frame. The length contraction equation can be derived from the LT but only by considering worldlines of the front and back of an object, and looking at *pairs* of events (one on each of the two worldlines) which are simultaneous in each frame--length in a given frame is just defined as the difference in position coordinate between the front and back of an object at a single time-coordinate in that frame, so it requires looking at a pair of events that are simultaneous in that frame. The result is that for any inertial object, it has its maximum length L in the frame where the object is at rest (the object's own 'rest frame'), and a shorter length L*sqrt(1 - v^2/c^2) in a different frame where the object has nonzero velocity v.

The t ---> t' equation is likewise not the same as the time dilation equation, it's just about the time coordinate assigned to a single event in each frame, although it has a simpler relation to time dilation since you can consider an event on the worldline that passes through the origin where both t and t' are equal to 0, and then the time coordinates t and t' assigned to some other event E on this worldline tell you the time elapsed in each frame between the origin and E. And the LT don't include any mass transformation equation.

Jesse

You're right of course. TY. I see the LT as giving appearances because, say for length contraction, the reduced length is not measured in the primed frame, but that is the length measurement from the pov of the unprimed or stationary frame.

In relativity one does not normally designate any particular frame to be the "stationary frame", since all concepts of motion and rest are defined in purely relative way; if one has two objects A and B in relative motion, one could talk about the frame where A is stationary (A's 'rest frame') or the frame where B is stationary (B's rest frame), but that's all. I'm not sure what you mean by "the reduced length is not measured in the primed frame"--which object's length are you talking about? If A's rest frame is the unprimed frame and B's rest frame is the primed frame, then the length of object A in the primed frame is reduced relative to its length in its own rest frame, i.e. the unprimed frame.

Let's consider a concrete example of a traveler moving at near light speed to Andromeda. From the traveler's frame, the distance to Andromeda is hugely reduced from its length of 2.5 MLY from the pov of a non-traveling observer. This seems to imply that the reduced length is only measured from the pov of the traveler, but not from the pov of the non-traveler, because of which I describe the measurement from the pov of the traveler as APPARENT. Do you agree that the traveler's measurement is apparent because the non-traveler measures the distance to Andromeda as unchanged? TY, AG  

I don't know what you mean by "apparent", but there is no asymmetry in the way Lorentz contraction works in each frame--if we assume there is a frame A where Milky Way and Andromeda are both at rest (ignoring the fact that in reality they have some motion relative to one another), and another frame B where the rocket ship of the traveler is at rest, then in frame B the Milky Way/Andromeda distance is shortened relative to the distance in their rest frame, and the rocket has its maximum length; in frame A the the rocket's length is shortened relative to its length in its rest frame, and the Milky Way/Andromeda distance has its maximum value. The only asymmetry here is in the choice of the two things to measure the length of (the distance between the Milky Way and Andromeda in their rest frame is obviously huge compared to the rest length of a rocket moving between them), the symmetry might be easier to see if we consider two rockets traveling towards each other (their noses facing each other), and each wants to know the distance it must traverse to get from the nose of the other rocket to its tail. Then for example if each rocket is 10 meters long in its rest frame, and the two rockets have a relative velocity of 0.8c, each will measure only a 6 meter distance between the nose and tail of the other rocket, and the time they each measure to cross that distance is just (6 meters)/(0.8c).

Jesse 

By apparent I just mean that the measurement the LT gives in this case, is not what is actually measured in the target frame. Moreover, this is differnt from the situation in the Twin Paradox as discussed in another recent post on this thread. AG 

What do you mean by target frame? If the unprimed frame is the frame where Milky Way/Andromeda are at rest and the primed frame is the frame where the rocket is at rest, are you saying the primed frame does not actually measure a shorter distance from Milky Way to Andromeda if we use the LT starting from the coordinates of everything in the unprimed frame? Or are you arguing something different? Are you using primed or unprimed as the "target frame"?

Jesse

The target frame is the primed frame, the result of the LT. The unprimed frame is the traveler's frame moving at some speed toward Andromeda. It's often claimed that the result of applying the LT will yield the actual measurement in the primed frame, but this isn't the case in this example. AG

OK, so you want the unprimed frame to be the frame where the rocket is at rest and the Milky Way/Andromeda are moving? In that case the unprimed frame will be the one where the distance between Milky Way/Andromeda is contracted according to the length contraction equation, since they are moving in that frame and at rest in the primed frame. And as I told you, the LT is not the same as the length contraction equation, if you apply the LT to the coordinates of the worldlines of Milky Way/Andromeda in the unprimed frame, you will get the correct answer that in the primed frame these worldlines have zero velocity (constant position as a function of time) and a greater coordinate distance between them than they did in the unprimed frame.

Jesse

 

 

 

 

 

About mass, since the measured mass grows exponentially to infinity as v --> c, isn't this derivable from the LT, but in which frame? AG 

The notion of a variable relativistic mass is just an alternate way of talking about relativistic momentum, often modern textbooks talk solely about the latter and the only mass concept they use is the rest mass. For example the page at https://courses.lumenlearning.com/suny-physics/chapter/28-5-relativistic-momentum has a box titled "Misconception alert: relativistic mass and momentum" which says the following (note that they are using u to denote velocity):

"The relativistically correct definition of momentum as p = γmu is sometimes taken to imply that mass varies with velocity: m_var = γm, particularly in older textbooks. However, note that m is the mass of the object as measured by a person at rest relative to the object. Thus, m is defined to be the rest mass, which could be measured at rest, perhaps using gravity. When a mass is moving relative to an observer, the only way that its mass can be determined is through collisions or other means in which momentum is involved. Since the mass of a moving object cannot be determined independently of momentum, the only meaningful mass is rest mass. Thus, when we use the term mass, assume it to be identical to rest mass."

I'd say there's nothing strictly incorrect about defining a variable relativistic mass, it's just a cosmetically different formalism, but it may be that part of the reason it was mostly abandoned is because for people learning relativity it can lead to misconceptions that there is more to the concept than just a difference in how momentum is calculated, whereas in fact there is no application of relativistic mass that does not involve relativistic momentum. Momentum is needed for situations like collisions or particle creation/annihilation where there's a change in which objects have which individual momenta, but total momentum must be conserved. It's also used in the more general form of the relation of energy to rest mass m and relativistic momentum p, given by the equation E^2 = (mc^2)^2 + (pc)^2, which reduces to the more well-known E=mc^2 in the special case where p=0. 

By the way, since relativistic momentum is given by p=mv/sqrt(1 - v^2/c^2), you can substitute this into the above equation to get E^2 = (m^2)(c^4) + (m^2)(v^2)(c^2)/(1 - v^2/c^2), and then if you take the first term on the right hand side, (m^2)(c^4), and multiply it by (1 - v^2/c^2)/(1 - v^2/c^2) and gather terms, you get E^2 = [(m^2)(c^4) - (m^2)(v^2)(c^2) + (m^2)(v^2)(c^2)]/(1 - v^2/c^2), and two terms cancel each other out so this simplifies to E^2 = (m^2)(c^4)/(1 - v^2/c^2), and then if you take the square root of both sides you get E = γmc^2. So the original equation for energy as a function fo rest mass m and relativistic momentum p can be rewritten as E=Mc^2 where M is the relativistic mass defined as M = γm, again showing that relativistic mass is only useful for rewriting equations involving relativistic momentum.

Jesse

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[Alan Grayson]

Firstly, copied below is what I posted earlier today. Because I wanted to contrast the Andromeda case with the TP, I used an SR solution for the latter, and you will note that the line segment paths of the inscribed polygon are inertial paths, and by infinitely refining the partition, I get the circular motion for the return path of the traveling twin. You will also note that the Earth-bound twin is at rest, and is analogous to the rest bound observer in the Andromeda case. In the TP, the Earth bound twin measures the traveling twin's clock running slower than his own clock, using the LT, but this effect is real for the traveling twin; otherwise he wouldn't see himself younger than his twin when they are juxtaposed upon his return. In contrast, you have the Andromeda traveler also at rest, and measuring the contracting distance in his frame, while the resting twin measures time dilation in his frame. However, in the frame of the moving rod representing the distance from Earth to Andromeda, according to your analysis the observer in that frame does NOT measure his length contracted;. only the rest frame measures the length contraction. IOW, the cases are similar except for the fact that one involves time dilation and other involves length contraction, but what is measured in the target frames is hugely different. This is the puzzle I am struggling with; namely, why is time dilation a measurable result for the traveling twin, but length contraction is NOT a measurable result for the frame of the moving rod in the Andromeda case, even though the frames doing the measuring in both cases have measurable results in their frames, but not in their respective moving frames? AG

PS for Clark; I am halfway through the video you posted. IMO, there can be several ways to solve a problem and acceleration is one legitimate way because there IS accelation for the traveling twin, and acceleration IS equivalent to gravity, and gravity DOES slow clocks. I will finish the video out of curiosity for the author's supposed solution, but it seems obvious that acceleration is one possible solution. AG

"Houston, we have a problem!" Now let's consider time dilation using SR in the Twin Paradox. Imagine the traveling twin moving in a circle and returning to Earth, and imagine the circle contains a polygon consisting of straight paths, which will later be infinitely partitioned, whose limit will be that circle. As measured by the stationary twin, the traveling twin's clock is dilated along each segment, so when the twins are juxtaposed, the traveling twin's elapsed time is LESS than clock readings for the stationary twin. If this is correct, it demostrates that what the stationary twin measures, is actually what the traveling twin's clock reads. IOW, what happens to time dilation in this case is OPPOSITE to what happens to the frames for the trip to Andromeda! Do you understand what I am alleging -- that length contraction acts in an opposite manner compared to time dilation, when I would expect them to behave similarly? AG 

 

About mass, since the measured mass grows exponentially to infinity as v --> c, isn't this derivable from the LT, but in which frame? AG 

[Jesse Mazer]

First, there's a habit in your writings that I find ambiguous: your use of the word "rest" and "moving" in an unqualified way, not clearly specifying that you are just speaking in a relative way about the rest frame of a particular object/observer in your thought-experiment. Sometimes this actually leaves me unclear on which frame you are actually talking about, at other times I can infer which one you're talking about but I wonder if you're trying to implicitly suggest that your thought-experiment shows that we must accept some notion of an objective truth about which observer is "really at rest", as opposed to the standard physicist's understanding that rest and motion can only be defined in relative terms.

So, can I ask that if you are just using these terms in a relative way, would you please always phrase it in a way that specifies whose rest frame you're talking about? (eg "the rocket's rest frame", "the Earth's rest frame" etc.) And second, if you *do* mean to make some argument that one of your thought-experiments suggests a concept of objective/absolute rest, in that case could you be explicit that you are making such an argument by talking about "absolute rest" or some similar term? Please let me know if you are willing to make this change to your way of writing before addressing my more specific questions below.

Now, when you say "the frame of the moving rod representing the distance from Earth to Andromeda" and "only the rest frame measures the length contraction", do you mean to introduce a rod at rest relative to Earth/Andromeda into the thought-experiment, and when you call this a "moving rod" you are talking about the perspective of the rest frame of the rocket which is in motion relative to Earth/Andromeda, and the rocket's rest frame is what you mean by "the rest frame"? If so, it is of course true that the rod would be contracted in the rocket's rest frame but not in the rod's own rest frame, and similarly true that the rocket would be contracted in the rod's rest frame but not in the rocket's own rest frame. This symmetry is similar to time dilation in that if you have two clocks A and B in inertial relative motion, in the rest frame of clock A it will be clock B that's running slow while clock A is running normally, and in the rest frame of clock B it will be clock A that's running slow while clock B is running normally.

Also when you say "You will also note that the Earth-bound twin is at rest" are you suggesting that the conclusion of this thought experiment is that the frame where Earth is at rest is more "correct" in some absolute sense, or just saying that this is what's true by convention if we analyze the problem from the perspective of the Earth rest frame? Would you agree or disagree that we could analyze the whole problem from the perspective of any other specific inertial frame, like a frame where the Earth is moving at 0.99c the whole time and the traveling twin is sometimes moving faster and sometimes moving slower during different sections of its non-inertial path, and we would get exactly the same answer to the question of how much each twin has aged at the moment they reunite at the same location (a 'local physical fact' in the sense I discussed before)?

Jesse

 

IOW, the cases are similar except for the fact that one involves time dilation and other involves length contraction, but what is measured in the target frames is hugely different. This is the puzzle I am struggling with; namely, why is time dilation a measurable result for the traveling twin, but length contraction is NOT a measurable result for the frame of the moving rod in the Andromeda case, even though the frames doing the measuring in both cases have measurable results in their frames, but not in their respective moving frames? AG

PS for Clark; I am halfway through the video you posted. IMO, there can be several ways to solve a problem and acceleration is one legitimate way because there IS accelation for the traveling twin, and acceleration IS equivalent to gravity, and gravity DOES slow clocks. I will finish the video out of curiosity for the author's supposed solution, but it seems obvious that acceleration is one possible solution. AG

"Houston, we have a problem!" Now let's consider time dilation using SR in the Twin Paradox. Imagine the traveling twin moving in a circle and returning to Earth, and imagine the circle contains a polygon consisting of straight paths, which will later be infinitely partitioned, whose limit will be that circle. As measured by the stationary twin, the traveling twin's clock is dilated along each segment, so when the twins are juxtaposed, the traveling twin's elapsed time is LESS than clock readings for the stationary twin. If this is correct, it demostrates that what the stationary twin measures, is actually what the traveling twin's clock reads. IOW, what happens to time dilation in this case is OPPOSITE to what happens to the frames for the trip to Andromeda! Do you understand what I am alleging -- that length contraction acts in an opposite manner compared to time dilation, when I would expect them to behave similarly? AG 

 

About mass, since the measured mass grows exponentially to infinity as v --> c, isn't this derivable from the LT, but in which frame? AG 

The notion of a variable relativistic mass is just an alternate way of talking about relativistic momentum, often modern textbooks talk solely about the latter and the only mass concept they use is the rest mass. For example the page at https://courses.lumenlearning.com/suny-physics/chapter/28-5-relativistic-momentum has a box titled "Misconception alert: relativistic mass and momentum" which says the following (note that they are using u to denote velocity):

"The relativistically correct definition of momentum as p = γmu is sometimes taken to imply that mass varies with velocity: m_var = γm, particularly in older textbooks. However, note that m is the mass of the object as measured by a person at rest relative to the object. Thus, m is defined to be the rest mass, which could be measured at rest, perhaps using gravity. When a mass is moving relative to an observer, the only way that its mass can be determined is through collisions or other means in which momentum is involved. Since the mass of a moving object cannot be determined independently of momentum, the only meaningful mass is rest mass. Thus, when we use the term mass, assume it to be identical to rest mass."

I'd say there's nothing strictly incorrect about defining a variable relativistic mass, it's just a cosmetically different formalism, but it may be that part of the reason it was mostly abandoned is because for people learning relativity it can lead to misconceptions that there is more to the concept than just a difference in how momentum is calculated, whereas in fact there is no application of relativistic mass that does not involve relativistic momentum. Momentum is needed for situations like collisions or particle creation/annihilation where there's a change in which objects have which individual momenta, but total momentum must be conserved. It's also used in the more general form of the relation of energy to rest mass m and relativistic momentum p, given by the equation E^2 = (mc^2)^2 + (pc)^2, which reduces to the more well-known E=mc^2 in the special case where p=0. 

By the way, since relativistic momentum is given by p=mv/sqrt(1 - v^2/c^2), you can substitute this into the above equation to get E^2 = (m^2)(c^4) + (m^2)(v^2)(c^2)/(1 - v^2/c^2), and then if you take the first term on the right hand side, (m^2)(c^4), and multiply it by (1 - v^2/c^2)/(1 - v^2/c^2) and gather terms, you get E^2 = [(m^2)(c^4) - (m^2)(v^2)(c^2) + (m^2)(v^2)(c^2)]/(1 - v^2/c^2), and two terms cancel each other out so this simplifies to E^2 = (m^2)(c^4)/(1 - v^2/c^2), and then if you take the square root of both sides you get E = γmc^2. So the original equation for energy as a function fo rest mass m and relativistic momentum p can be rewritten as E=Mc^2 where M is the relativistic mass defined as M = γm, again showing that relativistic mass is only useful for rewriting equations involving relativistic momentum.

Jesse

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[John Clark]

On Fri, Jan 17, 2025 at 12:31 PM Alan Grayson <agrays...@gmail.com> wrote:

>  I am halfway through the video you posted. IMO, there can be several ways to solve a problem and acceleration is one legitimate way because there IS accelation for the traveling twin, and acceleration IS equivalent to gravity, and gravity DOES slow clocks.

I may have missed a few but I think most of the diagrams posted so far in this thread have involved flat Minkowsky space, they show how clocks and measuring sticks would behave if you were in deep space far from any gravitational field, but gravity curves spacetime.  Special Relativity can handle the acceleration of a rocket if it's well beyond the orbit of Neptune, but not if it's close to the sun. Special relativity can't handle gravity because that involves curved spacetime, and for that you need General Relativity. Special Relativity has no explanation for why gravity causes things to move the way that they do, but General Relativity does.    

> I will finish the video out of curiosity for the author's supposed solution, but it seems obvious that acceleration is one possible solution.

If you watch the rest of the video you'll see a modification of the standard twin "paradox" that does not involve acceleration but produces the same odd situation. 

John K Clark    See what's on my new list at  Extropolis

7c1 

[Alan Grayson]

When I refer to the non-traveling twin in the TP, do you find this ambiguous? Do you find the traveling twin's frame ambiguous? If you do, I don't how I can be more specific. Concerning the Andromeda frames, there's a frame with a moving rod, representing the distance between the Earth and Andromeda, and the frame of an observer using this frame to determine the length contraction. I really don't know how I can be more specific. AG

[Alan Grayson]

On Friday, January 17, 2025 at 11:32:57 AM UTC-7 John Clark wrote:

On Fri, Jan 17, 2025 at 12:31 PM Alan Grayson <agrays...@gmail.com> wrote:

>  I am halfway through the video you posted. IMO, there can be several ways to solve a problem and acceleration is one legitimate way because there IS accelation for the traveling twin, and acceleration IS equivalent to gravity, and gravity DOES slow clocks.

I may have missed a few but I think most of the diagrams posted so far in this thread have involved flat Minkowsky space, they show how clocks and measuring sticks would behave if you were in deep space far from any gravitational field, but gravity curves spacetime.  Special Relativity can handle the acceleration of a rocket if it's well beyond the orbit of Neptune, but not if it's close to the sun. Special relativity can't handle gravity because that involves curved spacetime, and for that you need General Relativity. Special Relativity has no explanation for why gravity causes things to move the way that they do, but General Relativity does.    

IMO SR can handle curved spacetime. All one has to do is make the partitions very fine, so we're approximating inertial motion along very short paths. AG 

[Jesse Mazer]

On Fri, Jan 17, 2025 at 1:41 PM Alan Grayson <agrays...@gmail.com> wrote:

On Friday, January 17, 2025 at 11:32:57 AM UTC-7 John Clark wrote:

On Fri, Jan 17, 2025 at 12:31 PM Alan Grayson <agrays...@gmail.com> wrote:

>  I am halfway through the video you posted. IMO, there can be several ways to solve a problem and acceleration is one legitimate way because there IS accelation for the traveling twin, and acceleration IS equivalent to gravity, and gravity DOES slow clocks.

I may have missed a few but I think most of the diagrams posted so far in this thread have involved flat Minkowsky space, they show how clocks and measuring sticks would behave if you were in deep space far from any gravitational field, but gravity curves spacetime.  Special Relativity can handle the acceleration of a rocket if it's well beyond the orbit of Neptune, but not if it's close to the sun. Special relativity can't handle gravity because that involves curved spacetime, and for that you need General Relativity. Special Relativity has no explanation for why gravity causes things to move the way that they do, but General Relativity does.    

IMO SR can handle curved spacetime. All one has to do is make the partitions very fine, so we're approximating inertial motion along very short paths. AG 

Curved spacetime is different from curved worldlines on flat spacetime--SR can deal with the latter but not the former. Think of the difference between curved paths in a flat 2D plane (Euclidean geometry) vs. paths on the curved 2D surface of a sphere (non-Euclidean geometry).

Jesse

 

> I will finish the video out of curiosity for the author's supposed solution, but it seems obvious that acceleration is one possible solution.

If you watch the rest of the video you'll see a modification of the standard twin "paradox" that does not involve acceleration but produces the same odd situation. 

John K Clark    See what's on my new list at  Extropolis

7c1 

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[Jesse Mazer]

It's usually understood that "non-traveling twin" just means the twin that moves inertially between the two meetings, and "traveling twin" means the one that changed velocities at least once on its path between the meetings. If you mean something different, like the idea that there is some objective/absolute sense that the "non-traveling twin" is at rest rather than moving, then I would object to that. But as stated, those phrases don't involve the word "moving" or "at rest" without qualification, which is what I was objecting to in my comments above. Will you agree in future to specify what object/observer "rest" and "moving" are relative to if you mean them in a relative way (which can easily be specified with a phrase like 'moving relative to [some object]' or 'at rest relative to [some object]'), or else to explicitly use some phrase like "absolute rest" and "absolute movement" if you mean them in a non-relative way?

 

Concerning the Andromeda frames, there's a frame with a moving rod, representing the distance between the Earth and Andromeda,

"Moving" in an absolute or relative sense? If in a relative sense, moving relative to what? Are you talking about a rod which is at rest relative to Earth and Andromeda, and moving relative to the rocket?

 

and the frame of an observer using this frame to determine the length contraction.

Are you talking about an observer on the rocket which is moving relative to Earth/Andromeda, using his own rest frame to determine the length contraction of the rod which is at rest relative to Earth/Andromeda?

Jesse

 

 

IOW, the cases are similar except for the fact that one involves time dilation and other involves length contraction, but what is measured in the target frames is hugely different. This is the puzzle I am struggling with; namely, why is time dilation a measurable result for the traveling twin, but length contraction is NOT a measurable result for the frame of the moving rod in the Andromeda case, even though the frames doing the measuring in both cases have measurable results in their frames, but not in their respective moving frames? AG

PS for Clark; I am halfway through the video you posted. IMO, there can be several ways to solve a problem and acceleration is one legitimate way because there IS accelation for the traveling twin, and acceleration IS equivalent to gravity, and gravity DOES slow clocks. I will finish the video out of curiosity for the author's supposed solution, but it seems obvious that acceleration is one possible solution. AG

"Houston, we have a problem!" Now let's consider time dilation using SR in the Twin Paradox. Imagine the traveling twin moving in a circle and returning to Earth, and imagine the circle contains a polygon consisting of straight paths, which will later be infinitely partitioned, whose limit will be that circle. As measured by the stationary twin, the traveling twin's clock is dilated along each segment, so when the twins are juxtaposed, the traveling twin's elapsed time is LESS than clock readings for the stationary twin. If this is correct, it demostrates that what the stationary twin measures, is actually what the traveling twin's clock reads. IOW, what happens to time dilation in this case is OPPOSITE to what happens to the frames for the trip to Andromeda! Do you understand what I am alleging -- that length contraction acts in an opposite manner compared to time dilation, when I would expect them to behave similarly? AG 

 

About mass, since the measured mass grows exponentially to infinity as v --> c, isn't this derivable from the LT, but in which frame? AG 

The notion of a variable relativistic mass is just an alternate way of talking about relativistic momentum, often modern textbooks talk solely about the latter and the only mass concept they use is the rest mass. For example the page at https://courses.lumenlearning.com/suny-physics/chapter/28-5-relativistic-momentum has a box titled "Misconception alert: relativistic mass and momentum" which says the following (note that they are using u to denote velocity):

"The relativistically correct definition of momentum as p = γmu is sometimes taken to imply that mass varies with velocity: m_var = γm, particularly in older textbooks. However, note that m is the mass of the object as measured by a person at rest relative to the object. Thus, m is defined to be the rest mass, which could be measured at rest, perhaps using gravity. When a mass is moving relative to an observer, the only way that its mass can be determined is through collisions or other means in which momentum is involved. Since the mass of a moving object cannot be determined independently of momentum, the only meaningful mass is rest mass. Thus, when we use the term mass, assume it to be identical to rest mass."

I'd say there's nothing strictly incorrect about defining a variable relativistic mass, it's just a cosmetically different formalism, but it may be that part of the reason it was mostly abandoned is because for people learning relativity it can lead to misconceptions that there is more to the concept than just a difference in how momentum is calculated, whereas in fact there is no application of relativistic mass that does not involve relativistic momentum. Momentum is needed for situations like collisions or particle creation/annihilation where there's a change in which objects have which individual momenta, but total momentum must be conserved. It's also used in the more general form of the relation of energy to rest mass m and relativistic momentum p, given by the equation E^2 = (mc^2)^2 + (pc)^2, which reduces to the more well-known E=mc^2 in the special case where p=0. 

By the way, since relativistic momentum is given by p=mv/sqrt(1 - v^2/c^2), you can substitute this into the above equation to get E^2 = (m^2)(c^4) + (m^2)(v^2)(c^2)/(1 - v^2/c^2), and then if you take the first term on the right hand side, (m^2)(c^4), and multiply it by (1 - v^2/c^2)/(1 - v^2/c^2) and gather terms, you get E^2 = [(m^2)(c^4) - (m^2)(v^2)(c^2) + (m^2)(v^2)(c^2)]/(1 - v^2/c^2), and two terms cancel each other out so this simplifies to E^2 = (m^2)(c^4)/(1 - v^2/c^2), and then if you take the square root of both sides you get E = γmc^2. So the original equation for energy as a function fo rest mass m and relativistic momentum p can be rewritten as E=Mc^2 where M is the relativistic mass defined as M = γm, again showing that relativistic mass is only useful for rewriting equations involving relativistic momentum.

Jesse

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[Brent Meeker]

No I don't understand what you're alleging, nor what "moving in a circle and returning to Earth" refers to.  But yes length contraction and time dilation go together, that's what makes the speed of light the same in all frames.

Brent

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[Alan Grayson]

"Houston, we have a problem!" Now let's consider time dilation using SR in the Twin Paradox. Imagine the traveling twin moving in a circle and returning to Earth, and imagine the circle contains a polygon consisting of straight paths, which will later be infinitely partitioned, whose limitbse will be that circle. As measured by the stationary twin, the traveling twin's clock is dilated along each segment, so when the twins are juxtaposed, the traveling twin's elapsed time is LESS than clock readings for the stationary twin. If this is correct, it demostrates that what the stationary twin measures, is actually what the traveling twin's clock reads. IOW, what happens to time dilation in this case is OPPOSITE to what happens to the frames for the trip to Andromeda! Do you understand what I am alleging -- that length contraction acts in an opposite manner compared to time dilation, when I would expect them to behave similarly? AG

No I don't understand what you're alleging, nor what "moving in a circle and returning to Earth" refers to. 

It's a model of the path of the traveling twin in the TP. I wanted to use SR, so I needed the path to be composed of segments where there is only inertial motion. So I used a circle with an inscribed polygon, and then, as in calculus, I imagined this partition as infinitely fine, to approach the circle for the round-trip path. I then noted that from the pov of the stationary twin, time is dilated on those straight line inertial segments, so the traveling twin ages slower then the stationary twin. Note that in this situation not only does the stationary twin observe time dilation, but the traveling twin's clocks actually slows down. Otherwise the traveling twin won't be younger when the twins juxtaposed. But much more important, when considering the Andromeda case, the traveling observer (traveling with respect to the Earth) can be assumed to be at rest, and the frame of the rod representing the distance from Earth to Andromeda, can be assumed to be moving. So this situation mirrors the TP,  since now the moving frame containing the rod is analogous to the traveling twin, and the rest frame whose observer is observing the moving rod, is analogous to the stationary twin, the only difference is that now the Andromeda case is calculating length contraction, whereas the TP case is calculating time dilation. So what's the point of all this -- simply that the traveling twin's clock physically slows, whereas length contraction is NOT measured in the frame of the moving rod in the Andromeda case. I think this is a problem, that the frame containing the rod, does not manifest length contraction similar to the TP case, where the traveling twin's clock actually slows down. AG

 

But yes length contraction and time dilation go together, that's what makes the speed of light the same in all frames.

Brent

72bf93a9e508n%40googlegroups.com.

[Alan Grayson]

The non-traveling twin is at rest on the Earth throughout. I never heard of any other concept of the non-traveling twin. The traveling twin is moving with respect to the Earth. I never heard of any other concept of the traveling twin. AG

 

If you mean something different, like the idea that there is some objective/absolute sense that the "non-traveling twin" is at rest rather than moving, then I would object to that. But as stated, those phrases don't involve the word "moving" or "at rest" without qualification, which is what I was objecting to in my comments above. Will you agree in future to specify what object/observer "rest" and "moving" are relative to if you mean them in a relative way (which can easily be specified with a phrase like 'moving relative to [some object]' or 'at rest relative to [some object]'), or else to explicitly use some phrase like "absolute rest" and "absolute movement" if you mean them in a non-relative way?

No absolute anything. All motion is relative to something. AG 

 

Concerning the Andromeda frames, there's a frame with a moving rod, representing the distance between the Earth and Andromeda,

"Moving" in an absolute or relative sense? If in a relative sense, moving relative to what? Are you talking about a rod which is at rest relative to Earth and Andromeda, and moving relative to the rocket?

The observer in Andromeda case is traveling, moving with respect to the Earth. Then we can assume this observer is at rest, relative to a moving rod which represents the distance from Earth to Andromeda. AG  

 

and the frame of an observer using this frame to determine the length contraction.

Are you talking about an observer on the rocket which is moving relative to Earth/Andromeda,

Yes. AG

 

using his own rest frame to determine the length contraction of the rod which is at rest relative to Earth/Andromeda?

The rod is moving, the observer is stationary in his rest frame, from which he calculates the length contraction. AG 

[Jesse Mazer]

All that's important to the twin paradox is that one twin is inertial and the other is not. For example if twin A is moving away from Earth inertially the whole time, and twin B is at rest relative to Earth for a while and then accelerates to catch up with twin A, then twin A is the inertial twin and twin B is the non-inertial one during the time between their meetings, and so it's guaranteed that twin B will have aged less than twin A when they reunite.

 

 

If you mean something different, like the idea that there is some objective/absolute sense that the "non-traveling twin" is at rest rather than moving, then I would object to that. But as stated, those phrases don't involve the word "moving" or "at rest" without qualification, which is what I was objecting to in my comments above. Will you agree in future to specify what object/observer "rest" and "moving" are relative to if you mean them in a relative way (which can easily be specified with a phrase like 'moving relative to [some object]' or 'at rest relative to [some object]'), or else to explicitly use some phrase like "absolute rest" and "absolute movement" if you mean them in a non-relative way?

No absolute anything. All motion is relative to something. AG 

OK, then are you willing to alter your way of writing about these things to prevent ambiguity, to always use phrases like "moving relative to [object]" or "at rest relative to [object]" to specify the relative motion/rest you are thinking of?

 

 

Concerning the Andromeda frames, there's a frame with a moving rod, representing the distance between the Earth and Andromeda,

"Moving" in an absolute or relative sense? If in a relative sense, moving relative to what? Are you talking about a rod which is at rest relative to Earth and Andromeda, and moving relative to the rocket?

The observer in Andromeda case is traveling, moving with respect to the Earth. Then we can assume this observer is at rest, relative to a moving rod which represents the distance from Earth to Andromeda. AG  

 

and the frame of an observer using this frame to determine the length contraction.

Are you talking about an observer on the rocket which is moving relative to Earth/Andromeda,

Yes. AG

 

using his own rest frame to determine the length contraction of the rod which is at rest relative to Earth/Andromeda?

The rod is moving, the observer is stationary in his rest frame, from which he calculates the length contraction. AG 

The rod is moving relative to this observer A, and is thus contracted in A's frame. And likewise if we assume the observer A is on board a rocket as I suggested, the rocket is moving relative to an observer B who is at rest relative to Andromeda, and so the rocket is contracted in the frame of observer B. So length contraction is completely symmetric between inertial frames, as is time dilation--are you saying otherwise?

Jesse

 

 

 

IOW, the cases are similar except for the fact that one involves time dilation and other involves length contraction, but what is measured in the target frames is hugely different. This is the puzzle I am struggling with; namely, why is time dilation a measurable result for the traveling twin, but length contraction is NOT a measurable result for the frame of the moving rod in the Andromeda case, even though the frames doing the measuring in both cases have measurable results in their frames, but not in their respective moving frames? AG

PS for Clark; I am halfway through the video you posted. IMO, there can be several ways to solve a problem and acceleration is one legitimate way because there IS accelation for the traveling twin, and acceleration IS equivalent to gravity, and gravity DOES slow clocks. I will finish the video out of curiosity for the author's supposed solution, but it seems obvious that acceleration is one possible solution. AG

"Houston, we have a problem!" Now let's consider time dilation using SR in the Twin Paradox. Imagine the traveling twin moving in a circle and returning to Earth, and imagine the circle contains a polygon consisting of straight paths, which will later be infinitely partitioned, whose limit will be that circle. As measured by the stationary twin, the traveling twin's clock is dilated along each segment, so when the twins are juxtaposed, the traveling twin's elapsed time is LESS than clock readings for the stationary twin. If this is correct, it demostrates that what the stationary twin measures, is actually what the traveling twin's clock reads. IOW, what happens to time dilation in this case is OPPOSITE to what happens to the frames for the trip to Andromeda! Do you understand what I am alleging -- that length contraction acts in an opposite manner compared to time dilation, when I would expect them to behave similarly? AG 

 

About mass, since the measured mass grows exponentially to infinity as v --> c, isn't this derivable from the LT, but in which frame? AG 

The notion of a variable relativistic mass is just an alternate way of talking about relativistic momentum, often modern textbooks talk solely about the latter and the only mass concept they use is the rest mass. For example the page at https://courses.lumenlearning.com/suny-physics/chapter/28-5-relativistic-momentum has a box titled "Misconception alert: relativistic mass and momentum" which says the following (note that they are using u to denote velocity):

"The relativistically correct definition of momentum as p = γmu is sometimes taken to imply that mass varies with velocity: m_var = γm, particularly in older textbooks. However, note that m is the mass of the object as measured by a person at rest relative to the object. Thus, m is defined to be the rest mass, which could be measured at rest, perhaps using gravity. When a mass is moving relative to an observer, the only way that its mass can be determined is through collisions or other means in which momentum is involved. Since the mass of a moving object cannot be determined independently of momentum, the only meaningful mass is rest mass. Thus, when we use the term mass, assume it to be identical to rest mass."

I'd say there's nothing strictly incorrect about defining a variable relativistic mass, it's just a cosmetically different formalism, but it may be that part of the reason it was mostly abandoned is because for people learning relativity it can lead to misconceptions that there is more to the concept than just a difference in how momentum is calculated, whereas in fact there is no application of relativistic mass that does not involve relativistic momentum. Momentum is needed for situations like collisions or particle creation/annihilation where there's a change in which objects have which individual momenta, but total momentum must be conserved. It's also used in the more general form of the relation of energy to rest mass m and relativistic momentum p, given by the equation E^2 = (mc^2)^2 + (pc)^2, which reduces to the more well-known E=mc^2 in the special case where p=0. 

By the way, since relativistic momentum is given by p=mv/sqrt(1 - v^2/c^2), you can substitute this into the above equation to get E^2 = (m^2)(c^4) + (m^2)(v^2)(c^2)/(1 - v^2/c^2), and then if you take the first term on the right hand side, (m^2)(c^4), and multiply it by (1 - v^2/c^2)/(1 - v^2/c^2) and gather terms, you get E^2 = [(m^2)(c^4) - (m^2)(v^2)(c^2) + (m^2)(v^2)(c^2)]/(1 - v^2/c^2), and two terms cancel each other out so this simplifies to E^2 = (m^2)(c^4)/(1 - v^2/c^2), and then if you take the square root of both sides you get E = γmc^2. So the original equation for energy as a function fo rest mass m and relativistic momentum p can be rewritten as E=Mc^2 where M is the relativistic mass defined as M = γm, again showing that relativistic mass is only useful for rewriting equations involving relativistic momentum.

Jesse

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[Alan Grayson]

On and off over many years I've read about the TP. Never, not once, have I read it described as you do. Moreover, for the moving observer to leave and return, it's impossible for that observer to be totally inertial. If you model any observer leaving Earth, that observer cannot be inertial. AG  

 

If you mean something different, like the idea that there is some objective/absolute sense that the "non-traveling twin" is at rest rather than moving, then I would object to that. But as stated, those phrases don't involve the word "moving" or "at rest" without qualification, which is what I was objecting to in my comments above. Will you agree in future to specify what object/observer "rest" and "moving" are relative to if you mean them in a relative way (which can easily be specified with a phrase like 'moving relative to [some object]' or 'at rest relative to [some object]'), or else to explicitly use some phrase like "absolute rest" and "absolute movement" if you mean them in a non-relative way?

No absolute anything. All motion is relative to something. AG 

OK, then are you willing to alter your way of writing about these things to prevent ambiguity, to always use phrases like "moving relative to [object]" or "at rest relative to [object]" to specify the relative motion/rest you are thinking of?

 

 

Concerning the Andromeda frames, there's a frame with a moving rod, representing the distance between the Earth and Andromeda,

"Moving" in an absolute or relative sense? If in a relative sense, moving relative to what? Are you talking about a rod which is at rest relative to Earth and Andromeda, and moving relative to the rocket?

The observer in Andromeda case is traveling, moving with respect to the Earth. Then we can assume this observer is at rest, relative to a moving rod which represents the distance from Earth to Andromeda. AG  

 

and the frame of an observer using this frame to determine the length contraction.

Are you talking about an observer on the rocket which is moving relative to Earth/Andromeda,

Yes. AG

 

using his own rest frame to determine the length contraction of the rod which is at rest relative to Earth/Andromeda?

The rod is moving, the observer is stationary in his rest frame, from which he calculates the length contraction. AG 

The rod is moving relative to this observer A, and is thus contracted in A's frame.

 

But not, at that time, also contracted in B frame. This is different from the TP where time is dilated in frame B, the frame of traveling twin. IOW, using the LT in the TP, what is measured from the Earth or stationary frame, is what's actually measured in the moving frame. Not so in Andromeda problem. AG

[Alan Grayson]

Let me clarify the problem I'm trying to resolve; notice the T in LT. It stands for Transformation, presumably from one frame to another. It's claimed that the LT will produce the measured result in the target frame, based on parameters of the source frame for the transformation. And this seems to be the case in the TP; from the source frame, the frame at rest on the Earth, the LT tells us what will be measured in the traveling frame. And it seems to do just that, since the clock in the traveling frame actually ticks slower as predicted. If it didn't, the traveling observer would not age slower than the stationary observer. But when we consider the Andromeda problem, the LT seems NOT to predict what the frame of the moving rod will measure. Maybe it does, as you indicated, but only when a measurement is taken, unlike the case of the TP, where the result seems inherent, and not requiring a measurement.  AG 

[Alan Grayson]

The inconsistency I see is that measurements done in the frame of the moving rod do NOT show that the rod in its frame is contracted, even though this is ostensibly the prediction of the LT, from the pov of the stationary frame. This is different compared to what happens in the frames of the TP, where the predicted slowing of the traveling twin's clock is immediately manifested in the frame of the traveling twin. Do you understand what puzzles me?  AG

[Jesse Mazer]

You've agreed in the past that there's no problem analyzing these types of problems with idealizations like instantaneous acceleration (which contributes nothing to the elapsed time), so we can just imagine one twin instantaneously accelerating away from Earth and move inertially immediately after, and only analyze the problem from the moment of the instantaneous acceleration onwards, after which that twin moves inertially with constant velocity while the other twin is initially at rest on Earth and then later accelerates to catch up. 

For example say the inertial twin instantaneously accelerates to 0.6c relative to the Earth, then the non-inertial twin who's initially on Earth waits 4 years before accelerating instantaneously so that the inertial twin now measures them to be approaching at 0.6c, and the non-inertial twin experiences another 4 years before catching up. It makes no difference whether you analyze this problem from the perspective of a frame where the Earth is at rest and the inertial twin is moving at 0.6c, or a frame where the inertial twin is at rest and the Earth is moving at 0.6c--either way you get the same answer that when the twins reunite, the non-inertial twin has aged 8 years and the inertial one has aged 10. And of course it makes no difference to the problem if we keep all the relative velocities and times the same between the moment of departure and the moment they reunite, but change the wording of the problem so the inertial twin is at rest on Earth the whole time--this change in wording makes no difference to the math of the problem (the bottom of the page at https://www.askamathematician.com/2010/09/q-how-does-the-twin-paradox-work/comment-page-1/ includes spacetime diagrams drawn both in the frame where the inertial twin is at rest and drawn in the frame where the inertial twin is moving at 0.6c the whole time). Do you disagree?

 

 

If you mean something different, like the idea that there is some objective/absolute sense that the "non-traveling twin" is at rest rather than moving, then I would object to that. But as stated, those phrases don't involve the word "moving" or "at rest" without qualification, which is what I was objecting to in my comments above. Will you agree in future to specify what object/observer "rest" and "moving" are relative to if you mean them in a relative way (which can easily be specified with a phrase like 'moving relative to [some object]' or 'at rest relative to [some object]'), or else to explicitly use some phrase like "absolute rest" and "absolute movement" if you mean them in a non-relative way?

No absolute anything. All motion is relative to something. AG 

OK, then are you willing to alter your way of writing about these things to prevent ambiguity, to always use phrases like "moving relative to [object]" or "at rest relative to [object]" to specify the relative motion/rest you are thinking of?

Could you answer this question about whether you're willing to do this to prevent ambiguities?

 

 

 

Concerning the Andromeda frames, there's a frame with a moving rod, representing the distance between the Earth and Andromeda,

"Moving" in an absolute or relative sense? If in a relative sense, moving relative to what? Are you talking about a rod which is at rest relative to Earth and Andromeda, and moving relative to the rocket?

The observer in Andromeda case is traveling, moving with respect to the Earth. Then we can assume this observer is at rest, relative to a moving rod which represents the distance from Earth to Andromeda. AG  

 

and the frame of an observer using this frame to determine the length contraction.

Are you talking about an observer on the rocket which is moving relative to Earth/Andromeda,

Yes. AG

 

using his own rest frame to determine the length contraction of the rod which is at rest relative to Earth/Andromeda?

The rod is moving, the observer is stationary in his rest frame, from which he calculates the length contraction. AG 

The rod is moving relative to this observer A, and is thus contracted in A's frame.

 

But not, at that time, also contracted in B frame. This is different from the TP where time is dilated in frame B, the frame of traveling twin.

The traveling twin in the twin paradox does not have a single inertial frame, so you'll have to be more specific--in the case where the traveling twin's path consists of two inertial segments joined by an instantaneous acceleration, would frame B be the one where the traveling twin is at rest during the first ('outbound') segment, or would frame B be the one where the twin is at rest during the second ('inbound') segment? In the first case, the traveling twin's clock shows no time dilation in the B frame during the outbound segment, since the clock is at rest in that segment; in the second case, the traveling twin's clock shows no time dilation in the B frame during the inbound segment, since the clock is at rest in that segment. Either way, it's directly analogous to the case of the rod, since a rod shows no contraction in an inertial frame where it's at rest, and a clock shows no time dilation in an inertial frame where it's at rest.

 

IOW, using the LT in the TP, what is measured from the Earth or stationary frame, is what's actually measured in the moving frame. Not so in Andromeda problem. AG

Again you are using "moving frame" in an unqualified way, if you mean "the frame of the rod" can you just say that? If we start with the coordinates of everything in the rocket frame and use the LT to transform into the frame of the rod, we get correct answers about what's observed in the frame of the rod, including the fact that the rod is longer in this frame than it was in the rocket frame. Are you disagreeing with that?

Jesse

 

 

And likewise if we assume the observer A is on board a rocket as I suggested, the rocket is moving relative to an observer B who is at rest relative to Andromeda, and so the rocket is contracted in the frame of observer B. So length contraction is completely symmetric between inertial frames, as is time dilation--are you saying otherwise?

Jesse

 

 

 

IOW, the cases are similar except for the fact that one involves time dilation and other involves length contraction, but what is measured in the target frames is hugely different. This is the puzzle I am struggling with; namely, why is time dilation a measurable result for the traveling twin, but length contraction is NOT a measurable result for the frame of the moving rod in the Andromeda case, even though the frames doing the measuring in both cases have measurable results in their frames, but not in their respective moving frames? AG

PS for Clark; I am halfway through the video you posted. IMO, there can be several ways to solve a problem and acceleration is one legitimate way because there IS accelation for the traveling twin, and acceleration IS equivalent to gravity, and gravity DOES slow clocks. I will finish the video out of curiosity for the author's supposed solution, but it seems obvious that acceleration is one possible solution. AG

"Houston, we have a problem!" Now let's consider time dilation using SR in the Twin Paradox. Imagine the traveling twin moving in a circle and returning to Earth, and imagine the circle contains a polygon consisting of straight paths, which will later be infinitely partitioned, whose limit will be that circle. As measured by the stationary twin, the traveling twin's clock is dilated along each segment, so when the twins are juxtaposed, the traveling twin's elapsed time is LESS than clock readings for the stationary twin. If this is correct, it demostrates that what the stationary twin measures, is actually what the traveling twin's clock reads. IOW, what happens to time dilation in this case is OPPOSITE to what happens to the frames for the trip to Andromeda! Do you understand what I am alleging -- that length contraction acts in an opposite manner compared to time dilation, when I would expect them to behave similarly? AG 

 

About mass, since the measured mass grows exponentially to infinity as v --> c, isn't this derivable from the LT, but in which frame? AG 

The notion of a variable relativistic mass is just an alternate way of talking about relativistic momentum, often modern textbooks talk solely about the latter and the only mass concept they use is the rest mass. For example the page at https://courses.lumenlearning.com/suny-physics/chapter/28-5-relativistic-momentum has a box titled "Misconception alert: relativistic mass and momentum" which says the following (note that they are using u to denote velocity):

"The relativistically correct definition of momentum as p = γmu is sometimes taken to imply that mass varies with velocity: m_var = γm, particularly in older textbooks. However, note that m is the mass of the object as measured by a person at rest relative to the object. Thus, m is defined to be the rest mass, which could be measured at rest, perhaps using gravity. When a mass is moving relative to an observer, the only way that its mass can be determined is through collisions or other means in which momentum is involved. Since the mass of a moving object cannot be determined independently of momentum, the only meaningful mass is rest mass. Thus, when we use the term mass, assume it to be identical to rest mass."

I'd say there's nothing strictly incorrect about defining a variable relativistic mass, it's just a cosmetically different formalism, but it may be that part of the reason it was mostly abandoned is because for people learning relativity it can lead to misconceptions that there is more to the concept than just a difference in how momentum is calculated, whereas in fact there is no application of relativistic mass that does not involve relativistic momentum. Momentum is needed for situations like collisions or particle creation/annihilation where there's a change in which objects have which individual momenta, but total momentum must be conserved. It's also used in the more general form of the relation of energy to rest mass m and relativistic momentum p, given by the equation E^2 = (mc^2)^2 + (pc)^2, which reduces to the more well-known E=mc^2 in the special case where p=0. 

By the way, since relativistic momentum is given by p=mv/sqrt(1 - v^2/c^2), you can substitute this into the above equation to get E^2 = (m^2)(c^4) + (m^2)(v^2)(c^2)/(1 - v^2/c^2), and then if you take the first term on the right hand side, (m^2)(c^4), and multiply it by (1 - v^2/c^2)/(1 - v^2/c^2) and gather terms, you get E^2 = [(m^2)(c^4) - (m^2)(v^2)(c^2) + (m^2)(v^2)(c^2)]/(1 - v^2/c^2), and two terms cancel each other out so this simplifies to E^2 = (m^2)(c^4)/(1 - v^2/c^2), and then if you take the square root of both sides you get E = γmc^2. So the original equation for energy as a function fo rest mass m and relativistic momentum p can be rewritten as E=Mc^2 where M is the relativistic mass defined as M = γm, again showing that relativistic mass is only useful for rewriting equations involving relativistic momentum.

Jesse

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[Jesse Mazer]

See my point above about the need to specify which of the two segments of the non-inertial twin's path you mean when you refer to "the traveling frame"--the LT only deals with inertial frames, there is no inertial frame where the non-inertial twin is at rest during the entire journey! And if for example by "traveling frame" you mean the frame where the non-inertial clock is at rest during the outbound leg, you seem to have gotten things completely backwards here--the LT would *not* predict the non-inertial clock is ticking slower in this frame during the period where it's at rest in this frame, instead it would predict the non-inertial clock shows no time dilation during this leg of the trip, while the clock of the inertial twin is ticking slower.

Jesse

 

If it didn't, the traveling observer would not age slower than the stationary observer. But when we consider the Andromeda problem, the LT seems NOT to predict what the frame of the moving rod will measure. Maybe it does, as you indicated, but only when a measurement is taken, unlike the case of the TP, where the result seems inherent, and not requiring a measurement.  AG 

 

And likewise if we assume the observer A is on board a rocket as I suggested, the rocket is moving relative to an observer B who is at rest relative to Andromeda, and so the rocket is contracted in the frame of observer B. So length contraction is completely symmetric between inertial frames, as is time dilation--are you saying otherwise?

Jesse

 

 

 

IOW, the cases are similar except for the fact that one involves time dilation and other involves length contraction, but what is measured in the target frames is hugely different. This is the puzzle I am struggling with; namely, why is time dilation a measurable result for the traveling twin, but length contraction is NOT a measurable result for the frame of the moving rod in the Andromeda case, even though the frames doing the measuring in both cases have measurable results in their frames, but not in their respective moving frames? AG

PS for Clark; I am halfway through the video you posted. IMO, there can be several ways to solve a problem and acceleration is one legitimate way because there IS accelation for the traveling twin, and acceleration IS equivalent to gravity, and gravity DOES slow clocks. I will finish the video out of curiosity for the author's supposed solution, but it seems obvious that acceleration is one possible solution. AG

"Houston, we have a problem!" Now let's consider time dilation using SR in the Twin Paradox. Imagine the traveling twin moving in a circle and returning to Earth, and imagine the circle contains a polygon consisting of straight paths, which will later be infinitely partitioned, whose limit will be that circle. As measured by the stationary twin, the traveling twin's clock is dilated along each segment, so when the twins are juxtaposed, the traveling twin's elapsed time is LESS than clock readings for the stationary twin. If this is correct, it demostrates that what the stationary twin measures, is actually what the traveling twin's clock reads. IOW, what happens to time dilation in this case is OPPOSITE to what happens to the frames for the trip to Andromeda! Do you understand what I am alleging -- that length contraction acts in an opposite manner compared to time dilation, when I would expect them to behave similarly? AG 

 

About mass, since the measured mass grows exponentially to infinity as v --> c, isn't this derivable from the LT, but in which frame? AG 

The notion of a variable relativistic mass is just an alternate way of talking about relativistic momentum, often modern textbooks talk solely about the latter and the only mass concept they use is the rest mass. For example the page at https://courses.lumenlearning.com/suny-physics/chapter/28-5-relativistic-momentum has a box titled "Misconception alert: relativistic mass and momentum" which says the following (note that they are using u to denote velocity):

"The relativistically correct definition of momentum as p = γmu is sometimes taken to imply that mass varies with velocity: m_var = γm, particularly in older textbooks. However, note that m is the mass of the object as measured by a person at rest relative to the object. Thus, m is defined to be the rest mass, which could be measured at rest, perhaps using gravity. When a mass is moving relative to an observer, the only way that its mass can be determined is through collisions or other means in which momentum is involved. Since the mass of a moving object cannot be determined independently of momentum, the only meaningful mass is rest mass. Thus, when we use the term mass, assume it to be identical to rest mass."

I'd say there's nothing strictly incorrect about defining a variable relativistic mass, it's just a cosmetically different formalism, but it may be that part of the reason it was mostly abandoned is because for people learning relativity it can lead to misconceptions that there is more to the concept than just a difference in how momentum is calculated, whereas in fact there is no application of relativistic mass that does not involve relativistic momentum. Momentum is needed for situations like collisions or particle creation/annihilation where there's a change in which objects have which individual momenta, but total momentum must be conserved. It's also used in the more general form of the relation of energy to rest mass m and relativistic momentum p, given by the equation E^2 = (mc^2)^2 + (pc)^2, which reduces to the more well-known E=mc^2 in the special case where p=0. 

By the way, since relativistic momentum is given by p=mv/sqrt(1 - v^2/c^2), you can substitute this into the above equation to get E^2 = (m^2)(c^4) + (m^2)(v^2)(c^2)/(1 - v^2/c^2), and then if you take the first term on the right hand side, (m^2)(c^4), and multiply it by (1 - v^2/c^2)/(1 - v^2/c^2) and gather terms, you get E^2 = [(m^2)(c^4) - (m^2)(v^2)(c^2) + (m^2)(v^2)(c^2)]/(1 - v^2/c^2), and two terms cancel each other out so this simplifies to E^2 = (m^2)(c^4)/(1 - v^2/c^2), and then if you take the square root of both sides you get E = γmc^2. So the original equation for energy as a function fo rest mass m and relativistic momentum p can be rewritten as E=Mc^2 where M is the relativistic mass defined as M = γm, again showing that relativistic mass is only useful for rewriting equations involving relativistic momentum.

Jesse

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[Brent Meeker]

No it's not.  The traveling twin is like a point particle that traces  out a path thru spacetime.  Since the path is closed in space (a circle) it's endpoints can be joined by an inertial (Earth bound) twin.  None of this is true of a rod from Earth to Andromeda.  The rod doesn't go anywhere, much less circle back.

and the rest frame

Which is what?

whose observer is observing the moving rod, is analogous to the stationary twin,

If it's moving why isn't it analogous to the moving twin?

the only difference is that now the Andromeda case is calculating length contraction,

Andromeda is just one end of a long rod.  It doesn't have a "case".  What length is it calculating the contraction of?

whereas the TP case is calculating time dilation.

Of what clock?

So what's the point of all this -- simply that the traveling twin's clock physically slows,

This is the twin that made a big circle.

whereas length contraction is NOT measured in the frame of the moving rod in the Andromeda case.

Length contraction of what is not measured in the frame of the rod?  It's always the case that length contraction of X is not measured in the frame of X.  Remember it's called "relativity" theory.

I think this is a problem, that the frame containing the rod, does not manifest length contraction similar to the TP case

What length was contracted in the TP?

, where the traveling twin's clock actually slows down. AG

 

But yes length contraction and time dilation go together, that's what makes the speed of light the same in all frames.

Brent

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[Brent Meeker]

On 1/17/2025 3:00 PM, Jesse Mazer wrote:

The non-traveling twin is at rest on the Earth throughout. I never heard of any other concept of the non-traveling twin. The traveling twin is moving with respect to the Earth. I never heard of any other concept of the traveling twin. AG

All that's important to the twin paradox is that one twin is inertial and the other is not. For example if twin A is moving away from Earth inertially the whole time, and twin B is at rest relative to Earth for a while and then accelerates to catch up with twin A, then twin A is the inertial twin and twin B is the non-inertial one during the time between their meetings, and so it's guaranteed that twin B will have aged less than twin A when they reunite.

That's not exactly true.  All that's required for the twin paradox to eventuate is that between two events they take paths of different proper length.  Here's a few examples.

The Slingshot: The traveling twin uses the gravity of a distant star in a slingshot maneuver as beloved of NASA to return.  Both twins follow inertial paths.




Both twins accelerate with identical accelerations, just not at the same time.




And one in which the twin who accelerated ages more.




Part of my catalogue of twin paradoxes, explaining why it has nothing to do with acceleration, including the triplet version in which the outbound triplet just hands off his clock reading to the the inbound triplet (I'm not sure how to get his age handed off).



Brent

[Brent Meeker]

On 1/17/2025 5:40 PM, Alan Grayson wrote:

Let me clarify the problem I'm trying to resolve; notice the T in LT. It stands for Transformation, presumably from one frame to another. It's claimed that the LT will produce the measured result in the target frame,

See you've already introduced an ambiguity.  The measured result in the target frame is whatever it is in the target frame.  Is the "target frame" the one moving relative to the frame in which the measurement made?  What are we calling that frame?  Let's call it A and call the "target frame" B, the one moving relative to A with speed v.  Then things that measure length x in B will measure LT(v)x in A.

based on parameters of the source frame for the transformation.

That's just muddy.  What "parameters"?  Which is the "source" frame?

And this seems to be the case in the TP; from the source frame, the frame at rest on the Earth, the LT tells us what will be measured in the traveling frame. And it seems to do just that, since the clock in the traveling frame actually ticks slower as predicted. If it didn't, the traveling observer would not age slower than the stationary observer. But when we consider the Andromeda problem, the LT seems NOT to predict what the frame of the moving rod

What is the rod moving relative to?  Not the Earth and Andromeda.  So what?

Brent

[Brent Meeker]

On 1/17/2025 6:13 PM, Alan Grayson wrote:

The inconsistency I see is that measurements done in the frame of the moving rod do NOT show that the rod in its frame is contracted,

Nothing is ever Lorentz contracted in it's own frame because nothing is ever moving it it's own frame.

even though this is ostensibly the prediction of the LT, from the pov of the stationary frame.

No.  The prediction is that measurement of the rod from the stationary frame will find it to be shorter than it's length measured in it's own frame.  If you can't even keep straight that its RELATIVITY theory, there's no hope for you.

Brent

This is different compared to what happens in the frames of the TP, where the predicted slowing of the traveling twin's clock is immediately manifested in the frame of the traveling twin.

No it is not.  It is only slow relative to the other twin's clock.

Do you understand what puzzles me?  AG

You confuse yourself because in one case you measure an accumulated interval over a path length and because it's time you are surprised (see a paradox) that they are different.  In the other case you measure a distance at one instant (unrealistically) and compare it to the same distance measured in another frame.  You don't consider measuring along two different paths between two events because you already know that two paths between the same events can have very different lengths and so it doesn't seem paradoxical.

Brent

[John Clark]

On Fri, Jan 17, 2025 at 7:22 PM Alan Grayson <agrays...@gmail.com> wrote:

> If you model any observer leaving Earth, that observer cannot be inertial. AG  

You can if that observer had always been inertial and had passed the vicinity of earth at a given "proper time", the time he sees on his wristwatch.  And if X and Y are moving relative to each other they will each observe that the other's wristwatch is running slower than their own. If your trajectory is inertial (meaning you're not being pushed around by external forces) , you will always take the path that maximizes your proper time between two events. At first that may seem like a paradox but if you dig a little deeper you realize that it is not, it's just odd. You really should look at that video I recommended.

Richard Feynman gave a related anecdote in his book "Surely you're joking Mr. Feynman" when he posed this puzzle to an assistant of Einstein:  

"You blast off in a rocket which has a clock on board, and there's a clock on the ground. The idea is that you have to be back when the clock on the ground says one hour has passed. Now you want it so that when you come back, your clock is as far ahead as possible. According to Einstein, if you go very high, your clock will go faster, because the higher something is in a gravitational field, the faster its clock goes. But if you try to go too high, since you've only got an hour, you have to go so fast to get there that the speed slows your clock down. So you can't go too high. The question is, exactly what program of speed and height should you make so that you get the maximum time on your clock?"

"This assistant of Einstein worked on it for quite a bit before he realized that the answer is the real motion of matter. If you shoot something up in a normal way, so that the time it takes the shell to go up and come down is an hour, that's the correct motion. It's the fundamental principle of Einstein's gravity--that is, what's called the "proper time" is at a maximum for the actual curve."

John K Clark    See what's on my new list at  Extropolis

um8

[John Clark]

On Fri, Jan 17, 2025 at 1:41 PM Alan Grayson <agrays...@gmail.com> wrote:

> IMO SR can handle curved spacetime. All one has to do is make the partitions very fine, so we're approximating inertial motion along very short paths. AG 

All one has to do? Well yes but that's easier said than done, it took Einstein 10 years of grueling work to figure out exactly how to do it, and the effort nearly killed him, he got sick, lost 50 pounds and figured he would die soon. Fortunately he did not. One of the most difficult things he had to figure out was how to measure distance in 4D non-Euclidean spacetime that was curved in any given way that was useful and never produced self-contradictory results. Mathematicians insist that distance must have the following properties:

1) Non-negativity: d(x,y) ≥ 0
2) Identity of indiscernibles: d(x,y) = 0 if and only if x = y
3) Symmetry: d(x,y) = d(y,x)
4) Triangle inequality: d(x,z) ≤ d(x,y) + d(y,z)

After years of false starts and dead ends Einstein eventually found a measuring stick that worked, it's called the Metric Tensor. 

John K Clark    See what's on my new list at  Extropolis

tg1

[Alan Grayson]

Sorry to consume so much bandwidth here. I have one question about the LT and one about the Metric Tensor (MT) which will hopefully resolve most of my confusions.

I'll pose the LT question in the context of the TP. If the stationary twin at rest on the Earth uses the LT to calculate the clock reading at some time on the traveling twin's clock, or its clock rate using two or more time readings, what relationship, if any, does this have on what the traveler twin's clock actually reads, and what he observes as his clock rate? IOW, to what extent does the LT transform the parameters of one frame, to actual observations on another frame (previously referred to as the source and target frames respectively)?

Concerning the MT; it's defined as bilinear map to the real numbers on vectors in the tangent vector space at each point on a manifold. So, when trying to solve Einstein's field equation, the experts claim one must start by calculating the MT, presumably for some given distribution of matter and energy. Since, in general, there exists an infinite set of pairs of vectors in the tangent vector space at each point on the manifold, and assuming the MT has a unique real value at each point on the manifold, at each point on the manifold how do we choose which pair of vectors to do the calculation? 

TY, AG

tg1

[Alan Grayson]

On Thursday, January 16, 2025 at 5:52:52 PM UTC-7 Jesse Mazer wrote:

On Thu, Jan 16, 2025 at 7:33 PM Alan Grayson <agrays...@gmail.com> wrote:

On Thursday, January 16, 2025 at 2:39:55 PM UTC-7 Jesse Mazer wrote:

On Thu, Jan 16, 2025 at 2:43 PM Alan Grayson <agrays...@gmail.com> wrote:

On Thursday, January 16, 2025 at 11:36:48 AM UTC-7 Jesse Mazer wrote:

On Tue, Jan 14, 2025 at 12:02 AM Alan Grayson <agrays...@gmail.com> wrote:

Using the LT, we have the following transformations of Length, Time, and Mass, that is,

x --->x',  t ---> t',  m ---> m'

The length contraction equation is not part of the Lorentz transformation equations, the x --> x' equation in the LT is just about the position coordinate assigned to a *single* event in each frame. The length contraction equation can be derived from the LT but only by considering worldlines of the front and back of an object, and looking at *pairs* of events (one on each of the two worldlines) which are simultaneous in each frame--length in a given frame is just defined as the difference in position coordinate between the front and back of an object at a single time-coordinate in that frame, so it requires looking at a pair of events that are simultaneous in that frame. The result is that for any inertial object, it has its maximum length L in the frame where the object is at rest (the object's own 'rest frame'), and a shorter length L*sqrt(1 - v^2/c^2) in a different frame where the object has nonzero velocity v.

The t ---> t' equation is likewise not the same as the time dilation equation, it's just about the time coordinate assigned to a single event in each frame, although it has a simpler relation to time dilation since you can consider an event on the worldline that passes through the origin where both t and t' are equal to 0, and then the time coordinates t and t' assigned to some other event E on this worldline tell you the time elapsed in each frame between the origin and E. And the LT don't include any mass transformation equation.

Jesse

You're right of course. TY. I see the LT as giving appearances because, say for length contraction, the reduced length is not measured in the primed frame, but that is the length measurement from the pov of the unprimed or stationary frame.

In relativity one does not normally designate any particular frame to be the "stationary frame", since all concepts of motion and rest are defined in purely relative way; if one has two objects A and B in relative motion, one could talk about the frame where A is stationary (A's 'rest frame') or the frame where B is stationary (B's rest frame), but that's all. I'm not sure what you mean by "the reduced length is not measured in the primed frame"--which object's length are you talking about? If A's rest frame is the unprimed frame and B's rest frame is the primed frame, then the length of object A in the primed frame is reduced relative to its length in its own rest frame, i.e. the unprimed frame.

Let's consider a concrete example of a traveler moving at near light speed to Andromeda. From the traveler's frame, the distance to Andromeda is hugely reduced from its length of 2.5 MLY from the pov of a non-traveling observer. This seems to imply that the reduced length is only measured from the pov of the traveler, but not from the pov of the non-traveler, because of which I describe the measurement from the pov of the traveler as APPARENT. Do you agree that the traveler's measurement is apparent because the non-traveler measures the distance to Andromeda as unchanged? TY, AG  

I don't know what you mean by "apparent", but there is no asymmetry in the way Lorentz contraction works in each frame--

I mean, if one uses the LT, to transform from one frame to another frame, if the resultant parameters in the latter frame are not actually measured in the latter frame, I refer to those measurements as "apparent". What I'm stuggling with is what the LT actually results in. Does it tell us what is actually measured in the latter frame, or not? AG

 

if we assume there is a frame A where Milky Way and Andromeda are both at rest (ignoring the fact that in reality they have some motion relative to one another), and another frame B where the rocket ship of the traveler is at rest, then in frame B the Milky Way/Andromeda distance is shortened relative to the distance in their rest frame, and the rocket has its maximum length; in frame A the the rocket's length is shortened relative to its length in its rest frame, and the Milky Way/Andromeda distance has its maximum value. The only asymmetry here is in the choice of the two things to measure the length of (the distance between the Milky Way and Andromeda in their rest frame is obviously huge compared to the rest length of a rocket moving between them), the symmetry might be easier to see if we consider two rockets traveling towards each other (their noses facing each other), and each wants to know the distance it must traverse to get from the nose of the other rocket to its tail. Then for example if each rocket is 10 meters long in its rest frame, and the two rockets have a relative velocity of 0.8c, each will measure only a 6 meter distance between the nose and tail of the other rocket, and the time they each measure to cross that distance is just (6 meters)/(0.8c).

Jesse

 

 

About mass, since the measured mass grows exponentially to infinity as v --> c, isn't this derivable from the LT, but in which frame? AG 

The notion of a variable relativistic mass is just an alternate way of talking about relativistic momentum, often modern textbooks talk solely about the latter and the only mass concept they use is the rest mass. For example the page at https://courses.lumenlearning.com/suny-physics/chapter/28-5-relativistic-momentum has a box titled "Misconception alert: relativistic mass and momentum" which says the following (note that they are using u to denote velocity):

"The relativistically correct definition of momentum as p = γmu is sometimes taken to imply that mass varies with velocity: m_var = γm, particularly in older textbooks. However, note that m is the mass of the object as measured by a person at rest relative to the object. Thus, m is defined to be the rest mass, which could be measured at rest, perhaps using gravity. When a mass is moving relative to an observer, the only way that its mass can be determined is through collisions or other means in which momentum is involved. Since the mass of a moving object cannot be determined independently of momentum, the only meaningful mass is rest mass. Thus, when we use the term mass, assume it to be identical to rest mass."

I'd say there's nothing strictly incorrect about defining a variable relativistic mass, it's just a cosmetically different formalism, but it may be that part of the reason it was mostly abandoned is because for people learning relativity it can lead to misconceptions that there is more to the concept than just a difference in how momentum is calculated, whereas in fact there is no application of relativistic mass that does not involve relativistic momentum. Momentum is needed for situations like collisions or particle creation/annihilation where there's a change in which objects have which individual momenta, but total momentum must be conserved. It's also used in the more general form of the relation of energy to rest mass m and relativistic momentum p, given by the equation E^2 = (mc^2)^2 + (pc)^2, which reduces to the more well-known E=mc^2 in the special case where p=0. 

By the way, since relativistic momentum is given by p=mv/sqrt(1 - v^2/c^2), you can substitute this into the above equation to get E^2 = (m^2)(c^4) + (m^2)(v^2)(c^2)/(1 - v^2/c^2), and then if you take the first term on the right hand side, (m^2)(c^4), and multiply it by (1 - v^2/c^2)/(1 - v^2/c^2) and gather terms, you get E^2 = [(m^2)(c^4) - (m^2)(v^2)(c^2) + (m^2)(v^2)(c^2)]/(1 - v^2/c^2), and two terms cancel each other out so this simplifies to E^2 = (m^2)(c^4)/(1 - v^2/c^2), and then if you take the square root of both sides you get E = γmc^2. So the original equation for energy as a function fo rest mass m and relativistic momentum p can be rewritten as E=Mc^2 where M is the relativistic mass defined as M = γm, again showing that relativistic mass is only useful for rewriting equations involving relativistic momentum.

Jesse

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[Jesse Mazer]

On Sat, Jan 18, 2025 at 9:09 AM Alan Grayson <agrays...@gmail.com> wrote:

On Thursday, January 16, 2025 at 5:52:52 PM UTC-7 Jesse Mazer wrote:

On Thu, Jan 16, 2025 at 7:33 PM Alan Grayson <agrays...@gmail.com> wrote:

On Thursday, January 16, 2025 at 2:39:55 PM UTC-7 Jesse Mazer wrote:

On Thu, Jan 16, 2025 at 2:43 PM Alan Grayson <agrays...@gmail.com> wrote:

On Thursday, January 16, 2025 at 11:36:48 AM UTC-7 Jesse Mazer wrote:

On Tue, Jan 14, 2025 at 12:02 AM Alan Grayson <agrays...@gmail.com> wrote:

Using the LT, we have the following transformations of Length, Time, and Mass, that is,

x --->x',  t ---> t',  m ---> m'

The length contraction equation is not part of the Lorentz transformation equations, the x --> x' equation in the LT is just about the position coordinate assigned to a *single* event in each frame. The length contraction equation can be derived from the LT but only by considering worldlines of the front and back of an object, and looking at *pairs* of events (one on each of the two worldlines) which are simultaneous in each frame--length in a given frame is just defined as the difference in position coordinate between the front and back of an object at a single time-coordinate in that frame, so it requires looking at a pair of events that are simultaneous in that frame. The result is that for any inertial object, it has its maximum length L in the frame where the object is at rest (the object's own 'rest frame'), and a shorter length L*sqrt(1 - v^2/c^2) in a different frame where the object has nonzero velocity v.

The t ---> t' equation is likewise not the same as the time dilation equation, it's just about the time coordinate assigned to a single event in each frame, although it has a simpler relation to time dilation since you can consider an event on the worldline that passes through the origin where both t and t' are equal to 0, and then the time coordinates t and t' assigned to some other event E on this worldline tell you the time elapsed in each frame between the origin and E. And the LT don't include any mass transformation equation.

Jesse

You're right of course. TY. I see the LT as giving appearances because, say for length contraction, the reduced length is not measured in the primed frame, but that is the length measurement from the pov of the unprimed or stationary frame.

In relativity one does not normally designate any particular frame to be the "stationary frame", since all concepts of motion and rest are defined in purely relative way; if one has two objects A and B in relative motion, one could talk about the frame where A is stationary (A's 'rest frame') or the frame where B is stationary (B's rest frame), but that's all. I'm not sure what you mean by "the reduced length is not measured in the primed frame"--which object's length are you talking about? If A's rest frame is the unprimed frame and B's rest frame is the primed frame, then the length of object A in the primed frame is reduced relative to its length in its own rest frame, i.e. the unprimed frame.

Let's consider a concrete example of a traveler moving at near light speed to Andromeda. From the traveler's frame, the distance to Andromeda is hugely reduced from its length of 2.5 MLY from the pov of a non-traveling observer. This seems to imply that the reduced length is only measured from the pov of the traveler, but not from the pov of the non-traveler, because of which I describe the measurement from the pov of the traveler as APPARENT. Do you agree that the traveler's measurement is apparent because the non-traveler measures the distance to Andromeda as unchanged? TY, AG  

I don't know what you mean by "apparent", but there is no asymmetry in the way Lorentz contraction works in each frame--

I mean, if one uses the LT, to transform from one frame to another frame, if the resultant parameters in the latter frame are not actually measured in the latter frame, I refer to those measurements as "apparent". What I'm stuggling with is what the LT actually results in. Does it tell us what is actually measured in the latter frame, or not? AG

Yes, it always tells you what is actually measured in the frame that you're transforming into, using that frame's own rulers and clocks. You haven't made it at all clear why you suspect otherwise.

Jesse

 

 

if we assume there is a frame A where Milky Way and Andromeda are both at rest (ignoring the fact that in reality they have some motion relative to one another), and another frame B where the rocket ship of the traveler is at rest, then in frame B the Milky Way/Andromeda distance is shortened relative to the distance in their rest frame, and the rocket has its maximum length; in frame A the the rocket's length is shortened relative to its length in its rest frame, and the Milky Way/Andromeda distance has its maximum value. The only asymmetry here is in the choice of the two things to measure the length of (the distance between the Milky Way and Andromeda in their rest frame is obviously huge compared to the rest length of a rocket moving between them), the symmetry might be easier to see if we consider two rockets traveling towards each other (their noses facing each other), and each wants to know the distance it must traverse to get from the nose of the other rocket to its tail. Then for example if each rocket is 10 meters long in its rest frame, and the two rockets have a relative velocity of 0.8c, each will measure only a 6 meter distance between the nose and tail of the other rocket, and the time they each measure to cross that distance is just (6 meters)/(0.8c).

Jesse

 

 

About mass, since the measured mass grows exponentially to infinity as v --> c, isn't this derivable from the LT, but in which frame? AG 

The notion of a variable relativistic mass is just an alternate way of talking about relativistic momentum, often modern textbooks talk solely about the latter and the only mass concept they use is the rest mass. For example the page at https://courses.lumenlearning.com/suny-physics/chapter/28-5-relativistic-momentum has a box titled "Misconception alert: relativistic mass and momentum" which says the following (note that they are using u to denote velocity):

"The relativistically correct definition of momentum as p = γmu is sometimes taken to imply that mass varies with velocity: m_var = γm, particularly in older textbooks. However, note that m is the mass of the object as measured by a person at rest relative to the object. Thus, m is defined to be the rest mass, which could be measured at rest, perhaps using gravity. When a mass is moving relative to an observer, the only way that its mass can be determined is through collisions or other means in which momentum is involved. Since the mass of a moving object cannot be determined independently of momentum, the only meaningful mass is rest mass. Thus, when we use the term mass, assume it to be identical to rest mass."

I'd say there's nothing strictly incorrect about defining a variable relativistic mass, it's just a cosmetically different formalism, but it may be that part of the reason it was mostly abandoned is because for people learning relativity it can lead to misconceptions that there is more to the concept than just a difference in how momentum is calculated, whereas in fact there is no application of relativistic mass that does not involve relativistic momentum. Momentum is needed for situations like collisions or particle creation/annihilation where there's a change in which objects have which individual momenta, but total momentum must be conserved. It's also used in the more general form of the relation of energy to rest mass m and relativistic momentum p, given by the equation E^2 = (mc^2)^2 + (pc)^2, which reduces to the more well-known E=mc^2 in the special case where p=0. 

By the way, since relativistic momentum is given by p=mv/sqrt(1 - v^2/c^2), you can substitute this into the above equation to get E^2 = (m^2)(c^4) + (m^2)(v^2)(c^2)/(1 - v^2/c^2), and then if you take the first term on the right hand side, (m^2)(c^4), and multiply it by (1 - v^2/c^2)/(1 - v^2/c^2) and gather terms, you get E^2 = [(m^2)(c^4) - (m^2)(v^2)(c^2) + (m^2)(v^2)(c^2)]/(1 - v^2/c^2), and two terms cancel each other out so this simplifies to E^2 = (m^2)(c^4)/(1 - v^2/c^2), and then if you take the square root of both sides you get E = γmc^2. So the original equation for energy as a function fo rest mass m and relativistic momentum p can be rewritten as E=Mc^2 where M is the relativistic mass defined as M = γm, again showing that relativistic mass is only useful for rewriting equations involving relativistic momentum.

Jesse

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[Alan Grayson]

On Friday, January 17, 2025 at 11:17:37 PM UTC-7 Brent Meeker wrote:

On 1/17/2025 6:13 PM, Alan Grayson wrote:

The inconsistency I see is that measurements done in the frame of the moving rod do NOT show that the rod in its frame is contracted,

Nothing is ever Lorentz contracted in it's own frame because nothing is ever moving it it's own frame.

I am aware of that fact. AG 

even though this is ostensibly the prediction of the LT, from the pov of the stationary frame.

No.  The prediction is that measurement of the rod from the stationary frame will find it to be shorter than it's length measured in it's own frame.  If you can't even keep straight that its RELATIVITY theory, there's no hope for you.

 I am aware of that fact. AG 

Brent

This is different compared to what happens in the frames of the TP, where the predicted slowing of the traveling twin's clock is immediately manifested in the frame of the traveling twin.

No it is not.  It is only slow relative to the other twin's clock.

Yes, I am aware of that.  But what does the traveling twin actually measure his clock rate to be? Same as calculated by stationary twin on Earth? Still struggling with what the LT does for us. AG 

[Alan Grayson]

On Saturday, January 18, 2025 at 7:19:41 AM UTC-7 Jesse Mazer wrote:

On Sat, Jan 18, 2025 at 9:09 AM Alan Grayson <agrays...@gmail.com> wrote:

On Thursday, January 16, 2025 at 5:52:52 PM UTC-7 Jesse Mazer wrote:

On Thu, Jan 16, 2025 at 7:33 PM Alan Grayson <agrays...@gmail.com> wrote:

On Thursday, January 16, 2025 at 2:39:55 PM UTC-7 Jesse Mazer wrote:

On Thu, Jan 16, 2025 at 2:43 PM Alan Grayson <agrays...@gmail.com> wrote:

On Thursday, January 16, 2025 at 11:36:48 AM UTC-7 Jesse Mazer wrote:

On Tue, Jan 14, 2025 at 12:02 AM Alan Grayson <agrays...@gmail.com> wrote:

Using the LT, we have the following transformations of Length, Time, and Mass, that is,

x --->x',  t ---> t',  m ---> m'

The length contraction equation is not part of the Lorentz transformation equations, the x --> x' equation in the LT is just about the position coordinate assigned to a *single* event in each frame. The length contraction equation can be derived from the LT but only by considering worldlines of the front and back of an object, and looking at *pairs* of events (one on each of the two worldlines) which are simultaneous in each frame--length in a given frame is just defined as the difference in position coordinate between the front and back of an object at a single time-coordinate in that frame, so it requires looking at a pair of events that are simultaneous in that frame. The result is that for any inertial object, it has its maximum length L in the frame where the object is at rest (the object's own 'rest frame'), and a shorter length L*sqrt(1 - v^2/c^2) in a different frame where the object has nonzero velocity v.

The t ---> t' equation is likewise not the same as the time dilation equation, it's just about the time coordinate assigned to a single event in each frame, although it has a simpler relation to time dilation since you can consider an event on the worldline that passes through the origin where both t and t' are equal to 0, and then the time coordinates t and t' assigned to some other event E on this worldline tell you the time elapsed in each frame between the origin and E. And the LT don't include any mass transformation equation.

Jesse

You're right of course. TY. I see the LT as giving appearances because, say for length contraction, the reduced length is not measured in the primed frame, but that is the length measurement from the pov of the unprimed or stationary frame.

In relativity one does not normally designate any particular frame to be the "stationary frame", since all concepts of motion and rest are defined in purely relative way; if one has two objects A and B in relative motion, one could talk about the frame where A is stationary (A's 'rest frame') or the frame where B is stationary (B's rest frame), but that's all. I'm not sure what you mean by "the reduced length is not measured in the primed frame"--which object's length are you talking about? If A's rest frame is the unprimed frame and B's rest frame is the primed frame, then the length of object A in the primed frame is reduced relative to its length in its own rest frame, i.e. the unprimed frame.

Let's consider a concrete example of a traveler moving at near light speed to Andromeda. From the traveler's frame, the distance to Andromeda is hugely reduced from its length of 2.5 MLY from the pov of a non-traveling observer. This seems to imply that the reduced length is only measured from the pov of the traveler, but not from the pov of the non-traveler, because of which I describe the measurement from the pov of the traveler as APPARENT. Do you agree that the traveler's measurement is apparent because the non-traveler measures the distance to Andromeda as unchanged? TY, AG  

I don't know what you mean by "apparent", but there is no asymmetry in the way Lorentz contraction works in each frame--

I mean, if one uses the LT, to transform from one frame to another frame, if the resultant parameters in the latter frame are not actually measured in the latter frame, I refer to those measurements as "apparent". What I'm stuggling with is what the LT actually results in. Does it tell us what is actually measured in the latter frame, or not? AG

Yes, it always tells you what is actually measured in the frame that you're transforming into, using that frame's own rulers and clocks. You haven't made it at all clear why you suspect otherwise.

Jesse

I suspect otherwise because in the Andromeda problem, using the LT from the pov of the rest frame, at rest relative to the Earth, we get length contraction in the transformed frame (modeled as a rod moving toward the Earth), but never length contraction as what's observed IN the rod frame. I realize that length contraction requires something moving, so it's not reasonable to expect this, although it does show there's an exception to what you allege the LT does. AG 

[Jesse Mazer]

On Sat, Jan 18, 2025 at 9:36 AM Alan Grayson <agrays...@gmail.com> wrote:

On Saturday, January 18, 2025 at 7:19:41 AM UTC-7 Jesse Mazer wrote:

On Sat, Jan 18, 2025 at 9:09 AM Alan Grayson <agrays...@gmail.com> wrote:

On Thursday, January 16, 2025 at 5:52:52 PM UTC-7 Jesse Mazer wrote:

On Thu, Jan 16, 2025 at 7:33 PM Alan Grayson <agrays...@gmail.com> wrote:

On Thursday, January 16, 2025 at 2:39:55 PM UTC-7 Jesse Mazer wrote:

On Thu, Jan 16, 2025 at 2:43 PM Alan Grayson <agrays...@gmail.com> wrote:

On Thursday, January 16, 2025 at 11:36:48 AM UTC-7 Jesse Mazer wrote:

On Tue, Jan 14, 2025 at 12:02 AM Alan Grayson <agrays...@gmail.com> wrote:

Using the LT, we have the following transformations of Length, Time, and Mass, that is,

x --->x',  t ---> t',  m ---> m'

The length contraction equation is not part of the Lorentz transformation equations, the x --> x' equation in the LT is just about the position coordinate assigned to a *single* event in each frame. The length contraction equation can be derived from the LT but only by considering worldlines of the front and back of an object, and looking at *pairs* of events (one on each of the two worldlines) which are simultaneous in each frame--length in a given frame is just defined as the difference in position coordinate between the front and back of an object at a single time-coordinate in that frame, so it requires looking at a pair of events that are simultaneous in that frame. The result is that for any inertial object, it has its maximum length L in the frame where the object is at rest (the object's own 'rest frame'), and a shorter length L*sqrt(1 - v^2/c^2) in a different frame where the object has nonzero velocity v.

The t ---> t' equation is likewise not the same as the time dilation equation, it's just about the time coordinate assigned to a single event in each frame, although it has a simpler relation to time dilation since you can consider an event on the worldline that passes through the origin where both t and t' are equal to 0, and then the time coordinates t and t' assigned to some other event E on this worldline tell you the time elapsed in each frame between the origin and E. And the LT don't include any mass transformation equation.

Jesse

You're right of course. TY. I see the LT as giving appearances because, say for length contraction, the reduced length is not measured in the primed frame, but that is the length measurement from the pov of the unprimed or stationary frame.

In relativity one does not normally designate any particular frame to be the "stationary frame", since all concepts of motion and rest are defined in purely relative way; if one has two objects A and B in relative motion, one could talk about the frame where A is stationary (A's 'rest frame') or the frame where B is stationary (B's rest frame), but that's all. I'm not sure what you mean by "the reduced length is not measured in the primed frame"--which object's length are you talking about? If A's rest frame is the unprimed frame and B's rest frame is the primed frame, then the length of object A in the primed frame is reduced relative to its length in its own rest frame, i.e. the unprimed frame.

Let's consider a concrete example of a traveler moving at near light speed to Andromeda. From the traveler's frame, the distance to Andromeda is hugely reduced from its length of 2.5 MLY from the pov of a non-traveling observer. This seems to imply that the reduced length is only measured from the pov of the traveler, but not from the pov of the non-traveler, because of which I describe the measurement from the pov of the traveler as APPARENT. Do you agree that the traveler's measurement is apparent because the non-traveler measures the distance to Andromeda as unchanged? TY, AG  

I don't know what you mean by "apparent", but there is no asymmetry in the way Lorentz contraction works in each frame--

I mean, if one uses the LT, to transform from one frame to another frame, if the resultant parameters in the latter frame are not actually measured in the latter frame, I refer to those measurements as "apparent". What I'm stuggling with is what the LT actually results in. Does it tell us what is actually measured in the latter frame, or not? AG

Yes, it always tells you what is actually measured in the frame that you're transforming into, using that frame's own rulers and clocks. You haven't made it at all clear why you suspect otherwise.

Jesse

I suspect otherwise because in the Andromeda problem, using the LT from the pov of the rest frame, at rest relative to the Earth, we get length contraction in the transformed frame (modeled as a rod moving toward the Earth)

I thought the problem was supposed to involve the assumption of a traveler going from Earth to Andromeda (who I imagined as riding in a rocket), with a rod that is at rest relative to Earth and Andromeda and whose length defines the distance between them in each frame? But here you seem to be talking about a rod moving relative to the Earth? So now I'm not even clear about what physical scenario you are imagining--please lay it out clearly by specifying all the physical objects you are imagining in the problem (like rocket, rod, Earth, Andromeda), which of them have nonzero velocity in the Earth's frame, and the rest length for any object you want to do length calculations for.

Jesse

 

but never length contraction as what's observed IN the rod frame. I realize that length contraction requires something moving, so it's not reasonable to expect this, although it does show there's an exception to what you allege the LT does. AG 

 

 

if we assume there is a frame A where Milky Way and Andromeda are both at rest (ignoring the fact that in reality they have some motion relative to one another), and another frame B where the rocket ship of the traveler is at rest, then in frame B the Milky Way/Andromeda distance is shortened relative to the distance in their rest frame, and the rocket has its maximum length; in frame A the the rocket's length is shortened relative to its length in its rest frame, and the Milky Way/Andromeda distance has its maximum value. The only asymmetry here is in the choice of the two things to measure the length of (the distance between the Milky Way and Andromeda in their rest frame is obviously huge compared to the rest length of a rocket moving between them), the symmetry might be easier to see if we consider two rockets traveling towards each other (their noses facing each other), and each wants to know the distance it must traverse to get from the nose of the other rocket to its tail. Then for example if each rocket is 10 meters long in its rest frame, and the two rockets have a relative velocity of 0.8c, each will measure only a 6 meter distance between the nose and tail of the other rocket, and the time they each measure to cross that distance is just (6 meters)/(0.8c).

Jesse

 

 

About mass, since the measured mass grows exponentially to infinity as v --> c, isn't this derivable from the LT, but in which frame? AG 

The notion of a variable relativistic mass is just an alternate way of talking about relativistic momentum, often modern textbooks talk solely about the latter and the only mass concept they use is the rest mass. For example the page at https://courses.lumenlearning.com/suny-physics/chapter/28-5-relativistic-momentum has a box titled "Misconception alert: relativistic mass and momentum" which says the following (note that they are using u to denote velocity):

"The relativistically correct definition of momentum as p = γmu is sometimes taken to imply that mass varies with velocity: m_var = γm, particularly in older textbooks. However, note that m is the mass of the object as measured by a person at rest relative to the object. Thus, m is defined to be the rest mass, which could be measured at rest, perhaps using gravity. When a mass is moving relative to an observer, the only way that its mass can be determined is through collisions or other means in which momentum is involved. Since the mass of a moving object cannot be determined independently of momentum, the only meaningful mass is rest mass. Thus, when we use the term mass, assume it to be identical to rest mass."

I'd say there's nothing strictly incorrect about defining a variable relativistic mass, it's just a cosmetically different formalism, but it may be that part of the reason it was mostly abandoned is because for people learning relativity it can lead to misconceptions that there is more to the concept than just a difference in how momentum is calculated, whereas in fact there is no application of relativistic mass that does not involve relativistic momentum. Momentum is needed for situations like collisions or particle creation/annihilation where there's a change in which objects have which individual momenta, but total momentum must be conserved. It's also used in the more general form of the relation of energy to rest mass m and relativistic momentum p, given by the equation E^2 = (mc^2)^2 + (pc)^2, which reduces to the more well-known E=mc^2 in the special case where p=0. 

By the way, since relativistic momentum is given by p=mv/sqrt(1 - v^2/c^2), you can substitute this into the above equation to get E^2 = (m^2)(c^4) + (m^2)(v^2)(c^2)/(1 - v^2/c^2), and then if you take the first term on the right hand side, (m^2)(c^4), and multiply it by (1 - v^2/c^2)/(1 - v^2/c^2) and gather terms, you get E^2 = [(m^2)(c^4) - (m^2)(v^2)(c^2) + (m^2)(v^2)(c^2)]/(1 - v^2/c^2), and two terms cancel each other out so this simplifies to E^2 = (m^2)(c^4)/(1 - v^2/c^2), and then if you take the square root of both sides you get E = γmc^2. So the original equation for energy as a function fo rest mass m and relativistic momentum p can be rewritten as E=Mc^2 where M is the relativistic mass defined as M = γm, again showing that relativistic mass is only useful for rewriting equations involving relativistic momentum.

Jesse

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[Alan Grayson]

If you prefer, we can place the traveler to Andromeda in a spaceship moving with some velocity wrt the Earth, and now imagine a rod whose length is the distance from Earth to Andromeda. Since motion is relative, we can imagine the spaceship is at rest, and the rod moving toward Earth (since the spaceship is ..imagined as moving toward Andromeda). In the rest frame of the spaceship, the rod is contracted since it appears to be moving. We can use the LT to calculate how much its length contracts due to the velocity of the rod. However, as I pointed out earlier, this measurement is NOT possible IN the frame of the moving rod, since nothing is moving within this frame. This is an example of the LT NOT yielding a measurement in the target frame (in this case the frame containing the rod), which is an exception to the claim that the LT always predicts what a target frame will actually measure. I think time dilation is an example of the opposite; a measurement which is realized in the target frame. AG 

[Jesse Mazer]

So the rod is at rest relative to Earth and Andromeda, as I originally understood you to be saying?

 

Since motion is relative, we can imagine the spaceship is at rest, and the rod moving toward Earth (since the spaceship is ..imagined as moving toward Andromeda).

By "moving toward the Earth" do you mean moving *relative to the Earth* (i.e. distance between the Earth and either end of the rod is changing over time), or do you just mean that in the spaceship's frame, the rod is always moving in the direction of the Earth, but the Earth is moving at the same speed in the same direction so the end of the rod always coincides with the Earth (or remains at a fixed negligible distance from it)? If the latter, "the rod moving toward Earth" seems like confusing terminology for this, there would be much less room for confusion if you stuck to the language of talking about movement "relative to" a given object or observer as I suggested.

 

In the rest frame of the spaceship, the rod is contracted since it appears to be moving. We can use the LT to calculate how much its length contracts due to the velocity of the rod. However, as I pointed out earlier, this measurement is NOT possible IN the frame of the moving rod, since nothing is moving within this frame.

How much its length contracts in the frame of the spaceship, or in the frame of the rod (assuming the rod is at rest relative to Earth/Andromeda)?

 

This is an example of the LT NOT yielding a measurement in the target frame (in this case the frame containing the rod), which is an exception to the claim that the LT always predicts what a target frame will actually measure.

I don't know why you think that, but you'll have to clearly specify which frame you want to transform from, and which frame you want to use the LT to transform into (the 'target frame'). If you mean starting from the coordinates of the rod in the spaceship's rest frame, and using the LT to transform into the rest frame of the rod (so the rod's frame is the target frame), then if you actually do the algebra of the LT you will get the correct result that the rod is *longer* in its own rest frame than it was in the spaceship frame, not shorter. And if on the other hand you start with the coordinates in the rod's rest frame and use the LT to transform into the spaceship frame (so the spaceship frame is the target frame), you will get the correct result that the rod is shorter in the spaceship frame than in its own rest frame.

Jesse

 

I think time dilation is an example of the opposite; a measurement which is realized in the target frame. AG 

 

but never length contraction as what's observed IN the rod frame. I realize that length contraction requires something moving, so it's not reasonable to expect this, although it does show there's an exception to what you allege the LT does. AG 

 

 

if we assume there is a frame A where Milky Way and Andromeda are both at rest (ignoring the fact that in reality they have some motion relative to one another), and another frame B where the rocket ship of the traveler is at rest, then in frame B the Milky Way/Andromeda distance is shortened relative to the distance in their rest frame, and the rocket has its maximum length; in frame A the the rocket's length is shortened relative to its length in its rest frame, and the Milky Way/Andromeda distance has its maximum value. The only asymmetry here is in the choice of the two things to measure the length of (the distance between the Milky Way and Andromeda in their rest frame is obviously huge compared to the rest length of a rocket moving between them), the symmetry might be easier to see if we consider two rockets traveling towards each other (their noses facing each other), and each wants to know the distance it must traverse to get from the nose of the other rocket to its tail. Then for example if each rocket is 10 meters long in its rest frame, and the two rockets have a relative velocity of 0.8c, each will measure only a 6 meter distance between the nose and tail of the other rocket, and the time they each measure to cross that distance is just (6 meters)/(0.8c).

Jesse

 

 

About mass, since the measured mass grows exponentially to infinity as v --> c, isn't this derivable from the LT, but in which frame? AG 

The notion of a variable relativistic mass is just an alternate way of talking about relativistic momentum, often modern textbooks talk solely about the latter and the only mass concept they use is the rest mass. For example the page at https://courses.lumenlearning.com/suny-physics/chapter/28-5-relativistic-momentum has a box titled "Misconception alert: relativistic mass and momentum" which says the following (note that they are using u to denote velocity):

"The relativistically correct definition of momentum as p = γmu is sometimes taken to imply that mass varies with velocity: m_var = γm, particularly in older textbooks. However, note that m is the mass of the object as measured by a person at rest relative to the object. Thus, m is defined to be the rest mass, which could be measured at rest, perhaps using gravity. When a mass is moving relative to an observer, the only way that its mass can be determined is through collisions or other means in which momentum is involved. Since the mass of a moving object cannot be determined independently of momentum, the only meaningful mass is rest mass. Thus, when we use the term mass, assume it to be identical to rest mass."

I'd say there's nothing strictly incorrect about defining a variable relativistic mass, it's just a cosmetically different formalism, but it may be that part of the reason it was mostly abandoned is because for people learning relativity it can lead to misconceptions that there is more to the concept than just a difference in how momentum is calculated, whereas in fact there is no application of relativistic mass that does not involve relativistic momentum. Momentum is needed for situations like collisions or particle creation/annihilation where there's a change in which objects have which individual momenta, but total momentum must be conserved. It's also used in the more general form of the relation of energy to rest mass m and relativistic momentum p, given by the equation E^2 = (mc^2)^2 + (pc)^2, which reduces to the more well-known E=mc^2 in the special case where p=0. 

By the way, since relativistic momentum is given by p=mv/sqrt(1 - v^2/c^2), you can substitute this into the above equation to get E^2 = (m^2)(c^4) + (m^2)(v^2)(c^2)/(1 - v^2/c^2), and then if you take the first term on the right hand side, (m^2)(c^4), and multiply it by (1 - v^2/c^2)/(1 - v^2/c^2) and gather terms, you get E^2 = [(m^2)(c^4) - (m^2)(v^2)(c^2) + (m^2)(v^2)(c^2)]/(1 - v^2/c^2), and two terms cancel each other out so this simplifies to E^2 = (m^2)(c^4)/(1 - v^2/c^2), and then if you take the square root of both sides you get E = γmc^2. So the original equation for energy as a function fo rest mass m and relativistic momentum p can be rewritten as E=Mc^2 where M is the relativistic mass defined as M = γm, again showing that relativistic mass is only useful for rewriting equations involving relativistic momentum.

Jesse

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[Alan Grayson]

Let me try again. The spaceship is moving toward Andromeda from the Earth. Since motion is relative I can assume the spaceship is at rest, and a rod representing the distance from Earth to Andromeda is moving toward the Earth, since  it is assumed the spaceship was originally moving in the opposite direction, toward Andromeda. Now, from the pov of the rest frame, the spaceship, imagine a LT to determine the contraction of the rod. Note that in the frame of the moving rod, its contraction, predicted by the LT, cannot be measured because although the rod is moving, nothing within the rod frame is moving. This, IMO, contradicts the claim that the LT always yields what is actually measured in the target frame of the LT, in this case the frame of the rod. OK? AG

[Jesse Mazer]

You didn't address my question of whether "moving toward the Earth" means moving *relative* to the Earth (i.e. rod is moving in Earth's rest frame, and other frames like the spaceship frame see the distance between the end of the rod and the Earth as changing over time), or if it just means that in the rod is moving in the direction of the Earth at the same speed that the Earth itself is moving in this frame, so the distance between any point on the rod and the Earth is unchanging. If you mean the latter, this is confusing terminology, definitely not the sort of thing you would find in any relativity textbook. Either way, can you *please* stick to defining movement and rest "relative to" specific objects or observers to prevent this kind of verbal ambiguity?

 

Now, from the pov of the rest frame, the spaceship, imagine a LT to determine the contraction of the rod.

Contraction of the rod in what frame? Are you saying that if we start with the coordinates of the ends of the rod in the spaceship frame (I think the rod is supposed to be moving in this frame, correct me if I'm wrong), and then use the LT to find the coordinates of the ends of the rod in the rod's own rest frame, you think we'd get the false prediction that the rod is contracted in the rod's rest frame? If so this is simply WRONG, that's not what you would find if you actually did the algebra of the Lorentz transformation equations. If you mean something different you need to be more clear about what frame we are starting with, what is the "target frame" we are transforming into using the LT, and which of these frames you want to know the "contraction of the rod" in.

Jesse

 

Note that in the frame of the moving rod, its contraction, predicted by the LT, cannot be measured because although the rod is moving, nothing within the rod frame is moving. This, IMO, contradicts the claim that the LT always yields what is actually measured in the target frame of the LT, in this case the frame of the rod. OK? AG

 

I think time dilation is an example of the opposite; a measurement which is realized in the target frame. AG 

 

but never length contraction as what's observed IN the rod frame. I realize that length contraction requires something moving, so it's not reasonable to expect this, although it does show there's an exception to what you allege the LT does. AG 

 

 

if we assume there is a frame A where Milky Way and Andromeda are both at rest (ignoring the fact that in reality they have some motion relative to one another), and another frame B where the rocket ship of the traveler is at rest, then in frame B the Milky Way/Andromeda distance is shortened relative to the distance in their rest frame, and the rocket has its maximum length; in frame A the the rocket's length is shortened relative to its length in its rest frame, and the Milky Way/Andromeda distance has its maximum value. The only asymmetry here is in the choice of the two things to measure the length of (the distance between the Milky Way and Andromeda in their rest frame is obviously huge compared to the rest length of a rocket moving between them), the symmetry might be easier to see if we consider two rockets traveling towards each other (their noses facing each other), and each wants to know the distance it must traverse to get from the nose of the other rocket to its tail. Then for example if each rocket is 10 meters long in its rest frame, and the two rockets have a relative velocity of 0.8c, each will measure only a 6 meter distance between the nose and tail of the other rocket, and the time they each measure to cross that distance is just (6 meters)/(0.8c).

Jesse

 

 

About mass, since the measured mass grows exponentially to infinity as v --> c, isn't this derivable from the LT, but in which frame? AG 

The notion of a variable relativistic mass is just an alternate way of talking about relativistic momentum, often modern textbooks talk solely about the latter and the only mass concept they use is the rest mass. For example the page at https://courses.lumenlearning.com/suny-physics/chapter/28-5-relativistic-momentum has a box titled "Misconception alert: relativistic mass and momentum" which says the following (note that they are using u to denote velocity):

"The relativistically correct definition of momentum as p = γmu is sometimes taken to imply that mass varies with velocity: m_var = γm, particularly in older textbooks. However, note that m is the mass of the object as measured by a person at rest relative to the object. Thus, m is defined to be the rest mass, which could be measured at rest, perhaps using gravity. When a mass is moving relative to an observer, the only way that its mass can be determined is through collisions or other means in which momentum is involved. Since the mass of a moving object cannot be determined independently of momentum, the only meaningful mass is rest mass. Thus, when we use the term mass, assume it to be identical to rest mass."

I'd say there's nothing strictly incorrect about defining a variable relativistic mass, it's just a cosmetically different formalism, but it may be that part of the reason it was mostly abandoned is because for people learning relativity it can lead to misconceptions that there is more to the concept than just a difference in how momentum is calculated, whereas in fact there is no application of relativistic mass that does not involve relativistic momentum. Momentum is needed for situations like collisions or particle creation/annihilation where there's a change in which objects have which individual momenta, but total momentum must be conserved. It's also used in the more general form of the relation of energy to rest mass m and relativistic momentum p, given by the equation E^2 = (mc^2)^2 + (pc)^2, which reduces to the more well-known E=mc^2 in the special case where p=0. 

By the way, since relativistic momentum is given by p=mv/sqrt(1 - v^2/c^2), you can substitute this into the above equation to get E^2 = (m^2)(c^4) + (m^2)(v^2)(c^2)/(1 - v^2/c^2), and then if you take the first term on the right hand side, (m^2)(c^4), and multiply it by (1 - v^2/c^2)/(1 - v^2/c^2) and gather terms, you get E^2 = [(m^2)(c^4) - (m^2)(v^2)(c^2) + (m^2)(v^2)(c^2)]/(1 - v^2/c^2), and two terms cancel each other out so this simplifies to E^2 = (m^2)(c^4)/(1 - v^2/c^2), and then if you take the square root of both sides you get E = γmc^2. So the original equation for energy as a function fo rest mass m and relativistic momentum p can be rewritten as E=Mc^2 where M is the relativistic mass defined as M = γm, again showing that relativistic mass is only useful for rewriting equations involving relativistic momentum.

Jesse

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[Alan Grayson]

I assume, and you can as well, that the Earth is at rest, and the spaceship is moving toward Andromeda. Can you assume the Earth is at rest in this model and not allow us to get into a discussion of what "at rest" means? AG

 

Now, from the pov of the rest frame, the spaceship, imagine a LT to determine the contraction of the rod.

Contraction of the rod in what frame?

I was explicit; from the rest frame of spaceship, apply the LT to determine the contraction of the rod, assuming you know its original length. Now ask yourself this question; in the rod frame, does the contraction information you've received from the LT correspond to any measurement result in the rod frame? (Answer; of course NOT!). This is what I've been trying to demonstrate; results given by the LT do NOT always give us what can be measured in the target frame of the LT, in this case the rod frame. AG (PS; I sent you a message on Facebook. You can reply here if you want.)

 

Are you saying that if we start with the coordinates of the ends of the rod in the spaceship frame (I think the rod is supposed to be moving in this frame, correct me if I'm wrong), and then use the LT to find the coordinates of the ends of the rod in the rod's own rest frame, you think we'd get the false prediction that the rod is contracted in the rod's rest frame?

 

The spaceship is in one frame, at relative rest WRT the rod; the rod is in another frame, in relative motion WRT to the spaceship, at the velocity which the spaceship was originally moving toward Andromeda. We know the rod's initial length and we calculate its contraction from the spaceship's frame, using the rod's relative velocity. AG

 

If so this is simply WRONG, that's not what you would find if you actually did the algebra of the Lorentz transformation equations. If you mean something different you need to be more clear about what frame we are starting with, what is the "target frame" we are transforming into using the LT, and which of these frames you want to know the "contraction of the rod" in.

Jesse

 

Note that in the frame of the moving rod, its contraction, predicted by the LT, cannot be measured because although the rod is moving, nothing within the rod frame is moving. This, IMO, contradicts the claim that the LT always yields what is actually measured in the target frame of the LT, in this case the frame of the rod. OK? AG

I think time dilation is an example of the opposite; a measurement which is realized in the target frame. AG 

but never length contraction as what's observed IN the rod frame. I realize that length contraction requires something moving, so it's not reasonable to expect this, although it does show there's an exception to what you allege the LT does. AG 

if we assume there is a frame A where Milky Way and Andromeda are both at rest (ignoring the fact that in reality they have some motion relative to one another), and another frame B where the rocket ship of the traveler is at rest, then in frame B the Milky Way/Andromeda distance is shortened relative to the distance in their rest frame, and the rocket has its maximum length; in frame A the the rocket's length is shortened relative to its length in its rest frame, and the Milky Way/Andromeda distance has its maximum value. The only asymmetry here is in the choice of the two things to measure the length of (the distance between the Milky Way and Andromeda in their rest frame is obviously huge compared to the rest length of a rocket moving between them), the symmetry might be easier to see if we consider two rockets traveling towards each other (their noses facing each other), and each wants to know the distance it must traverse to get from the nose of the other rocket to its tail. Then for example if each rocket is 10 meters long in its rest frame, and the two rockets have a relative velocity of 0.8c, each will measure only a 6 meter distance between the nose and tail of the other rocket, and the time they each measure to cross that distance is just (6 meters)/(0.8c).

Jesse

 

 

About mass, since the measured mass grows exponentially to infinity as v --> c, isn't this derivable from the LT, but in which frame? AG 

The notion of a variable relativistic mass is just an alternate way of talking about relativistic momentum, often modern textbooks talk solely about the latter and the only mass concept they use is the rest mass. For example the page at https://courses.lumenlearning.com/suny-physics/chapter/28-5-relativistic-momentum has a box titled "Misconception alert: relativistic mass and momentum" which says the following (note that they are using u to denote velocity):

"The relativistically correct definition of momentum as p = γmu is sometimes taken to imply that mass varies with velocity: m_var = γm, particularly in older textbooks. However, note that m is the mass of the object as measured by a person at rest relative to the object. Thus, m is defined to be the rest mass, which could be measured at rest, perhaps using gravity. When a mass is moving relative to an observer, the only way that its mass can be determined is through collisions or other means in which momentum is involved. Since the mass of a moving object cannot be determined independently of momentum, the only meaningful mass is rest mass. Thus, when we use the term mass, assume it to be identical to rest mass."

I'd say there's nothing strictly incorrect about defining a variable relativistic mass, it's just a cosmetically different formalism, but it may be that part of the reason it was mostly abandoned is because for people learning relativity it can lead to misconceptions that there is more to the concept than just a difference in how momentum is calculated, whereas in fact there is no application of relativistic mass that does not involve relativistic momentum. Momentum is needed for situations like collisions or particle creation/annihilation where there's a change in which objects have which individual momenta, but total momentum must be conserved. It's also used in the more general form of the relation of energy to rest mass m and relativistic momentum p, given by the equation E^2 = (mc^2)^2 + (pc)^2, which reduces to the more well-known E=mc^2 in the special case where p=0. 

By the way, since relativistic momentum is given by p=mv/sqrt(1 - v^2/c^2), you can substitute this into the above equation to get E^2 = (m^2)(c^4) + (m^2)(v^2)(c^2)/(1 - v^2/c^2), and then if you take the first term on the right hand side, (m^2)(c^4), and multiply it by (1 - v^2/c^2)/(1 - v^2/c^2) and gather terms, you get E^2 = [(m^2)(c^4) - (m^2)(v^2)(c^2) + (m^2)(v^2)(c^2)]/(1 - v^2/c^2), and two terms cancel each other out so this simplifies to E^2 = (m^2)(c^4)/(1 - v^2/c^2), and then if you take the square root of both sides you get E = γmc^2. So the original equation for energy as a function fo rest mass m and relativistic momentum p can be rewritten as E=Mc^2 where M is the relativistic mass defined as M = γm, again showing that relativistic mass is only useful for rewriting equations involving relativistic momentum.

Jesse.

[Jesse Mazer]

I think this way of speaking just leads to confusion, which is why I'd prefer that you not designate any particular frame as being "at rest" and talk exclusively in a relative way about "the rest frame of the Earth" and "the rest frame of the spaceship". Even if you are stubborn about this and really want to designate one frame as being "at rest" and one as "moving", you are not even being consistent in your designation, since in your post immediately before this you said "Since motion is relative I can assume the spaceship is at rest". There would be much less room for confusion if you would just do as I suggest and use "rest" and "moving" in a relative way ('the rest frame of X', 'X is moving relative to Y' etc.), would you be willing to do this for my sake or do you absolutely refuse? Please answer this question directly, I've asked before and you've simply not responded.

Also, note that my question above was specifically about the *rod", which you didn't mention at all in your response above--I was asking about the meaning of your statement that the rod is "moving toward the Earth", can you please just answer yes or no if the rod is meant to be at rest relative to the Earth, and therefore moving at the same velocity as the Earth in the spaceship rest frame?

 

 

Now, from the pov of the rest frame, the spaceship, imagine a LT to determine the contraction of the rod.

Contraction of the rod in what frame?

I was explicit; from the rest frame of spaceship, apply the LT to determine the contraction of the rod, assuming you know its original length. Now ask yourself this question; in the rod frame, does the contraction information you've received from the LT correspond to any measurement result in the rod frame? (Answer; of course NOT!).

Assuming we start from the coordinates in the spaceship's rest frame and transform into the Earth's rest frame (and assuming the rod is at rest relative to the Earth), you will get the conclusion that the rod is LONGER in the Earth's rest frame than in the spaceship's rest frame, so I don't know what you mean by "the contraction information you've received from the LT".

 

This is what I've been trying to demonstrate; results given by the LT do NOT always give us what can be measured in the target frame of the LT, in this case the rod frame. AG (PS; I sent you a message on Facebook. You can reply here if you want.)

 

Are you saying that if we start with the coordinates of the ends of the rod in the spaceship frame (I think the rod is supposed to be moving in this frame, correct me if I'm wrong), and then use the LT to find the coordinates of the ends of the rod in the rod's own rest frame, you think we'd get the false prediction that the rod is contracted in the rod's rest frame?

 

The spaceship is in one frame, at relative rest WRT the rod; the rod is in another frame, in relative motion WRT to the spaceship,

How can the spaceship be at rest relative to the rod while the rod is in motion relative to the spaceship? Rest/motion are always symmetric in relativity as in classical mechanics--if X has a speed v in Y's rest frame (including v=0 in the case where X is at rest relative to Y), then Y always has the same speed v in X's rest frame. Did you just type this out wrong or are you really confused on this point?

Jesse

 

at the velocity which the spaceship was originally moving toward Andromeda. We know the rod's initial length and we calculate its contraction from the spaceship's frame, using the rod's relative velocity. AG

 

If so this is simply WRONG, that's not what you would find if you actually did the algebra of the Lorentz transformation equations. If you mean something different you need to be more clear about what frame we are starting with, what is the "target frame" we are transforming into using the LT, and which of these frames you want to know the "contraction of the rod" in.

Jesse

 

Note that in the frame of the moving rod, its contraction, predicted by the LT, cannot be measured because although the rod is moving, nothing within the rod frame is moving. This, IMO, contradicts the claim that the LT always yields what is actually measured in the target frame of the LT, in this case the frame of the rod. OK? AG

 

I think time dilation is an example of the opposite; a measurement which is realized in the target frame. AG 

 

but never length contraction as what's observed IN the rod frame. I realize that length contraction requires something moving, so it's not reasonable to expect this, although it does show there's an exception to what you allege the LT does. AG 

 

 

if we assume there is a frame A where Milky Way and Andromeda are both at rest (ignoring the fact that in reality they have some motion relative to one another), and another frame B where the rocket ship of the traveler is at rest, then in frame B the Milky Way/Andromeda distance is shortened relative to the distance in their rest frame, and the rocket has its maximum length; in frame A the the rocket's length is shortened relative to its length in its rest frame, and the Milky Way/Andromeda distance has its maximum value. The only asymmetry here is in the choice of the two things to measure the length of (the distance between the Milky Way and Andromeda in their rest frame is obviously huge compared to the rest length of a rocket moving between them), the symmetry might be easier to see if we consider two rockets traveling towards each other (their noses facing each other), and each wants to know the distance it must traverse to get from the nose of the other rocket to its tail. Then for example if each rocket is 10 meters long in its rest frame, and the two rockets have a relative velocity of 0.8c, each will measure only a 6 meter distance between the nose and tail of the other rocket, and the time they each measure to cross that distance is just (6 meters)/(0.8c).

Jesse

 

 

About mass, since the measured mass grows exponentially to infinity as v --> c, isn't this derivable from the LT, but in which frame? AG 

The notion of a variable relativistic mass is just an alternate way of talking about relativistic momentum, often modern textbooks talk solely about the latter and the only mass concept they use is the rest mass. For example the page at https://courses.lumenlearning.com/suny-physics/chapter/28-5-relativistic-momentum has a box titled "Misconception alert: relativistic mass and momentum" which says the following (note that they are using u to denote velocity):

"The relativistically correct definition of momentum as p = γmu is sometimes taken to imply that mass varies with velocity: m_var = γm, particularly in older textbooks. However, note that m is the mass of the object as measured by a person at rest relative to the object. Thus, m is defined to be the rest mass, which could be measured at rest, perhaps using gravity. When a mass is moving relative to an observer, the only way that its mass can be determined is through collisions or other means in which momentum is involved. Since the mass of a moving object cannot be determined independently of momentum, the only meaningful mass is rest mass. Thus, when we use the term mass, assume it to be identical to rest mass."

I'd say there's nothing strictly incorrect about defining a variable relativistic mass, it's just a cosmetically different formalism, but it may be that part of the reason it was mostly abandoned is because for people learning relativity it can lead to misconceptions that there is more to the concept than just a difference in how momentum is calculated, whereas in fact there is no application of relativistic mass that does not involve relativistic momentum. Momentum is needed for situations like collisions or particle creation/annihilation where there's a change in which objects have which individual momenta, but total momentum must be conserved. It's also used in the more general form of the relation of energy to rest mass m and relativistic momentum p, given by the equation E^2 = (mc^2)^2 + (pc)^2, which reduces to the more well-known E=mc^2 in the special case where p=0. 

By the way, since relativistic momentum is given by p=mv/sqrt(1 - v^2/c^2), you can substitute this into the above equation to get E^2 = (m^2)(c^4) + (m^2)(v^2)(c^2)/(1 - v^2/c^2), and then if you take the first term on the right hand side, (m^2)(c^4), and multiply it by (1 - v^2/c^2)/(1 - v^2/c^2) and gather terms, you get E^2 = [(m^2)(c^4) - (m^2)(v^2)(c^2) + (m^2)(v^2)(c^2)]/(1 - v^2/c^2), and two terms cancel each other out so this simplifies to E^2 = (m^2)(c^4)/(1 - v^2/c^2), and then if you take the square root of both sides you get E = γmc^2. So the original equation for energy as a function fo rest mass m and relativistic momentum p can be rewritten as E=Mc^2 where M is the relativistic mass defined as M = γm, again showing that relativistic mass is only useful for rewriting equations involving relativistic momentum.

Jesse.

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[Brent Meeker]

It doesn't matter whether your path is inertial or not.  You age by the proper time measured along that path.

Brent

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[Brent Meeker]

Well he didn't discover it.  Bernard Riemann did many years earlier.  Einstein's friend Marcel Grossman put Einstein onto it.  Most of those years of trial and error were spent trying figure out how to represent gravity in the equation and what function of the metric corresponded to what representation of gravity.

Brent

John K Clark    See what's on my new list at  Extropolis

tg1

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[Brent Meeker]

On 1/18/2025 5:42 AM, Alan Grayson wrote:

On Saturday, January 18, 2025 at 6:13:27 AM UTC-7 John Clark wrote:

On Fri, Jan 17, 2025 at 1:41 PM Alan Grayson <agrays...@gmail.com> wrote:

> IMO SR can handle curved spacetime. All one has to do is make the partitions very fine, so we're approximating inertial motion along very short paths. AG 

All one has to do? Well yes but that's easier said than done, it took Einstein 10 years of grueling work to figure out exactly how to do it, and the effort nearly killed him, he got sick, lost 50 pounds and figured he would die soon. Fortunately he did not. One of the most difficult things he had to figure out was how to measure distance in 4D non-Euclidean spacetime that was curved in any given way that was useful and never produced self-contradictory results. Mathematicians insist that distance must have the following properties:

1) Non-negativity: d(x,y) ≥ 0
2) Identity of indiscernibles: d(x,y) = 0 if and only if x = y
3) Symmetry: d(x,y) = d(y,x)
4) Triangle inequality: d(x,z) ≤ d(x,y) + d(y,z)

After years of false starts and dead ends Einstein eventually found a measuring stick that worked, it's called the Metric Tensor. 

John K Clark    See what's on my new list at  Extropolis

Sorry to consume so much bandwidth here. I have one question about the LT and one about the Metric Tensor (MT) which will hopefully resolve most of my confusions.

I'll pose the LT question in the context of the TP. If the stationary twin at rest on the Earth uses the LT to calculate the clock reading at some time on the traveling twin's clock, or its clock rate using two or more time readings, what relationship, if any, does this have on what the traveler twin's clock actually reads,

Actually reads when?

and what he observes as his clock rate?

He observes his clock rate to be one second per second.

IOW, to what extent does the LT transform the parameters of one frame, to actual observations on another frame

It transforms spatial and temporal intervals in one frame, say Earth's frame, into intervals in a frame moving relative to Earth.

(previously referred to as the source and target frames respectively)?

Concerning the MT; it's defined as bilinear map to the real numbers on vectors in the tangent vector space at each point on a manifold. So, when trying to solve Einstein's field equation, the experts claim one must start by calculating the MT, presumably for some given distribution of matter and energy.

That's for general relativity in which spacetime is not approximately flat, but it is for the twin paradox.

Since, in general, there exists an infinite set of pairs of vectors in the tangent vector space at each point on the manifold, and assuming the MT has a unique real value at each point on the manifold, at each point on the manifold how do we choose which pair of vectors to do the calculation?

If you have a pair of vectors you just get the product of projected lengths.  So for a path you're just interested in two copies of the vector along the path.  Then multiplying them with the metric tensor gives you the squared length of the vector.

Brent

TY, AG

tg1

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[Alan Grayson]

On Saturday, January 18, 2025 at 4:28:06 PM UTC-7 Brent Meeker wrote:

On 1/18/2025 5:42 AM, Alan Grayson wrote:

On Saturday, January 18, 2025 at 6:13:27 AM UTC-7 John Clark wrote:

On Fri, Jan 17, 2025 at 1:41 PM Alan Grayson <agrays...@gmail.com> wrote:

> IMO SR can handle curved spacetime. All one has to do is make the partitions very fine, so we're approximating inertial motion along very short paths. AG 

All one has to do? Well yes but that's easier said than done, it took Einstein 10 years of grueling work to figure out exactly how to do it, and the effort nearly killed him, he got sick, lost 50 pounds and figured he would die soon. Fortunately he did not. One of the most difficult things he had to figure out was how to measure distance in 4D non-Euclidean spacetime that was curved in any given way that was useful and never produced self-contradictory results. Mathematicians insist that distance must have the following properties:

1) Non-negativity: d(x,y) ≥ 0
2) Identity of indiscernibles: d(x,y) = 0 if and only if x = y
3) Symmetry: d(x,y) = d(y,x)
4) Triangle inequality: d(x,z) ≤ d(x,y) + d(y,z)

After years of false starts and dead ends Einstein eventually found a measuring stick that worked, it's called the Metric Tensor. 

John K Clark    See what's on my new list at  Extropolis

Sorry to consume so much bandwidth here. I have one question about the LT and one about the Metric Tensor (MT) which will hopefully resolve most of my confusions.

I'll pose the LT question in the context of the TP. If the stationary twin at rest on the Earth uses the LT to calculate the clock reading at some time on the traveling twin's clock, or its clock rate using two or more time readings, what relationship, if any, does this have on what the traveler twin's clock actually reads,

Actually reads when?

and what he observes as his clock rate?

He observes his clock rate to be one second per second.

IOW, using the LT, the stationary twin knows precisely what the traveling twin will measure for his clock rate, but the traveling twin detects nothing. You gotta luv it. AG

[Brent Meeker]

On 1/18/2025 4:32 PM, Alan Grayson wrote:

On Saturday, January 18, 2025 at 4:28:06 PM UTC-7 Brent Meeker wrote:

On 1/18/2025 5:42 AM, Alan Grayson wrote:

On Saturday, January 18, 2025 at 6:13:27 AM UTC-7 John Clark wrote:

On Fri, Jan 17, 2025 at 1:41 PM Alan Grayson <agrays...@gmail.com> wrote:

> IMO SR can handle curved spacetime. All one has to do is make the partitions very fine, so we're approximating inertial motion along very short paths. AG 

All one has to do? Well yes but that's easier said than done, it took Einstein 10 years of grueling work to figure out exactly how to do it, and the effort nearly killed him, he got sick, lost 50 pounds and figured he would die soon. Fortunately he did not. One of the most difficult things he had to figure out was how to measure distance in 4D non-Euclidean spacetime that was curved in any given way that was useful and never produced self-contradictory results. Mathematicians insist that distance must have the following properties:

1) Non-negativity: d(x,y) ≥ 0
2) Identity of indiscernibles: d(x,y) = 0 if and only if x = y
3) Symmetry: d(x,y) = d(y,x)
4) Triangle inequality: d(x,z) ≤ d(x,y) + d(y,z)

After years of false starts and dead ends Einstein eventually found a measuring stick that worked, it's called the Metric Tensor. 

John K Clark    See what's on my new list at  Extropolis

Sorry to consume so much bandwidth here. I have one question about the LT and one about the Metric Tensor (MT) which will hopefully resolve most of my confusions.

I'll pose the LT question in the context of the TP. If the stationary twin at rest on the Earth uses the LT to calculate the clock reading at some time on the traveling twin's clock, or its clock rate using two or more time readings, what relationship, if any, does this have on what the traveler twin's clock actually reads,

Actually reads when?

and what he observes as his clock rate?

He observes his clock rate to be one second per second.

IOW, using the LT, the stationary twin knows precisely what the traveling twin will measure for his clock rate, but the traveling twin detects nothing. You gotta luv it. AG

It's the same as length contraction, nobody ever measures time dilation with their own clock.

Brent

[Alan Grayson]

So the claim that the LT yields what the target frame -- in this case the frame of the traveling twin  -- will measure, is false, and why I called these LT measuring results APPARENT. AG

[Alan Grayson]

I was thinking the spaceship is relatively at rest, the rod is moving toward the Earth, and the LT is used by the spaceship to calculate the contraction of the rod. I figure the rod moving toward the Earth, with the spaceship in relative rest, is equivalent to a spaceship moving toward Andromeda. AG 

I tried to make it clear, very clear, but obviously I failed. Let's try this; since there's general agreement in the physics community that traveling very fast to Andromeda causes its distance from Earth to contract. So you tell me; from which frame would you'd like to measure the contraction? And since, as Brent just remarked, in the contracted frame, nothing that the LT determined, is measurable in the contracted frame, my claim is proven; that the results of the LT do NOT tell us what the target frame (in this case the frame of the contracted length) will measure. QED! AG  

[Brent Meeker]

That doesn't even parse.  Frames doesn't measure anything.  Relations for inertial observers are symmetric.  So A measures B's clocks are slow and his rulers are short, while B measures that A's clocks are slow and her rulers are short.

Brent

  -- will measure, is false, and why I called these LT measuring results APPARENT. AG

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[Alan Grayson]

You can nitpik my words, but the fact is that it's misleading to claim that the results of the LT are what observers in the target frame will measure. AG 

[Jesse Mazer]

"Relatively at rest" is a meaningless phrase unless you specify what *specific object* it is at rest relative to. Are you so attached to your own non-standard way of talking that you refuse to just use the formulations "at rest relative to X" and "moving relative to X" (where X is some object in the problem) as a small concession towards being understood by others? I have asked if you are willing to do this a bunch of times, last time I even asked you "please answer this question directly", but you continue to just ignore it. I'm not going to continue the discussion with you if you refuse to do me this simple courtesy, without it I genuinely can't understand what scenario you're envisioning.

 

the rod is moving toward the Earth,

And I likewise asked several times to specify if you mean the rod is moving *relative* to the Earth (i.e. the distance between the rod and the Earth is changing over time), or if you just mean it's moving in the direction of the Earth but at a fixed distance (i.e. it's at rest relative to the Earth). My question last time was "can you please just answer yes or no if the rod is meant to be at rest relative to the Earth, and therefore moving at the same velocity as the Earth in the spaceship rest frame?" If you aren't willing to answer simple yes-or-no questions like this, then again I'm not going to continue the discussion.

 

and the LT is used by the spaceship to calculate the contraction of the rod.

Contraction of the rod in which object's rest frame?

 

I tried to make it clear, very clear, but obviously I failed. Let's try this; since there's general agreement in the physics community that traveling very fast to Andromeda causes its distance from Earth to contract. So you tell me; from which frame would you'd like to measure the contraction?

In the spaceship frame, obviously.

 

And since, as Brent just remarked, in the contracted frame, nothing that the LT determined, is measurable in the contracted frame, my claim is proven;

"Contracted frame" is another unclear phrase, the distance from Earth to Andromeda is contracted in the spaceship frame but the spaceship itself is contracted in the Earth/Andromeda frame. *If* by "contracted frame" you mean the spaceship frame where the distance from Earth to Andromeda is contracted relative to the distance between them in their rest frame, then you need to understand that what is measured in the spaceship frame (using a system of rulers and clocks at rest in that frame, the clocks synchronized using the Einstein convention) is exactly what would be predicted if you started with the coordinates in some other frame and use the LT to transform into the spaceship frame. If you think Brent is saying otherwise, I can guarantee you've just misunderstood him.

 

that the results of the LT do NOT tell us what the target frame (in this case the frame of the contracted length) will measure. QED! AG  

 

Now, from the pov of the rest frame, the spaceship, imagine a LT to determine the contraction of the rod.

Contraction of the rod in what frame?

I was explicit; from the rest frame of spaceship, apply the LT to determine the contraction of the rod, assuming you know its original length. Now ask yourself this question; in the rod frame, does the contraction information you've received from the LT correspond to any measurement result in the rod frame? (Answer; of course NOT!).

Assuming we start from the coordinates in the spaceship's rest frame and transform into the Earth's rest frame (and assuming the rod is at rest relative to the Earth), you will get the conclusion that the rod is LONGER in the Earth's rest frame than in the spaceship's rest frame, so I don't know what you mean by "the contraction information you've received from the LT".

 

This is what I've been trying to demonstrate; results given by the LT do NOT always give us what can be measured in the target frame of the LT, in this case the rod frame. AG (PS; I sent you a message on Facebook. You can reply here if you want.)

 

Are you saying that if we start with the coordinates of the ends of the rod in the spaceship frame (I think the rod is supposed to be moving in this frame, correct me if I'm wrong), and then use the LT to find the coordinates of the ends of the rod in the rod's own rest frame, you think we'd get the false prediction that the rod is contracted in the rod's rest frame?

 

The spaceship is in one frame, at relative rest WRT the rod; the rod is in another frame, in relative motion WRT to the spaceship,

How can the spaceship be at rest relative to the rod while the rod is in motion relative to the spaceship? Rest/motion are always symmetric in relativity as in classical mechanics--if X has a speed v in Y's rest frame (including v=0 in the case where X is at rest relative to Y), then Y always has the same speed v in X's rest frame. Did you just type this out wrong or are you really confused on this point?

You also did not answer my question above--that point about two objects always having symmetrical judgments about whether the other object is at rest or moving (and each judging the other's speed to be identical) is a very basic and important one, if you don't understand this it's likely to lead to endless confusion. So if you do want to continue the discussion and are willing to answer yes-or-no questions, then please answer yes or no if you understand that if the spaceship is at rest relative to the rod, that guarantees that the rod is at rest relative to the spaceship.

Jesse

 

at the velocity which the spaceship was originally moving toward Andromeda. We know the rod's initial length and we calculate its contraction from the spaceship's frame, using the rod's relative velocity. AG

 

If so this is simply WRONG, that's not what you would find if you actually did the algebra of the Lorentz transformation equations. If you mean something different you need to be more clear about what frame we are starting with, what is the "target frame" we are transforming into using the LT, and which of these frames you want to know the "contraction of the rod" in.

Jesse

 

Note that in the frame of the moving rod, its contraction, predicted by the LT, cannot be measured because although the rod is moving, nothing within the rod frame is moving. This, IMO, contradicts the claim that the LT always yields what is actually measured in the target frame of the LT, in this case the frame of the rod. OK? AG

 

I think time dilation is an example of the opposite; a measurement which is realized in the target frame. AG 

 

but never length contraction as what's observed IN the rod frame. I realize that length contraction requires something moving, so it's not reasonable to expect this, although it does show there's an exception to what you allege the LT does. AG 

 

 

if we assume there is a frame A where Milky Way and Andromeda are both at rest (ignoring the fact that in reality they have some motion relative to one another), and another frame B where the rocket ship of the traveler is at rest, then in frame B the Milky Way/Andromeda distance is shortened relative to the distance in their rest frame, and the rocket has its maximum length; in frame A the the rocket's length is shortened relative to its length in its rest frame, and the Milky Way/Andromeda distance has its maximum value. The only asymmetry here is in the choice of the two things to measure the length of (the distance between the Milky Way and Andromeda in their rest frame is obviously huge compared to the rest length of a rocket moving between them), the symmetry might be easier to see if we consider two rockets traveling towards each other (their noses facing each other), and each wants to know the distance it must traverse to get from the nose of the other rocket to its tail. Then for example if each rocket is 10 meters long in its rest frame, and the two rockets have a relative velocity of 0.8c, each will measure only a 6 meter distance between the nose and tail of the other rocket, and the time they each measure to cross that distance is just (6 meters)/(0.8c).

Jesse

 

 

About mass, since the measured mass grows exponentially to infinity as v --> c, isn't this derivable from the LT, but in which frame? AG 

The notion of a variable relativistic mass is just an alternate way of talking about relativistic momentum, often modern textbooks talk solely about the latter and the only mass concept they use is the rest mass. For example the page at https://courses.lumenlearning.com/suny-physics/chapter/28-5-relativistic-momentum has a box titled "Misconception alert: relativistic mass and momentum" which says the following (note that they are using u to denote velocity):

"The relativistically correct definition of momentum as p = γmu is sometimes taken to imply that mass varies with velocity: m_var = γm, particularly in older textbooks. However, note that m is the mass of the object as measured by a person at rest relative to the object. Thus, m is defined to be the rest mass, which could be measured at rest, perhaps using gravity. When a mass is moving relative to an observer, the only way that its mass can be determined is through collisions or other means in which momentum is involved. Since the mass of a moving object cannot be determined independently of momentum, the only meaningful mass is rest mass. Thus, when we use the term mass, assume it to be identical to rest mass."

I'd say there's nothing strictly incorrect about defining a variable relativistic mass, it's just a cosmetically different formalism, but it may be that part of the reason it was mostly abandoned is because for people learning relativity it can lead to misconceptions that there is more to the concept than just a difference in how momentum is calculated, whereas in fact there is no application of relativistic mass that does not involve relativistic momentum. Momentum is needed for situations like collisions or particle creation/annihilation where there's a change in which objects have which individual momenta, but total momentum must be conserved. It's also used in the more general form of the relation of energy to rest mass m and relativistic momentum p, given by the equation E^2 = (mc^2)^2 + (pc)^2, which reduces to the more well-known E=mc^2 in the special case where p=0. 

By the way, since relativistic momentum is given by p=mv/sqrt(1 - v^2/c^2), you can substitute this into the above equation to get E^2 = (m^2)(c^4) + (m^2)(v^2)(c^2)/(1 - v^2/c^2), and then if you take the first term on the right hand side, (m^2)(c^4), and multiply it by (1 - v^2/c^2)/(1 - v^2/c^2) and gather terms, you get E^2 = [(m^2)(c^4) - (m^2)(v^2)(c^2) + (m^2)(v^2)(c^2)]/(1 - v^2/c^2), and two terms cancel each other out so this simplifies to E^2 = (m^2)(c^4)/(1 - v^2/c^2), and then if you take the square root of both sides you get E = γmc^2. So the original equation for energy as a function fo rest mass m and relativistic momentum p can be rewritten as E=Mc^2 where M is the relativistic mass defined as M = γm, again showing that relativistic mass is only useful for rewriting equations involving relativistic momentum.

Jesse.

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[Jesse Mazer]

I assume you'd agree an inertial frame's coordinates are often defined in intro textbooks in terms of local measurements on a system of rulers with clocks attached to different markings, all mutually at rest (and with the clocks synchronized in their rest frame using the Einstein convention), as in the illustration at http://www.upscale.utoronto.ca/GeneralInterest/Harrison/SpecRel/SpecRel.html#Exploring -- even if such systems are not constructed in practice, this is a way to conceive of an inertial frame's coordinates as physical "measurements" which could be done in principle. And if we know the coordinates assigned to events by the ruler/clock system corresponding to some frame A, and want to know the coordinates that'd be assigned to the same events by a ruler/clock system corresponding to some different frame B, we can just apply the Lorentz transformation to the coordinates in A to get the right answer for the coordinates in B--Alan's confusion here (one of them, anyway) is that he thinks there are situations where the Lorentz transformation would give the wrong answer, and it seems like he misinterpreted your comment above as supporting that. 

When you said "nobody ever measures time dilation with their own clock", did you mean that if there's a clock B in motion relative to me, I can't measure its time dilation with a *single* clock A at rest relative to me? But to clarify for Alan, presumably you'd agree I could in principle measure the time dilation of clock B in my frame with local measurements on a system of multiple clocks of the kind I mentioned, which are all at rest relative to me and pre-synchronized by the Einstein convention?

Jesse

[John Clark]

On Sat, Jan 18, 2025 at 8:42 AM Alan Grayson <agrays...@gmail.com> wrote:

> I'll pose the LT question in the context of the TP. If the stationary twin at rest on the Earth uses the LT to calculate the clock reading at some time on the traveling twin's clock, or its clock rate using two or more time readings, what relationship, if any, does this have on what the traveler twin's clock actually reads, and what he observes as his clock rate? IOW, to what extent does the LT transform the parameters of one frame, to actual observations on another frame (previously referred to as the source and target frames respectively)?

I'm not sure I understand your questions but the Lorentz transformations allow the earthbound twin to calculate what reading his spaceship traveling twin is reading on his clock right "now", and he can even calculate how much different his "now" is from his twins "now". And when he returns back to earth his earthbound twin is not at all surprised to see that his spaceship brother has aged much less than he has.

Although the components of the metric tensor do change under coordinate transformations, like when we change the origin of our coordinate system between observers that are moving relative to each other, the components of the metric tensor change in a way that preserves the invariant spacetime interval; and this is true even if it's not just a question of coordinates but even if there is a real physical difference between the environment of two observers, such as when spacetime is curved.

John K Clark    See what's on my new list at  Extropolis

3ma

 

[Alan Grayson]

The spaceship is moving toward Andromeda at some velocity v wrt the Earth. The rod is moving with velocity -v toward the Earth, while the spaceship is now at rest wrt the Earth. I am replacing a moving spacecraft with a moving rod of known initial length, the distance between Andromeda the the Earth, moving in the opposite direction. Then I am doing a LT from the stationary frame of the spacecraft to the moving frame of the rod, to calculate the rod's contraction from the pov of the frame of the spacecraft. I hope I have met your criterion for defining these frames of reference clearly.

But all this is unnecessary to prove my point; that in the frame of the rod, observers CANNOT measure its length contraction (or time dilation). This FACT is universally known and not in dispute, yet at the same time many people who claim to understand relativity assert that in the target frame of the LT, that is, in this case, in the frame of the rod, observers CAN measure what the LT implies. ThIs is FALSE! Of course, as you point out, from the pov of the rod frame, the spacecraft is contracted and its clock is dilated, but this fact is irrelevant to what I am alleging. Finally, I have NOT misunderstood Brent's recent comments on this very thread. You can read them yourself to verify my claim. Look at two or three of his recent short posts. I don't get it. Why do people who understand relativity keep affirming something -- what can be MEASURED in the target frame of the LT -- which isn't true? 

Another thing worth considering about the parking paradox: it's clear, as Clark pointed out not so long ago, that the paradox is caused by the assumption of universal time, specifically that the car fits and doesn't fit, AT THE SAME TIME. Using the disagreement about simultaneity this error is corrected and allegedly the paradox goes away. Yet Brent claims his plots show that the fitting and not fitting occur a the same time. I asked him more than once to explain this, but he hasn't replied. Do you know what's going on on this issue? 

AG

[Jesse Mazer]

Here you have again ignored the general question I ask about whether you are willing to phrase all your comments about rest/motion in a way that makes explicit what frame the statement is relative to (like 'moving relative to the Earth' which makes clear we are talking about motion in the Earth's rest frame). Are you willing to do this, yes or no? If you won't give me an answer, I'm not going to continue this discussion with you.

 

 

the rod is moving toward the Earth,

And I likewise asked several times to specify if you mean the rod is moving *relative* to the Earth (i.e. the distance between the rod and the Earth is changing over time), or if you just mean it's moving in the direction of the Earth but at a fixed distance (i.e. it's at rest relative to the Earth). My question last time was "can you please just answer yes or no if the rod is meant to be at rest relative to the Earth, and therefore moving at the same velocity as the Earth in the spaceship rest frame?" If you aren't willing to answer simple yes-or-no questions like this, then again I'm not going to continue the discussion.

 

and the LT is used by the spaceship to calculate the contraction of the rod.

Contraction of the rod in which object's rest frame?

 

I tried to make it clear, very clear, but obviously I failed. Let's try this; since there's general agreement in the physics community that traveling very fast to Andromeda causes its distance from Earth to contract. So you tell me; from which frame would you'd like to measure the contraction?

In the spaceship frame, obviously.

 

And since, as Brent just remarked, in the contracted frame, nothing that the LT determined, is measurable in the contracted frame, my claim is proven;

"Contracted frame" is another unclear phrase, the distance from Earth to Andromeda is contracted in the spaceship frame but the spaceship itself is contracted in the Earth/Andromeda frame. *If* by "contracted frame" you mean the spaceship frame where the distance from Earth to Andromeda is contracted relative to the distance between them in their rest frame, then you need to understand that what is measured in the spaceship frame (using a system of rulers and clocks at rest in that frame, the clocks synchronized using the Einstein convention) is exactly what would be predicted if you started with the coordinates in some other frame and use the LT to transform into the spaceship frame. If you think Brent is saying otherwise, I can guarantee you've just misunderstood him.

The spaceship is moving toward Andromeda at some velocity v wrt the Earth. The rod is moving with velocity -v toward the Earth,

While "velocity v wrt the Earth" is the sort of phrase I asked for, I have already commented in three previous posts about the ambiguity of your statement that the rod is moving "toward the Earth" and asked you to clarify, which you still have not done, instead you're just repeating the same ambiguous phrase. As I said before, "moving toward the Earth" could either mean "moving relative to the Earth" or it could mean "moving in the direction of the Earth's position, as seen in the spaceship's frame where both rod and Earth have the same velocity -v" (in the latter case where they both have the same velocity in the spaceship frame, this would imply the rod is at rest in the Earth's own frame). Will you please answer my question about whether the rod is moving *relative to* (or wrt) the Earth, yes or no?

In general it's frustrating that when I ask you several pointed questions, especially ones where I ask for a simple yes-or-no answer, you just try to restate your overall scenario in a way that you perhaps vaguely think addresses everything, without actually quoting my questions and giving individual responses to them. Can you please answer each question individually? I note that you also completely ignored the final question in my last post about the symmetry of rest vs. moving between frames (see my last comment below).

[BTW, an alternative way you could avoid all this verbal ambiguity would be to give an actual numerical example where you state the position as a function of time for each object in the scenario. For example, say the spaceship and Earth/Andromeda have a relative speed of 0.6c so that the distance from Earth to Andromeda is contracted by a factor of 0.8 in the spaceship rest frame, meaning if the distance is 2.5 Gly in the Earth/Andromeda rest frame (Gly = Giga-light-years, so 2.5 Gly = 2.5 million light years), then the distance between them would be 2 Gly in the ship rest frame. In that case, if we also use units of Giga-years for time so that c=1, then you could say something like "in the ship's rest frame, at t=0 the initial conditions are that the ship is at position x = 0, the Earth is also at x = 0, and Andromeda is at position x = -2, and the ship's position as a function of time in this frame is x(t) = 0, the Earth's position as a function of time is x(t) = 0.6*t, and the Andromeda galaxy's position as a function of time is x(t) = 0.6*t - 2". Then if you also gave the corresponding equations of motion for the front and back of the rod in the ship's rest frame (I'm still not clear on whether they'd be identical to the equations of motion for Earth and Andromeda, so that one end of the rod always coincides with Earth and the other end always coincides with Andromeda, or if they'd be different) then there would be no ambiguity about what the rod is supposed to be at rest relative to and what it is moving relative to.]

 

while the spaceship is now at rest wrt the Earth. I am replacing a moving spacecraft with a moving rod of known initial length, the distance between Andromeda the the Earth, moving in the opposite direction. Then I am doing a LT from the stationary frame of the spacecraft to the moving frame of the rod, to calculate the rod's contraction from the pov of the frame of the spacecraft. I hope I have met your criterion for defining these frames of reference clearly.

But all this is unnecessary to prove my point; that in the frame of the rod, observers CANNOT measure its length contraction (or time dilation). This FACT is universally known and not in dispute, yet at the same time many people who claim to understand relativity assert that in the target frame of the LT, that is, in this case, in the frame of the rod, observers CAN measure what the LT implies. ThIs is FALSE!

The LT simply doesn't "imply" there would be any length contraction of the rod "in the frame of the rod", so your premise here is completely wrong (if you had an actual numerical example with the equations of motion x(t) initially stated in the spaceship rest frame, as I suggested above, you could plug them into the LT equations directly and see what they predict about the equations of motion x'(t') in the rod's frame--I assume you have not actually done such an algebraic exercise and are just relying on some confused verbal argument to get the wrong idea that the LT would predict contraction of the rod in the rod's own frame). Whatever the LT implies about lengths/times in a specific inertial frame, it always corresponds exactly to what would actually be measured using a system of rulers and clocks which are at rest in that frame (the clocks synchronized by the Einstein convention), no exceptions.

 

Of course, as you point out, from the pov of the rod frame, the spacecraft is contracted and its clock is dilated, but this fact is irrelevant to what I am alleging. Finally, I have NOT misunderstood Brent's recent comments on this very thread. You can read them yourself to verify my claim. Look at two or three of his recent short posts. I don't get it. Why do people who understand relativity keep affirming something -- what can be MEASURED in the target frame of the LT -- which isn't true? 

Another thing worth considering about the parking paradox: it's clear, as Clark pointed out not so long ago, that the paradox is caused by the assumption of universal time, specifically that the car fits and doesn't fit, AT THE SAME TIME. Using the disagreement about simultaneity this error is corrected and allegedly the paradox goes away. Yet Brent claims his plots show that the fitting and not fitting occur a the same time. I asked him more than once to explain this, but he hasn't replied. Do you know what's going on on this issue? 

AG

 

that the results of the LT do NOT tell us what the target frame (in this case the frame of the contracted length) will measure. QED! AG  

 

Now, from the pov of the rest frame, the spaceship, imagine a LT to determine the contraction of the rod.

Contraction of the rod in what frame?

I was explicit; from the rest frame of spaceship, apply the LT to determine the contraction of the rod, assuming you know its original length. Now ask yourself this question; in the rod frame, does the contraction information you've received from the LT correspond to any measurement result in the rod frame? (Answer; of course NOT!).

Assuming we start from the coordinates in the spaceship's rest frame and transform into the Earth's rest frame (and assuming the rod is at rest relative to the Earth), you will get the conclusion that the rod is LONGER in the Earth's rest frame than in the spaceship's rest frame, so I don't know what you mean by "the contraction information you've received from the LT".

 

This is what I've been trying to demonstrate; results given by the LT do NOT always give us what can be measured in the target frame of the LT, in this case the rod frame. AG (PS; I sent you a message on Facebook. You can reply here if you want.)

 

Are you saying that if we start with the coordinates of the ends of the rod in the spaceship frame (I think the rod is supposed to be moving in this frame, correct me if I'm wrong), and then use the LT to find the coordinates of the ends of the rod in the rod's own rest frame, you think we'd get the false prediction that the rod is contracted in the rod's rest frame?

 

The spaceship is in one frame, at relative rest WRT the rod; the rod is in another frame, in relative motion WRT to the spaceship,

How can the spaceship be at rest relative to the rod while the rod is in motion relative to the spaceship? Rest/motion are always symmetric in relativity as in classical mechanics--if X has a speed v in Y's rest frame (including v=0 in the case where X is at rest relative to Y), then Y always has the same speed v in X's rest frame. Did you just type this out wrong or are you really confused on this point?

You also did not answer my question above--that point about two objects always having symmetrical judgments about whether the other object is at rest or moving (and each judging the other's speed to be identical) is a very basic and important one, if you don't understand this it's likely to lead to endless confusion. So if you do want to continue the discussion and are willing to answer yes-or-no questions, then please answer yes or no if you understand that if the spaceship is at rest relative to the rod, that guarantees that the rod is at rest relative to the spaceship.

Can you please address the individual yes-or-no question above, not as part of a longer statement mushing all your comments together but as a distinct answer to what I ask here?

Jesse

 

 

at the velocity which the spaceship was originally moving toward Andromeda. We know the rod's initial length and we calculate its contraction from the spaceship's frame, using the rod's relative velocity. AG

 

If so this is simply WRONG, that's not what you would find if you actually did the algebra of the Lorentz transformation equations. If you mean something different you need to be more clear about what frame we are starting with, what is the "target frame" we are transforming into using the LT, and which of these frames you want to know the "contraction of the rod" in.

Jesse

 

Note that in the frame of the moving rod, its contraction, predicted by the LT, cannot be measured because although the rod is moving, nothing within the rod frame is moving. This, IMO, contradicts the claim that the LT always yields what is actually measured in the target frame of the LT, in this case the frame of the rod. OK? AG

 

I think time dilation is an example of the opposite; a measurement which is realized in the target frame. AG 

 

but never length contraction as what's observed IN the rod frame. I realize that length contraction requires something moving, so it's not reasonable to expect this, although it does show there's an exception to what you allege the LT does. AG 

 

 

if we assume there is a frame A where Milky Way and Andromeda are both at rest (ignoring the fact that in reality they have some motion relative to one another), and another frame B where the rocket ship of the traveler is at rest, then in frame B the Milky Way/Andromeda distance is shortened relative to the distance in their rest frame, and the rocket has its maximum length; in frame A the the rocket's length is shortened relative to its length in its rest frame, and the Milky Way/Andromeda distance has its maximum value. The only asymmetry here is in the choice of the two things to measure the length of (the distance between the Milky Way and Andromeda in their rest frame is obviously huge compared to the rest length of a rocket moving between them), the symmetry might be easier to see if we consider two rockets traveling towards each other (their noses facing each other), and each wants to know the distance it must traverse to get from the nose of the other rocket to its tail. Then for example if each rocket is 10 meters long in its rest frame, and the two rockets have a relative velocity of 0.8c, each will measure only a 6 meter distance between the nose and tail of the other rocket, and the time they each measure to cross that distance is just (6 meters)/(0.8c).

Jesse

 

 

About mass, since the measured mass grows exponentially to infinity as v --> c, isn't this derivable from the LT, but in which frame? AG 

The notion of a variable relativistic mass is just an alternate way of talking about relativistic momentum, often modern textbooks talk solely about the latter and the only mass concept they use is the rest mass. For example the page at https://courses.lumenlearning.com/suny-physics/chapter/28-5-relativistic-momentum has a box titled "Misconception alert: relativistic mass and momentum" which says the following (note that they are using u to denote velocity):

"The relativistically correct definition of momentum as p = γmu is sometimes taken to imply that mass varies with velocity: m_var = γm, particularly in older textbooks. However, note that m is the mass of the object as measured by a person at rest relative to the object. Thus, m is defined to be the rest mass, which could be measured at rest, perhaps using gravity. When a mass is moving relative to an observer, the only way that its mass can be determined is through collisions or other means in which momentum is involved. Since the mass of a moving object cannot be determined independently of momentum, the only meaningful mass is rest mass. Thus, when we use the term mass, assume it to be identical to rest mass."

I'd say there's nothing strictly incorrect about defining a variable relativistic mass, it's just a cosmetically different formalism, but it may be that part of the reason it was mostly abandoned is because for people learning relativity it can lead to misconceptions that there is more to the concept than just a difference in how momentum is calculated, whereas in fact there is no application of relativistic mass that does not involve relativistic momentum. Momentum is needed for situations like collisions or particle creation/annihilation where there's a change in which objects have which individual momenta, but total momentum must be conserved. It's also used in the more general form of the relation of energy to rest mass m and relativistic momentum p, given by the equation E^2 = (mc^2)^2 + (pc)^2, which reduces to the more well-known E=mc^2 in the special case where p=0. 

By the way, since relativistic momentum is given by p=mv/sqrt(1 - v^2/c^2), you can substitute this into the above equation to get E^2 = (m^2)(c^4) + (m^2)(v^2)(c^2)/(1 - v^2/c^2), and then if you take the first term on the right hand side, (m^2)(c^4), and multiply it by (1 - v^2/c^2)/(1 - v^2/c^2) and gather terms, you get E^2 = [(m^2)(c^4) - (m^2)(v^2)(c^2) + (m^2)(v^2)(c^2)]/(1 - v^2/c^2), and two terms cancel each other out so this simplifies to E^2 = (m^2)(c^4)/(1 - v^2/c^2), and then if you take the square root of both sides you get E = γmc^2. So the original equation for energy as a function fo rest mass m and relativistic momentum p can be rewritten as E=Mc^2 where M is the relativistic mass defined as M = γm, again showing that relativistic mass is only useful for rewriting equations involving relativistic momentum.

Jesse

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[Brent Meeker]

No I meant that one doesn't measure time dilation in their own frame.  It's always relative.

Personally I don't like to invoke ruler/clock systems.  I think they muddy explanations because they introduce simultaneity, as measured in the ruler/clock system; while it's important to realize there is no physical significance to "simultaneous" for spacelike events.

Brent

But to clarify for Alan, presumably you'd agree I could in principle measure the time dilation of clock B in my frame with local measurements on a system of multiple clocks of the kind I mentioned, which are all at rest relative to me and pre-synchronized by the Einstein convention?

Jesse

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[Alan Grayson]

The rod is moving relative to the Earth, toward the Earth at some velocity v. The Earth is at rest and the rod is moving. And I've done those LT's in the past, and I've reviewed those done by Brent on his plots, so there's no need to do them again. Also, in all discussions about the Parking Paradox, we have one frame moving and the other at rest. So I don't see why  I can't have the rocket at rest, and the rod moving. You claim that is wrong, but it's done repeatedly in the Parking Paradox. But we don't need a rocket ship. We can just calculate using the LT from the Earth to determine the length contraction. Do you have a better method?  I surmise you haven't reviewed Brent's short posts on this thread. He concedes that one cannot MEASURE length contraction and time dilution in the target frame of the LT, because, as he says, nothing is moving WITHIN that frame. So I am not misinterpreting his words. AG

 

the rod is moving toward the Earth,

And I likewise asked several times to specify if you mean the rod is moving *relative* to the Earth (i.e. the distance between the rod and the Earth is changing over time), or if you just mean it's moving in the direction of the Earth but at a fixed distance (i.e. it's at rest relative to the Earth). My question last time was "can you please just answer yes or no if the rod is meant to be at rest relative to the Earth, and therefore moving at the same velocity as the Earth in the spaceship rest frame?" If you aren't willing to answer simple yes-or-no questions like this, then again I'm not going to continue the discussion.

 

and the LT is used by the spaceship to calculate the contraction of the rod.

Contraction of the rod in which object's rest frame?

 

I tried to make it clear, very clear, but obviously I failed. Let's try this; since there's general agreement in the physics community that traveling very fast to Andromeda causes its distance from Earth to contract. So you tell me; from which frame would you'd like to measure the contraction?

In the spaceship frame, obviously.

 

And since, as Brent just remarked, in the contracted frame, nothing that the LT determined, is measurable in the contracted frame, my claim is proven;

"Contracted frame" is another unclear phrase, the distance from Earth to Andromeda is contracted in the spaceship frame but the spaceship itself is contracted in the Earth/Andromeda frame. *If* by "contracted frame" you mean the spaceship frame where the distance from Earth to Andromeda is contracted relative to the distance between them in their rest frame, then you need to understand that what is measured in the spaceship frame (using a system of rulers and clocks at rest in that frame, the clocks synchronized using the Einstein convention) is exactly what would be predicted if you started with the coordinates in some other frame and use the LT to transform into the spaceship frame. If you think Brent is saying otherwise, I can guarantee you've just misunderstood him.

The spaceship is moving toward Andromeda at some velocity v wrt the Earth. The rod is moving with velocity -v toward the Earth,

While "velocity v wrt the Earth" is the sort of phrase I asked for, I have already commented in three previous posts about the ambiguity of your statement that the rod is moving "toward the Earth" and asked you to clarify, which you still have not done, instead you're just repeating the same ambiguous phrase. As I said before, "moving toward the Earth" could either mean "moving relative to the Earth" or it could mean "moving in the direction of the Earth's position, as seen in the spaceship's frame where both rod and Earth have the same velocity -v" (in the latter case where they both have the same velocity in the spaceship frame, this would imply the rod is at rest in the Earth's own frame). Will you please answer my question about whether the rod is moving *relative to* (or wrt) the Earth, yes or no?

In general it's frustrating that when I ask you several pointed questions, especially ones where I ask for a simple yes-or-no answer, you just try to restate your overall scenario in a way that you perhaps vaguely think addresses everything, without actually quoting my questions and giving individual responses to them. Can you please answer each question individually? I note that you also completely ignored the final question in my last post about the symmetry of rest vs. moving between frames (see my last comment below).

As previously stated, the LT is done from a frame at rest. This is consistently done by Brent and in all discussions of the Parking Paradox. So the symmetry you refer to, refers to how each frame can view the other from assuming it's at rest. IOW, from the car frame assuming the car is at rest, we can use the LT to determine what happens in the garage frame which we assume is moving and contracted, and vice-versa. AG

[BTW, an alternative way you could avoid all this verbal ambiguity would be to give an actual numerical example where you state the position as a function of time for each object in the scenario. For example, say the spaceship and Earth/Andromeda have a relative speed of 0.6c so that the distance from Earth to Andromeda is contracted by a factor of 0.8 in the spaceship rest frame, meaning if the distance is 2.5 Gly in the Earth/Andromeda rest frame (Gly = Giga-light-years, so 2.5 Gly = 2.5 million light years), then the distance between them would be 2 Gly in the ship rest frame. In that case, if we also use units of Giga-years for time so that c=1, then you could say something like "in the ship's rest frame, at t=0 the initial conditions are that the ship is at position x = 0, the Earth is also at x = 0, and Andromeda is at position x = -2, and the ship's position as a function of time in this frame is x(t) = 0, the Earth's position as a function of time is x(t) = 0.6*t, and the Andromeda galaxy's position as a function of time is x(t) = 0.6*t - 2". Then if you also gave the corresponding equations of motion for the front and back of the rod in the ship's rest frame (I'm still not clear on whether they'd be identical to the equations of motion for Earth and Andromeda, so that one end of the rod always coincides with Earth and the other end always coincides with Andromeda, or if they'd be different) then there would be no ambiguity about what the rod is supposed to be at rest relative to and what it is moving relative to.]

 

while the spaceship is now at rest wrt the Earth. I am replacing a moving spacecraft with a moving rod of known initial length, the distance between Andromeda the the Earth, moving in the opposite direction. Then I am doing a LT from the stationary frame of the spacecraft to the moving frame of the rod, to calculate the rod's contraction from the pov of the frame of the spacecraft. I hope I have met your criterion for defining these frames of reference clearly.

But all this is unnecessary to prove my point; that in the frame of the rod, observers CANNOT measure its length contraction (or time dilation). This FACT is universally known and not in dispute, yet at the same time many people who claim to understand relativity assert that in the target frame of the LT, that is, in this case, in the frame of the rod, observers CAN measure what the LT implies. ThIs is FALSE!

The LT simply doesn't "imply" there would be any length contraction of the rod "in the frame of the rod",

Then what does the LT do? AG

 

so your premise here is completely wrong (if you had an actual numerical example with the equations of motion x(t) initially stated in the spaceship rest frame, as I suggested above, you could plug them into the LT equations directly and see what they predict about the equations of motion x'(t') in the rod's frame--I assume you have not actually done such an algebraic exercise and are just relying on some confused verbal argument to get the wrong idea that the LT would predict contraction of the rod in the rod's own frame).

That's NOT my claim but what those allegedly knowledgeable about SR claim; that the results of the LT give us what is actually MEASURED in the target frame. In fact, that's what you claimed in some post on this subject. AG

 

Whatever the LT implies about lengths/times in a specific inertial frame, it always corresponds exactly to what would actually be measured using a system of rulers and clocks which are at rest in that frame (the clocks synchronized by the Einstein convention), no exceptions.

Brent was explicit; one cannot MEASURE length contraction or time dilation in the target frame, because there is no motion WITHIN that frame. AG

[Jesse Mazer]

Do you just mean one doesn't measure any time dilation of a clock in that clock's own rest frame? I'd of course agree with that, but I was making the different point that an observer measuring a clock B in motion relative to themselves (using their own clocks at rest relative to themselves to measure it) can measure the time dilation of clock B in their frame, by comparing readings on B with readings on their own clocks as B passes by multiple clocks. And naturally this does not mean that B's time is dilated in any objective frame-independent sense, it's a frame-dependent measurement.

Personally I don't like to invoke ruler/clock systems.  I think they muddy explanations because they introduce simultaneity, as measured in the ruler/clock system; while it's important to realize there is no physical significance to "simultaneous" for spacelike events.

Sure, assigning coordinates in an inertial frame using such a ruler/clock system always depends on a purely conventional definition of simultaneity in that frame (the Einstein synchronization convention), but it would still be a type of physical measurement (and other measurements of frame-dependent quantities, like velocity, also depend on conventions). Alan seems to interpret you as making a broader claim that the coordinates obtained by the Lorentz transform can't be assigned meaning in terms of any kind of physical measurement whatsoever, but I assume that's not what you're saying.

Jesse

But to clarify for Alan, presumably you'd agree I could in principle measure the time dilation of clock B in my frame with local measurements on a system of multiple clocks of the kind I mentioned, which are all at rest relative to me and pre-synchronized by the Einstein convention?

Jesse

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[Jesse Mazer]

OK, so both the rod and the spaceship are moving relative to the Earth? Do they both have the same velocity in the Earth rest frame (so they are at rest relative to each other), or different velocities? If different, please give some sort of numerical example so we can actually do some calculations--for example if the spaceship is moving at 0.6c in the -x direction in the Earth rest frame, how fast is the rod moving in this frame, and in what direction?

Your original problem involved the distance from Earth to Andromeda galaxy, I thought the rod was just supposed to measure that distance by stretching from Earth to Andromeda--if not, what's the point of adding it to the problem? Why not just talk about how the distance from Earth to Andromeda looks in the spaceship frame vs. the Earth frame, why do we also need to add the idea of a rod whose length is completely unrelated to the Earth/Andromeda distance?

 

The Earth is at rest and the rod is moving. And I've done those LT's in the past, and I've reviewed those done by Brent on his plots, so there's no need to do them again. Also, in all discussions about the Parking Paradox, we have one frame moving and the other at rest. So I don't see why  I can't have the rocket at rest, and the rod moving. You claim that is wrong,

No, I have never claimed it's wrong to use a frame where the rocket is at rest and the rod moving, I just said you're wrong to say that when you start from this frame and use the LT to transform into the rod's frame, the rod will be CONTRACTED--in fact, if you actually do the math of the LT in such an example, you will find that the length of the rod in its own rest frame is predicted to be EXPANDED relative to its length in the rocket frame where the rod is moving.

 

but it's done repeatedly in the Parking Paradox. But we don't need a rocket ship. We can just calculate using the LT from the Earth to determine the length contraction. Do you have a better method?  I surmise you haven't reviewed Brent's short posts on this thread. He concedes that one cannot MEASURE length contraction and time dilution in the target frame of the LT, because, as he says, nothing is moving WITHIN that frame. So I am not misinterpreting his words. AG

I have been asking him to clarify, we'll see what he says, but I still think you're misunderstanding. In any case, *I* am definitely saying that the coordinates of any frame can be understood in terms of measurements on a ruler/clock system at rest in that frame, and this is a standard notion in relativity textbooks.

 

 

the rod is moving toward the Earth,

And I likewise asked several times to specify if you mean the rod is moving *relative* to the Earth (i.e. the distance between the rod and the Earth is changing over time), or if you just mean it's moving in the direction of the Earth but at a fixed distance (i.e. it's at rest relative to the Earth). My question last time was "can you please just answer yes or no if the rod is meant to be at rest relative to the Earth, and therefore moving at the same velocity as the Earth in the spaceship rest frame?" If you aren't willing to answer simple yes-or-no questions like this, then again I'm not going to continue the discussion.

 

and the LT is used by the spaceship to calculate the contraction of the rod.

Contraction of the rod in which object's rest frame?

 

I tried to make it clear, very clear, but obviously I failed. Let's try this; since there's general agreement in the physics community that traveling very fast to Andromeda causes its distance from Earth to contract. So you tell me; from which frame would you'd like to measure the contraction?

In the spaceship frame, obviously.

 

And since, as Brent just remarked, in the contracted frame, nothing that the LT determined, is measurable in the contracted frame, my claim is proven;

"Contracted frame" is another unclear phrase, the distance from Earth to Andromeda is contracted in the spaceship frame but the spaceship itself is contracted in the Earth/Andromeda frame. *If* by "contracted frame" you mean the spaceship frame where the distance from Earth to Andromeda is contracted relative to the distance between them in their rest frame, then you need to understand that what is measured in the spaceship frame (using a system of rulers and clocks at rest in that frame, the clocks synchronized using the Einstein convention) is exactly what would be predicted if you started with the coordinates in some other frame and use the LT to transform into the spaceship frame. If you think Brent is saying otherwise, I can guarantee you've just misunderstood him.

The spaceship is moving toward Andromeda at some velocity v wrt the Earth. The rod is moving with velocity -v toward the Earth,

While "velocity v wrt the Earth" is the sort of phrase I asked for, I have already commented in three previous posts about the ambiguity of your statement that the rod is moving "toward the Earth" and asked you to clarify, which you still have not done, instead you're just repeating the same ambiguous phrase. As I said before, "moving toward the Earth" could either mean "moving relative to the Earth" or it could mean "moving in the direction of the Earth's position, as seen in the spaceship's frame where both rod and Earth have the same velocity -v" (in the latter case where they both have the same velocity in the spaceship frame, this would imply the rod is at rest in the Earth's own frame). Will you please answer my question about whether the rod is moving *relative to* (or wrt) the Earth, yes or no?

In general it's frustrating that when I ask you several pointed questions, especially ones where I ask for a simple yes-or-no answer, you just try to restate your overall scenario in a way that you perhaps vaguely think addresses everything, without actually quoting my questions and giving individual responses to them. Can you please answer each question individually? I note that you also completely ignored the final question in my last post about the symmetry of rest vs. moving between frames (see my last comment below).

As previously stated, the LT is done from a frame at rest. This is consistently done by Brent and in all discussions of the Parking Paradox. So the symmetry you refer to, refers to how each frame can view the other from assuming it's at rest.

My question about symmetry was responding to your previous statement "The spaceship is in one frame, at relative rest WRT the rod; the rod is in another frame, in relative motion WRT to the spaceship". Do you agree or disagree that if the spaceship is at rest WRT to the rod, then automatically that must mean the rod is at rest WRT the spaceship?

 

IOW, from the car frame assuming the car is at rest, we can use the LT to determine what happens in the garage frame which we assume is moving and contracted, and vice-versa. AG

[BTW, an alternative way you could avoid all this verbal ambiguity would be to give an actual numerical example where you state the position as a function of time for each object in the scenario. For example, say the spaceship and Earth/Andromeda have a relative speed of 0.6c so that the distance from Earth to Andromeda is contracted by a factor of 0.8 in the spaceship rest frame, meaning if the distance is 2.5 Gly in the Earth/Andromeda rest frame (Gly = Giga-light-years, so 2.5 Gly = 2.5 million light years), then the distance between them would be 2 Gly in the ship rest frame. In that case, if we also use units of Giga-years for time so that c=1, then you could say something like "in the ship's rest frame, at t=0 the initial conditions are that the ship is at position x = 0, the Earth is also at x = 0, and Andromeda is at position x = -2, and the ship's position as a function of time in this frame is x(t) = 0, the Earth's position as a function of time is x(t) = 0.6*t, and the Andromeda galaxy's position as a function of time is x(t) = 0.6*t - 2". Then if you also gave the corresponding equations of motion for the front and back of the rod in the ship's rest frame (I'm still not clear on whether they'd be identical to the equations of motion for Earth and Andromeda, so that one end of the rod always coincides with Earth and the other end always coincides with Andromeda, or if they'd be different) then there would be no ambiguity about what the rod is supposed to be at rest relative to and what it is moving relative to.]

 

while the spaceship is now at rest wrt the Earth. I am replacing a moving spacecraft with a moving rod of known initial length, the distance between Andromeda the the Earth, moving in the opposite direction. Then I am doing a LT from the stationary frame of the spacecraft to the moving frame of the rod, to calculate the rod's contraction from the pov of the frame of the spacecraft. I hope I have met your criterion for defining these frames of reference clearly.

But all this is unnecessary to prove my point; that in the frame of the rod, observers CANNOT measure its length contraction (or time dilation). This FACT is universally known and not in dispute, yet at the same time many people who claim to understand relativity assert that in the target frame of the LT, that is, in this case, in the frame of the rod, observers CAN measure what the LT implies. ThIs is FALSE!

The LT simply doesn't "imply" there would be any length contraction of the rod "in the frame of the rod",

Then what does the LT do? AG

It transforms coordinates from one frame to another. If you use these coordinates to define the length of an object in both frames, then depending on the object's velocity in the starting frame, sometimes a given object will be shorter in the target frame than it was in the starting frame, and sometimes it will be longer in the target frame than it was in the starting frame. Only if the object is at rest in the starting frame can we be confident without any further details that the object will be shorter in the target frame where it's moving.

 

 

so your premise here is completely wrong (if you had an actual numerical example with the equations of motion x(t) initially stated in the spaceship rest frame, as I suggested above, you could plug them into the LT equations directly and see what they predict about the equations of motion x'(t') in the rod's frame--I assume you have not actually done such an algebraic exercise and are just relying on some confused verbal argument to get the wrong idea that the LT would predict contraction of the rod in the rod's own frame).

That's NOT my claim but what those allegedly knowledgeable about SR claim; that the results of the LT give us what is actually MEASURED in the target frame. In fact, that's what you claimed in some post on this subject. AG

Yes, that's what I still claim. The premise of yours that is wrong is that when we apply the LT we predict the rod is CONTRACTED in its own rest frame.

 

 

Whatever the LT implies about lengths/times in a specific inertial frame, it always corresponds exactly to what would actually be measured using a system of rulers and clocks which are at rest in that frame (the clocks synchronized by the Einstein convention), no exceptions.

Brent was explicit; one cannot MEASURE length contraction or time dilation in the target frame, because there is no motion WITHIN that frame. AG

See my last question to him, I think he was only talking about a ruler or clock at rest in the frame we're talking about, in which case I agree you won't measure any length contraction or time dilation of an object in its own rest frame.

Jesse

 

 

Of course, as you point out, from the pov of the rod frame, the spacecraft is contracted and its clock is dilated, but this fact is irrelevant to what I am alleging. Finally, I have NOT misunderstood Brent's recent comments on this very thread. You can read them yourself to verify my claim. Look at two or three of his recent short posts. I don't get it. Why do people who understand relativity keep affirming something -- what can be MEASURED in the target frame of the LT -- which isn't true? 

Another thing worth considering about the parking paradox: it's clear, as Clark pointed out not so long ago, that the paradox is caused by the assumption of universal time, specifically that the car fits and doesn't fit, AT THE SAME TIME. Using the disagreement about simultaneity this error is corrected and allegedly the paradox goes away. Yet Brent claims his plots show that the fitting and not fitting occur a the same time. I asked him more than once to explain this, but he hasn't replied. Do you know what's going on on this issue? 

AG

 

that the results of the LT do NOT tell us what the target frame (in this case the frame of the contracted length) will measure. QED! AG  

 

Now, from the pov of the rest frame, the spaceship, imagine a LT to determine the contraction of the rod.

Contraction of the rod in what frame?

I was explicit; from the rest frame of spaceship, apply the LT to determine the contraction of the rod, assuming you know its original length. Now ask yourself this question; in the rod frame, does the contraction information you've received from the LT correspond to any measurement result in the rod frame? (Answer; of course NOT!).

Assuming we start from the coordinates in the spaceship's rest frame and transform into the Earth's rest frame (and assuming the rod is at rest relative to the Earth), you will get the conclusion that the rod is LONGER in the Earth's rest frame than in the spaceship's rest frame, so I don't know what you mean by "the contraction information you've received from the LT".

 

This is what I've been trying to demonstrate; results given by the LT do NOT always give us what can be measured in the target frame of the LT, in this case the rod frame. AG (PS; I sent you a message on Facebook. You can reply here if you want.)

 

Are you saying that if we start with the coordinates of the ends of the rod in the spaceship frame (I think the rod is supposed to be moving in this frame, correct me if I'm wrong), and then use the LT to find the coordinates of the ends of the rod in the rod's own rest frame, you think we'd get the false prediction that the rod is contracted in the rod's rest frame?

 

The spaceship is in one frame, at relative rest WRT the rod; the rod is in another frame, in relative motion WRT to the spaceship,

How can the spaceship be at rest relative to the rod while the rod is in motion relative to the spaceship? Rest/motion are always symmetric in relativity as in classical mechanics--if X has a speed v in Y's rest frame (including v=0 in the case where X is at rest relative to Y), then Y always has the same speed v in X's rest frame. Did you just type this out wrong or are you really confused on this point?

You also did not answer my question above--that point about two objects always having symmetrical judgments about whether the other object is at rest or moving (and each judging the other's speed to be identical) is a very basic and important one, if you don't understand this it's likely to lead to endless confusion. So if you do want to continue the discussion and are willing to answer yes-or-no questions, then please answer yes or no if you understand that if the spaceship is at rest relative to the rod, that guarantees that the rod is at rest relative to the spaceship.

Can you please address the individual yes-or-no question above, not as part of a longer statement mushing all your comments together but as a distinct answer to what I ask here?

Jesse

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[Alan Grayson]

You got my position totally wrong! That wasn't my position. It is contracted in its own frame, and we know that from necessary corrections for GR and SR effects on GPS clocks. What I claimed is that you can't MEASURE the contraction in its own frame. You claim it's possible. Brent says it's not possible, because there is no motion WITHIN that frame. Further, our discussion about travel to Andromeda is totally unnessary.  I was just trying to develop a model for measuring the length contraction to Andromeda. We don't need a spacecraft. We just need the Earth at rest, and a rod moving toward the Earth, wrt the Earth, from which we use the LT. The moving rod is equivalent to a spaceship traveling to Andromeda, so the rod moves toward the Earth since, as seen by the traveler it moves in the opposite direction. Finally, about symmetry; what's used in the Parking Paradox is how symmetry exists between frames. The source frame, from which the LT is applied, is always at relative rest wrt the moving frame, which is always length contracted, and vice-versa. AG 

[Jesse Mazer]

No it isn't! "It is contracted in its own frame" is exactly the position of yours I'm saying is completely wrong, assuming we are talking about inertial frames. If you want to actually look at the math so I can prove to you that the LT does *not* predict the rod is contracted in its own inertial frame, you'll need to fill in some specific numbers for things like the length of the rod and its speed and direction relative to the Earth. 

 

and we know that from necessary corrections for GR and SR effects on GPS clocks. What I claimed is that you can't MEASURE the contraction in its own frame. You claim it's possible.

No, I never claimed you could "measure the contraction in its own frame", I claim it is *not* contracted in its own inertial frame according to the LT, so there is no predicted contraction to measure. GPS clocks are irrelevant here because the GPS system does not use an inertial frame for defining the motion and time dilation of the clocks, it uses a different kind of non-inertial coordinate system (inertial coordinate systems are not even possible in large regions where the shape of spacetime has been curved by mass/energy as GR predicts, you can only use them in the flat spacetime of SR which has no curvature). The LT only deals with the relationship between different inertial frames.

 

Brent says it's not possible, because there is no motion WITHIN that frame. Further, our discussion about travel to Andromeda is totally unnessary.  I was just trying to develop a model for measuring the length contraction to Andromeda. We don't need a spacecraft. We just need the Earth at rest, and a rod moving toward the Earth, wrt the Earth, from which we use the LT. The moving rod is equivalent to a spaceship traveling to Andromeda, so the rod moves toward the Earth since, as seen by the traveler it moves in the opposite direction.

OK, give me a specific number for the rod's speed in the Earth's inertial rest frame, its direction (in the +x or -x direction), and the initial position of each end of the rod at t=0 in the Earth frame. I can then show you the math proving that according to the LT the rod is *longer* in its own inertial rest frame than it is in the Earth frame, it isn't predicted to be contracted in its own rest frame.

Finally, about symmetry; what's used in the Parking Paradox is how symmetry exists between frames. The source frame, from which the LT is applied, is always at relative rest wrt the moving frame,

Are you using "relative rest" in some bizarre new way (for example, do you just mean that you want to verbally designate the source frame as the one which is 'at rest'?), or are you talking about the standard meaning, where it's just talking about the velocity of one thing relative to the other? For example, if we label what you call "the source frame" with A, and label the frame you call "the moving frame" with B, the usual meaning of "A and B are at relative rest" would be that B has a velocity of 0 in the A frame. If that's *not* what you meant, you really need to learn the standard way of talking about these things and not keep inventing your own confusing terminology, you shouldn't expect other people to be mind-readers.

 

which is always length contracted, and vice-versa. AG 

This still doesn't answer my question about your prior statement "The spaceship is in one frame, at relative rest WRT the rod; the rod is in another frame, in relative motion WRT to the spaceship". I asked if you AGREE or DISAGREE with my statement "if the spaceship is at rest WRT to the rod, then automatically that must mean the rod is at rest WRT the spaceship". Don't give me a meandering answer, just tell me if you AGREE or DISAGREE with that statement!

Jesse

[Alan Grayson]

??? Really? But you seem to deny it in subsequent statements. AG

 

The premise of yours that is wrong is that when we apply the LT we predict the rod is CONTRACTED in its own rest frame.

You got my position totally wrong! That wasn't my position. It is contracted in its own frame,

No it isn't! "It is contracted in its own frame" is exactly the position of yours I'm saying is completely wrong, assuming we are talking about inertial frames.

By parsing my statement you misconstrued my meaning. AG

 

If you want to actually look at the math so I can prove to you that the LT does *not* predict the rod is contracted in its own inertial frame, you'll need to fill in some specific numbers for things like the length of the rod and its speed and direction relative to the Earth. 

 

and we know that from necessary corrections for GR and SR effects on GPS clocks. What I claimed is that you can't MEASURE the contraction in its own frame. You claim it's possible.

No, I never claimed you could "measure the contraction in its own frame", I claim it is *not* contracted in its own inertial frame according to the LT, so there is no predicted contraction to measure.

As I clearly recall, you claimed the LT gives us what is actually MEASURED in the target frame, and I objected. If there's no physical contraction that can be MEASURED in the target frame, then it's reasonable to say the LT gives how the situation APPEARS from the pov of the source frame. Nonetheless, I think the result is physically real in the target frame even if it can't be measured.  AG

 

GPS clocks are irrelevant here because the GPS system does not use an inertial frame for defining the motion and time dilation of the clocks, it uses a different kind of non-inertial coordinate system (inertial coordinate systems are not even possible in large regions where the shape of spacetime has been curved by mass/energy as GR predicts, you can only use them in the flat spacetime of SR which has no curvature). The LT only deals with the relationship between different inertial frames.

Satellites in orbit are in freefall, which is inertial motion, and I'm pretty sure clock corrections are done for GR and SR effects. AG 

 

Brent says it's not possible, because there is no motion WITHIN that frame. Further, our discussion about travel to Andromeda is totally unnessary.  I was just trying to develop a model for measuring the length contraction to Andromeda. We don't need a spacecraft. We just need the Earth at rest, and a rod moving toward the Earth, wrt the Earth, from which we use the LT. The moving rod is equivalent to a spaceship traveling to Andromeda, so the rod moves toward the Earth since, as seen by the traveler it moves in the opposite direction.

OK, give me a specific number for the rod's speed in the Earth's inertial rest frame, its direction (in the +x or -x direction), and the initial position of each end of the rod at t=0 in the Earth frame. I can then show you the math proving that according to the LT the rod is *longer* in its own inertial rest frame than it is in the Earth frame, it isn't predicted to be contracted in its own rest frame.

I never claimed that the source frame is contracted in its own rest frame. If the Earth is at rest and the rod moving wrt the Earth, the LT says the length of rod is contracted from the pov of source frame, as shown by Brent's calculation text in his plots. See Brent's post on Dec 10, 2024, 11:15:16 PM, on thread Length Contraction in SR (again). What happens in target frame is what's under discussion. I claim the contraction is not measurable in the target frame but likely is a real physical occurrence. He does specific calculations so I suggest you view his text on the calculations, which show contraction of length in target frame. I don't see anything gained if I produce any calculations. AG

Finally, about symmetry; what's used in the Parking Paradox is how symmetry exists between frames. The source frame, from which the LT is applied, is always at relative rest wrt the moving frame,

Are you using "relative rest" in some bizarre new way (for example, do you just mean that you want to verbally designate the source frame as the one which is 'at rest'?),

 

Yes, but not bizarre. This is the frame which applies the LT.  AG

 

or are you talking about the standard meaning, where it's just talking about the velocity of one thing relative to the other? For example, if we label what you call "the source frame" with A, and label the frame you call "the moving frame" with B, the usual meaning of "A and B are at relative rest" would be that B has a velocity of 0 in the A frame. If that's *not* what you meant, you really need to learn the standard way of talking about these things and not keep inventing your own confusing terminology, you shouldn't expect other people to be mind-readers.

You seem to be splitting hairs. I'm not doing anything bizarre; I am doing what Brent did in the Parking Paradox and you didn't object. The LT is applied from a frame at rest, or if you prefer relatively at rest, wrt a moving frame. AG 

 

which is always length contracted, and vice-versa. AG 

This still doesn't answer my question about your prior statement "The spaceship is in one frame, at relative rest WRT the rod; the rod is in another frame, in relative motion WRT to the spaceship". I asked if you AGREE or DISAGREE with my statement "if the spaceship is at rest WRT to the rod, then automatically that must mean the rod is at rest WRT the spaceship". Don't give me a meandering answer, just tell me if you AGREE or DISAGREE with that statement!

I definitely disagree. AG 

Jesse

[Jesse Mazer]

No I don't. What statements of mine are you talking about?   

 

The premise of yours that is wrong is that when we apply the LT we predict the rod is CONTRACTED in its own rest frame.

You got my position totally wrong! That wasn't my position. It is contracted in its own frame,

No it isn't! "It is contracted in its own frame" is exactly the position of yours I'm saying is completely wrong, assuming we are talking about inertial frames.

By parsing my statement you misconstrued my meaning. AG

How so? You said the rod is predicted to be "contracted in its own frame" using the LT, didn't you? If so, that's precisely the claim I deny.

 

 

If you want to actually look at the math so I can prove to you that the LT does *not* predict the rod is contracted in its own inertial frame, you'll need to fill in some specific numbers for things like the length of the rod and its speed and direction relative to the Earth. 

 

and we know that from necessary corrections for GR and SR effects on GPS clocks. What I claimed is that you can't MEASURE the contraction in its own frame. You claim it's possible.

No, I never claimed you could "measure the contraction in its own frame", I claim it is *not* contracted in its own inertial frame according to the LT, so there is no predicted contraction to measure.

As I clearly recall, you claimed the LT gives us what is actually MEASURED in the target frame, and I objected. If there's no physical contraction that can be MEASURED in the target frame, then it's reasonable to say the LT gives how the situation APPEARS from the pov of the source frame. Nonetheless, I think the result is physically real in the target frame even if it can't be measured.  AG

No, what is measured matches exactly with what is predicted by the LT. I say that (1) The LT predicts there is *no* physical contraction of the rod in the rod's frame (i.e. the inertial frame where the rod has a velocity of 0), and (2) measurements show there is no physical contraction of the rod in the rod's frame. You apparently assume (1a) the LT predicts there *is* physical contraction of the rod in the rod's frame, and that's why you think there is a conflict with (2). If so, the problem is just that (1a) is false as a statement about what the LT predict.

 

 

GPS clocks are irrelevant here because the GPS system does not use an inertial frame for defining the motion and time dilation of the clocks, it uses a different kind of non-inertial coordinate system (inertial coordinate systems are not even possible in large regions where the shape of spacetime has been curved by mass/energy as GR predicts, you can only use them in the flat spacetime of SR which has no curvature). The LT only deals with the relationship between different inertial frames.

Satellites in orbit are in freefall, which is inertial motion, and I'm pretty sure clock corrections are done for GR and SR effects. AG 

In GR there is the concept of a "local inertial frame", which deals only with measurements in an infinitesimal neighborhood of spacetime around a single point, where curvature can be neglected (see for example the article at https://www.einstein-online.info/en/spotlight/equivalence_principle/ which says 'Within an infinitely small (“infinitesimal”) spacetime region, one can always find a reference frame – an infinitely small elevator cabin, observed over an infinitely brief period of time – in which the laws of physics are the same as in special relativity'). In this sense a satellite in free-fall can be said to be moving "inertially" as measured in a local inertial frame in the infinitesimal neighborhood of any given point on the satellite's worldline. But any coordinate system large enough to cover a non-infinitesimal region of curved GR spacetime, like the spacetime around the Earth, cannot itself be an inertial coordinate system (unless we are talking about an approximation scheme where GR effects are ignored), so the coordinate system used in GPS calculations is non-inertial.

 

 

Brent says it's not possible, because there is no motion WITHIN that frame. Further, our discussion about travel to Andromeda is totally unnessary.  I was just trying to develop a model for measuring the length contraction to Andromeda. We don't need a spacecraft. We just need the Earth at rest, and a rod moving toward the Earth, wrt the Earth, from which we use the LT. The moving rod is equivalent to a spaceship traveling to Andromeda, so the rod moves toward the Earth since, as seen by the traveler it moves in the opposite direction.

OK, give me a specific number for the rod's speed in the Earth's inertial rest frame, its direction (in the +x or -x direction), and the initial position of each end of the rod at t=0 in the Earth frame. I can then show you the math proving that according to the LT the rod is *longer* in its own inertial rest frame than it is in the Earth frame, it isn't predicted to be contracted in its own rest frame.

I never claimed that the source frame is contracted in its own rest frame.

I don't know what it means to say a "frame" is contracted—only *objects* can be contracted as measured in a given frame, like the rod as measured in the Earth frame. Are you not claiming that according to the LT, the rod is predicted to be contracted in the rod's own frame (which I think you are calling the 'target frame')?

And what about my request for a concrete example with specific numbers? All I'm asking is that you provide a few basic numbers in the Earth frame, so that we can then use the LT to explicitly transform into the rod frame and see what we get. Once again: "give me a specific number for the rod's speed in the Earth's inertial rest frame, its direction (in the +x or -x direction), and the initial position of each end of the rod at t=0 in the Earth frame" This should not be difficult for you to do, and it would be one simple way of cutting through all the linguistic confusion that's caused this discussion to drag out so much.

 

If the Earth is at rest and the rod moving wrt the Earth, the LT says the length of rod is contracted from the pov of source frame, as shown by Brent's calculation text in his plots. See Brent's post on Dec 10, 2024, 11:15:16 PM, on thread Length Contraction in SR (again). What happens in target frame is what's under discussion. I claim the contraction is not measurable in the target frame but likely is a real physical occurrence. He does specific calculations so I suggest you view his text on the calculations, which show contraction of length in target frame. I don't see anything gained if I produce any calculations. AG

Finally, about symmetry; what's used in the Parking Paradox is how symmetry exists between frames. The source frame, from which the LT is applied, is always at relative rest wrt the moving frame,

Are you using "relative rest" in some bizarre new way (for example, do you just mean that you want to verbally designate the source frame as the one which is 'at rest'?),

 

Yes, but not bizarre. This is the frame which applies the LT.  AG

If my statement above in parentheses is indeed what you meant, that definitely would be bizarre according to the standard way of talking in relativity, even if it makes sense in your own head. In relativity, saying two objects/frames A and B are in a state of "relative rest" or "relative motion" always refers to to the velocity of one as measured in the other one's frame--if that velocity is 0 they are in a state of relative rest, if it's nonzero they are in a state of relative motion. Absolutely NO physicist would say something like "B has a velocity of 0.6c in A's frame, but I am going to designate B's frame 'the rest frame', and since I understand that all talk of rest and motion is relative, this means that B is in a state of 'relative rest' wrt A". And Brent did *not* talk this way in the post at https://groups.google.com/g/everything-list/c/gbOE5B-7a6g/m/ZUPNbWYQAQAJ which you referenced.

So, once again, will you please agree to adopt the standard way of talking about motion and rest, only saying "A is at rest wrt B" if you mean that A has a velocity of 0 in B's frame, or equivalent that B has a velocity of 0 in A's frame? Since you do sometimes talk about relative motion in this way but freely mix it with your own non-standard way of talking without clearly specifying which is which, it's very difficult for me to parse what you're trying to say at any given time. People agree on terminological conventions for a reason, to prevent the fog of verbal confusion that would arise if everyone came up with their own idiosyncratic lingo for talking about the same subject.

 

 

or are you talking about the standard meaning, where it's just talking about the velocity of one thing relative to the other? For example, if we label what you call "the source frame" with A, and label the frame you call "the moving frame" with B, the usual meaning of "A and B are at relative rest" would be that B has a velocity of 0 in the A frame. If that's *not* what you meant, you really need to learn the standard way of talking about these things and not keep inventing your own confusing terminology, you shouldn't expect other people to be mind-readers.

You seem to be splitting hairs. I'm not doing anything bizarre; I am doing what Brent did in the Parking Paradox and you didn't object. The LT is applied from a frame at rest, or if you prefer relatively at rest, wrt a moving frame. AG 

Brent never designated one frame "the rest frame" and the other "the moving frame", and he definitely never used phrases like "A and B are relatively at rest" or "A is at rest relative to B" to mean anything other than the standard meaning I discussed above: that A has a velocity of 0 in B's frame, and B has a velocity of 0 in A's frame. I'm not just splitting hairs, as I said before I genuinely can't follow what scenario you are describing or what claims you are making about the LT half the time because you don't stick to the standard way of talking about relative motion and rest.

 

 

which is always length contracted, and vice-versa. AG 

This still doesn't answer my question about your prior statement "The spaceship is in one frame, at relative rest WRT the rod; the rod is in another frame, in relative motion WRT to the spaceship". I asked if you AGREE or DISAGREE with my statement "if the spaceship is at rest WRT to the rod, then automatically that must mean the rod is at rest WRT the spaceship". Don't give me a meandering answer, just tell me if you AGREE or DISAGREE with that statement!

I definitely disagree. AG 

Do you disagree because you are using your non-standard way of defining the phrase "at rest WRT to", or would you continue to disagree even if I clarified that we are using standard terminology where "spaceship is at rest WRT to the rod" means "spaceship has a velocity of 0 in the rod's frame" and "rod is at rest WRT the spaceship" means "rod has a velocity of 0 in the spaceship's frame"?

Jesse

[Alan Grayson]

Below is what Brent wrote to describe his plots: notice that he uses the car moving at 0.8c. But respect to what? He doesn't say. And then he contracts the car from the garage frame. Is garage frame moving or at rest? He doesn't say. So much presumably ambiguous non standard terminology and not a peep out of you. Let's do this; since this discussion has reached the point of tedious worthlessness, let's terminate it. AG

"First note by comparing the two diagrams that the car is longer than the garage, 12' vs 10'.  So the car doesn't fit at small relative speed.  What does "fit" mean?  It means that the event of the front of the car coinciding with the right-hand end of the garage is after or at the same time as the rear of the car coinciding with the left-had end of the garage.  In both diagrams the car is moving to the right at 0.8c so \gamma=sqrt{1-0.8^2}=0.6.  Consequently, in the car's reference frame, the garage is contracted to 6' length and when the rear of the car is just entering the garage, the front is simultaneously, in the car's reference frame, already 6' beyond the right-hand end of the garage.  Then in the garage's reference frame the car's length is contracted to 0.6*12'=7.2' so at the moment the front of the car coincides with the right end of the garage, the rear of the car will simultaneously, in the garage reference system, be 2.8' inside the garage as shown below."

[Jesse Mazer]

On Mon, Jan 20, 2025 at 3:30 PM Alan Grayson <agrays...@gmail.com> wrote:

Below is what Brent wrote to describe his plots: notice that he uses the car moving at 0.8c. But respect to what? He doesn't say.

I would actually quibble with his statement "In both diagrams the car is moving to the right at 0.8c", since in the diagram for the car's frame the car isn't moving at all, but I'm sure if asked to clarify he'd say he just meant that both diagrams show a scenario where the car has a velocity of 0.8c relative to the garage, so this doesn't actually lead me to be confused about his scenario the way some of your statements do (in part because you refuse to give a straightforward numerical example like Brent did). And everything else in his quote does clearly specify whether he is talking about what's true in "the car's reference frame" or in "the garage's reference frame" (naming the frame of a specific object as I have asked you to do), he uses variations of those phrases several times.

 

And then he contracts the car from the garage frame. Is garage frame moving or at rest? He doesn't say. So much presumably ambiguous non standard terminology and not a peep out of you.

Here you seem confused about what I am saying is "non standard terminology"--I think it's GOOD that Brent doesn't declare one frame to be "moving" and the other to be "at rest", because that is precisely the sort of confusing non-standard terminology YOU use that I'm objecting to! In the standard terminology, one wouldn't say a given frame is "at rest" or "moving" except as part of a longer phrase that specifies some other object or frame those words are supposed to be relative to, like "the car is moving relative to the garage frame" or "the garage is moving relative to the car frame" or "the car is at rest relative to the car frame", with these phrases just telling you something about the velocity of the named object in the named frame.

 

Let's do this; since this discussion has reached the point of tedious worthlessness, let's terminate it. AG

As I said, one easy way to avoid terminological confusion would be to answer my simple request for a numerical example: "give me a specific number for the rod's speed in the Earth's inertial rest frame, its direction (in the +x or -x direction), and the initial position of each end of the rod at t=0 in the Earth frame". 

Is there some reason you are unwilling to give me a few numbers for velocity and initial position of the rod in the Earth frame to work with? If you did, I could then show you with a little simple algebra what happens when we use the LT to transform these numbers into the rod's frame, proving that when we do this the length of the rod is predicted to be EXPANDED rather than contracted, compared to its length in the Earth frame.

Jesse

 

"First note by comparing the two diagrams that the car is longer than the garage, 12' vs 10'.  So the car doesn't fit at small relative speed.  What does "fit" mean?  It means that the event of the front of the car coinciding with the right-hand end of the garage is after or at the same time as the rear of the car coinciding with the left-had end of the garage.  In both diagrams the car is moving to the right at 0.8c so \gamma=sqrt{1-0.8^2}=0.6.  Consequently, in the car's reference frame, the garage is contracted to 6' length and when the rear of the car is just entering the garage, the front is simultaneously, in the car's reference frame, already 6' beyond the right-hand end of the garage.  Then in the garage's reference frame the car's length is contracted to 0.6*12'=7.2' so at the moment the front of the car coincides with the right end of the garage, the rear of the car will simultaneously, in the garage reference system, be 2.8' inside the garage as shown below."

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[Jesse Mazer]

On Mon, Jan 20, 2025 at 4:13 PM Jesse Mazer <laser...@gmail.com> wrote:

On Mon, Jan 20, 2025 at 3:30 PM Alan Grayson <agrays...@gmail.com> wrote:

Below is what Brent wrote to describe his plots: notice that he uses the car moving at 0.8c. But respect to what? He doesn't say.

I would actually quibble with his statement "In both diagrams the car is moving to the right at 0.8c", since in the diagram for the car's frame the car isn't moving at all, but I'm sure if asked to clarify he'd say he just meant that both diagrams show a scenario where the car has a velocity of 0.8c relative to the garage, so this doesn't actually lead me to be confused about his scenario the way some of your statements do (in part because you refuse to give a straightforward numerical example like Brent did). And everything else in his quote does clearly specify whether he is talking about what's true in "the car's reference frame" or in "the garage's reference frame" (naming the frame of a specific object as I have asked you to do), he uses variations of those phrases several times.

 

And then he contracts the car from the garage frame. Is garage frame moving or at rest? He doesn't say. So much presumably ambiguous non standard terminology and not a peep out of you.

Here you seem confused about what I am saying is "non standard terminology"--I think it's GOOD that Brent doesn't declare one frame to be "moving" and the other to be "at rest", because that is precisely the sort of confusing non-standard terminology YOU use that I'm objecting to! In the standard terminology, one wouldn't say a given frame is "at rest" or "moving" except as part of a longer phrase that specifies some other object or frame those words are supposed to be relative to, like "the car is moving relative to the garage frame" or "the garage is moving relative to the car frame" or "the car is at rest relative to the car frame", with these phrases just telling you something about the velocity of the named object in the named frame.

 

Let's do this; since this discussion has reached the point of tedious worthlessness, let's terminate it. AG

As I said, one easy way to avoid terminological confusion would be to answer my simple request for a numerical example: "give me a specific number for the rod's speed in the Earth's inertial rest frame, its direction (in the +x or -x direction), and the initial position of each end of the rod at t=0 in the Earth frame". 

Is there some reason you are unwilling to give me a few numbers for velocity and initial position of the rod in the Earth frame to work with? If you did, I could then show you with a little simple algebra what happens when we use the LT to transform these numbers into the rod's frame, proving that when we do this the length of the rod is predicted to be EXPANDED rather than contracted, compared to its length in the Earth frame.

Since I gave a similar numerical example in my recent comment at https://groups.google.com/g/everything-list/c/QgVdhXi3Hdc/m/SG_JbWiYDQAJ I'm going to re-use it with a few modifications to prove my point above: if we start with the equations for the worldlines of the back and front end of a rod which is moving in the Earth frame, and then use the LT to calculate the equations for the worldlines of the ends of the rod in the rod's own frame, we will see the its length in the rod frame is EXPANDED compared to the Earth frame, not contracted.

Say that relative to the Earth frame, the rod is moving at 0.6c in the +x direction and is 8 light-seconds long as measured in this frame, so the equation for the back of the rod could be x = 0.6c*t and the equation for the front of the rod could be x = 8 + 0.6c*t. Now if we want to know the equations in the rod's own frame, we can substitute those expressions for x into the Lorentz transformation's position transformation equation, x' = gamma*(x - v*t). Since the rod frame has v=0.6c as measured in the Earth frame, we have gamma=1/sqrt(1 - 0.6^2) = 1.25, so the LT equation can be written as x' = 1.25*(x - 0.6c*t).

Now, if you take the equation for the back of the rod in the unprimed (Earth) frame, x=0.6c*t, and substitute that in for x in the LT equation x' = 1.25*(x - 0.6c*t), you get x' = 1.25*(0.6c*t - 0.6c*t) = 0, meaning in the primed (rod) frame the back end of the rod has a fixed position x' = 0 which doesn't change with time (the rod is at rest in the primed frame).

And if you take the equation for the front of the rod in the unprimed (Earth) frame, x = 8 + 0.6c*t and similarly substitute it into x' = 1.25*(x - 0.6c*t), you get x' = 1.25*(8 + 0.6c*t - 0.6c*t) = 1.25*8 = 10, meaning in the primed (rod) frame the front end of the rod is fixed at x' = 10.

So you can see that if we start with the coordinates for a rod that in the unprimed (Earth) frame has a length of 8 light-second and is moving at 0.6c, and then we apply the LT equations, we end up with the coordinates for a rod that's 10 light-seconds long and at rest in the primed (rod) frame.

Jesse

[Brent Meeker]

Just so you know, the way I make the diagrams is I start by entering the loci of the object (e.g. the exit door of the garage) as stationary at equal time intervals (e.g. nano-seconds) and then I Lorentz transform using the matrix for L(v).

Brent

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[Alan Grayson]

On Monday, January 20, 2025 at 2:13:21 PM UTC-7 Jesse Mazer wrote:

On Mon, Jan 20, 2025 at 3:30 PM Alan Grayson <agrays...@gmail.com> wrote:

Below is what Brent wrote to describe his plots: notice that he uses the car moving at 0.8c. But respect to what? He doesn't say.

I would actually quibble with his statement "In both diagrams the car is moving to the right at 0.8c", since in the diagram for the car's frame the car isn't moving at all, but I'm sure if asked to clarify he'd say he just meant that both diagrams show a scenario where the car has a velocity of 0.8c relative to the garage, so this doesn't actually lead me to be confused about his scenario the way some of your statements do (in part because you refuse to give a straightforward numerical example like Brent did). And everything else in his quote does clearly specify whether he is talking about what's true in "the car's reference frame" or in "the garage's reference frame" (naming the frame of a specific object as I have asked you to do), he uses variations of those phrases several times.

 

And then he contracts the car from the garage frame. Is garage frame moving or at rest? He doesn't say. So much presumably ambiguous non standard terminology and not a peep out of you.

Here you seem confused about what I am saying is "non standard terminology"--I think it's GOOD that Brent doesn't declare one frame to be "moving" and the other to be "at rest", because that is precisely the sort of confusing non-standard terminology YOU use that I'm objecting to! In the standard terminology, one wouldn't say a given frame is "at rest" or "moving" except as part of a longer phrase that specifies some other object or frame those words are supposed to be relative to, like "the car is moving relative to the garage frame" or "the garage is moving relative to the car frame" or "the car is at rest relative to the car frame", with these phrases just telling you something about the velocity of the named object in the named frame.

 

Let's do this; since this discussion has reached the point of tedious worthlessness, let's terminate it. AG

As I said, one easy way to avoid terminological confusion would be to answer my simple request for a numerical example: "give me a specific number for the rod's speed in the Earth's inertial rest frame, its direction (in the +x or -x direction), and the initial position of each end of the rod at t=0 in the Earth frame". 

Is there some reason you are unwilling to give me a few numbers for velocity and initial position of the rod in the Earth frame to work with? If you did, I could then show you with a little simple algebra what happens when we use the LT to transform these numbers into the rod's frame, proving that when we do this the length of the rod is predicted to be EXPANDED rather than contracted, compared to its length in the Earth frame.

Yes, two reasons. Firstly, you can use Brent's numbers which he uses in his plots; and secondly, it seems as if your results contradict length contraction, so they are suspect. As for standard terminology, since motion is relative, what's wrong with saying the rod is moving relative to the Earth, towad the Earth, at some speed v, so the Earth can be imagined as at rest? AG 

[Jesse Mazer]

On Sat, Jan 25, 2025 at 4:03 AM Alan Grayson <agrays...@gmail.com> wrote:

On Monday, January 20, 2025 at 2:13:21 PM UTC-7 Jesse Mazer wrote:

On Mon, Jan 20, 2025 at 3:30 PM Alan Grayson <agrays...@gmail.com> wrote:

Below is what Brent wrote to describe his plots: notice that he uses the car moving at 0.8c. But respect to what? He doesn't say.

I would actually quibble with his statement "In both diagrams the car is moving to the right at 0.8c", since in the diagram for the car's frame the car isn't moving at all, but I'm sure if asked to clarify he'd say he just meant that both diagrams show a scenario where the car has a velocity of 0.8c relative to the garage, so this doesn't actually lead me to be confused about his scenario the way some of your statements do (in part because you refuse to give a straightforward numerical example like Brent did). And everything else in his quote does clearly specify whether he is talking about what's true in "the car's reference frame" or in "the garage's reference frame" (naming the frame of a specific object as I have asked you to do), he uses variations of those phrases several times.

 

And then he contracts the car from the garage frame. Is garage frame moving or at rest? He doesn't say. So much presumably ambiguous non standard terminology and not a peep out of you.

Here you seem confused about what I am saying is "non standard terminology"--I think it's GOOD that Brent doesn't declare one frame to be "moving" and the other to be "at rest", because that is precisely the sort of confusing non-standard terminology YOU use that I'm objecting to! In the standard terminology, one wouldn't say a given frame is "at rest" or "moving" except as part of a longer phrase that specifies some other object or frame those words are supposed to be relative to, like "the car is moving relative to the garage frame" or "the garage is moving relative to the car frame" or "the car is at rest relative to the car frame", with these phrases just telling you something about the velocity of the named object in the named frame.

 

Let's do this; since this discussion has reached the point of tedious worthlessness, let's terminate it. AG

As I said, one easy way to avoid terminological confusion would be to answer my simple request for a numerical example: "give me a specific number for the rod's speed in the Earth's inertial rest frame, its direction (in the +x or -x direction), and the initial position of each end of the rod at t=0 in the Earth frame". 

Is there some reason you are unwilling to give me a few numbers for velocity and initial position of the rod in the Earth frame to work with? If you did, I could then show you with a little simple algebra what happens when we use the LT to transform these numbers into the rod's frame, proving that when we do this the length of the rod is predicted to be EXPANDED rather than contracted, compared to its length in the Earth frame.

Yes, two reasons. Firstly, you can use Brent's numbers which he uses in his plots;

OK, looking at Brent's original post with the diagrams at https://groups.google.com/g/everything-list/c/_O3VOdJ-KUU/m/24R401SnBAAJ he shows the car to have a length of 12 in its own frame, and to be moving at 0.8c in the +x direction in the garage frame (so gamma = 1/sqrt(1 - 0.8^2) = 1/0.6, and the length contraction equation says length is contracted by 1/gamma so length of the car in the garage frame is 0.6*12 = 7.2), and the worldline of the back end of the car passes through the point x=0, t=0 in the garage frame. So, you similarly want the rod to have a length of 12 in its frame, to be moving at 0.8c in the +x direction in the Earth frame so it has a contracted length of 7.2 in the Earth frame, and the back end passing through x=0, t=0 in the Earth frame. If so, I'll just make some small modifications to the other numerical example I have you a couple posts back at https://groups.google.com/g/everything-list/c/ykkIYDL3mTg/m/hH75rNWuDQAJ --

Let the unprimed frame be the Earth frame, and the primed frame be the rod's frame. In the Earth frame, the BACK end of the rod has velocity 0.8c and passes through x=0, t=0, so its equation for position as a function of time for the BACK must be x = 0.8c*t. And in the Earth frame the FRONT of the rod is also moving at 0.8c and is always in front of the back by 7.2 (the rod's length in this frame), so its equation for position as a function of time for the FRONT must be x = 7.2 + 0.8c*t

Now if we want to know the equations in the rod's own (primed) frame, we can substitute those expressions for x into the Lorentz transformation's position transformation equation, x' = gamma*(x - v*t). Since the rod frame has v=0.8c as measured in the Earth frame, we have gamma=1/0.6, so the LT equation can be written as x' = (x - 0.8c*t)/0.6

Now, if you take the equation for the BACK of the rod in the unprimed (Earth) frame, x = 0.8c*t, and substitute that in for x in the LT equation x' = (x - 0.8c*t)/0.6, you get x' = (0.8c*t - 0.8c*t)/0.6 = 0, meaning in the primed (rod) frame the BACK end of the rod has a fixed position x' = 0 which doesn't change with time (the rod is at rest in the primed frame).

And if you take the equation for the FRONT of the rod in the unprimed (Earth) frame, x = 7.2 + 0.8c*t, and similarly substitute it into x' = (x - 0.8c*t)/0.6, you get x' = (7.2 + 0.8c*t - 0.8c*t)/0.6 = 7.2/0.6 = 12, meaning in the primed (rod) frame the FRONT end of the rod is fixed at x' = 12. If the BACK has fixed position x' = 0 in the rod frame and the FRONT has fixed position x' = 12, that means the rod has a length of 12 in this frame.

So you can see that if we start with the coordinates for a rod that in the unprimed (Earth) frame has a length of 7.2 light-second and is moving at 0.8c, and then we apply the LT equations, we end up with the coordinates for a rod that's 12 light-seconds long and at rest in the primed (rod) frame.

 

and secondly, it seems as if your results contradict length contraction, so they are suspect.

No, those results agree with length contraction. The length contraction is written as a relation between the object's "proper length" L_0 in the frame where the object is at rest (in this case, the rod's frame), and the contracted length L in some other frame where the object is in motion (in this case the Earth's frame)--the equation is just L = L_0/gamma, and gamma > 1 so this always means L is smaller than L_0. So, if you start from the frame where the rod is moving and the length is L, and use the LT to transform into the frame where the rod is at rest (using the rod's frame as what you called the 'target frame' for the LT), in order to agree with the length contraction equation the LT should predict a LONGER length L_0 in the target frame.

 

As for standard terminology, since motion is relative, what's wrong with saying the rod is moving relative to the Earth, towad the Earth, at some speed v, so the Earth can be imagined as at rest? AG 

The standard terminology is to define length and rest relative in terms of the coordinates of whichever frame you are explicitly talking about. So it's fine to say something like "in the Earth frame, the rod is moving and the Earth is at rest" as long as you would likewise say "in the rod frame, the rod is at rest and the Earth is moving". But what is non-standard is to say something like "I am going to define the Earth as at rest and the rod as moving throughout this discussion, and will continue to define them this way even when I am talking about how things are relative to the rod--for example, I will say 'the Earth is at relative rest wrt the rod' even though in the rod frame, the Earth has a nonzero coordinate velocity". If I'm understanding correctly, something like that was the logic behind your statement in the post at https://groups.google.com/g/everything-list/c/ykkIYDL3mTg/m/cZ_dWIcwDAAJ that "The spaceship is in one frame, at relative rest WRT the rod; the rod is in another frame, in relative motion WRT to the spaceship" (the Earth later took the place of the spaceship in how you described the example). So if I do understand the logic of that statement correctly, what's wrong with it is just that physicists don't talk about rest and motion that way, there are agreed-upon conventions and inventing your own alternate conventions creates unneeded confusion.

Jesse