Question on SR

[Alan Grayson]

In this scenario, is there any contradition with the principles of SR? Suppose there exist two inertial frames, moving in opposite directions with velocity v < c along the x-axis, where one clock of each frame is initially located one unit, positively and negatively respectively from the origin, and when these clocks are juxtaposed at the origin, the multiple set of clocks in both frames can be synchronized? Does this scenario imply an unwarranted affirmation of simultaneity?

TY, AG

[Brent Meeker]

The fact that you never specify whether "synchronized" means "set to the same time" or "caused to run at the same rate" or both, makes me think you don't understand your own question.

Brent

On 10/23/2024 6:00 AM, Alan Grayson wrote:

In this scenario, is there any contradition with the principles of SR? Suppose there exist two inertial frames, moving in opposite directions with velocity v < c along the x-axis, where one clock of each frame is initially located one unit, positively and negatively respectively from the origin, and when these clocks are juxtaposed at the origin, the multiple set of clocks in both frames can be synchronized? Does this scenario imply an unwarranted affirmation of simultaneity?

TY, AG

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[Alan Grayson]

On Wednesday, October 23, 2024 at 1:55:13 PM UTC-6 Brent Meeker wrote:

The fact that you never specify whether "synchronized" means "set to the same time" or "caused to run at the same rate" or both, makes me think you don't understand your own question.

Brent

I meant when juxtaposted, to set at the two clocks at the same time, and then synchronized throughout each frame. Then I expect, but am not certain, that the rates in the two frames will be the same. AG

[Jesse Mazer]

On Wed, Oct 23, 2024 at 6:31 PM Alan Grayson <agrays...@gmail.com> wrote:

On Wednesday, October 23, 2024 at 1:55:13 PM UTC-6 Brent Meeker wrote:

The fact that you never specify whether "synchronized" means "set to the same time" or "caused to run at the same rate" or both, makes me think you don't understand your own question.

Brent

I meant when juxtaposted, to set at the two clocks at the same time, and then synchronized throughout each frame. Then I expect, but am not certain, that the rates in the two frames will be the same. AG

"Synchronized" only has meaning relative to a particular frame's definition of simultaneity--since the frames disagree on simulataneity, you can momentarily set all clocks so that they read the same time at the same moment relative to one frame, but you can't do this in both frames. And whichever frame you pick, unless you artificially adjust the ticking rate of the clocks moving relative to that frame to "correct" for time dilation, the moving clocks won't stay synchronized with the clocks at rest in that frame.

Jesse

 

On 10/23/2024 6:00 AM, Alan Grayson wrote:

In this scenario, is there any contradition with the principles of SR? Suppose there exist two inertial frames, moving in opposite directions with velocity v < c along the x-axis, where one clock of each frame is initially located one unit, positively and negatively respectively from the origin, and when these clocks are juxtaposed at the origin, the multiple set of clocks in both frames can be synchronized? Does this scenario imply an unwarranted affirmation of simultaneity?

TY, AG

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[Alan Grayson]

On Wednesday, October 23, 2024 at 4:47:13 PM UTC-6 Jesse Mazer wrote:

On Wed, Oct 23, 2024 at 6:31 PM Alan Grayson <agrays...@gmail.com> wrote:

On Wednesday, October 23, 2024 at 1:55:13 PM UTC-6 Brent Meeker wrote:

The fact that you never specify whether "synchronized" means "set to the same time" or "caused to run at the same rate" or both, makes me think you don't understand your own question.

Brent

I meant when juxtaposted, to set at the two clocks at the same time, and then synchronized throughout each frame. Then I expect, but am not certain, that the rates in the two frames will be the same. AG

"Synchronized" only has meaning relative to a particular frame's definition of simultaneity--since the frames disagree on simulataneity, you can momentarily set all clocks so that they read the same time at the same moment relative to one frame, but you can't do this in both frames. And whichever frame you pick, unless you artificially adjust the ticking rate of the clocks moving relative to that frame to "correct" for time dilation, the moving clocks won't stay synchronized with the clocks at rest in that frame.

Jesse

As I see it, when the clocks are juxtaposed, a comparison of any clock in one frame, will read the same time as the corresponding clock in the other frame, that is, corresponding with position as they pass each other. And since the frames are moving with the same velocity wrt each other, I don't see the role of simultaneity in changing the rate of any clock in any frame. What I think this scenario shows, is that time dilation doesn't exist. AG 

[Jesse Mazer]

On Wed, Oct 23, 2024 at 7:10 PM Alan Grayson <agrays...@gmail.com> wrote:

On Wednesday, October 23, 2024 at 4:47:13 PM UTC-6 Jesse Mazer wrote:

On Wed, Oct 23, 2024 at 6:31 PM Alan Grayson <agrays...@gmail.com> wrote:

On Wednesday, October 23, 2024 at 1:55:13 PM UTC-6 Brent Meeker wrote:

The fact that you never specify whether "synchronized" means "set to the same time" or "caused to run at the same rate" or both, makes me think you don't understand your own question.

Brent

I meant when juxtaposted, to set at the two clocks at the same time, and then synchronized throughout each frame. Then I expect, but am not certain, that the rates in the two frames will be the same. AG

"Synchronized" only has meaning relative to a particular frame's definition of simultaneity--since the frames disagree on simulataneity, you can momentarily set all clocks so that they read the same time at the same moment relative to one frame, but you can't do this in both frames. And whichever frame you pick, unless you artificially adjust the ticking rate of the clocks moving relative to that frame to "correct" for time dilation, the moving clocks won't stay synchronized with the clocks at rest in that frame.

Jesse

As I see it, when the clocks are juxtaposed, a comparison of any clock in one frame, will read the same time as the corresponding clock in the other frame, that is, corresponding with position as they pass each other. And since the frames are moving with the same velocity wrt each other, I don't see the role of simultaneity in changing the rate of any clock in any frame. What I think this scenario shows, is that time dilation doesn't exist. AG 

But that's wrong according to relativity, and the Lorentz coordinate transformation is mathematically/logically consistent, and the prediction that the laws of physics work symmetrically in these different frames (so that readings on natural physical clocks at different locations will align with coordinate time in their rest frame, assuming they are synchronized according to the Einstein convention at https://en.wikipedia.org/wiki/Einstein_synchronisation ) has held up experimentally. I once made a diagram showing two rows of clocks in motion relative to each other, synchronized according to Einstein's convention, so people can see how it works--see https://physics.stackexchange.com/a/155016/59406

Jesse

 

 

On 10/23/2024 6:00 AM, Alan Grayson wrote:

In this scenario, is there any contradition with the principles of SR? Suppose there exist two inertial frames, moving in opposite directions with velocity v < c along the x-axis, where one clock of each frame is initially located one unit, positively and negatively respectively from the origin, and when these clocks are juxtaposed at the origin, the multiple set of clocks in both frames can be synchronized? Does this scenario imply an unwarranted affirmation of simultaneity?

TY, AG

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[ilsa]

Could you please send me a private email so Alan and I could talk I tried to send him up personal email but it bounced back and said it was refused that I didn't have permission to write to him so I'd like to talk to him about something he did for me in 2012

On Wed, Oct 23, 2024, 6:00 AM Alan Grayson <agrays...@gmail.com> wrote:

In this scenario, is there any contradition with the principles of SR? Suppose there exist two inertial frames, moving in opposite directions with velocity v < c along the x-axis, where one clock of each frame is initially located one unit, positively and negatively respectively from the origin, and when these clocks are juxtaposed at the origin, the multiple set of clocks in both frames can be synchronized? Does this scenario imply an unwarranted affirmation of simultaneity?

TY, AG

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[Alan Grayson]

On Wednesday, October 23, 2024 at 6:03:58 PM UTC-6 Jesse Mazer wrote:

On Wed, Oct 23, 2024 at 7:10 PM Alan Grayson <agrays...@gmail.com> wrote:

On Wednesday, October 23, 2024 at 4:47:13 PM UTC-6 Jesse Mazer wrote:

On Wed, Oct 23, 2024 at 6:31 PM Alan Grayson <agrays...@gmail.com> wrote:

On Wednesday, October 23, 2024 at 1:55:13 PM UTC-6 Brent Meeker wrote:

The fact that you never specify whether "synchronized" means "set to the same time" or "caused to run at the same rate" or both, makes me think you don't understand your own question.

Brent

I meant when juxtaposted, to set at the two clocks at the same time, and then synchronized throughout each frame. Then I expect, but am not certain, that the rates in the two frames will be the same. AG

"Synchronized" only has meaning relative to a particular frame's definition of simultaneity--since the frames disagree on simulataneity, you can momentarily set all clocks so that they read the same time at the same moment relative to one frame, but you can't do this in both frames. And whichever frame you pick, unless you artificially adjust the ticking rate of the clocks moving relative to that frame to "correct" for time dilation, the moving clocks won't stay synchronized with the clocks at rest in that frame.

Jesse

As I see it, when the clocks are juxtaposed, a comparison of any clock in one frame, will read the same time as the corresponding clock in the other frame, that is, corresponding with position as they pass each other. And since the frames are moving with the same velocity wrt each other, I don't see the role of simultaneity in changing the rate of any clock in any frame. What I think this scenario shows, is that time dilation doesn't exist. AG 

But that's wrong according to relativity, and the Lorentz coordinate transformation is mathematically/logically consistent, and the prediction that the laws of physics work symmetrically in these different frames (so that readings on natural physical clocks at different locations will align with coordinate time in their rest frame, assuming they are synchronized according to the Einstein convention at https://en.wikipedia.org/wiki/Einstein_synchronisation ) has held up experimentally. I once made a diagram showing two rows of clocks in motion relative to each other, synchronized according to Einstein's convention, so people can see how it works--see https://physics.stackexchange.com/a/155016/59406

Jesse

Jesse; I'll check out your links, for sure. I will just say now that time dilation can be established using a rest frame and moving frame, but in my model there is no rest frame; both frames are moving. AG 

[Brent Meeker]

I think it's confusing to introduce "frames" that are spacially extended (although Einstein did).   Better to recognize that all physical times are proper times of some clock and they can be set to the same time when they are at the same place and as ideal clock's they are assumed to keep perfect time along their world lines.

Brent

To view this discussion visit https://groups.google.com/d/msgid/everything-list/CAPCWU3%2BKVsZRW5MBLQCYG_NwXcOgrk%2BEZEW%2BpmtPKpwv%3D%2BqO%2BA%40mail.gmail.com.

[Alan Grayson]

On Wednesday, October 23, 2024 at 6:30:44 PM UTC-6 ilsa wrote:

Could you please send me a private email so Alan and I could talk I tried to send him up personal email but it bounced back and said it was refused that I didn't have permission to write to him so I'd like to talk to him about something he did for me in 2012

You may have confused me with Alan Grayson, who was a US representative from FL.  However, if you want to contact ME, you may use this email address;  anonym...@protonmail.com .  AG

[ilsa]

Wasn't there a time art project It was just for a weekend or something and it had the the clocks and they were always changing and they were always different You could probably look up a look it up and see if it's an artistic description of what you're talking about It was at the San Francisco Museum of modern art you know 15 years ago or something

To view this discussion visit https://groups.google.com/d/msgid/everything-list/baf16371-0457-4f98-81df-89f53cbe9f7a%40gmail.com.

[Alan Grayson]

On Wednesday, October 23, 2024 at 7:18:34 PM UTC-6 ilsa wrote:

Wasn't there a time art project It was just for a weekend or something and it had the the clocks and they were always changing and they were always different You could probably look up a look it up and see if it's an artistic description of what you're talking about It was at the San Francisco Museum of modern art you know 15 years ago or something

I don't recall that event. I probably wasn't there. The LHS in the address is anonymous67. AG

[Alan Grayson]

On Wednesday, October 23, 2024 at 7:10:36 PM UTC-6 Brent Meeker wrote:

I think it's confusing to introduce "frames" that are spacially extended (although Einstein did).   Better to recognize that all physical times are proper times of some clock and they can be set to the same time when they are at the same place and as ideal clock's they are assumed to keep perfect time along their world lines.

Brent

How do we get time dilation from world lines if both represent the same path in the sense that neither is considered at rest? AG 

[Brent Meeker]

You bring them together at the same place.  The only other useful frame is the cosmic frame in which all the matter (stars, galaxies, CMB) is on average at rest.  This can be determined locally without any reference to any other distant objects.

Brent

To view this discussion visit https://groups.google.com/d/msgid/everything-list/5ff13d7e-0470-4e84-a86c-70887fab6973n%40googlegroups.com.

[Jesse Mazer]

On Wed, Oct 23, 2024 at 9:03 PM Alan Grayson <agrays...@gmail.com> wrote:

On Wednesday, October 23, 2024 at 6:03:58 PM UTC-6 Jesse Mazer wrote:

On Wed, Oct 23, 2024 at 7:10 PM Alan Grayson <agrays...@gmail.com> wrote:

On Wednesday, October 23, 2024 at 4:47:13 PM UTC-6 Jesse Mazer wrote:

On Wed, Oct 23, 2024 at 6:31 PM Alan Grayson <agrays...@gmail.com> wrote:

On Wednesday, October 23, 2024 at 1:55:13 PM UTC-6 Brent Meeker wrote:

The fact that you never specify whether "synchronized" means "set to the same time" or "caused to run at the same rate" or both, makes me think you don't understand your own question.

Brent

I meant when juxtaposted, to set at the two clocks at the same time, and then synchronized throughout each frame. Then I expect, but am not certain, that the rates in the two frames will be the same. AG

"Synchronized" only has meaning relative to a particular frame's definition of simultaneity--since the frames disagree on simulataneity, you can momentarily set all clocks so that they read the same time at the same moment relative to one frame, but you can't do this in both frames. And whichever frame you pick, unless you artificially adjust the ticking rate of the clocks moving relative to that frame to "correct" for time dilation, the moving clocks won't stay synchronized with the clocks at rest in that frame.

Jesse

As I see it, when the clocks are juxtaposed, a comparison of any clock in one frame, will read the same time as the corresponding clock in the other frame, that is, corresponding with position as they pass each other. And since the frames are moving with the same velocity wrt each other, I don't see the role of simultaneity in changing the rate of any clock in any frame. What I think this scenario shows, is that time dilation doesn't exist. AG 

But that's wrong according to relativity, and the Lorentz coordinate transformation is mathematically/logically consistent, and the prediction that the laws of physics work symmetrically in these different frames (so that readings on natural physical clocks at different locations will align with coordinate time in their rest frame, assuming they are synchronized according to the Einstein convention at https://en.wikipedia.org/wiki/Einstein_synchronisation ) has held up experimentally. I once made a diagram showing two rows of clocks in motion relative to each other, synchronized according to Einstein's convention, so people can see how it works--see https://physics.stackexchange.com/a/155016/59406

Jesse

Jesse; I'll check out your links, for sure. I will just say now that time dilation can be established using a rest frame and moving frame, but in my model there is no rest frame; both frames are moving. AG 

In relativity "rest frame" is only used in a relative sense, there is no objective truth about which frame is called the "rest frame" and which frame is "moving". For example if you have two clocks A and B in relative motion, you can calculate things from the perspective of the coordinate system where A's coordinate position doesn't change with time which is called "A's rest frame", but you can equally well calculate things from the perspective of the coordinate system where B's coordinate position doesn't change with time which is called "B's rest frame", and both calculations should agree in their predictions about all local events eg what readings show on both clocks at the moment they pass next to each other (I illustrated this with a few clocks in the diagrams from my link).

Jesse

 

On 10/23/2024 6:00 AM, Alan Grayson wrote:

In this scenario, is there any contradition with the principles of SR? Suppose there exist two inertial frames, moving in opposite directions with velocity v < c along the x-axis, where one clock of each frame is initially located one unit, positively and negatively respectively from the origin, and when these clocks are juxtaposed at the origin, the multiple set of clocks in both frames can be synchronized? Does this scenario imply an unwarranted affirmation of simultaneity?

TY, AG

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[Alan Grayson]

On Thursday, October 24, 2024 at 12:01:30 AM UTC-6 Jesse Mazer wrote:

On Wed, Oct 23, 2024 at 9:03 PM Alan Grayson <agrays...@gmail.com> wrote:

On Wednesday, October 23, 2024 at 6:03:58 PM UTC-6 Jesse Mazer wrote:

On Wed, Oct 23, 2024 at 7:10 PM Alan Grayson <agrays...@gmail.com> wrote:

On Wednesday, October 23, 2024 at 4:47:13 PM UTC-6 Jesse Mazer wrote:

On Wed, Oct 23, 2024 at 6:31 PM Alan Grayson <agrays...@gmail.com> wrote:

On Wednesday, October 23, 2024 at 1:55:13 PM UTC-6 Brent Meeker wrote:

The fact that you never specify whether "synchronized" means "set to the same time" or "caused to run at the same rate" or both, makes me think you don't understand your own question.

Brent

I meant when juxtaposted, to set at the two clocks at the same time, and then synchronized throughout each frame. Then I expect, but am not certain, that the rates in the two frames will be the same. AG

"Synchronized" only has meaning relative to a particular frame's definition of simultaneity--since the frames disagree on simulataneity, you can momentarily set all clocks so that they read the same time at the same moment relative to one frame, but you can't do this in both frames. And whichever frame you pick, unless you artificially adjust the ticking rate of the clocks moving relative to that frame to "correct" for time dilation, the moving clocks won't stay synchronized with the clocks at rest in that frame.

Jesse

As I see it, when the clocks are juxtaposed, a comparison of any clock in one frame, will read the same time as the corresponding clock in the other frame, that is, corresponding with position as they pass each other. And since the frames are moving with the same velocity wrt each other, I don't see the role of simultaneity in changing the rate of any clock in any frame. What I think this scenario shows, is that time dilation doesn't exist. AG 

But that's wrong according to relativity, and the Lorentz coordinate transformation is mathematically/logically consistent, and the prediction that the laws of physics work symmetrically in these different frames (so that readings on natural physical clocks at different locations will align with coordinate time in their rest frame, assuming they are synchronized according to the Einstein convention at https://en.wikipedia.org/wiki/Einstein_synchronisation ) has held up experimentally. I once made a diagram showing two rows of clocks in motion relative to each other, synchronized according to Einstein's convention, so people can see how it works--see https://physics.stackexchange.com/a/155016/59406

Jesse

Jesse; I'll check out your links, for sure. I will just say now that time dilation can be established using a rest frame and moving frame, but in my model there is no rest frame; both frames are moving. AG 

In relativity "rest frame" is only used in a relative sense, there is no objective truth about which frame is called the "rest frame" and which frame is "moving". For example if you have two clocks A and B in relative motion, you can calculate things from the perspective of the coordinate system where A's coordinate position doesn't change with time which is called "A's rest frame", but you can equally well calculate things from the perspective of the coordinate system where B's coordinate position doesn't change with time which is called "B's rest frame", and both calculations should agree in their predictions about all local events eg what readings show on both clocks at the moment they pass next to each other (I illustrated this with a few clocks in the diagrams from my link).

Jesse

TY. Yes, I am aware of what you wrote. That's why I tried to come up with an example where there is no frame identified as the rest frame. I will look at your diagrams today. I think Brent makes the same point with world lines. AG

[Jesse Mazer]

On Thu, Oct 24, 2024 at 8:06 AM Alan Grayson <agrays...@gmail.com> wrote:

On Thursday, October 24, 2024 at 12:01:30 AM UTC-6 Jesse Mazer wrote:

On Wed, Oct 23, 2024 at 9:03 PM Alan Grayson <agrays...@gmail.com> wrote:

On Wednesday, October 23, 2024 at 6:03:58 PM UTC-6 Jesse Mazer wrote:

On Wed, Oct 23, 2024 at 7:10 PM Alan Grayson <agrays...@gmail.com> wrote:

On Wednesday, October 23, 2024 at 4:47:13 PM UTC-6 Jesse Mazer wrote:

On Wed, Oct 23, 2024 at 6:31 PM Alan Grayson <agrays...@gmail.com> wrote:

On Wednesday, October 23, 2024 at 1:55:13 PM UTC-6 Brent Meeker wrote:

The fact that you never specify whether "synchronized" means "set to the same time" or "caused to run at the same rate" or both, makes me think you don't understand your own question.

Brent

I meant when juxtaposted, to set at the two clocks at the same time, and then synchronized throughout each frame. Then I expect, but am not certain, that the rates in the two frames will be the same. AG

"Synchronized" only has meaning relative to a particular frame's definition of simultaneity--since the frames disagree on simulataneity, you can momentarily set all clocks so that they read the same time at the same moment relative to one frame, but you can't do this in both frames. And whichever frame you pick, unless you artificially adjust the ticking rate of the clocks moving relative to that frame to "correct" for time dilation, the moving clocks won't stay synchronized with the clocks at rest in that frame.

Jesse

As I see it, when the clocks are juxtaposed, a comparison of any clock in one frame, will read the same time as the corresponding clock in the other frame, that is, corresponding with position as they pass each other. And since the frames are moving with the same velocity wrt each other, I don't see the role of simultaneity in changing the rate of any clock in any frame. What I think this scenario shows, is that time dilation doesn't exist. AG 

But that's wrong according to relativity, and the Lorentz coordinate transformation is mathematically/logically consistent, and the prediction that the laws of physics work symmetrically in these different frames (so that readings on natural physical clocks at different locations will align with coordinate time in their rest frame, assuming they are synchronized according to the Einstein convention at https://en.wikipedia.org/wiki/Einstein_synchronisation ) has held up experimentally. I once made a diagram showing two rows of clocks in motion relative to each other, synchronized according to Einstein's convention, so people can see how it works--see https://physics.stackexchange.com/a/155016/59406

Jesse

Jesse; I'll check out your links, for sure. I will just say now that time dilation can be established using a rest frame and moving frame, but in my model there is no rest frame; both frames are moving. AG 

In relativity "rest frame" is only used in a relative sense, there is no objective truth about which frame is called the "rest frame" and which frame is "moving". For example if you have two clocks A and B in relative motion, you can calculate things from the perspective of the coordinate system where A's coordinate position doesn't change with time which is called "A's rest frame", but you can equally well calculate things from the perspective of the coordinate system where B's coordinate position doesn't change with time which is called "B's rest frame", and both calculations should agree in their predictions about all local events eg what readings show on both clocks at the moment they pass next to each other (I illustrated this with a few clocks in the diagrams from my link).

Jesse

TY. Yes, I am aware of what you wrote. That's why I tried to come up with an example where there is no frame identified as the rest frame. I will look at your diagrams today. I think Brent makes the same point with world lines. AG

But when I referred to the calculations with A and B, no frame would be identified as "the" rest frame, there are just two frames which as a matter of verbal convention are called "A's rest frame" (you could also just call it 'the frame where A has a velocity of 0') and "B's rest frame" (or 'the frame where B has a velocity of 0'). You could also do the calculations from the perspective of a frame where both A and B have nonzero velocity if you wish, the point in special relativity is that the same equations of physics apply in all inertial frames so you're free to use whichever one you find convenient. But you do need to pick *some* spacetime coordinate system to define how the coordinate position of each object you're analyzing changes with coordinate time, since the equations for the laws of physics are generally written in terms of such coordinates.

Jesse

 

 

On 10/23/2024 6:00 AM, Alan Grayson wrote:

In this scenario, is there any contradition with the principles of SR? Suppose there exist two inertial frames, moving in opposite directions with velocity v < c along the x-axis, where one clock of each frame is initially located one unit, positively and negatively respectively from the origin, and when these clocks are juxtaposed at the origin, the multiple set of clocks in both frames can be synchronized? Does this scenario imply an unwarranted affirmation of simultaneity?

TY, AG

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[ilsa]

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[Alan Grayson]

On Thursday, October 24, 2024 at 12:03:22 PM UTC-6 Jesse Mazer wrote:

On Thu, Oct 24, 2024 at 8:06 AM Alan Grayson <agrays...@gmail.com> wrote:

On Thursday, October 24, 2024 at 12:01:30 AM UTC-6 Jesse Mazer wrote:

On Wed, Oct 23, 2024 at 9:03 PM Alan Grayson <agrays...@gmail.com> wrote:

On Wednesday, October 23, 2024 at 6:03:58 PM UTC-6 Jesse Mazer wrote:

On Wed, Oct 23, 2024 at 7:10 PM Alan Grayson <agrays...@gmail.com> wrote:

On Wednesday, October 23, 2024 at 4:47:13 PM UTC-6 Jesse Mazer wrote:

On Wed, Oct 23, 2024 at 6:31 PM Alan Grayson <agrays...@gmail.com> wrote:

On Wednesday, October 23, 2024 at 1:55:13 PM UTC-6 Brent Meeker wrote:

The fact that you never specify whether "synchronized" means "set to the same time" or "caused to run at the same rate" or both, makes me think you don't understand your own question.

Brent

I meant when juxtaposted, to set at the two clocks at the same time, and then synchronized throughout each frame. Then I expect, but am not certain, that the rates in the two frames will be the same. AG

"Synchronized" only has meaning relative to a particular frame's definition of simultaneity--since the frames disagree on simulataneity, you can momentarily set all clocks so that they read the same time at the same moment relative to one frame, but you can't do this in both frames. And whichever frame you pick, unless you artificially adjust the ticking rate of the clocks moving relative to that frame to "correct" for time dilation, the moving clocks won't stay synchronized with the clocks at rest in that frame.

Jesse

As I see it, when the clocks are juxtaposed, a comparison of any clock in one frame, will read the same time as the corresponding clock in the other frame, that is, corresponding with position as they pass each other. And since the frames are moving with the same velocity wrt each other, I don't see the role of simultaneity in changing the rate of any clock in any frame. What I think this scenario shows, is that time dilation doesn't exist. AG 

But that's wrong according to relativity, and the Lorentz coordinate transformation is mathematically/logically consistent, and the prediction that the laws of physics work symmetrically in these different frames (so that readings on natural physical clocks at different locations will align with coordinate time in their rest frame, assuming they are synchronized according to the Einstein convention at https://en.wikipedia.org/wiki/Einstein_synchronisation ) has held up experimentally. I once made a diagram showing two rows of clocks in motion relative to each other, synchronized according to Einstein's convention, so people can see how it works--see https://physics.stackexchange.com/a/155016/59406

Jesse

Jesse; I'll check out your links, for sure. I will just say now that time dilation can be established using a rest frame and moving frame, but in my model there is no rest frame; both frames are moving. AG 

In relativity "rest frame" is only used in a relative sense, there is no objective truth about which frame is called the "rest frame" and which frame is "moving". For example if you have two clocks A and B in relative motion, you can calculate things from the perspective of the coordinate system where A's coordinate position doesn't change with time which is called "A's rest frame", but you can equally well calculate things from the perspective of the coordinate system where B's coordinate position doesn't change with time which is called "B's rest frame", and both calculations should agree in their predictions about all local events eg what readings show on both clocks at the moment they pass next to each other (I illustrated this with a few clocks in the diagrams from my link).

Jesse

TY. Yes, I am aware of what you wrote. That's why I tried to come up with an example where there is no frame identified as the rest frame. I will look at your diagrams today. I think Brent makes the same point with world lines. AG

But when I referred to the calculations with A and B, no frame would be identified as "the" rest frame, there are just two frames which as a matter of verbal convention are called "A's rest frame" (you could also just call it 'the frame where A has a velocity of 0') and "B's rest frame" (or 'the frame where B has a velocity of 0'). You could also do the calculations from the perspective of a frame where both A and B have nonzero velocity if you wish, the point in special relativity is that the same equations of physics apply in all inertial frames so you're free to use whichever one you find convenient. But you do need to pick *some* spacetime coordinate system to define how the coordinate position of each object you're analyzing changes with coordinate time, since the equations for the laws of physics are generally written in terms of such coordinates.

Jesse

In my scenario for analyzing the Clock Paradox, can you identify where, EXACTLY, I mistakenly assumed simultaneity, which presumably led to the wrong conclusion? TY, AG

[Brent Meeker]

Here's  how a light-clock ticks in when in motion.  A light-clock is just two perfect mirrors a fixed distance apart with a photon bouncing back an forth between them.  It's a hypothetical ideal clock for which the effect of motion is easily visualized.



These are the spacetime diagrams of three identical light-clocks moving at +c relative to the blue one.  Because the speed of light is invariant the photon paths are at unit slope inside all three clocks, so it is easily seen why the relative motion makes the clock seem slow although each clock is ticking at the same rate in it's own reference frame.  The red diagram is just the blue diagram Lorentz transformed as it would be seen in a frame moving the left at 0.5c, and the black diagram as it would be seen from a frame moving to the right.

Brent

To view this discussion visit https://groups.google.com/d/msgid/everything-list/CAPCWU3LoxbKXSvYZ1pJCOd2fVqSbQxbQRQxCB83%3DQnkEAo7Ang%40mail.gmail.com.

[Jesse Mazer]

Your scenario just seems unclear to me, you said "multiple set of clocks in both frames can be synchronized" without spelling out what technique you want to use to synchronize them, I was just pointing out the ambiguity. If you have two sets of clocks A' and B' where all the clocks in a given set are at rest relative to one another, and all the clocks in A' are supposed to be synchronized with one another, likewise all the clocks in B' are supposed to be synchronized with one another, how are you proposing to do this when each member of a given set is at a different location so they can't be compared in a local way? If you synchronize them using the Einstein synchronization convention involving the assumption that light signals travel at the same speed in both directions, then the clocks in A' cannot be synchronized with the clocks in B'. If you want some other method of synchronizing clocks that are at rest relative to one another, you need to spell out the method.

Jesse

 

On 10/23/2024 6:00 AM, Alan Grayson wrote:

In this scenario, is there any contradition with the principles of SR? Suppose there exist two inertial frames, moving in opposite directions with velocity v < c along the x-axis, where one clock of each frame is initially located one unit, positively and negatively respectively from the origin, and when these clocks are juxtaposed at the origin, the multiple set of clocks in both frames can be synchronized? Does this scenario imply an unwarranted affirmation of simultaneity?

TY, AG

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[Alan Grayson]

If the clocks in both frames are identical, and I set the readings of the juxtaposed clocks to some identical value, and then sychronize all clocks in each frame to the value of its respective juxtaposed clock, there is no reason to think that the clock rates would change and differ when the frames are compared. You argue that the clock rates would appear different as one frame views the other, but what's your reason for this assumption? AG 

[Alan Grayson]

On Thursday, October 24, 2024 at 1:30:32 PM UTC-6 Brent Meeker wrote:

Here's  how a light-clock ticks in when in motion.  A light-clock is just two perfect mirrors a fixed distance apart with a photon bouncing back an forth between them.  It's a hypothetical ideal clock for which the effect of motion is easily visualized.



These are the spacetime diagrams of three identical light-clocks moving at +c relative to the blue one. 

Three clocks?  Black diagram? If only this was as clear as you claim. TY, AG

[Jesse Mazer]

By juxtaposed do you mean right next to each other spatially, i.e. you set a clock from group A' to read the same thing as a clock from group B' at the moment they pass next to each other? But that wasn't what I was asking you, I was asking how you set different clocks from the *same* set so that they are "synchronized" with one another according to you. Clocks from the same set are at rest relative to one another and at different spatial locations, so they will never be next to each other unless you moved them around beforehand (and if you did that, you have to worry about time dilation due to motion when you move them apart). Do you use the Einstein clock synchronization convention, or something different?

Jesse

 

 

On 10/23/2024 6:00 AM, Alan Grayson wrote:

In this scenario, is there any contradition with the principles of SR? Suppose there exist two inertial frames, moving in opposite directions with velocity v < c along the x-axis, where one clock of each frame is initially located one unit, positively and negatively respectively from the origin, and when these clocks are juxtaposed at the origin, the multiple set of clocks in both frames can be synchronized? Does this scenario imply an unwarranted affirmation of simultaneity?

TY, AG

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[Alan Grayson]

I think I used Einstein's convention. Not sure I recall what it was. I assumed the clocks exist everywhere in any frame and aren't moved. And they could be set to the same time as the juxtaposed clock in that frame, probably using a light signal. AG 

[Alan Grayson]

I think I only need one clock other than the juxtaposed clock, with a known distance between them. Knowing that distance, I can synchronize these two clocks using a light signal. AG 

[Jesse Mazer]

So like two clocks A1 and A2 that are at rest relative to one another and synchronized by light signal (the Einstein convention), and then just one other clock B1 that's moving relative to them, so it first passes A1 and then passes A2 a bit later? In this case, if you set B1 to read the same time as A1 at the moment they pass next to each other, then because of time dilation due to motion in the rest frame of the A clocks, by the time B1 passes A2 its reading will be behind that of A2.

Jesse

 

 

 

On 10/23/2024 6:00 AM, Alan Grayson wrote:

In this scenario, is there any contradition with the principles of SR? Suppose there exist two inertial frames, moving in opposite directions with velocity v < c along the x-axis, where one clock of each frame is initially located one unit, positively and negatively respectively from the origin, and when these clocks are juxtaposed at the origin, the multiple set of clocks in both frames can be synchronized? Does this scenario imply an unwarranted affirmation of simultaneity?

TY, AG

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[Brent Meeker]

On 10/24/2024 5:46 PM, Alan Grayson wrote:

On Thursday, October 24, 2024 at 1:30:32 PM UTC-6 Brent Meeker wrote:

Here's  how a light-clock ticks in when in motion.  A light-clock is just two perfect mirrors a fixed distance apart with a photon bouncing back an forth between them.  It's a hypothetical ideal clock for which the effect of motion is easily visualized.



These are the spacetime diagrams of three identical light-clocks moving at +c relative to the blue one. 

Three clocks?  Black diagram? If only this was as clear as you claim. TY, AG

You can't handle more than two?  The left clock is black with a red photon.  Is that hard to comprehend?  Didn't they teach spacetime diagrams at your kindergarten?
               
Brent

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[Alan Grayson]

On Thursday, October 24, 2024 at 11:07:18 PM UTC-6 Brent Meeker wrote:

On 10/24/2024 5:46 PM, Alan Grayson wrote:

On Thursday, October 24, 2024 at 1:30:32 PM UTC-6 Brent Meeker wrote:

Here's  how a light-clock ticks in when in motion.  A light-clock is just two perfect mirrors a fixed distance apart with a photon bouncing back an forth between them.  It's a hypothetical ideal clock for which the effect of motion is easily visualized.



These are the spacetime diagrams of three identical light-clocks moving at +c relative to the blue one. 

Three clocks?  Black diagram? If only this was as clear as you claim. TY, AG

You can't handle more than two?  The left clock is black with a red photon.  Is that hard to comprehend?  Didn't they teach spacetime diagrams at your kindergarten?
               
Brent

If you could cease behaving like an arrogant a'hole, likely beyond your maturity and capability, maybe we could get somewhere on this problem. I was thinking about your diagrams, and concluded you could prove your point with the simple observation and simpler diagram, that with your photon clock, it would be easy to show that from the pov of a rest clock, the moving clock would appear to have a slower rate. And since inertial frames are equivalent in SR, the same result would be evident for the rest frame, when considered as the moving frame. BUT, what I have shown, with an arbitrary clock, that all clocks in both frames can be synchronized, ostensibly showing the absense of time dilation. AG

[Alan Grayson]

And using this method of synchronization, one could synchronize any set of clocks in either frame, provided we know the distance of each from the juxtapose clocks. AG 

[Alan Grayson]

On Thursday, October 24, 2024 at 11:07:18 PM UTC-6 Brent Meeker wrote:

On 10/24/2024 5:46 PM, Alan Grayson wrote:

On Thursday, October 24, 2024 at 1:30:32 PM UTC-6 Brent Meeker wrote:

Here's  how a light-clock ticks in when in motion.  A light-clock is just two perfect mirrors a fixed distance apart with a photon bouncing back an forth between them.  It's a hypothetical ideal clock for which the effect of motion is easily visualized.



These are the spacetime diagrams of three identical light-clocks moving at +c relative to the blue one. 

Three clocks?  Black diagram? If only this was as clear as you claim. TY, AG

You can't handle more than two?  The left clock is black with a red photon.  Is that hard to comprehend?  Didn't they teach spacetime diagrams at your kindergarten?
               
Brent

What makes you think you can teach? I can handle dozens of clocks. I know what a spacetime diagram. It was taught in pre-school. Why did you introduce a red photon? A joke perhaps? How can a clock move at light speed? It's a real muddle. I think you meant well, but you don't have the maturity to contain your temper. Nonetheless, the photon clock gave me a good idea, which I just wrote about. AG 

[Brent Meeker]

On 10/24/2024 11:26 PM, Alan Grayson wrote:

On Thursday, October 24, 2024 at 11:07:18 PM UTC-6 Brent Meeker wrote:

On 10/24/2024 5:46 PM, Alan Grayson wrote:

On Thursday, October 24, 2024 at 1:30:32 PM UTC-6 Brent Meeker wrote:

Here's  how a light-clock ticks in when in motion.  A light-clock is just two perfect mirrors a fixed distance apart with a photon bouncing back an forth between them.  It's a hypothetical ideal clock for which the effect of motion is easily visualized.



These are the spacetime diagrams of three identical light-clocks moving at +c relative to the blue one. 

Three clocks?  Black diagram? If only this was as clear as you claim. TY, AG

You can't handle more than two?  The left clock is black with a red photon.  Is that hard to comprehend?  Didn't they teach spacetime diagrams at your kindergarten?
               
Brent

If you could cease behaving like an arrogant a'hole, likely beyond your maturity and capability, maybe we could get somewhere on this problem. I was thinking about your diagrams, and concluded you could prove your point with the simple observation and simpler diagram, that with your photon clock, it would be easy to show that from the pov of a rest clock, the moving clock would appear to have a slower rate. And since inertial frames are equivalent in SR, the same result would be evident for the rest frame, when considered as the moving frame. BUT, what I have shown, with an arbitrary clock, that all clocks in both frames can be synchronized, ostensibly showing the absense of time dilation. AG

If you've shown it, where's your diagram?

Brent

[Brent Meeker]

On 10/25/2024 1:36 AM, Alan Grayson wrote:

On Thursday, October 24, 2024 at 11:07:18 PM UTC-6 Brent Meeker wrote:

On 10/24/2024 5:46 PM, Alan Grayson wrote:

On Thursday, October 24, 2024 at 1:30:32 PM UTC-6 Brent Meeker wrote:

Here's  how a light-clock ticks in when in motion.  A light-clock is just two perfect mirrors a fixed distance apart with a photon bouncing back an forth between them.  It's a hypothetical ideal clock for which the effect of motion is easily visualized.



These are the spacetime diagrams of three identical light-clocks moving at +c relative to the blue one. 

Three clocks?  Black diagram? If only this was as clear as you claim. TY, AG

You can't handle more than two?  The left clock is black with a red photon.  Is that hard to comprehend?  Didn't they teach spacetime diagrams at your kindergarten?
               
Brent

What makes you think you can teach?

That I have taught and my students came back for more.

I can handle dozens of clocks. I know what a spacetime diagram. It was taught in pre-school. Why did you introduce a red photon? A joke perhaps? How can a clock move at light speed?

None of the clocks in the diagram are moving at light speed.  The black one and the red one are moving at 0.5c as the label says.  What is it you don't understand about this diagram?

Brent

It's a real muddle. I think you meant well, but you don't have the maturity to contain your temper. Nonetheless, the photon clock gave me a good idea, which I just wrote about. AG 

 

Because the speed of light is invariant the photon paths are at unit slope inside all three clocks, so it is easily seen why the relative motion makes the clock seem slow although each clock is ticking at the same rate in it's own reference frame.  The red diagram is just the blue diagram Lorentz transformed as it would be seen in a frame moving the left at 0.5c, and the black diagram as it would be seen from a frame moving to the right.

Brent

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[Alan Grayson]

On Friday, October 25, 2024 at 2:44:06 AM UTC-6 Brent Meeker wrote:

On 10/25/2024 1:36 AM, Alan Grayson wrote:

On Thursday, October 24, 2024 at 11:07:18 PM UTC-6 Brent Meeker wrote:

On 10/24/2024 5:46 PM, Alan Grayson wrote:

On Thursday, October 24, 2024 at 1:30:32 PM UTC-6 Brent Meeker wrote:

Here's  how a light-clock ticks in when in motion.  A light-clock is just two perfect mirrors a fixed distance apart with a photon bouncing back an forth between them.  It's a hypothetical ideal clock for which the effect of motion is easily visualized.



These are the spacetime diagrams of three identical light-clocks moving at +c relative to the blue one. 

Three clocks?  Black diagram? If only this was as clear as you claim. TY, AG

You can't handle more than two?  The left clock is black with a red photon.  Is that hard to comprehend?  Didn't they teach spacetime diagrams at your kindergarten?
               
Brent

What makes you think you can teach?

That I have taught and my students came back for more.

Maybe you're now having an extended "senior moment"? AG 

I can handle dozens of clocks. I know what a spacetime diagram. It was taught in pre-school. Why did you introduce a red photon? A joke perhaps? How can a clock move at light speed?

None of the clocks in the diagram are moving at light speed. 

Then why do you write "These are the spacetime diagrams of three identical light-clocks moving at +c relative to the blue one."? AG

[Alan Grayson]

On Friday, October 25, 2024 at 2:40:26 AM UTC-6 Brent Meeker wrote:

On 10/24/2024 11:26 PM, Alan Grayson wrote:

On Thursday, October 24, 2024 at 11:07:18 PM UTC-6 Brent Meeker wrote:

On 10/24/2024 5:46 PM, Alan Grayson wrote:

On Thursday, October 24, 2024 at 1:30:32 PM UTC-6 Brent Meeker wrote:

Here's  how a light-clock ticks in when in motion.  A light-clock is just two perfect mirrors a fixed distance apart with a photon bouncing back an forth between them.  It's a hypothetical ideal clock for which the effect of motion is easily visualized.



These are the spacetime diagrams of three identical light-clocks moving at +c relative to the blue one. 

Three clocks?  Black diagram? If only this was as clear as you claim. TY, AG

You can't handle more than two?  The left clock is black with a red photon.  Is that hard to comprehend?  Didn't they teach spacetime diagrams at your kindergarten?
               
Brent

If you could cease behaving like an arrogant a'hole, likely beyond your maturity and capability, maybe we could get somewhere on this problem. I was thinking about your diagrams, and concluded you could prove your point with the simple observation and simpler diagram, that with your photon clock, it would be easy to show that from the pov of a rest clock, the moving clock would appear to have a slower rate. And since inertial frames are equivalent in SR, the same result would be evident for the rest frame, when considered as the moving frame. BUT, what I have shown, with an arbitrary clock, that all clocks in both frames can be synchronized, ostensibly showing the absence (sp) of time dilation. AG

If you've shown it, where's your diagram?

Brent

"We don't need no stinkin' badges." -- a famous line from a great movie. I showed that all clocks in both frames can be sychronized, so that establishes (IMO), or at least strongly suggests, that there is no time dilation. Why would it exist if all clocks in both frames are in synch, and since they're identical, they tick at the same rate? AG 

[Alan Grayson]

On Friday, October 25, 2024 at 2:44:06 AM UTC-6 Brent Meeker wrote:

On 10/25/2024 1:36 AM, Alan Grayson wrote:

On Thursday, October 24, 2024 at 11:07:18 PM UTC-6 Brent Meeker wrote:

On 10/24/2024 5:46 PM, Alan Grayson wrote:

On Thursday, October 24, 2024 at 1:30:32 PM UTC-6 Brent Meeker wrote:

Here's  how a light-clock ticks in when in motion.  A light-clock is just two perfect mirrors a fixed distance apart with a photon bouncing back an forth between them.  It's a hypothetical ideal clock for which the effect of motion is easily visualized.



These are the spacetime diagrams of three identical light-clocks moving at +c relative to the blue one. 

Three clocks?  Black diagram? If only this was as clear as you claim. TY, AG

You can't handle more than two?  The left clock is black with a red photon.  Is that hard to comprehend?  Didn't they teach spacetime diagrams at your kindergarten?
               
Brent

What makes you think you can teach?

That I have taught and my students came back for more.

I can handle dozens of clocks. I know what a spacetime diagram. It was taught in pre-school. Why did you introduce a red photon? A joke perhaps? How can a clock move at light speed?

None of the clocks in the diagram are moving at light speed.  The black one and the red one are moving at 0.5c as the label says.  What is it you don't understand about this diagram?

Brent

One thing among several that I don't understand is how the LT is applied. For example, if we transform from one frame to another, say in E&M, IIUC we get what the fields will actually be measured by an observer in the target or primed frame. (I assume we're transferring from frame S to frame S'). But when we use it to establish time dilation say, we don't get what's actually measured in the target frame, but rather how it appears from the pov of the source or unprimed frame. Presumably, that's why you say that after a LT, the internal situation in each transformed frame remains unchanged (or something to that effect). AG

[Jesse Mazer]

On Fri, Oct 25, 2024 at 5:44 AM Alan Grayson <agrays...@gmail.com> wrote:

On Friday, October 25, 2024 at 2:44:06 AM UTC-6 Brent Meeker wrote:

On 10/25/2024 1:36 AM, Alan Grayson wrote:

On Thursday, October 24, 2024 at 11:07:18 PM UTC-6 Brent Meeker wrote:

On 10/24/2024 5:46 PM, Alan Grayson wrote:

On Thursday, October 24, 2024 at 1:30:32 PM UTC-6 Brent Meeker wrote:

Here's  how a light-clock ticks in when in motion.  A light-clock is just two perfect mirrors a fixed distance apart with a photon bouncing back an forth between them.  It's a hypothetical ideal clock for which the effect of motion is easily visualized.



These are the spacetime diagrams of three identical light-clocks moving at +c relative to the blue one. 

Three clocks?  Black diagram? If only this was as clear as you claim. TY, AG

You can't handle more than two?  The left clock is black with a red photon.  Is that hard to comprehend?  Didn't they teach spacetime diagrams at your kindergarten?
               
Brent

What makes you think you can teach?

That I have taught and my students came back for more.

I can handle dozens of clocks. I know what a spacetime diagram. It was taught in pre-school. Why did you introduce a red photon? A joke perhaps? How can a clock move at light speed?

None of the clocks in the diagram are moving at light speed.  The black one and the red one are moving at 0.5c as the label says.  What is it you don't understand about this diagram?

Brent

One thing among several that I don't understand is how the LT is applied. For example, if we transform from one frame to another, say in E&M, IIUC we get what the fields will actually be measured by an observer in the target or primed frame. (I assume we're transferring from frame S to frame S'). But when we use it to establish time dilation say, we don't get what's actually measured in the target frame, but rather how it appears from the pov of the source or unprimed frame. Presumably, that's why you say that after a LT, the internal situation in each transformed frame remains unchanged (or something to that effect). AG

Can you give a concrete example? If you some coordinate-based facts in frame S (source frame) and use the Lorentz transformation to get to frame S' (target frame), the result should be exactly what is measured in the target frame S' using their own system of rulers and clocks at rest relative to themselves (with their own clocks synchronized by the Einstein synchronization convention).

Jesse

 

It's a real muddle. I think you meant well, but you don't have the maturity to contain your temper. Nonetheless, the photon clock gave me a good idea, which I just wrote about. AG 

 

Because the speed of light is invariant the photon paths are at unit slope inside all three clocks, so it is easily seen why the relative motion makes the clock seem slow although each clock is ticking at the same rate in it's own reference frame.  The red diagram is just the blue diagram Lorentz transformed as it would be seen in a frame moving the left at 0.5c, and the black diagram as it would be seen from a frame moving to the right.

Brent

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[Brent Meeker]

On 10/25/2024 1:58 AM, Alan Grayson wrote:

On Friday, October 25, 2024 at 2:44:06 AM UTC-6 Brent Meeker wrote:

On 10/25/2024 1:36 AM, Alan Grayson wrote:

On Thursday, October 24, 2024 at 11:07:18 PM UTC-6 Brent Meeker wrote:

On 10/24/2024 5:46 PM, Alan Grayson wrote:

On Thursday, October 24, 2024 at 1:30:32 PM UTC-6 Brent Meeker wrote:

Here's  how a light-clock ticks in when in motion.  A light-clock is just two perfect mirrors a fixed distance apart with a photon bouncing back an forth between them.  It's a hypothetical ideal clock for which the effect of motion is easily visualized.



These are the spacetime diagrams of three identical light-clocks moving at +c relative to the blue one. 

Three clocks?  Black diagram? If only this was as clear as you claim. TY, AG

You can't handle more than two?  The left clock is black with a red photon.  Is that hard to comprehend?  Didn't they teach spacetime diagrams at your kindergarten?
               
Brent

What makes you think you can teach?

That I have taught and my students came back for more.

Maybe you're now having an extended "senior moment"? AG

And maybe you're too interested in snark to learn anything.

I can handle dozens of clocks. I know what a spacetime diagram. It was taught in pre-school. Why did you introduce a red photon? A joke perhaps? How can a clock move at light speed?

None of the clocks in the diagram are moving at light speed. 

Then why do you write "These are the spacetime diagrams of three identical light-clocks moving at +c relative to the blue one."? AG

Just a typo; I Ieft out the 0.5, which is clearly displayed on the diagram.

Brent

 

The black one and the red one are moving at 0.5c as the label says.  What is it you don't understand about this diagram?

Brent

It's a real muddle. I think you meant well, but you don't have the maturity to contain your temper. Nonetheless, the photon clock gave me a good idea, which I just wrote about. AG 

 

Because the speed of light is invariant the photon paths are at unit slope inside all three clocks, so it is easily seen why the relative motion makes the clock seem slow although each clock is ticking at the same rate in it's own reference frame.  The red diagram is just the blue diagram Lorentz transformed as it would be seen in a frame moving the left at 0.5c, and the black diagram as it would be seen from a frame moving to the right.

Brent

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[Brent Meeker]

On 10/25/2024 2:15 AM, Alan Grayson wrote:

On Friday, October 25, 2024 at 2:40:26 AM UTC-6 Brent Meeker wrote:

On 10/24/2024 11:26 PM, Alan Grayson wrote:

On Thursday, October 24, 2024 at 11:07:18 PM UTC-6 Brent Meeker wrote:

On 10/24/2024 5:46 PM, Alan Grayson wrote:

On Thursday, October 24, 2024 at 1:30:32 PM UTC-6 Brent Meeker wrote:

Here's  how a light-clock ticks in when in motion.  A light-clock is just two perfect mirrors a fixed distance apart with a photon bouncing back an forth between them.  It's a hypothetical ideal clock for which the effect of motion is easily visualized.



These are the spacetime diagrams of three identical light-clocks moving at +c relative to the blue one. 

Three clocks?  Black diagram? If only this was as clear as you claim. TY, AG

You can't handle more than two?  The left clock is black with a red photon.  Is that hard to comprehend?  Didn't they teach spacetime diagrams at your kindergarten?
               
Brent

If you could cease behaving like an arrogant a'hole, likely beyond your maturity and capability, maybe we could get somewhere on this problem. I was thinking about your diagrams, and concluded you could prove your point with the simple observation and simpler diagram, that with your photon clock, it would be easy to show that from the pov of a rest clock, the moving clock would appear to have a slower rate. And since inertial frames are equivalent in SR, the same result would be evident for the rest frame, when considered as the moving frame. BUT, what I have shown, with an arbitrary clock, that all clocks in both frames can be synchronized, ostensibly showing the absence (sp) of time dilation. AG

If you've shown it, where's your diagram?

Brent

"We don't need no stinkin' badges." -- a famous line from a great movie.

Treasure of Sierra Madre

I showed that all clocks in both frames can be sychronized, so that establishes (IMO), or at least strongly suggests, that there is no time dilation.

So where is this revelation written down.  And what do you mean by "can be synchronized"?  Nobody doubts that two moving clocks, while in close proximity, can be set to the same time; but what they do after that is generally called exhibiting time dilation...although I think that's misleading when their relationship is symmetric.

Why would it exist if all clocks in both frames are in synch, and since they're identical, they tick at the same rate? AG

Each ticks at the the given rate in it's own frame.  But it doesn't follow that they tick the same as measured different frame moving relative to theirs.  That's the point of the diagram.  All three clocks tick at the same rate in their own frame, the black and red clocks click slower in the frame of the blue clock.

Brent

 

Because the speed of light is invariant the photon paths are at unit slope inside all three clocks, so it is easily seen why the relative motion makes the clock seem slow although each clock is ticking at the same rate in it's own reference frame.  The red diagram is just the blue diagram Lorentz transformed as it would be seen in a frame moving the left at 0.5c, and the black diagram as it would be seen from a frame moving to the right.

Brent

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[Brent Meeker]

On 10/25/2024 2:44 AM, Alan Grayson wrote:

On Friday, October 25, 2024 at 2:44:06 AM UTC-6 Brent Meeker wrote:

On 10/25/2024 1:36 AM, Alan Grayson wrote:

On Thursday, October 24, 2024 at 11:07:18 PM UTC-6 Brent Meeker wrote:

On 10/24/2024 5:46 PM, Alan Grayson wrote:

On Thursday, October 24, 2024 at 1:30:32 PM UTC-6 Brent Meeker wrote:

Here's  how a light-clock ticks in when in motion.  A light-clock is just two perfect mirrors a fixed distance apart with a photon bouncing back an forth between them.  It's a hypothetical ideal clock for which the effect of motion is easily visualized.



These are the spacetime diagrams of three identical light-clocks moving at +c relative to the blue one. 

Three clocks?  Black diagram? If only this was as clear as you claim. TY, AG

You can't handle more than two?  The left clock is black with a red photon.  Is that hard to comprehend?  Didn't they teach spacetime diagrams at your kindergarten?
               
Brent

What makes you think you can teach?

That I have taught and my students came back for more.

I can handle dozens of clocks. I know what a spacetime diagram. It was taught in pre-school. Why did you introduce a red photon? A joke perhaps? How can a clock move at light speed?

None of the clocks in the diagram are moving at light speed.  The black one and the red one are moving at 0.5c as the label says.  What is it you don't understand about this diagram?

Brent

One thing among several that I don't understand is how the LT is applied.

The set of points (t,x) representing a stationary light clock, two vertical lines with 45deg "photon" lines bouncing between them, is generated.  This is the blue one.  Then a Lorentz transform is applied to the that set of points.  Here's the LISP code I used:

(defun gammma (v) (/ (sqrt (- 1.0 (* v v)))))

(defun lorentz-2d (v)
"Returns a function that takes a point (t,x) and
returns the transformed point (t',x')"
   (lambda (p)
      (let ((t0 (car p))
            (x0 (cadr p))
            (g (gammma v)))
      (list (+ (* g t0) (* g (/ x0 v))) ;this is t'
            (+ (* g x0) (* g v t0)))))) ;this is x'

(defun lorentz-trans (v points)
"Transform a set of points, e.g. a world-line"
      (translate (mapcar (lorentz-2d v) points) (car points)))

I don't know if you read LISP but in this case it's very simple.  The first line is just defining the gamma value for a given speed, in this case v=0.5.  The second function lorentz-2d(v) takes a speed and returns a function lambda(p) that takes a point and returns the point's coordinates in the frame moving at speed v.  The third function just applies that transform to every point on the given list points.  The translate function just moves the graph timewise so it lines up with a given starting point.

For example, if we transform from one frame to another, say in E&M, IIUC we get what the fields will actually be measured by an observer in the target or primed frame. (I assume we're transferring from frame S to frame S'). But when we use it to establish time dilation say, we don't get what's actually measured in the target frame, but rather how it appears from the pov of the source or unprimed frame.

Right.

Presumably, that's why you say that after a LT, the internal situation in each transformed frame remains unchanged (or something to that effect). AG

Right.  A coordinate transform shouldn't change anything local.  In this case it is just distances and times that are different than one expects from Euclidean space+time.  Everyone in their own local inertial is ex hypothesi looking at an identical perfect clock.

Brent

It's a real muddle. I think you meant well, but you don't have the maturity to contain your temper. Nonetheless, the photon clock gave me a good idea, which I just wrote about. AG 

 

Because the speed of light is invariant the photon paths are at unit slope inside all three clocks, so it is easily seen why the relative motion makes the clock seem slow although each clock is ticking at the same rate in it's own reference frame.  The red diagram is just the blue diagram Lorentz transformed as it would be seen in a frame moving the left at 0.5c, and the black diagram as it would be seen from a frame moving to the right.

Brent

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[Alan Grayson]

On Friday, October 25, 2024 at 11:34:13 AM UTC-6 Jesse Mazer wrote:

On Fri, Oct 25, 2024 at 5:44 AM Alan Grayson <agrays...@gmail.com> wrote:

On Friday, October 25, 2024 at 2:44:06 AM UTC-6 Brent Meeker wrote:

On 10/25/2024 1:36 AM, Alan Grayson wrote:

On Thursday, October 24, 2024 at 11:07:18 PM UTC-6 Brent Meeker wrote:

On 10/24/2024 5:46 PM, Alan Grayson wrote:

On Thursday, October 24, 2024 at 1:30:32 PM UTC-6 Brent Meeker wrote:

Here's  how a light-clock ticks in when in motion.  A light-clock is just two perfect mirrors a fixed distance apart with a photon bouncing back an forth between them.  It's a hypothetical ideal clock for which the effect of motion is easily visualized.



These are the spacetime diagrams of three identical light-clocks moving at +c relative to the blue one. 

Three clocks?  Black diagram? If only this was as clear as you claim. TY, AG

You can't handle more than two?  The left clock is black with a red photon.  Is that hard to comprehend?  Didn't they teach spacetime diagrams at your kindergarten?
               
Brent

What makes you think you can teach?

That I have taught and my students came back for more.

I can handle dozens of clocks. I know what a spacetime diagram. It was taught in pre-school. Why did you introduce a red photon? A joke perhaps? How can a clock move at light speed?

None of the clocks in the diagram are moving at light speed.  The black one and the red one are moving at 0.5c as the label says.  What is it you don't understand about this diagram?

Brent

One thing among several that I don't understand is how the LT is applied. For example, if we transform from one frame to another, say in E&M, IIUC we get what the fields will actually be measured by an observer in the target or primed frame. (I assume we're transferring from frame S to frame S'). But when we use it to establish time dilation say, we don't get what's actually measured in the target frame, but rather how it appears from the pov of the source or unprimed frame. Presumably, that's why you say that after a LT, the internal situation in each transformed frame remains unchanged (or something to that effect). AG

Can you give a concrete example? If you some coordinate-based facts in frame S (source frame) and use the Lorentz transformation to get to frame S' (target frame), the result should be exactly what is measured in the target frame S' using their own system of rulers and clocks at rest relative to themselves (with their own clocks synchronized by the Einstein synchronization convention).

Jesse

Glad you asked that question. Yes, this is what I expect when we use the LT. We measure some observable in S, use the LT to calculate its value in S', and this what an observer in S' will measure. But notice this, say for length contraction. Whereas from the pov of S, a moving rod shrinks as calculated and viewed from S, the observer in S' doesn't measure the rod as shortened! This is why I claim that the LT sometimes just tells how things appear in the source frame S, but not what an observer in S' actually measures. AG

On another point concerning time dilation; I demonstrated that given two inertial frames with relative velocity v < c, it's easy to synchronize clocks in both frames provided we know the distance of clocks from the location of juxtaposition, but I was mistaken in concluding this alone shows time dilation doesn't exist. It does, because we insist on using the LT as the only transformation between these frames, and the reason we do this is because the LT is presumably the only transformation that guarantees the invariance of the velocity of light. So time dilation is, so to speak, the price we pay for imposing the invariance of the velocity of light on our frame transformation. But I remain unclear how a breakdown in simultaneity resolves the apparent paradox of two frames viewing a passing clock in another frame, as running slower than its own clock. AG

Finally, for Brent, a word about "snarky". You get snarky when I don't understand something, like your "kindergarten" reference in one of your recent replies. And occasionally I am correct in my criticisms. Moreover, if you have typos in your explanation of your graph, you shouldn't be surprised if they make it hard to understand your graphical explanation of time dilation. AG

[Jesse Mazer]

On Fri, Oct 25, 2024 at 5:49 PM Alan Grayson <agrays...@gmail.com> wrote:

On Friday, October 25, 2024 at 11:34:13 AM UTC-6 Jesse Mazer wrote:

On Fri, Oct 25, 2024 at 5:44 AM Alan Grayson <agrays...@gmail.com> wrote:

On Friday, October 25, 2024 at 2:44:06 AM UTC-6 Brent Meeker wrote:

On 10/25/2024 1:36 AM, Alan Grayson wrote:

On Thursday, October 24, 2024 at 11:07:18 PM UTC-6 Brent Meeker wrote:

On 10/24/2024 5:46 PM, Alan Grayson wrote:

On Thursday, October 24, 2024 at 1:30:32 PM UTC-6 Brent Meeker wrote:

Here's  how a light-clock ticks in when in motion.  A light-clock is just two perfect mirrors a fixed distance apart with a photon bouncing back an forth between them.  It's a hypothetical ideal clock for which the effect of motion is easily visualized.



These are the spacetime diagrams of three identical light-clocks moving at +c relative to the blue one. 

Three clocks?  Black diagram? If only this was as clear as you claim. TY, AG

You can't handle more than two?  The left clock is black with a red photon.  Is that hard to comprehend?  Didn't they teach spacetime diagrams at your kindergarten?
               
Brent

What makes you think you can teach?

That I have taught and my students came back for more.

I can handle dozens of clocks. I know what a spacetime diagram. It was taught in pre-school. Why did you introduce a red photon? A joke perhaps? How can a clock move at light speed?

None of the clocks in the diagram are moving at light speed.  The black one and the red one are moving at 0.5c as the label says.  What is it you don't understand about this diagram?

Brent

One thing among several that I don't understand is how the LT is applied. For example, if we transform from one frame to another, say in E&M, IIUC we get what the fields will actually be measured by an observer in the target or primed frame. (I assume we're transferring from frame S to frame S'). But when we use it to establish time dilation say, we don't get what's actually measured in the target frame, but rather how it appears from the pov of the source or unprimed frame. Presumably, that's why you say that after a LT, the internal situation in each transformed frame remains unchanged (or something to that effect). AG

Can you give a concrete example? If you some coordinate-based facts in frame S (source frame) and use the Lorentz transformation to get to frame S' (target frame), the result should be exactly what is measured in the target frame S' using their own system of rulers and clocks at rest relative to themselves (with their own clocks synchronized by the Einstein synchronization convention).

Jesse

Glad you asked that question. Yes, this is what I expect when we use the LT. We measure some observable in S, use the LT to calculate its value in S', and this what an observer in S' will measure. But notice this, say for length contraction. Whereas from the pov of S, a moving rod shrinks as calculated and viewed from S, the observer in S' doesn't measure the rod as shortened! This is why I claim that the LT sometimes just tells how things appear in the source frame S, but not what an observer in S' actually measures. AG

But the length contraction equation isn't one of the Lorentz transformation equations that map coordinates in S to coordinates in S'. It is derivable from them, but the intended use of the length contraction formula is for an observer at rest in S is if they want to predict how long a rod will be in their own rest frame, given knowledge of its length in the rod's rest frame S'.

If you use the actual Lorentz coordinate transformation to derive it, one way to do it would be to start in frame S' where the rod is at rest (so position coordinates of its two ends are unchanging), then pick two events on the worldline of either end which are not simultaneous in S' but are simultaneous in S, and then use the inverse Lorentz transformation from S' to S (equations given in the second box at https://en.wikipedia.org/wiki/Lorentz_transformation#Coordinate_transformation) to find the position coordinates of each end at a single moment in S (since "length" in a given frame is understood as the distance between two ends of an object at a single moment in time in that frame).

For example suppose S' is moving at v=0.6c relative to S, so the gamma factor in those equations is 1.25. And suppose in S' the rod's left end is at rest at position coordinate x'=0 light years, and the right end is at rest at position coordinate x'=12 light years. So for the event on the worldline of the left end, pick x'=0 light years, t'=0 years; and for the event on the worldline of the right end, pick x'=10 light years, t'=-6 seconds. In this case if you use the inverse Lorentz transformation to find the coordinates of these events in frame S, the event at the left end would be x=0 light years, t=0 years (the origins of the two frames coincide according to the transformation), and the event on the right end would work out to:

t = gamma*(t' + v*x') = 1.25 * (-6 + 0.6*10) = 1.25 * (-6 + 6) = 0

x = gamma*(x' + v*t') = 1.25 * (10 + 0.6*-6) = 1.25 * (6.4) = 8

So you get the conclusion that for a rod with rest length 10 light years in its own frame S', in the S frame at time t=0 the left end is at position x=0 light years and the right end is at position x=8 light years, so its length is 8 light years in the S frame, which is exactly what's predicted by the Lorentz contraction equation.

 

On another point concerning time dilation; I demonstrated that given two inertial frames with relative velocity v < c, it's easy to synchronize clocks in both frames provided we know the distance of clocks from the location of juxtaposition, but I was mistaken in concluding this alone shows time dilation doesn't exist.

I don't think you've shown that, or at least you haven't clearly explained what you mean--you didn't answer my question about your procedure that I asked right before your back-and-forth exchange with Brent (the message prior to the one where I asked for a concrete example).

Jesse

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[Brent Meeker]

On 10/25/2024 2:49 PM, Alan Grayson wrote:

On Friday, October 25, 2024 at 11:34:13 AM UTC-6 Jesse Mazer wrote:

On Fri, Oct 25, 2024 at 5:44 AM Alan Grayson <agrays...@gmail.com> wrote:

On Friday, October 25, 2024 at 2:44:06 AM UTC-6 Brent Meeker wrote:

On 10/25/2024 1:36 AM, Alan Grayson wrote:

On Thursday, October 24, 2024 at 11:07:18 PM UTC-6 Brent Meeker wrote:

On 10/24/2024 5:46 PM, Alan Grayson wrote:

On Thursday, October 24, 2024 at 1:30:32 PM UTC-6 Brent Meeker wrote:

Here's  how a light-clock ticks in when in motion.  A light-clock is just two perfect mirrors a fixed distance apart with a photon bouncing back an forth between them.  It's a hypothetical ideal clock for which the effect of motion is easily visualized.



These are the spacetime diagrams of three identical light-clocks moving at +c relative to the blue one. 

Three clocks?  Black diagram? If only this was as clear as you claim. TY, AG

You can't handle more than two?  The left clock is black with a red photon.  Is that hard to comprehend?  Didn't they teach spacetime diagrams at your kindergarten?
               
Brent

What makes you think you can teach?

That I have taught and my students came back for more.

I can handle dozens of clocks. I know what a spacetime diagram. It was taught in pre-school. Why did you introduce a red photon? A joke perhaps? How can a clock move at light speed?

None of the clocks in the diagram are moving at light speed.  The black one and the red one are moving at 0.5c as the label says.  What is it you don't understand about this diagram?

Brent

One thing among several that I don't understand is how the LT is applied. For example, if we transform from one frame to another, say in E&M, IIUC we get what the fields will actually be measured by an observer in the target or primed frame. (I assume we're transferring from frame S to frame S'). But when we use it to establish time dilation say, we don't get what's actually measured in the target frame, but rather how it appears from the pov of the source or unprimed frame. Presumably, that's why you say that after a LT, the internal situation in each transformed frame remains unchanged (or something to that effect). AG

Can you give a concrete example? If you some coordinate-based facts in frame S (source frame) and use the Lorentz transformation to get to frame S' (target frame), the result should be exactly what is measured in the target frame S' using their own system of rulers and clocks at rest relative to themselves (with their own clocks synchronized by the Einstein synchronization convention).

Jesse

Glad you asked that question. Yes, this is what I expect when we use the LT. We measure some observable in S, use the LT to calculate its value in S', and this what an observer in S' will measure. But notice this, say for length contraction. Whereas from the pov of S, a moving rod shrinks as calculated and viewed from S, the observer in S' doesn't measure the rod as shortened! This is why I claim that the LT sometimes just tells how things appear in the source frame S, but not what an observer in S' actually measures. AG

Yes, although "appear" can be misleading when you consider things moving near light speed.  More accurate is "measure", using the invariant speed of light.

On another point concerning time dilation; I demonstrated that given two inertial frames with relative velocity v < c, it's easy to synchronize clocks in both frames provided we know the distance of clocks from the location of juxtaposition, but I was mistaken in concluding this alone shows time dilation doesn't exist. It does, because we insist on using the LT as the only transformation between these frames, and the reason we do this is because the LT is presumably the only transformation that guarantees the invariance of the velocity of light. So time dilation is, so to speak, the price we pay for imposing the invariance of the velocity of light on our frame transformation. But I remain unclear how a breakdown in simultaneity resolves the apparent paradox of two frames viewing a passing clock in another frame, as running slower than its own clock. AG

Look at the diagram I provided.  At the bottom (t=0) the three clocks are passing by one another.  The blue clock sees the other two as running slower.

Finally, for Brent, a word about "snarky". You get snarky when I don't understand something, like your "kindergarten" reference in one of your recent replies. And occasionally I am correct in my criticisms. Moreover, if you have typos in your explanation of your graph, you shouldn't be surprised if they make it hard to understand your graphical explanation of time dilation. AG

So that one typo, which was correct elsewhere made it muddled for you?

Brent

[Brent Meeker]

If you look at the diagram I posted, you can see that the distance between the mirrors of the blue light clock is about 2.4 units while for the moving light clocks it is about 2.0 divisions.  That's the Lorentz contraction.  Interestingly Lorentz derived the equation for this contraction assuming the the motion of a body consisting of atoms held in place by their relative electrical forces and moving thru the aether.



Brent

If you use the actual Lorentz coordinate transformation to derive it, one way to do it would be to start in frame S' where the rod is at rest (so position coordinates of its two ends are unchanging), then pick two events on the worldline of either end which are not simultaneous in S' but are simultaneous in S, and then use the inverse Lorentz transformation from S' to S (equations given in the second box at https://en.wikipedia.org/wiki/Lorentz_transformation#Coordinate_transformation) to find the position coordinates of each end at a single moment in S (since "length" in a given frame is understood as the distance between two ends of an object at a single moment in time in that frame).

For example suppose S' is moving at v=0.6c relative to S, so the gamma factor in those equations is 1.25. And suppose in S' the rod's left end is at rest at position coordinate x'=0 light years, and the right end is at rest at position coordinate x'=12 light years. So for the event on the worldline of the left end, pick x'=0 light years, t'=0 years; and for the event on the worldline of the right end, pick x'=10 light years, t'=-6 seconds. In this case if you use the inverse Lorentz transformation to find the coordinates of these events in frame S, the event at the left end would be x=0 light years, t=0 years (the origins of the two frames coincide according to the transformation), and the event on the right end would work out to:

t = gamma*(t' + v*x') = 1.25 * (-6 + 0.6*10) = 1.25 * (-6 + 6) = 0

x = gamma*(x' + v*t') = 1.25 * (10 + 0.6*-6) = 1.25 * (6.4) = 8

So you get the conclusion that for a rod with rest length 10 light years in its own frame S', in the S frame at time t=0 the left end is at position x=0 light years and the right end is at position x=8 light years, so its length is 8 light years in the S frame, which is exactly what's predicted by the Lorentz contraction equation.

 

On another point concerning time dilation; I demonstrated that given two inertial frames with relative velocity v < c, it's easy to synchronize clocks in both frames provided we know the distance of clocks from the location of juxtaposition, but I was mistaken in concluding this alone shows time dilation doesn't exist.

I don't think you've shown that, or at least you haven't clearly explained what you mean--you didn't answer my question about your procedure that I asked right before your back-and-forth exchange with Brent (the message prior to the one where I asked for a concrete example).

Jesse

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[Jesse Mazer]

Sorry, that should say that "the right end is at rest at position coordinate x'=10 light years", I changed the numbers as I was typing out this example but forgot to edit that one.

[Alan Grayson]

Well, one could set the two juxtaposed clocks to both read t=0, and then set another clock, in either frame, at some known measured distance removed, and send a signal, say sound waves, of known velocity, and then set the time on this clock, advanced from the juxtaposed clock, depending on how long it takes to signal to reach its destination.  And one could do this repeatedly for other clocks in either frame. AG

[Alan Grayson]

In part yes. When I think an author doesn't know what he's expounding about, I lose interest. Also, although I was a software engineer at JPL, I don't know LISP,  so it would be hard to see what assumptions you made in generating the plot. And the plot is claimed to establish time dilation, and I'm not sure how you developed the width of the blue path say, to show time passes more rapidly compared to the other plots.  AG

Brent

[Jesse Mazer]

OK, suppose you have two clocks in relative motion A1 and B1, and at the moment B1 passes next to A1 (i.e. is spatially juxtaposed with it), they both read t=0. Then you have another clock A2 at rest relative to A1, and you used the Einstein synchronization convention in the A frame to set A2 based on the reading on A1, assuming light moves at constant speed in both directions in the A frame. You also have another clock B2 at rest relative to B1, and you used the Einstein synchronization convention in the B frame to set B2 based on the reading on B1, assuming light moves at constant speed in both directions in the B frame. In this case, if the distances are such that B2 is passing next to A2 when A2 reads t=0, then B2 will *not* read t=0 at that moment, because using the Einstein synchronization convention for different sets of clocks in different frames results in different definitions of simultaneity for each frame.

Of course, if you preferred you could just use the Einstein synchronization convention only for the two A clocks, and set both B clocks to read 0 at the moment they passed an A clock. But this would be an arbitrary choice to favor the A frame's definition of simultaneity--you could just as easily have done the opposite and used the Einstein synchronization convention only for the two B clocks, and set both A clocks to read 0 at the moment they passed a B clock. So you haven't given a procedure that gives a unique definition of simultaneity.

Jesse

 

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[Alan Grayson]

I don't follow. A2 never reads t=0. It starts at some t > 0, accounting for the elapsed time on A1 when the signal reaches A2. And B2 is similarly synchronized and never reads t=0. AG 

[Brent Meeker]

I just assumed a width for the blue path.  All that determines is how fast the light clock ticks.  Then the other two light clock world lines were generated by point-by-point application of the given Lorentz transform.  So I showed the two clocks moving relative to blue ticked more slowly, not the other way around.  Do you not see that the bouncing photon hits the mirror less often in red's clock as measured in blue's frame.

Brent

Brent

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[Alan Grayson]

Yes, so that implies tics are less frequent in red's clock, compared to blue's clock, so the time rate for red is less than blue, which is what I in effect posted -- that blue clock tics more rapidly than red clock. Why do you fail to understand what I wrote? AG 

Brent

[Jesse Mazer]

I was assuming A2 was synchronized with A1 at some moment before either read t=0, for example A could have transmitted the signal when it read t=-20. The details aren't important but my point was about what would happen assuming the synchronization for each pair of clocks (A1/A2 as well as B1/B2) happened prior to the moment when B1 passed A1, and prior to the moment the B2 passed A2. You could equally well assume the synchronization signal was sent at t=0 but the crossing of paths of the A's and the B's didn't happen until a later time, with the synchronization of each planned in such a way that A1 and B2 would read exactly the same time (say, t=10) at the moment they crossed paths.

Jesse

 

[Brent Meeker]

I understood it, but it read as if you didn't realize red was just the transform of blue and it is in the clock's own frame it runs fastest.  You wrote as though I "developed the width of the blue path say, to show time passes more rapidly"  whereas I chose it arbitrarily and derived the other two.

Brent

Brent

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[Alan Grayson]

So that one typo, which was correct elsewhere made it muddled for you?

In part yes. When I think an author doesn't know what he's expounding about, I lose interest. Also, although I was a software engineer at JPL, I don't know LISP,  so it would be hard to see what assumptions you made in generating the plot. And the plot is claimed to establish time dilation, and I'm not sure how you developed the width of the blue path say, to show time passes more rapidly compared to the other plots.  AG

I just assumed a width for the blue path.  All that determines is how fast the light clock ticks.  Then the other two light clock world lines were generated by point-by-point application of the given Lorentz transform.  So I showed the two clocks moving relative to blue ticked more slowly, not the other way around.  Do you not see that the bouncing photon hits the mirror less often in red's clock as measured in blue's frame.

Yes, so that implies tics are less frequent in red's clock, compared to blue's clock, so the time rate for red is less than blue, which is what I in effect posted -- that blue clock tics more rapidly than red clock. Why do you fail to understand what I wrote? AG

I understood it, but it read as if you didn't realize red was just the transform of blue and it is in the clock's own frame it runs fastest.  You wrote as though I "developed the width of the blue path say, to show time passes more rapidly"  whereas I chose it arbitrarily and derived the other two.

Brent

 

Are you saying the red clock is in the same frame as the blue clock? I missed that point. Why did you model it this way, instead of just using two frames, one at rest, the other moving? Why does the red clock's photon cross at right angles, but this isn't so for the blue clock? Were they arbitrary choices? AG

This discussion began with my claim that there could be a clock paradox, defined by two clocks, each running slower than the other. If such a paradox existed, it would be impossible to produce a plot which would show it. So, what exactly does your plot show; that the LT establishes that a moving clock runs slower than a stationary clock? This is not something I disputed. I don't see how your plot resolves a possible paradox. AG

I thought that if I could synchronize clocks in two inertial frames without the LT, I could establish the paradox. But now I don't think this is true. What is true, is that the LT causes time dilation, and is, so to speak, the price we pay to guarantee frame invariance of the SoL. AG

For Jesse; I looked up Einstein's method for determining simultaneous events. IIUC, it involves two clocks and a light source midway between them to produce simultaneous events, with the conclusion that simultaneity exists in the rest frame of the clocks, but not in a moving frame. I didn't use it to establish that clocks in two inertial frames can be synchronized. Neither did I deny it. I don't see why you think there's something awry that I didn't use it. AG

[Brent Meeker]

On 10/26/2024 12:06 AM, Alan Grayson wrote:

So that one typo, which was correct elsewhere made it muddled for you?

In part yes. When I think an author doesn't know what he's expounding about, I lose interest. Also, although I was a software engineer at JPL, I don't know LISP,  so it would be hard to see what assumptions you made in generating the plot. And the plot is claimed to establish time dilation, and I'm not sure how you developed the width of the blue path say, to show time passes more rapidly compared to the other plots.  AG

I just assumed a width for the blue path.  All that determines is how fast the light clock ticks.  Then the other two light clock world lines were generated by point-by-point application of the given Lorentz transform.  So I showed the two clocks moving relative to blue ticked more slowly, not the other way around.  Do you not see that the bouncing photon hits the mirror less often in red's clock as measured in blue's frame.

Yes, so that implies tics are less frequent in red's clock, compared to blue's clock, so the time rate for red is less than blue, which is what I in effect posted -- that blue clock tics more rapidly than red clock. Why do you fail to understand what I wrote? AG

I understood it, but it read as if you didn't realize red was just the transform of blue and it is in the clock's own frame it runs fastest.  You wrote as though I "developed the width of the blue path say, to show time passes more rapidly"  whereas I chose it arbitrarily and derived the other two.

Brent

 

Are you saying the red clock is in the same frame as the blue clock? I missed that point.

No, it's obviously moving relative to the blue clock.

Why did you model it this way, instead of just using two frames, one at rest, the other moving?

Because I was addressing a different question.  I didn't create it just for you.

Why does the red clock's photon cross at right angles, but this isn't so for the blue clock? Were they arbitrary choices? AG

All the photon lines are at 45deg because the speed of light is invariant; its 1.0c in the diagram for blue.  If I transformed to red's frame then red would look just like blue does in this diagram.

Brent

This discussion began with my claim that there could be a clock paradox, defined by two clocks, each running slower than the other. If such a paradox existed, it would be impossible to produce a plot which would show it. So, what exactly does your plot show; that the LT establishes that a moving clock runs slower than a stationary clock? This is not something I disputed. I don't see how your plot resolves a possible paradox. AG

A failure of imagination.  As drawn in blues frame, red is obviously running slower than blue.  But suppose we transform to red's frame.  Then red would go straight up the diagram taking the place of blue.  We know this because red was generated from blue by the inverse transform.  Then blue will be transformed to black.  And in that picture blue will obviously be slower than red.

Brent

I thought that if I could synchronize clocks in two inertial frames without the LT, I could establish the paradox. But now I don't think this is true. What is true, is that the LT causes time dilation, and is, so to speak, the price we pay to guarantee frame invariance of the SoL. AG

For Jesse; I looked up Einstein's method for determining simultaneous events. IIUC, it involves two clocks and a light source midway between them to produce simultaneous events, with the conclusion that simultaneity exists in the rest frame of the clocks, but not in a moving frame. I didn't use it to establish that clocks in two inertial frames can be synchronized. Neither did I deny it. I don't see why you think there's something awry that I didn't use it. AG

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[Alan Grayson]

On Saturday, October 26, 2024 at 2:38:23 AM UTC-6 Brent Meeker wrote:

On 10/26/2024 12:06 AM, Alan Grayson wrote:

So that one typo, which was correct elsewhere made it muddled for you?

In part yes. When I think an author doesn't know what he's expounding about, I lose interest. Also, although I was a software engineer at JPL, I don't know LISP,  so it would be hard to see what assumptions you made in generating the plot. And the plot is claimed to establish time dilation, and I'm not sure how you developed the width of the blue path say, to show time passes more rapidly compared to the other plots.  AG

I just assumed a width for the blue path.  All that determines is how fast the light clock ticks.  Then the other two light clock world lines were generated by point-by-point application of the given Lorentz transform.  So I showed the two clocks moving relative to blue ticked more slowly, not the other way around.  Do you not see that the bouncing photon hits the mirror less often in red's clock as measured in blue's frame.

Yes, so that implies tics are less frequent in red's clock, compared to blue's clock, so the time rate for red is less than blue, which is what I in effect posted -- that blue clock tics more rapidly than red clock. Why do you fail to understand what I wrote? AG

I understood it, but it read as if you didn't realize red was just the transform of blue and it is in the clock's own frame it runs fastest.  You wrote as though I "developed the width of the blue path say, to show time passes more rapidly"  whereas I chose it arbitrarily and derived the other two.

Brent

 

Are you saying the red clock is in the same frame as the blue clock? I missed that point.

No, it's obviously moving relative to the blue clock.

Not obvious since your first sentence above is unintelligible. AG

Why did you model it this way, instead of just using two frames, one at rest, the other moving?

Because I was addressing a different question.  I didn't create it just for you.

Why does the red clock's photon cross at right angles, but this isn't so for the blue clock? Were they arbitrary choices? AG

All the photon lines are at 45deg because the speed of light is invariant; its 1.0c in the diagram for blue.  If I transformed to red's frame then red would look just like blue does in this diagram.

Brent

This discussion began with my claim that there could be a clock paradox, defined by two clocks, each running slower than the other. If such a paradox existed, it would be impossible to produce a plot which would show it. So, what exactly does your plot show; that the LT establishes that a moving clock runs slower than a stationary clock? This is not something I disputed. I don't see how your plot resolves a possible paradox. AG

A failure of imagination. 

Best to avoid the snark. AG

 

   |   As drawn in blues frame, red is obviously running slower than blue.

But you replied above that red is not in blue's frame. It's hard to see has this resolves the issue of a possible clock paradox. AG

[Alan Grayson]

On Saturday, October 26, 2024 at 8:05:45 AM UTC-6 Alan Grayson wrote:

On Saturday, October 26, 2024 at 2:38:23 AM UTC-6 Brent Meeker wrote:

On 10/26/2024 12:06 AM, Alan Grayson wrote:

So that one typo, which was correct elsewhere made it muddled for you?

In part yes. When I think an author doesn't know what he's expounding about, I lose interest. Also, although I was a software engineer at JPL, I don't know LISP,  so it would be hard to see what assumptions you made in generating the plot. And the plot is claimed to establish time dilation, and I'm not sure how you developed the width of the blue path say, to show time passes more rapidly compared to the other plots.  AG

I just assumed a width for the blue path.  All that determines is how fast the light clock ticks.  Then the other two light clock world lines were generated by point-by-point application of the given Lorentz transform.  So I showed the two clocks moving relative to blue ticked more slowly, not the other way around.  Do you not see that the bouncing photon hits the mirror less often in red's clock as measured in blue's frame.

Yes, so that implies tics are less frequent in red's clock, compared to blue's clock, so the time rate for red is less than blue, which is what I in effect posted -- that blue clock tics more rapidly than red clock. Why do you fail to understand what I wrote? AG

I understood it, but it read as if you didn't realize red was just the transform of blue and it is in the clock's own frame it runs fastest.  You wrote as though I "developed the width of the blue path say, to show time passes more rapidly"  whereas I chose it arbitrarily and derived the other two.

Brent

 

Are you saying the red clock is in the same frame as the blue clock? I missed that point.

No, it's obviously moving relative to the blue clock.

Not obvious since your first sentence above is unintelligible. AG

Why did you model it this way, instead of just using two frames, one at rest, the other moving?

Because I was addressing a different question.  I didn't create it just for you.

Why does the red clock's photon cross at right angles, but this isn't so for the blue clock? Were they arbitrary choices? AG

All the photon lines are at 45deg because the speed of light is invariant; its 1.0c in the diagram for blue.  If I transformed to red's frame then red would look just like blue does in this diagram.

Brent

This discussion began with my claim that there could be a clock paradox, defined by two clocks, each running slower than the other. If such a paradox existed, it would be impossible to produce a plot which would show it. So, what exactly does your plot show; that the LT establishes that a moving clock runs slower than a stationary clock? This is not something I disputed. I don't see how your plot resolves a possible paradox. AG

A failure of imagination. 

Best to avoid the snark. AG

 

   |   As drawn in blues frame, red is obviously running slower than blue.

 But you replied above that red is not in blue's frame. When you refer to a frame transformation, you're in a different frame AFAIK. It's hard to see how this resolves the issue of a possible clock paradox. You're just using the LT and its inverse, and this just shows what we already know from the LT. Incidentally, you have two typos related to the SoL. AG

[Jesse Mazer]

Again, the problem is that you simply haven't clearly laid out what your procedure is for synchronizing different clocks at rest in the *same* frame, so your summary of the experiment you want to set up is too vague without that information. Are all the A clocks synchronized with one another using the Einstein synchronization procedure in the A frame, and then the B clocks set with reference to whichever A clock they are next to at some moment? Or is just one B clock set by reference to the A clock it's next to, and the other B clocks synchronized with that first B clock using the Einstein synchronization procedure in the B frame? Or some other option?

Jesse

[Alan Grayson]

I assumed that if the two juxtaposed clocks were set to t=0, and I specified how any other clock in either frame could be synchronized to those two clocks, one can infer how to synchronize any other clock, in any known distance to any previously synchronized clock, and synchronize that clock. I was trying to imagine a scheme for synchronizing all hypothetical clocks in both frames. If I could do this, I was thinking I'd be able to solve the apparent clock paradox. But I now realize that even if I could do this, the clocks will NOT remain synchronized because the LT won't allow it, and we must use the LT since it's presumably the only frame transformation that constrains the SoL to be frame independent. Therefore, I've come to the conclusion that the problem I'm trying to understand, can only be solved via the breakdown of simultaneity after more clearly defining the problem. AG

Finally, I note that Brent was correct in his initial response, that I failed to define the problem clearly. This might account for the fact that IMO his plot fails to offer a solution to my initially ill-posed question. AFAICT, his plot just shows that the LT implies a clock in a spatially moving frame, his red clock, tics at a slower rate than the clock in spatially fixed frame, his blue clock. This we already knew, and the problem I seek to understand, is what happens when the red and blue clocks become juxtaposed, but there's insufficient resolution in his plot, at this point of interest, to shed any light on my problem. I now intend to go back one of your previous posts where you allegedly showed that the LT is, indeed, a transformation, in the sense that it tells an observer in the spatially fixed frame, the rest frame, what the observer in the moving frame will actually measure. This issue puzzles me because of the contraction of a moving rod. In which frame does the shrinking rod reside? TY, AG 

[smitra]

On 27-10-2024 07:56, Alan Grayson wrote:
> On Saturday, October 26, 2024 at 8:44:51 PM UTC-6 Jesse Mazer wrote:
>
>> On Sat, Oct 26, 2024 at 3:06 AM Alan Grayson <agrays...@gmail.com>
>> wrote:
>>
>> So that one typo, which was correct elsewhere made it muddled for
>> you?
>

> IN PART YES. WHEN I THINK AN AUTHOR DOESN'T KNOW WHAT HE'S EXPOUNDING
> ABOUT, I LOSE INTEREST. ALSO, ALTHOUGH I WAS A SOFTWARE ENGINEER AT
> JPL, I DON'T KNOW LISP, SO IT WOULD BE HARD TO SEE WHAT ASSUMPTIONS
> YOU MADE IN GENERATING THE PLOT. AND THE PLOT IS CLAIMED TO ESTABLISH
> TIME DILATION, AND I'M NOT SURE HOW YOU DEVELOPED THE WIDTH OF THE
> BLUE PATH SAY, TO SHOW TIME PASSES MORE RAPIDLY COMPARED TO THE OTHER
> PLOTS. AG

> I just assumed a width for the blue path. All that determines is how
> fast the light clock ticks. Then the other two light clock world
> lines were generated by point-by-point application of the given
> Lorentz transform. So I showed the two clocks moving relative to blue
> ticked more slowly, not the other way around. Do you not see that the
> bouncing photon hits the mirror less often in red's clock as measured
> in blue's frame.
>
> Yes, so that implies tics are less frequent in red's clock, compared
> to blue's clock, so the time rate for red is less than blue, which is
> what I in effect posted -- that blue clock tics more rapidly than red
> clock. Why do you fail to understand what I wrote? AG
>
> I understood it, but it read as if you didn't realize red was just the
> transform of blue and it is in the clock's own frame it runs fastest.
> You wrote as though I "developed the width of the blue path say, to
> show time passes more rapidly" whereas I chose it arbitrarily and
> derived the other two.
>
> Brent
>

> ARE YOU SAYING THE RED CLOCK IS IN THE SAME FRAME AS THE BLUE CLOCK? I
> MISSED THAT POINT. WHY DID YOU MODEL IT THIS WAY, INSTEAD OF JUST
> USING TWO FRAMES, ONE AT REST, THE OTHER MOVING? WHY DOES THE RED
> CLOCK'S PHOTON CROSS AT RIGHT ANGLES, BUT THIS ISN'T SO FOR THE BLUE
> CLOCK? WERE THEY ARBITRARY CHOICES? AG
>
> THIS DISCUSSION BEGAN WITH MY CLAIM THAT THERE COULD BE A CLOCK
> PARADOX, DEFINED BY TWO CLOCKS, EACH RUNNING SLOWER THAN THE OTHER. IF
> SUCH A PARADOX EXISTED, IT WOULD BE IMPOSSIBLE TO PRODUCE A PLOT WHICH
> WOULD SHOW IT. SO, WHAT EXACTLY DOES YOUR PLOT SHOW; THAT THE LT
> ESTABLISHES THAT A MOVING CLOCK RUNS SLOWER THAN A STATIONARY CLOCK?
> THIS IS NOT SOMETHING I DISPUTED. I DON'T SEE HOW YOUR PLOT RESOLVES A
> POSSIBLE PARADOX. AG
>
> I THOUGHT THAT IF I COULD SYNCHRONIZE CLOCKS IN TWO INERTIAL FRAMES
> WITHOUT THE LT, I COULD ESTABLISH THE PARADOX. BUT NOW I DON'T THINK
> THIS IS TRUE. WHAT IS TRUE, IS THAT THE LT CAUSES TIME DILATION, AND
> IS, SO TO SPEAK, THE PRICE WE PAY TO GUARANTEE FRAME INVARIANCE OF THE
> SOL. AG
>
> FOR JESSE; I LOOKED UP EINSTEIN'S METHOD FOR DETERMINING SIMULTANEOUS
> EVENTS. IIUC, IT INVOLVES TWO CLOCKS AND A LIGHT SOURCE MIDWAY BETWEEN
> THEM TO PRODUCE SIMULTANEOUS EVENTS, WITH THE CONCLUSION THAT
> SIMULTANEITY EXISTS IN THE REST FRAME OF THE CLOCKS, BUT NOT IN A
> MOVING FRAME. I DIDN'T USE IT TO ESTABLISH THAT CLOCKS IN TWO INERTIAL
> FRAMES CAN BE SYNCHRONIZED. NEITHER DID I DENY IT. I DON'T SEE WHY YOU
> THINK THERE'S SOMETHING AWRY THAT I DIDN'T USE IT. AG

>
> Again, the problem is that you simply haven't clearly laid out what
> your procedure is for synchronizing different clocks at rest in the
> *same* frame, so your summary of the experiment you want to set up is
> too vague without that information. Are all the A clocks synchronized
> with one another using the Einstein synchronization procedure in the A
> frame, and then the B clocks set with reference to whichever A clock
> they are next to at some moment? Or is just one B clock set by
> reference to the A clock it's next to, and the other B clocks
> synchronized with that first B clock using the Einstein
> synchronization procedure in the B frame? Or some other option?
>
> Jesse
>

> I ASSUMED THAT IF THE TWO JUXTAPOSED CLOCKS WERE SET TO T=0, AND I
> SPECIFIED HOW ANY OTHER CLOCK IN EITHER FRAME COULD BE SYNCHRONIZED TO
> THOSE TWO CLOCKS, ONE CAN INFER HOW TO SYNCHRONIZE ANY OTHER CLOCK, IN
> ANY KNOWN DISTANCE TO ANY PREVIOUSLY SYNCHRONIZED CLOCK, AND
> SYNCHRONIZE THAT CLOCK. I WAS TRYING TO IMAGINE A SCHEME FOR
> SYNCHRONIZING ALL HYPOTHETICAL CLOCKS IN BOTH FRAMES. IF I COULD DO
> THIS, I WAS THINKING I'D BE ABLE TO SOLVE THE APPARENT CLOCK PARADOX.
> BUT I NOW REALIZE THAT EVEN IF I COULD DO THIS, THE CLOCKS WILL NOT
> REMAIN SYNCHRONIZED BECAUSE THE LT WON'T ALLOW IT, AND WE MUST USE THE
> LT SINCE IT'S PRESUMABLY THE ONLY FRAME TRANSFORMATION THAT CONSTRAINS
> THE SOL TO BE FRAME INDEPENDENT. THEREFORE, I'VE COME TO THE
> CONCLUSION THAT THE PROBLEM I'M TRYING TO UNDERSTAND, CAN ONLY BE
> SOLVED VIA THE BREAKDOWN OF SIMULTANEITY AFTER MORE CLEARLY DEFINING
> THE PROBLEM. AG
>
> FINALLY, I NOTE THAT BRENT WAS CORRECT IN HIS INITIAL RESPONSE, THAT I
> FAILED TO DEFINE THE PROBLEM CLEARLY. THIS MIGHT ACCOUNT FOR THE FACT
> THAT IMO HIS PLOT FAILS TO OFFER A SOLUTION TO MY INITIALLY ILL-POSED
> QUESTION. AFAICT, HIS PLOT JUST SHOWS THAT THE LT IMPLIES A CLOCK IN A
> SPATIALLY MOVING FRAME, HIS RED CLOCK, TICS AT A SLOWER RATE THAN THE
> CLOCK IN SPATIALLY FIXED FRAME, HIS BLUE CLOCK. THIS WE ALREADY KNEW,
> AND THE PROBLEM I SEEK TO UNDERSTAND, IS WHAT HAPPENS WHEN THE RED AND
> BLUE CLOCKS BECOME JUXTAPOSED, BUT THERE'S INSUFFICIENT RESOLUTION IN
> HIS PLOT, AT THIS POINT OF INTEREST, TO SHED ANY LIGHT ON MY PROBLEM.
> I NOW INTEND TO GO BACK ONE OF YOUR PREVIOUS POSTS WHERE YOU ALLEGEDLY
> SHOWED THAT THE LT IS, INDEED, A TRANSFORMATION, IN THE SENSE THAT IT
> TELLS AN OBSERVER IN THE SPATIALLY FIXED FRAME, THE REST FRAME, WHAT
> THE OBSERVER IN THE MOVING FRAME WILL ACTUALLY MEASURE. THIS ISSUE
> PUZZLES ME BECAUSE OF THE CONTRACTION OF A MOVING ROD. IN WHICH FRAME
> DOES THE SHRINKING ROD RESIDE? TY, AG
>
http://insti.physics.sunysb.edu/~siegel/sr.html

Saibal

[Alan Grayson]

From link: 

 A famous "paradox" is trying to park a relativistic car in a garage: From the point of view of the car, the garage has "Lorentz contracted", and the car will no longer fit. But from the point of view of the garage, the car is now shorter, and so will fit even better. The resolution of the paradox is that if the front end of the car stops simultaneously to the back end from one "reference frame", that will not be true in the other. If both ends do not stop at the same time, the car changes length. (This has often been observed nonrelativistically, for cars stopped by trees or other cars.)

So how exactly is the paradox resolved? Not sure I get it. BTW, I don't post using all CAPS. AG

[Alan Grayson]

Maybe the shrinking rod is not in either frame, but viewed from the rest frame and the moving frame. Is it measured as shrinking from both frames, say on a trip to Andromeda? AG 

[Alan Grayson]

Is the paradox resolved by assuming that if simultaneity is broken from the pov of the garage frame, the car length is undefined?  AG 

[Jesse Mazer]

By "simultaneity is broken" do you just mean the idea from the link that if both ends of the car stop simultaneously in the car's frame, these events are non-simultaneous in the garage frame? In that scenario, the car always has a well-defined length at a given moment in the garage frame, but the back end will stop first while the front end continues to move forward for a while before it also stops (since the events of each end stopping are non-simultaneous in this frame), so the car will stretch out during the period between each end stopping. Note though that this is just one possibility, you could also have a scenario where both ends stop simultaneously in the garage frame, or where both frames say one end stopped before the other (for example, if the front end stops because it crashes into the back wall of the garage, the back end will not begin to react to the crash until a sound wave has traveled from front to back, which takes a finite time in both frames).

Jesse

 

[Alan Grayson]

The link just says that the apparent paradox is resolved by a breakdown in simultaneity, but doesn't specify exactly what that means. I notice that an apparent paradox can be defined for length contraction, whereas I was trying to resolve it for time dilation, but so far I cannot define the problem with clarity. Do you have any suggestion in this regard? AG

[John Clark]

On Mon, Oct 28, 2024 at 9:19 AM Alan Grayson <agrays...@gmail.com> wrote:

> The link just says that the apparent paradox is resolved by a breakdown in simultaneity, but doesn't specify exactly what that means. I notice that an apparent paradox can be defined for length contraction, whereas I was trying to resolve it for time dilation, but so far I cannot define the problem with clarity. Do you have any suggestion in this regard? AG

The garage is 9 feet deep and has doors on the front and back that can be closed and locked, but the car is 10 feet long so apparently it can never fit in the garage. However from the point of view of somebody standing still next to the garage the car is moving so fast that, due to Lorentz contraction, the car is now only 8 feet long. And to prove that the contraction is real and not just an optical illusion, as soon as the back of the car has fully entered the garage the man quickly closes and locks the front of the garage, at that exact instant from the garage man's point of view, the car is in the garage AND simultaneously it is between BOTH of two closed and locked doors. The man then quickly runs to the back of the garage and unlocks and opens the back door which allows the card to continue on at nearly the speed of light. So there is no paradox.

But how would this look from the driver of the car's point of view? He would see the car as being stationary and therefore 10 feet long, but the garage is moving so fast due to Lorentz contraction the garage is now only 8 feet deep not 9, and apparently making things even worse. However, what the car driver sees is that as soon as the front of the car enters the garage the garage man runs around to the back and opens the back door of the garage. From the car driver's point of view at NO time is the car simultaneously between  BOTH of two closed and locked doors. So there is no paradox, although the car driver and the garage man do not agree what is "simultaneous" and what is not.

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

cal

    

 

[Jesse Mazer]

The link doesn't use the word "breakdown", do you use that term just to mean different frames have different definitions of simultaneity, even if they both can make definite statements about which pairs of events on the worldlines of the front and back of the car are simultaneous (and therefore the distance in their frame between any such pair of simultaneous events)? Your comment about the length being "undefined" seemed to suggest you had a different idea in mind.

Time dilation and length contraction are not really symmetrical quantities, length contraction depends on having *two* worldlines with identical velocities (the front and back of the object) and looking at the distance between simultaneous points on them, time dilation is based on comparing proper time between two points on a *single* worldline with the coordinate time between those points in some inertial frame. So I don't think you could come up with anything exactly analogous to the garage paradox for time dilation.

Jesse

[Alan Grayson]

Two doors. Doors locked and then unlocked. Or whatever. You seem to have an inclination for overly complicated analyses. Why not just say the car driver knows the length of his car because he can simultaneously measure its endpoints, and due to contraction of the garage's length, he knows his car won't fit inside. In the garage frame, the car's length cannot be measured due to a breakdown in simultaneity. So this observer hasn't a valid opinion whether or not the car will fit inside. So, in this analysis the paradox is solved, and the car won't fit inside the garage. What do you find insufficient about this analysis? AGca

[Jesse Mazer]

It's simply not true that there is a "breakdown in simultaneity" leading the car's length to be unmeasurable in the garage frame, the garage frame just has a *different* view of simultaneity than the car frame but they are both perfectly well-defined, and in relativity you can't say one is "true" and the other is wrong. 

John Clark's version makes things simpler by avoiding the need for the car to move non-inertially (decelerate), if both front and back doors of the garage are open at the moment the car passes through them, then the car can just sail right in one door and out the other. Then the two frames disagree about whether the car ever "fit in the garage" because they disagree about whether the event "front of car exits the open back door of the garage" happened before or after the event "back of car passes through the open front door of the garage". If the front of the car passing through the back door happened *before* the back of the car passing the front door, then the car was never fully inside the garage because the front end was starting to poke out before the back end was fully inside, whereas if the former event happened *after* the latter event, then the car was fully inside the garage for some time. So, disagreement over simultaneity is equivalent to disagreement over the answer to the question "was the car ever fully inside the garage at any moment?"

Jesse

[Alan Grayson]

I'm not looking for something exactly analogous to the garage paradox for time dilation; just any model the captures what the problem is. I don't see that it can involve two points on a single world line. Rather, it's comparing clocks in different frames, on different world lines, and explaining how each can apparently run slower than the other. AG

[Alan Grayson]

Initially you claim there is no breakdown in simultaneity, but you conclude by claiming there is simultaneity and what it implies. Doesn't the article from the posted link claim that the breakdown is what apparently causes the paradox? AG 

[Jesse Mazer]

I don't understand your language, what does "breakdown in simultaneity" mean and what does "there is simultaneity" mean? Neither phrase is used in the link or in any text on relativity I've ever seen, and the meaning isn't self-evident at all, I asked you to explain "breakdown" earlier but you didn't respond. All I'm saying is that each frame has their own well-defined definition of simultaneity, the two frames' definitions disagree, and neither is objectively more correct than the other (analogous to how different inertial frames have their own definitions of what the 'velocity' of different objects is and none is preferred, velocity is an inherently coordinate-dependent quantity). And I'm also saying that if you are using the phrase "breakdown in simultaneity" in a way that has something to do with your claim that length is undefined in some frame, that's not how things work in relativity, each frame can assign the front and back of the object well-defined position coordinates at each moment in coordinate time, and can use that to define the object's "length" at each moment.

Jesse

[Alan Grayson]

On Monday, October 28, 2024 at 11:13:18 PM UTC-6 Jesse Mazer wrote:

On Mon, Oct 28, 2024 at 9:47 PM Alan Grayson <agrays...@gmail.com> wrote:

On Monday, October 28, 2024 at 6:44:18 PM UTC-6 Jesse Mazer wrote:

On Mon, Oct 28, 2024 at 7:26 PM Alan Grayson <agrays...@gmail.com> wrote:

On Monday, October 28, 2024 at 12:01:33 PM UTC-6 John Clark wrote:

On Mon, Oct 28, 2024 at 9:19 AM Alan Grayson <agrays...@gmail.com> wrote:

> The link just says that the apparent paradox is resolved by a breakdown in simultaneity, but doesn't specify exactly what that means. I notice that an apparent paradox can be defined for length contraction, whereas I was trying to resolve it for time dilation, but so far I cannot define the problem with clarity. Do you have any suggestion in this regard? AG

The garage is 9 feet deep and has doors on the front and back that can be closed and locked, but the car is 10 feet long so apparently it can never fit in the garage. However from the point of view of somebody standing still next to the garage the car is moving so fast that, due to Lorentz contraction, the car is now only 8 feet long. And to prove that the contraction is real and not just an optical illusion, as soon as the back of the car has fully entered the garage the man quickly closes and locks the front of the garage, at that exact instant from the garage man's point of view, the car is in the garage AND simultaneously it is between BOTH of two closed and locked doors. The man then quickly runs to the back of the garage and unlocks and opens the back door which allows the card to continue on at nearly the speed of light. So there is no paradox.

But how would this look from the driver of the car's point of view? He would see the car as being stationary and therefore 10 feet long, but the garage is moving so fast due to Lorentz contraction the garage is now only 8 feet deep not 9, and apparently making things even worse. However, what the car driver sees is that as soon as the front of the car enters the garage the garage man runs around to the back and opens the back door of the garage. From the car driver's point of view at NO time is the car simultaneously between  BOTH of two closed and locked doors. So there is no paradox, although the car driver and the garage man do not agree what is "simultaneous" and what is not.

Two doors. Doors locked and then unlocked. Or whatever. You seem to have an inclination for overly complicated analyses. Why not just say the car driver knows the length of his car because he can simultaneously measure its endpoints, and due to contraction of the garage's length, he knows his car won't fit inside. In the garage frame, the car's length cannot be measured due to a breakdown in simultaneity. So this observer hasn't a valid opinion whether or not the car will fit inside. So, in this analysis the paradox is solved, and the car won't fit inside the garage. What do you find insufficient about this analysis? AGca

It's simply not true that there is a "breakdown in simultaneity" leading the car's length to be unmeasurable in the garage frame, the garage frame just has a *different* view of simultaneity than the car frame but they are both perfectly well-defined, and in relativity you can't say one is "true" and the other is wrong. 

John Clark's version makes things simpler by avoiding the need for the car to move non-inertially (decelerate), if both front and back doors of the garage are open at the moment the car passes through them, then the car can just sail right in one door and out the other. Then the two frames disagree about whether the car ever "fit in the garage" because they disagree about whether the event "front of car exits the open back door of the garage" happened before or after the event "back of car passes through the open front door of the garage". If the front of the car passing through the back door happened *before* the back of the car passing the front door, then the car was never fully inside the garage because the front end was starting to poke out before the back end was fully inside, whereas if the former event happened *after* the latter event, then the car was fully inside the garage for some time. So, disagreement over simultaneity is equivalent to disagreement over the answer to the question "was the car ever fully inside the garage at any moment?"

Jesse

Initially you claim there is no breakdown in simultaneity, but you conclude by claiming there is simultaneity and what it implies.

I don't understand your language, what does "breakdown in simultaneity" mean and what does "there is simultaneity" mean?

The meaning of simultaneity is fairly straight-forward. If, for example, an observer who is situated between two mirrors and sends a beam of light toward both, and receives their reflections at the same instant, knows their locations by making a simple calculation and correction for the time required for the round trip, and knows they are equidistant. Or, if the beams don't return at the same time, the observer knows they are not equidistant, but he can calculate how far away each one is. That occurs in some rest frame. But an observer in a relatively moving frame, will not see both reflections at the same time, even if they are equidistant in the rest frame, which is the definition of breakdown of simultaneity. IOW, events which are simultaneous in one frame, the rest frame, will not be simultaneous in the moving frame. AG

 

Neither phrase is used in the link or in any text on relativity I've ever seen, and the meaning isn't self-evident at all, I asked you to explain "breakdown" earlier but you didn't respond. All I'm saying is that each frame has their own well-defined definition of simultaneity, the two frames' definitions disagree,

I doubt the definitions disagree. AG

 

and neither is objectively more correct than the other (analogous to how different inertial frames have their own definitions of what the 'velocity' of different objects is and none is preferred, velocity is an inherently coordinate-dependent quantity).

None is preferred, but both frames use the same coordinate system. AG

 

And I'm also saying that if you are using the phrase "breakdown in simultaneity" in a way that has something to do with your claim that length is undefined in some frame,

IMO,  length defined in a rest frame, and depends on simultaneity?  AG

[Jesse Mazer]

On Tue, Oct 29, 2024 at 2:26 AM Alan Grayson <agrays...@gmail.com> wrote:

On Monday, October 28, 2024 at 11:13:18 PM UTC-6 Jesse Mazer wrote:

On Mon, Oct 28, 2024 at 9:47 PM Alan Grayson <agrays...@gmail.com> wrote:

On Monday, October 28, 2024 at 6:44:18 PM UTC-6 Jesse Mazer wrote:

On Mon, Oct 28, 2024 at 7:26 PM Alan Grayson <agrays...@gmail.com> wrote:

On Monday, October 28, 2024 at 12:01:33 PM UTC-6 John Clark wrote:

On Mon, Oct 28, 2024 at 9:19 AM Alan Grayson <agrays...@gmail.com> wrote:

> The link just says that the apparent paradox is resolved by a breakdown in simultaneity, but doesn't specify exactly what that means. I notice that an apparent paradox can be defined for length contraction, whereas I was trying to resolve it for time dilation, but so far I cannot define the problem with clarity. Do you have any suggestion in this regard? AG

The garage is 9 feet deep and has doors on the front and back that can be closed and locked, but the car is 10 feet long so apparently it can never fit in the garage. However from the point of view of somebody standing still next to the garage the car is moving so fast that, due to Lorentz contraction, the car is now only 8 feet long. And to prove that the contraction is real and not just an optical illusion, as soon as the back of the car has fully entered the garage the man quickly closes and locks the front of the garage, at that exact instant from the garage man's point of view, the car is in the garage AND simultaneously it is between BOTH of two closed and locked doors. The man then quickly runs to the back of the garage and unlocks and opens the back door which allows the card to continue on at nearly the speed of light. So there is no paradox.

But how would this look from the driver of the car's point of view? He would see the car as being stationary and therefore 10 feet long, but the garage is moving so fast due to Lorentz contraction the garage is now only 8 feet deep not 9, and apparently making things even worse. However, what the car driver sees is that as soon as the front of the car enters the garage the garage man runs around to the back and opens the back door of the garage. From the car driver's point of view at NO time is the car simultaneously between  BOTH of two closed and locked doors. So there is no paradox, although the car driver and the garage man do not agree what is "simultaneous" and what is not.

Two doors. Doors locked and then unlocked. Or whatever. You seem to have an inclination for overly complicated analyses. Why not just say the car driver knows the length of his car because he can simultaneously measure its endpoints, and due to contraction of the garage's length, he knows his car won't fit inside. In the garage frame, the car's length cannot be measured due to a breakdown in simultaneity. So this observer hasn't a valid opinion whether or not the car will fit inside. So, in this analysis the paradox is solved, and the car won't fit inside the garage. What do you find insufficient about this analysis? AGca

It's simply not true that there is a "breakdown in simultaneity" leading the car's length to be unmeasurable in the garage frame, the garage frame just has a *different* view of simultaneity than the car frame but they are both perfectly well-defined, and in relativity you can't say one is "true" and the other is wrong. 

John Clark's version makes things simpler by avoiding the need for the car to move non-inertially (decelerate), if both front and back doors of the garage are open at the moment the car passes through them, then the car can just sail right in one door and out the other. Then the two frames disagree about whether the car ever "fit in the garage" because they disagree about whether the event "front of car exits the open back door of the garage" happened before or after the event "back of car passes through the open front door of the garage". If the front of the car passing through the back door happened *before* the back of the car passing the front door, then the car was never fully inside the garage because the front end was starting to poke out before the back end was fully inside, whereas if the former event happened *after* the latter event, then the car was fully inside the garage for some time. So, disagreement over simultaneity is equivalent to disagreement over the answer to the question "was the car ever fully inside the garage at any moment?"

Jesse

Initially you claim there is no breakdown in simultaneity, but you conclude by claiming there is simultaneity and what it implies.

I don't understand your language, what does "breakdown in simultaneity" mean and what does "there is simultaneity" mean?

The meaning of simultaneity is fairly straight-forward. If, for example, an observer who is situated between two mirrors and sends a beam of light toward both, and receives their reflections at the same instant, knows their locations by making a simple calculation and correction for the time required for the round trip, and knows they are equidistant. Or, if the beams don't return at the same time, the observer knows they are not equidistant, but he can calculate how far away each one is. That occurs in some rest frame. But an observer in a relatively moving frame, will not see both reflections at the same time, even if they are equidistant in the rest frame, which is the definition of breakdown of simultaneity. IOW, events which are simultaneous in one frame, the rest frame, will not be simultaneous in the moving frame. AG

I agree with all this, but what was confusing me was your connecting it to the notion of length being undefined in some frame--see below

 

Neither phrase is used in the link or in any text on relativity I've ever seen, and the meaning isn't self-evident at all, I asked you to explain "breakdown" earlier but you didn't respond. All I'm saying is that each frame has their own well-defined definition of simultaneity, the two frames' definitions disagree,

I doubt the definitions disagree. AG

Maybe you mean something more abstract by a "definition" of simultaneity (for example, the general notion that in every frame it's defined by local readings on clocks that are at rest in the frame and synchronized by the Einstein convention), but all I mean is that for a given pair of events, if one frame defines them as simultaneous, a different frame defines them as non-simultaneous.

 

and neither is objectively more correct than the other (analogous to how different inertial frames have their own definitions of what the 'velocity' of different objects is and none is preferred, velocity is an inherently coordinate-dependent quantity).

None is preferred, but both frames use the same coordinate system. AG

They both use the same *type* of coordinate system, but if they don't assign the same position and time coordinates to each event, that means they use "different coordinate systems" in the way physicists talk.

 

 

And I'm also saying that if you are using the phrase "breakdown in simultaneity" in a way that has something to do with your claim that length is undefined in some frame,

IMO,  length defined in a rest frame, and depends on simultaneity?  AG

Length is not specifically defined in the object's rest frame, no (though the 'rest length' is). Length in relativity simply refers to the distance between the front and back end of an object at a single moment in time, so it can be defined for moving objects--for example if the back end's position as a function of time in your frame is x_b = 0.8c*t and the position of the front end is x_f = 0.8c*t + 2 light years, then for any given value of t, say t=0, the distance between the front and back end is always 2 light years (for example at x=0 we have x_b = 0 and x_f = 2 light years), so the length of the object is 2 light years in this frame.

Jesse

[John Clark]

On Mon, Oct 28, 2024 at 7:26 PM Alan Grayson <agrays...@gmail.com> wrote:

>> ME: The garage is 9 feet deep and has doors on the front and back that can be closed and locked, but the car is 10 feet long so apparently it can never fit in the garage. However from the point of view of somebody standing still next to the garage the car is moving so fast that, due to Lorentz contraction, the car is now only 8 feet long. And to prove that the contraction is real and not just an optical illusion, as soon as the back of the car has fully entered the garage the man quickly closes and locks the front of the garage, at that exact instant from the garage man's point of view, the car is in the garage AND simultaneously it is between BOTH of two closed and locked doors. The man then quickly runs to the back of the garage and unlocks and opens the back door which allows the card to continue on at nearly the speed of light. So there is no paradox.  But how would this look from the driver of the car's point of view? He would see the car as being stationary and therefore 10 feet long, but the garage is moving so fast due to Lorentz contraction the garage is now only 8 feet deep not 9, and apparently making things even worse. However, what the car driver sees is that as soon as the front of the car enters the garage the garage man runs around to the back and opens the back door of the garage. From the car driver's point of view at NO time is the car simultaneously between  BOTH of two closed and locked doors. So there is no paradox, although the car driver and the garage man do not agree what is "simultaneous" and what is not.

> Two doors. Doors locked and then unlocked. Or whatever. You seem to have an inclination for overly complicated analyses. Why not just say the car driver knows the length of his car because he can simultaneously measure its endpoints,

The car driver knows the length of his car because he knows what the speed of light is and he sent 2 intense but very short bursts of light to mirrors placed at the front and back of his car; the driver used the time he observed on his wristwatch, which in physics is called "Proper Time",  to determine how long it took him to see the beams reflected in the mirrors. "Proper Time" may be poorly named, it's not really "proper", it's not the one true time, it's just the time measured by a clock that is stationary relative to an observer.

The car driver used a similar method to determine how deep the garage was. He also used his wristwatch to determine if things happen at the same time or not. The garage man used the exact same method to determine the length of the car and the depth of the garage EXCEPT that he used his "proper time", the time the garage man  saw on his wristwatch, which was different from the car driver's "proper time".  

 

 > In the garage frame, the car's length cannot be measured due to a breakdown in simultaneity. So this observer hasn't a valid opinion whether or not the car will fit inside.

If the garage man knew Special Relativity then he WOULD have a valid opinion, the car WILL fit inside the garage. And if he didn't know Special Relativity but actually performed the experiment then he wouldn't need Einstein, he would have enough information to derive Special Relativity for himself, in fact he'd have to if he didn't want to live with a logical paradox.

> So, in this analysis the paradox is solved, and the car won't fit inside the garage. What do you find insufficient about this analysis? AGc

 

The thing that makes your analysis unsatisfactory is that it contradicts experimental results, if the garage man followed your line of thought he would conclude that the car could never fit inside his garage, but when he actually performed the experiment he found that he was wrong, the car COULD fit inside his garage with both doors closed. 

 

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

dbc

[Alan Grayson]

On Tuesday, October 29, 2024 at 1:47:36 AM UTC-6 Jesse Mazer wrote:

On Tue, Oct 29, 2024 at 2:26 AM Alan Grayson <agrays...@gmail.com> wrote:

On Monday, October 28, 2024 at 11:13:18 PM UTC-6 Jesse Mazer wrote:

On Mon, Oct 28, 2024 at 9:47 PM Alan Grayson <agrays...@gmail.com> wrote:

On Monday, October 28, 2024 at 6:44:18 PM UTC-6 Jesse Mazer wrote:

On Mon, Oct 28, 2024 at 7:26 PM Alan Grayson <agrays...@gmail.com> wrote:

On Monday, October 28, 2024 at 12:01:33 PM UTC-6 John Clark wrote:

On Mon, Oct 28, 2024 at 9:19 AM Alan Grayson <agrays...@gmail.com> wrote:

> The link just says that the apparent paradox is resolved by a breakdown in simultaneity, but doesn't specify exactly what that means. I notice that an apparent paradox can be defined for length contraction, whereas I was trying to resolve it for time dilation, but so far I cannot define the problem with clarity. Do you have any suggestion in this regard? AG

The garage is 9 feet deep and has doors on the front and back that can be closed and locked, but the car is 10 feet long so apparently it can never fit in the garage. However from the point of view of somebody standing still next to the garage the car is moving so fast that, due to Lorentz contraction, the car is now only 8 feet long. And to prove that the contraction is real and not just an optical illusion, as soon as the back of the car has fully entered the garage the man quickly closes and locks the front of the garage, at that exact instant from the garage man's point of view, the car is in the garage AND simultaneously it is between BOTH of two closed and locked doors. The man then quickly runs to the back of the garage and unlocks and opens the back door which allows the card to continue on at nearly the speed of light. So there is no paradox.

But how would this look from the driver of the car's point of view? He would see the car as being stationary and therefore 10 feet long, but the garage is moving so fast due to Lorentz contraction the garage is now only 8 feet deep not 9, and apparently making things even worse. However, what the car driver sees is that as soon as the front of the car enters the garage the garage man runs around to the back and opens the back door of the garage. From the car driver's point of view at NO time is the car simultaneously between  BOTH of two closed and locked doors. So there is no paradox, although the car driver and the garage man do not agree what is "simultaneous" and what is not.

Two doors. Doors locked and then unlocked. Or whatever. You seem to have an inclination for overly complicated analyses. Why not just say the car driver knows the length of his car because he can simultaneously measure its endpoints, and due to contraction of the garage's length, he knows his car won't fit inside. In the garage frame, the car's length cannot be measured due to a breakdown in simultaneity. So this observer hasn't a valid opinion whether or not the car will fit inside. So, in this analysis the paradox is solved, and the car won't fit inside the garage. What do you find insufficient about this analysis? AGca

It's simply not true that there is a "breakdown in simultaneity" leading the car's length to be unmeasurable in the garage frame, the garage frame just has a *different* view of simultaneity than the car frame but they are both perfectly well-defined, and in relativity you can't say one is "true" and the other is wrong. 

John Clark's version makes things simpler by avoiding the need for the car to move non-inertially (decelerate), if both front and back doors of the garage are open at the moment the car passes through them, then the car can just sail right in one door and out the other. Then the two frames disagree about whether the car ever "fit in the garage" because they disagree about whether the event "front of car exits the open back door of the garage" happened before or after the event "back of car passes through the open front door of the garage". If the front of the car passing through the back door happened *before* the back of the car passing the front door, then the car was never fully inside the garage because the front end was starting to poke out before the back end was fully inside, whereas if the former event happened *after* the latter event, then the car was fully inside the garage for some time. So, disagreement over simultaneity is equivalent to disagreement over the answer to the question "was the car ever fully inside the garage at any moment?"

Jesse

Initially you claim there is no breakdown in simultaneity, but you conclude by claiming there is simultaneity and what it implies.

I don't understand your language, what does "breakdown in simultaneity" mean and what does "there is simultaneity" mean?

The meaning of simultaneity is fairly straight-forward. If, for example, an observer who is situated between two mirrors and sends a beam of light toward both, and receives their reflections at the same instant, knows their locations by making a simple calculation and correction for the time required for the round trip, and knows they are equidistant. Or, if the beams don't return at the same time, the observer knows they are not equidistant, but he can calculate how far away each one is. That occurs in some rest frame. But an observer in a relatively moving frame, will not see both reflections at the same time, even if they are equidistant in the rest frame, which is the definition of breakdown of simultaneity. IOW, events which are simultaneous in one frame, the rest frame, will not be simultaneous in the moving frame. AG

I agree with all this, but what was confusing me was your connecting it to the notion of length being undefined in some frame--see below

 

Neither phrase is used in the link or in any text on relativity I've ever seen, and the meaning isn't self-evident at all, I asked you to explain "breakdown" earlier but you didn't respond. All I'm saying is that each frame has their own well-defined definition of simultaneity, the two frames' definitions disagree,

I doubt the definitions disagree. AG

Maybe you mean something more abstract by a "definition" of simultaneity (for example, the general notion that in every frame it's defined by local readings on clocks that are at rest in the frame and synchronized by the Einstein convention), but all I mean is that for a given pair of events, if one frame defines them as simultaneous, a different frame defines them as non-simultaneous.

OK. AG

and neither is objectively more correct than the other (analogous to how different inertial frames have their own definitions of what the 'velocity' of different objects is and none is preferred, velocity is an inherently coordinate-dependent quantity).

None is preferred, but both frames use the same coordinate system. AG

They both use the same *type* of coordinate system, but if they don't assign the same position and time coordinates to each event, that means they use "different coordinate systems" in the way physicists talk. 

 

And I'm also saying that if you are using the phrase "breakdown in simultaneity" in a way that has something to do with your claim that length is undefined in some frame,

IMO,  length defined in a rest frame, and depends on simultaneity?  AG

Length is not specifically defined in the object's rest frame, no (though the 'rest length' is). Length in relativity simply refers to the distance between the front and back end of an object at a single moment in time, so it can be defined for moving objects--for example if the back end's position as a function of time in your frame is x_b = 0.8c*t and the position of the front end is x_f = 0.8c*t + 2 light years, then for any given value of t, say t=0, the distance between the front and back end is always 2 light years (for example at x=0 we have x_b = 0 and x_f = 2 light years), so the length of the object is 2 light years in this frame.

Jesse

I see I have been making a mistake on some of these issues. For example, many of the peculiar results of SR are due to the fact that we must use the LT in order for the SoL to be frame independent. So, the length of a rod can be measured differently in different frames, and the breakdown in simultaneity just means that events which are simultaneous in one frame, will not be simultaneous in another frame. IOW, just like the E field in the S frame will have different values in the S' frame, E', there's no inherent contradiction with different values of length say, in different frames. In the car garage problem, the car length and other variables could be frame dependent, but the only contradiction would be if the frames disagreed on whether the car could fit in the garage. BTW, the link which started this discussion does refer to simultaneous measurements and infers, but doesn't explicitly state that the resolution of the apparent paradox depends on the differing results of simultaneity between the frames, which I referred to as a breakdown in simultaneity. AG

[Alan Grayson]

My analysis above is incorrect. Without analyzing simultaneity, the frames disagree on whether the car will fit in the garage. From the frame of the car, its length remains what it is, but if traveling fast enough, the garage length is seen as smaller than the car, and it won't fit inside. OTOH, from the pov of the garage frame, the length of the garage is unchanged, but the car shrinks small enough if going sufficiently fast, and the car will fit inside the garage. So, I will re-read JC's analysis, and expect that when simultaneity is properly analyzed, both frames will agree on the outcome and the paradox is no more. AG 

ca

[Quentin Anciaux]

The paradox is resolved because from the pov of the cars, the doors are never simultaneously closed and from the pov of the garage they are, from the pov of the cars it will fit because it passes through the doors are not simultaneously closed, from the pov it fit in between the closed door because the car lenght is shorter and the closing are simultaneous. 

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[Brent Meeker]

Just to make this clear I prepared a couple of diagrams for my relativity lecture.  Here's the car passing thru the garage at 0.8 which is 10units long, the red vertical lines, in this the garage frame.  The long image is how the driver sees his car, note that the image extended in time as well as space.  The short image is the car with the two ends measured simultaneously, notice it's plenty short enough to fit in the 10u garage.



Now here's the same spacetime points transformed to the cars reference frame and the garage moving right to left at 0.8.  Here we see that in it's own frame the car is 12unit long, 2units more than the garage length of 10u, which the above diagram show it fits into at 0.8.




Brent

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