Twin Paradox / Clark

[Alan Grayson]

If you recall, you recently posted some links to the TP and accused me of not reading them. Well, I certainly intended to read them and I explained why. But then I got involved in other  discussions here and put that temporarily aside. But now those links are, from my pov, lost in a myriad of discussion threads. So, please be so kind as to post them again here. TY, AG

[John Clark]

On Sat, Sep 6, 2025 at 1:10 AM Alan Grayson <agrays...@gmail.com> wrote:

> If you recall, you recently posted some links to the TP and accused me of not reading them. Well, I certainly intended to read them and I explained why. But then I got involved in other  discussions here and put that temporarily aside. But now those links are, from my pov, lost in a myriad of discussion threads. So, please be so kind as to post them again here. TY, AG

Start with this video:  

I Never Understood How To Intuitively Solve The Twin's Paradox ... Until Now!

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

fgt

[Alan Grayson]

I've encountered that fellow before. He speaks too fast, way too fast. It seems to me that GR solves the problem, and without frame jumping. Just imagine several changes in velocity, each spread out, so not instantaneous. During those changes, time for the traveler slows compared to rest frame on Earth, so when he returns to meet his twin, he is younger. Time slows for the traveler due to Equivalence Principle, since gravity is locally equivalent to acceleration. Do you see anything wrong with this analysis? AG 

fgt

[John Clark]

On Sat, Sep 6, 2025 at 11:01 AM Alan Grayson <agrays...@gmail.com> wrote:

>>Start with this video:  

I Never Understood How To Intuitively Solve The Twin's Paradox ... Until Now!

> I've encountered that fellow before. He speaks too fast,

Try it again. Any YouTube video can be slowed down or sped up with no change in pitch with just a few clicks of a mouse button, and he provides the clearest explanation of how to resolve the twin paradox that I know of, so if he can't give you an intuitive understanding of it then I'm not going to be able to either.  

> It seems to me that GR solves the problem,

You need to have a good feel for Special Relativity before you jump to General Relativity because it is far more complex and even less intuitive.  

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

wl1

 

fgt

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[Alan Grayson]

On Saturday, September 6, 2025 at 12:16:21 PM UTC-6 John Clark wrote:

On Sat, Sep 6, 2025 at 11:01 AM Alan Grayson <agrays...@gmail.com> wrote:

>>Start with this video:  

I Never Understood How To Intuitively Solve The Twin's Paradox ... Until Now!

> I've encountered that fellow before. He speaks too fast,

Try it again. Any YouTube video can be slowed down or sped up with no change in pitch with just a few clicks of a mouse button, and he provides the clearest explanation of how to resolve the twin paradox that I know of, so if he can't give you an intuitive understanding of it then I'm not going to be able to either.  

> It seems to me that GR solves the problem,

You need to have a good feel for Special Relativity before you jump to General Relativity because it is far more complex and even less intuitive.  

While GR isn't the most intuitive way to resolve the TP, it just relies on the results of GR and recognition of lack of symmetry. The TP can also be resolved using SR by imaging the traveling twin moves in a circle so as to return, and imaging a polygon inscribed within that circle, and calculating the time contraction along each leg as seen from the rest frame. Then, like in calculus, partitioning the path infinitely, with the resultant path exactly circular, the reason being I don't know what happens to time when going from one leg to another, when there are only finitely many paths. (Actually, circular motion isn't necessary but can be used for convenience.) So, in summary, I see the TP solved using SR or GR. Do you see any flaws in either approach? AG 

[Brent Meeker]

No. You're over complicating the problem.  It's as simple as the fact that two different thru spacetime are different lengths.  Because the spatial coordinate distance, X, appears with a minus sign relative to the coordinate time, T, the proper time, S (which is what a clock measures).  So the rocket, which takes the longer spatial path, experiences less proper time lapse.



This result has nothing to do with accelerations experience by the traveler.  This a common specious  "explanation" trying to connect it to general relativity.  This is most easily seen by the triplet version of the paradox.  In this version the one triplet stays home, one travelers away from Earth, and one who has been far away returns to Earth.  When the outbound triplet passes the inbound triplet he hands off the time to the inbound one, so together they measure the same path as the turnaround twin.



Notice that I have also avoided any acceleration at the beginning and end, so no triplet ever accelerates.

Brent

On 9/6/2025 8:01 AM, Alan Grayson wrote:

On Saturday, September 6, 2025 at 5:39:18 AM UTC-6 John Clark wrote:

I've encountered that fellow before. He speaks too fast, way too fast. It seems to me that GR solves the problem, and without frame jumping. Just imagine several changes in velocity, each spread out, so not instantaneous. During those changes, time for the traveler slows compared to rest frame on Earth, so when he returns to meet his twin, he is younger. Do you see anything wrong with this analysis? AG 

fgt

[Alan Grayson]

On Saturday, September 6, 2025 at 5:56:36 PM UTC-6 Brent Meeker wrote:

No. You're over complicating the problem.  It's as simple as the fact that two different thru spacetime are different lengths.  Because the spatial coordinate distance, X, appears with a minus sign relative to the coordinate time, T, the proper time, S (which is what a clock measures).  So the rocket, which takes the longer spatial path, experiences less proper time lapse.

Your solution is essentially no different than the two I explained, one using GR and other SR. All proposed solutions rely on some accepted result from relativity theory. AG 

This result has nothing to do with accelerations experience by the traveler.  This a common specious  "explanation" trying to connect it to general relativity. 

Why "specious"? I used an established result in GR to show the traveling twin is younger when he returns and compares clock readings with the stationary twin. The fact that you prefer your solution to mine, doesn't mean using GR is wrong. AG

 

This is most easily seen by the triplet version of the paradox.  In this version the one triplet stays home, one travelers away from Earth, and one who has been far away returns to Earth.  When the outbound triplet passes the inbound triplet he hands off the time to the inbound one, so together they measure the same path as the turnaround twin.

Without acceleration, there can be no comparison of clock readings when the inbound twin compares his clock readings with stationary twin. And I don't like the handing off of clock readings. AG 

Notice that I have also avoided any acceleration at the beginning and end, so no triplet ever accelerates.

Brent

If there's no acceleration, then you've misstated the TP, where both twins start at rest, juxtaposed so their clocks are synched, and the traveler leaves and returns at rest to compare clocks. AG 

[Brent Meeker]

On 9/6/2025 11:20 PM, Alan Grayson wrote:

On Saturday, September 6, 2025 at 5:56:36 PM UTC-6 Brent Meeker wrote:

No. You're over complicating the problem.  It's as simple as the fact that two different thru spacetime are different lengths.  Because the spatial coordinate distance, X, appears with a minus sign relative to the coordinate time, T, the proper time, S (which is what a clock measures).  So the rocket, which takes the longer spatial path, experiences less proper time lapse.

Your solution is essentially no different than the two I explained, one using GR and other SR. All proposed solutions rely on some accepted result from relativity theory. AG 

This result has nothing to do with accelerations experience by the traveler.  This a common specious  "explanation" trying to connect it to general relativity. 

Why "specious"? I used an established result in GR to show the traveling twin is younger when he returns and compares clock readings with the stationary twin. The fact that you prefer your solution to mine, doesn't mean using GR is wrong. AG

 

This is most easily seen by the triplet version of the paradox.  In this version the one triplet stays home, one travelers away from Earth, and one who has been far away returns to Earth.  When the outbound triplet passes the inbound triplet he hands off the time to the inbound one, so together they measure the same path as the turnaround twin.

Without acceleration, there can be no comparison of clock readings when the inbound twin compares his clock readings with stationary twin. 

That's flat wrong.

And I don't like the handing off of clock readings. AG 

Notice that I have also avoided any acceleration at the beginning and end, so no triplet ever accelerates.

Brent

If there's no acceleration, then you've misstated the TP, where both twins start at rest, juxtaposed so their clocks are synched, and the traveler leaves and returns at rest to compare clocks. AG 

You may not like it, but it shows the twin paradox has nothing to do with general relativity or acceleration.  It's simply the geometric fact that some paths are shorter than others.  There is no reason to require that the clocks are set (not synched) equal at rest.  I've not "misstated" anything.   You have apparently not understood the twins paradox.

Brent

On 9/6/2025 8:01 AM, Alan Grayson wrote:

On Saturday, September 6, 2025 at 5:39:18 AM UTC-6 John Clark wrote:

On Sat, Sep 6, 2025 at 1:10 AM Alan Grayson <agrays...@gmail.com> wrote:

> If you recall, you recently posted some links to the TP and accused me of not reading them. Well, I certainly intended to read them and I explained why. But then I got involved in other  discussions here and put that temporarily aside. But now those links are, from my pov, lost in a myriad of discussion threads. So, please be so kind as to post them again here. TY, AG

Start with this video:  

I Never Understood How To Intuitively Solve The Twin's Paradox ... Until Now!

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

I've encountered that fellow before. He speaks too fast, way too fast. It seems to me that GR solves the problem, and without frame jumping. Just imagine several changes in velocity, each spread out, so not instantaneous. During those changes, time for the traveler slows compared to rest frame on Earth, so when he returns to meet his twin, he is younger. Do you see anything wrong with this analysis? AG 

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[Alan Grayson]

On Sunday, September 7, 2025 at 11:17:42 AM UTC-6 Brent Meeker wrote:

On 9/6/2025 11:20 PM, Alan Grayson wrote:

On Saturday, September 6, 2025 at 5:56:36 PM UTC-6 Brent Meeker wrote:

No. You're over complicating the problem.  It's as simple as the fact that two different thru spacetime are different lengths.  Because the spatial coordinate distance, X, appears with a minus sign relative to the coordinate time, T, the proper time, S (which is what a clock measures).  So the rocket, which takes the longer spatial path, experiences less proper time lapse.

Your solution is essentially no different than the two I explained, one using GR and other SR. All proposed solutions rely on some accepted result from relativity theory. AG 

This result has nothing to do with accelerations experience by the traveler.  This a common specious  "explanation" trying to connect it to general relativity. 

Why "specious"? I used an established result in GR to show the traveling twin is younger when he returns and compares clock readings with the stationary twin. The fact that you prefer your solution to mine, doesn't mean using GR is wrong. AG

 

This is most easily seen by the triplet version of the paradox.  In this version the one triplet stays home, one travelers away from Earth, and one who has been far away returns to Earth.  When the outbound triplet passes the inbound triplet he hands off the time to the inbound one, so together they measure the same path as the turnaround twin.

Without acceleration, there can be no comparison of clock readings when the inbound twin compares his clock readings with stationary twin. 

That's flat wrong.

And I don't like the handing off of clock readings. AG 

Notice that I have also avoided any acceleration at the beginning and end, so no triplet ever accelerates.

Brent

If there's no acceleration, then you've misstated the TP, where both twins start at rest, juxtaposed so their clocks are synched, and the traveler leaves and returns at rest to compare clocks. AG 

You may not like it, but it shows the twin paradox has nothing to do with general relativity or acceleration.  It's simply the geometric fact that some paths are shorter than others.  There is no reason to require that the clocks are set (not synched) equal at rest. 

If the clocks aren't synched, how can you know which twin is younger when they compare clock reading? That is, how can you know which path is shorter or longer? AG

 

I've not "misstated" anything.   You have apparently not understood the twins paradox.

 

Brent

I can read English. In virtually EVERY description of the TP, the twins starts from rest on Earth, the traveling twin leaves and returns, and they compare clock readings. That's how it's stated! I used GR (and SR) to show the traveling twin ages more slowly. You have a different solution, but that doesn't mean that mine is wrong. In fact, the "paradox" is the mistaken assumption that the twins are in symmetric situations. Neither of us assumes this to resolve the alleged paradox. And No, I don't dislike your solution, but like mine, there are underlying results of relativity that yield NON-UNIQUE solutions. AG

[Alan Grayson]

On Sunday, September 7, 2025 at 11:17:42 AM UTC-6 Brent Meeker wrote:

On 9/6/2025 11:20 PM, Alan Grayson wrote:

On Saturday, September 6, 2025 at 5:56:36 PM UTC-6 Brent Meeker wrote:

No. You're over complicating the problem.  It's as simple as the fact that two different thru spacetime are different lengths.  Because the spatial coordinate distance, X, appears with a minus sign relative to the coordinate time, T, the proper time, S (which is what a clock measures).  So the rocket, which takes the longer spatial path, experiences less proper time lapse.

Your solution is essentially no different than the two I explained, one using GR and other SR. All proposed solutions rely on some accepted result from relativity theory. AG 

This result has nothing to do with accelerations experience by the traveler.  This a common specious  "explanation" trying to connect it to general relativity. 

Why "specious"? I used an established result in GR to show the traveling twin is younger when he returns and compares clock readings with the stationary twin. The fact that you prefer your solution to mine, doesn't mean using GR is wrong. AG 

This is most easily seen by the triplet version of the paradox.  In this version the one triplet stays home, one travelers away from Earth, and one who has been far away returns to Earth.  When the outbound triplet passes the inbound triplet he hands off the time to the inbound one, so together they measure the same path as the turnaround twin.

Without acceleration, there can be no comparison of clock readings when the inbound twin compares his clock readings with stationary twin. 

That's flat wrong.

In reality, there surely IS acceleration, even though it might not be necessary to use it to solve the problem. Can the traveling twin return without acceleration? Of course not! AG

And I don't like the handing off of clock readings. AG 

Notice that I have also avoided any acceleration at the beginning and end, so no triplet ever accelerates.

Brent

If there's no acceleration, then you've misstated the TP, where both twins start at rest, juxtaposed so their clocks are synched, and the traveler leaves and returns at rest to compare clocks. AG 

[John Clark]

On Sun, Sep 7, 2025 at 2:56 PM Alan Grayson <agrays...@gmail.com> wrote:

Here is another video on the twin paradox by the same guy that I recommended before, he explains it in a slightly different way but it's still crystal clear at least in my mind. The guy is really good. 

I wish I was taught the Twin's Paradox this way!

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

mmm

[Alan Grayson]

Thanks. I'll view it, but I am satisfied I understand its resolution despite what Brent says. He thinks there is a unique solution, his, but that's not true. AG 

mmm

[Brent Meeker]

On 9/7/2025 12:12 PM, Alan Grayson wrote:

On Sunday, September 7, 2025 at 11:17:42 AM UTC-6 Brent Meeker wrote:

On 9/6/2025 11:20 PM, Alan Grayson wrote:

On Saturday, September 6, 2025 at 5:56:36 PM UTC-6 Brent Meeker wrote:

No. You're over complicating the problem.  It's as simple as the fact that two different thru spacetime are different lengths.  Because the spatial coordinate distance, X, appears with a minus sign relative to the coordinate time, T, the proper time, S (which is what a clock measures).  So the rocket, which takes the longer spatial path, experiences less proper time lapse.

Your solution is essentially no different than the two I explained, one using GR and other SR. All proposed solutions rely on some accepted result from relativity theory. AG 

This result has nothing to do with accelerations experience by the traveler.  This a common specious  "explanation" trying to connect it to general relativity. 

Why "specious"? I used an established result in GR to show the traveling twin is younger when he returns and compares clock readings with the stationary twin. The fact that you prefer your solution to mine, doesn't mean using GR is wrong. AG 

This is most easily seen by the triplet version of the paradox.  In this version the one triplet stays home, one travelers away from Earth, and one who has been far away returns to Earth.  When the outbound triplet passes the inbound triplet he hands off the time to the inbound one, so together they measure the same path as the turnaround twin.

Without acceleration, there can be no comparison of clock readings when the inbound twin compares his clock readings with stationary twin. 

That's flat wrong.

In reality, there surely IS acceleration, even though it might not be necessary to use it to solve the problem. Can the traveling twin return without acceleration? Of course not! AG

Actually he can.  All he has to do it slingshot around a distant planet in order to head back to Earth:



You may object that he accelerated in turning around.  But general relativity teaches us that force free motion in a gravitational field is geodesic and there is no acceleration.

Brent

And I don't like the handing off of clock readings. AG 

Notice that I have also avoided any acceleration at the beginning and end, so no triplet ever accelerates.

Brent

If there's no acceleration, then you've misstated the TP, where both twins start at rest, juxtaposed so their clocks are synched, and the traveler leaves and returns at rest to compare clocks. AG 

You may not like it, but it shows the twin paradox has nothing to do with general relativity or acceleration.  It's simply the geometric fact that some paths are shorter than others.  There is no reason to require that the clocks are set (not synched) equal at rest.  I've not "misstated" anything.   You have apparently not understood the twins paradox.

Brent

On 9/6/2025 8:01 AM, Alan Grayson wrote:

On Saturday, September 6, 2025 at 5:39:18 AM UTC-6 John Clark wrote:

On Sat, Sep 6, 2025 at 1:10 AM Alan Grayson <agrays...@gmail.com> wrote:

> If you recall, you recently posted some links to the TP and accused me of not reading them. Well, I certainly intended to read them and I explained why. But then I got involved in other  discussions here and put that temporarily aside. But now those links are, from my pov, lost in a myriad of discussion threads. So, please be so kind as to post them again here. TY, AG

Start with this video:  

I Never Understood How To Intuitively Solve The Twin's Paradox ... Until Now!

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

I've encountered that fellow before. He speaks too fast, way too fast. It seems to me that GR solves the problem, and without frame jumping. Just imagine several changes in velocity, each spread out, so not instantaneous. During those changes, time for the traveler slows compared to rest frame on Earth, so when he returns to meet his twin, he is younger. Do you see anything wrong with this analysis? AG 

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[Brent Meeker]

A complicated explanation of the triplet paradox.  Length contraction is consistent, but it's not necessary to understand the effect.  AG will reject it because he doesn't "believe in" handing off clock readings.

Brent

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[Alan Grayson]

On Sunday, September 7, 2025 at 2:38:40 PM UTC-6 Brent Meeker wrote:

A complicated explanation of the triplet paradox.  Length contraction is consistent, but it's not necessary to understand the effect.  AG will reject it because he doesn't "believe in" handing off clock readings.

Brent

No, that's not it. Rather, I am uncomfortable with de-facto frame-jumping because I am unsure what happens to time when this is included in a solution. And if the twins are at rest and juxtaposed as the scenario begins -- which, BTW, is how the TP is habitually DEFINED -- the traveling twin MUST accelerate to begin his journey. But in the final analysis it's "your way or the highway", meaning that alternate solutions are unacceptable for you. So, if there is acceleration, there is also gravity by applying the Equivalence Principle, and clocks in gravitational fields slow down, and this applies solely to the traveling twin. Notice, I never used or applied the concept of force, so claiming I did so, shows you didn't understand my solution (using GR!). AG

[Alan Grayson]

On Sunday, September 7, 2025 at 2:29:11 PM UTC-6 Brent Meeker wrote:

On 9/7/2025 12:12 PM, Alan Grayson wrote:

On Sunday, September 7, 2025 at 11:17:42 AM UTC-6 Brent Meeker wrote:

On 9/6/2025 11:20 PM, Alan Grayson wrote:

On Saturday, September 6, 2025 at 5:56:36 PM UTC-6 Brent Meeker wrote:

No. You're over complicating the problem.  It's as simple as the fact that two different thru spacetime are different lengths.  Because the spatial coordinate distance, X, appears with a minus sign relative to the coordinate time, T, the proper time, S (which is what a clock measures).  So the rocket, which takes the longer spatial path, experiences less proper time lapse.

Your solution is essentially no different than the two I explained, one using GR and other SR. All proposed solutions rely on some accepted result from relativity theory. AG 

This result has nothing to do with accelerations experience by the traveler.  This a common specious  "explanation" trying to connect it to general relativity. 

Why "specious"? I used an established result in GR to show the traveling twin is younger when he returns and compares clock readings with the stationary twin. The fact that you prefer your solution to mine, doesn't mean using GR is wrong. AG 

This is most easily seen by the triplet version of the paradox.  In this version the one triplet stays home, one travelers away from Earth, and one who has been far away returns to Earth.  When the outbound triplet passes the inbound triplet he hands off the time to the inbound one, so together they measure the same path as the turnaround twin.

Without acceleration, there can be no comparison of clock readings when the inbound twin compares his clock readings with stationary twin. 

That's flat wrong.

In reality, there surely IS acceleration, even though it might not be necessary to use it to solve the problem. Can the traveling twin return without acceleration? Of course not! AG

Actually he can.  All he has to do it slingshot around a distant planet in order to head back to Earth:

But that's NOT how the TP is defined! AG 

You may object that he accelerated in turning around.  But general relativity teaches us that force free motion in a gravitational field is geodesic and there is no acceleration.

So the traveling twin turns around without acceleration? AG 

[Brent Meeker]

On 9/7/2025 5:52 PM, Alan Grayson wrote:

On Sunday, September 7, 2025 at 2:29:11 PM UTC-6 Brent Meeker wrote:

On 9/7/2025 12:12 PM, Alan Grayson wrote:

In reality, there surely IS acceleration, even though it might not be necessary to use it to solve the problem. Can the traveling twin return without acceleration? Of course not! AG

Actually he can.  All he has to do it slingshot around a distant planet in order to head back to Earth:

But that's NOT how the TP is defined! AG 

What is this "defined"?  It's not defined anywhere.  It's just a thought experiment that was paradoxical in Newtonian mechanics.  Every version I've shown you is paradoxical in Newtonian mechanics in exactly the same way.  If you'd open you eyes and mind, you'd see that they give an intuitive grasp on why they all give the same answer in relativity and so resolve the same paradox

You may object that he accelerated in turning around.  But general relativity teaches us that force free motion in a gravitational field is geodesic and there is no acceleration.

So the traveling twin turns around without acceleration? AG 

Read my last sentence above over again a few times.

Brent

[Alan Grayson]

On Sunday, September 7, 2025 at 7:36:32 PM UTC-6 Brent Meeker wrote:

On 9/7/2025 5:52 PM, Alan Grayson wrote:

On Sunday, September 7, 2025 at 2:29:11 PM UTC-6 Brent Meeker wrote:

On 9/7/2025 12:12 PM, Alan Grayson wrote:

In reality, there surely IS acceleration, even though it might not be necessary to use it to solve the problem. Can the traveling twin return without acceleration? Of course not! AG

Actually he can.  All he has to do it slingshot around a distant planet in order to head back to Earth:

But that's NOT how the TP is defined! AG 

What is this "defined"?  It's not defined anywhere. 

That's how it's described in almost any text one can find. You have a private definition. AG

 

It's just a thought experiment that was paradoxical in Newtonian mechanics.  Every version I've shown you is paradoxical in Newtonian mechanics in exactly the same way.  If you'd open you eyes and mind, you'd see that they give an intuitive grasp on why they all give the same answer in relativity and so resolve the same paradox

You may object that he accelerated in turning around.  But general relativity teaches us that force free motion in a gravitational field is geodesic and there is no acceleration.

So the traveling twin turns around without acceleration? AG 

Read my last sentence above over again a few times.

Brent

I did, initially. During the turnaround the motion is NOT force free, which GR allows, and one can apply the Equivalence Principle, and then time dilation. AG 

[Brent Meeker]

On 9/7/2025 5:44 PM, Alan Grayson wrote:

On Sunday, September 7, 2025 at 2:38:40 PM UTC-6 Brent Meeker wrote:

A complicated explanation of the triplet paradox.  Length contraction is consistent, but it's not necessary to understand the effect.  AG will reject it because he doesn't "believe in" handing off clock readings.

Brent

No, that's not it. Rather, I am uncomfortable with de-facto frame-jumping because I am unsure what happens to time when this is included in a solution. And if the twins are at rest and juxtaposed as the scenario begins -- which, BTW, is how the TP is habitually DEFINED -- the traveling twin MUST accelerate to begin his journey. But in the final analysis it's "your way or the highway", meaning that alternate solutions are unacceptable for you. 

Not at all.  You think it depends on acceleration.  Fine, then here's an alternate version with acceleration.  The twins each accelerates exactly the same level for exactly the same duration.  But Red is still younger than Blue for exactly the same reason; his path is longer in space and therefore shorter in spacetime.

So, if there is acceleration, there is also gravity by applying the Equivalence Principle, 

So did you apply gravitational time dilation to each twin above?

and clocks in gravitational fields slow down, and this applies solely to the traveling twin. Notice, I never used or applied the concept of force, 

Above you seem to think the equivalence principle means acceleration implies gravity

Brent

[Brent Meeker]

You seemed to have missed GR 101 where it is explained that gravity is NOT a force.

Brent

[Alan Grayson]

It's your way or the highway. Or shall we say a touch of arrogance? Haven't you ever heard of the EP? Let me remind you. Gravity is locally equivalent to acceleration, so when the traveling twin accelerates, it's equivalent to being in a gravity field, where clock rates are slower compared to rest frames. Where did I use the word "force"? AG 

[Alan Grayson]

The traveler accelerates by applying a force, not by the force being applied by the gravitational field. Do you get it now? AG 

[Alan Grayson]

How does the traveler do that; by firing a rocket attached to his butt. Same way he left the stationary observer! GR allows the traveler do that. AG 

[John Clark]

On Sun, Sep 7, 2025 at 4:05 PM Alan Grayson <agrays...@gmail.com> wrote:

>> Here is another video on the twin paradox by the same guy that I recommended before, he explains it in a slightly different way but it's still crystal clear at least in my mind. The guy is really good. 

I wish I was taught the Twin's Paradox this way!

> Thanks. I'll view it, but I am satisfied I understand its resolution despite what Brent says. He thinks there is a unique solution, his, but that's not true. AG 

You really should look at that video because it shows a way to state the twin "paradox" such that no acceleration is involved, but even in that case it can be resolved and he demonstrates it's not a paradox at all, it's just a strange situation.  People call it a "paradox" because, although they remember time dilation and length contraction, they forget a third equally important thing special relativity tells us about the universe, the impossibility of absolute simultaneously except for the case of two events occurring at the same place and at the same time. So even if you can run faster than me I can still beat you to the finish line if I hear the starting gun before you do.

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

kf9

[Alan Grayson]

I see a problem with this GR scenario. If the traveler is partly coasting on a geodesic path, his clock will be running faster than than his stationary twin, so he will age at a greater rate. OTOH, during periods when a force is applied to accelerate (not by any gravity field), his clock will slow down, compared to its rate while he's in geodesic motion. So for the traveler to return to Earth younger than his twin, his slower clock while accelerating, must be large enough overall, to cause his clock to fall behind his stationary twin. The traveler could apply a continous but changing acceleration, say by traveling in a circle, but whether his clock will slow enough to make him age less than his stationary twin I don't know. AG

[Alan Grayson]

I'm not sure the impossibility of absolute simultaneity solves the problem, if it's caused by the fallacious assumption of symmetry. I haven't yet viewed the video you just posted, but I will. AG 

[John Clark]

On Mon, Sep 8, 2025 at 7:00 AM Alan Grayson <agrays...@gmail.com> wrote:

> I'm not sure the impossibility of absolute simultaneity solves the problem,

Watch the video! If you follow what he does step-by-step you will see that he is right. It's not difficult. 

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

erx

On

>> Here is another video on the twin paradox by the same guy that I recommended before, he explains it in a slightly different way but it's still crystal clear at least in my mind. The guy is really good. 

I wish I was taught the Twin's Paradox this way!

> Thanks. I'll view it, but I am satisfied I understand its resolution despite what Brent says. He thinks there is a unique solution, his, but that's not true. AG 

You really should look at that video because it shows a way to state the twin "paradox" such that no acceleration is involved, but even in that case it can be resolved and he demonstrates it's not a paradox at all, it's just a strange situation.  People call it a "paradox" because, although they remember time dilation and length contraction, they forget a third equally important thing special relativity tells us about the universe, the impossibility of absolute simultaneously except for the case of two events occurring at the same place and at the same time. So even if you can run faster than me I can still beat you to the finish line if I hear the starting gun before you do.

I'm not sure the impossibility of absolute simultaneity solves the problem, if it's caused by the fallacious assumption of symmetry. I haven't yet viewed the video you just posted, but I will. AG 

kf9

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[Alan Grayson]

On Monday, September 8, 2025 at 5:06:36 AM UTC-6 John Clark wrote:

On Mon, Sep 8, 2025 at 7:00 AM Alan Grayson <agrays...@gmail.com> wrote:

> I'm not sure the impossibility of absolute simultaneity solves the problem,

Watch the video! If you follow what he does step-by-step you will see that he is right. It's not difficult. 

I'll definitely watch it, very soon, but a-priori the impossibility of absolute simultaneity can't solve the paradox because it's not its cause. Can you succinctly state the cause of the paradox? It's the application of time dilation in SR, under the mistaken assumption that the twins take symmetric paths; that their situations are symmetric. This results in the situation that when they meet and compare clock readings, each concludes the other is younger. Do you agree that I've correctly stated the paradox? AG

[Brent Meeker]

But in the turnaround the time in the gravity field can be arbitrarily short compared to the the coasting phase before and after.

Where did I use the word "force"? AG 

" During the turnaround the motion is NOT force free."

Brent

[Brent Meeker]

No.  The traveler uses the gravitational field of a planet to slingshot back around toward Earth.  No force, in the GR sense, at all.
Get it now?...no of course not becuase you didn't spend a moments thought on the diagram I posted.

Brent

[Brent Meeker]

That's confused on several points.  First, nobodies clock runs fast or slow because he's coasting...in a straight line in flat space in this problem.  Second, the traveler and the stay home person each see the other's clock as running slow.

Brent

OTOH, during periods when a force is applied to accelerate (not by any gravity field), his clock will slow down, compared to its rate while he's in geodesic motion. So for the traveler to return to Earth younger than his twin, his slower clock while accelerating, must be large enough overall, to cause his clock to fall behind his stationary twin. The traveler could apply a continous but changing acceleration, say by traveling in a circle, but whether his clock will slow enough to make him age less than his stationary twin I don't know. AG

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[Alan Grayson]

That's your modification or interpretation of the TP. There are different ways to return to Earth, and the slingshot is just one example. Another is strapping a rocket to your butt and lighting it. AG

[Alan Grayson]

Wrong. In flat space there is no gravity, and a clock coasting in flat space runs faster than the clock residing on the Earth in its gravity field. The gravity field in GR slows the clock rate compared to an observer in free fall -- one of the effects on GPS clocks. AG

Second, the traveler and the stay home person each see the other's clock as running slow.

Yes, that's the paradox, and it's not caused by assuming absolute simultaneity (as I think the video recommended by Clark alleges, which I have not yet viewed), but the erroneous assumption that the twins are in symmetric situations. One twin accelerates and the other does NOT. AG

[Alan Grayson]

Not true. I considered your spacetime plot for two paths and found it instructive.  More proper time elapses for stationary twin. I have to check if the invariance of (ds)^2 is true in GR as well as SR. AG

[Brent Meeker]

On 9/8/2025 11:19 AM, Alan Grayson wrote:

On Monday, September 8, 2025 at 5:06:36 AM UTC-6 John Clark wrote:

On Mon, Sep 8, 2025 at 7:00 AM Alan Grayson <agrays...@gmail.com> wrote:

> I'm not sure the impossibility of absolute simultaneity solves the problem,

Watch the video! If you follow what he does step-by-step you will see that he is right. It's not difficult. 

I'll definitely watch it, very soon, but a-priori the impossibility of absolute simultaneity can't solve the paradox because it's not its cause. Can you succinctly state the cause of the paradox? It's the application of time dilation in SR, under the mistaken assumption that the twins take symmetric paths; that their situations are symmetric. This results in the situation that when they meet and compare clock readings, each concludes the other is younger. 

No that's wrong.  The stay at home twin has a clock that indicates a longer interval than the traveling twins clock.   They agree that the traveling twin is younger.

Brent

Do you agree that I've correctly stated the paradox? AG

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

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[Alan Grayson]

On Monday, September 8, 2025 at 9:35:09 PM UTC-6 Brent Meeker wrote:

On 9/8/2025 11:19 AM, Alan Grayson wrote:

On Monday, September 8, 2025 at 5:06:36 AM UTC-6 John Clark wrote:

On Mon, Sep 8, 2025 at 7:00 AM Alan Grayson <agrays...@gmail.com> wrote:

> I'm not sure the impossibility of absolute simultaneity solves the problem,

Watch the video! If you follow what he does step-by-step you will see that he is right. It's not difficult. 

I'll definitely watch it, very soon, but a-priori the impossibility of absolute simultaneity can't solve the paradox because it's not its cause. Can you succinctly state the cause of the paradox? It's the application of time dilation in SR, under the mistaken assumption that the twins take symmetric paths; that their situations are symmetric. This results in the situation that when they meet and compare clock readings, each concludes the other is younger. 

No that's wrong.  The stay at home twin has a clock that indicates a longer interval than the traveling twins clock.   They agree that the traveling twin is younger.

Brent

Can't you understand English? I was stating the paradox and its cause. With an accurate analysis, the traveling twin is younger. Also, FWIW, for the traveling twin to return for the clock comparison, some acceleration is necessary, although it can be minimized if the comparison is done by fly-by. a AG 

[Alan Grayson]

On Sunday, September 7, 2025 at 7:49:01 PM UTC-6 Brent Meeker wrote:

On 9/7/2025 5:44 PM, Alan Grayson wrote:

On Sunday, September 7, 2025 at 2:38:40 PM UTC-6 Brent Meeker wrote:

A complicated explanation of the triplet paradox.  Length contraction is consistent, but it's not necessary to understand the effect.  AG will reject it because he doesn't "believe in" handing off clock readings.

Brent

No, that's not it. Rather, I am uncomfortable with de-facto frame-jumping because I am unsure what happens to time when this is included in a solution. And if the twins are at rest and juxtaposed as the scenario begins -- which, BTW, is how the TP is habitually DEFINED -- the traveling twin MUST accelerate to begin his journey. But in the final analysis it's "your way or the highway", meaning that alternate solutions are unacceptable for you. 

Not at all.  You think it depends on acceleration.  Fine, then here's an alternate version with acceleration.  The twins each accelerates exactly the same level for exactly the same duration.  But Red is still younger than Blue for exactly the same reason; his path is longer in space and therefore shorter in spacetime.

So, if there is acceleration, there is also gravity by applying the Equivalence Principle, 

So did you apply gravitational time dilation to each twin above?

Later I posted why my GR model doesn't work. There's no obvious way for the twins to compare clock and determine their relative ages. It might depend on the paths taken, and I don't see how to do a calculation for any particular path for the traveling twin. Nonetheless, your denial of acceleration is mistaken. In your diagram with two spacetime paths, the proper times differ because along one path all the spatial derivatives are zero, unlike along the other path of the traveling twin. This is your de-facto admission that differences in accelerations is the key to solving the paradox. Your solution is ostensibly simpler because you fail to state exactly why the proper times are different along the two paths. AG

and clocks in gravitational fields slow down, and this applies solely to the traveling twin. Notice, I never used or applied the concept of force, 

Above you seem to think the equivalence principle means acceleration implies gravity

Brent

When you're accelerating, it seems as if you're in a local gravitational field; that is, you cannot distinguish your acceleration from local gravity field. If that's not what the EP is, what's your take? AG

[Alan Grayson]

It's your way or the highway. Or shall we say a touch of arrogance? Haven't you ever heard of the EP? Let me remind you. Gravity is locally equivalent to acceleration, so when the traveling twin accelerates, it's equivalent to being in a gravity field, where clock rates are slower compared to rest frames. AG

This is where I made a mistake. When accelerating, the traveling twin's clock is ticking slower than a comparable clock in inertial space where there is no gravity, but what its rate is compared to the stationary twin's clock on the Earth I cannot determine for comparison. In GPS, for example, the clocks are in free fall and ticking faster than ground clocks, but slower than clocks in inertial space. So I think the paradox is solved in SR. It might be possible to solve the paradox in GR, but the calculation to do so is likely rather difficult.  AG 

But in the turnaround the time in the gravity field can be arbitrarily short compared to the the coasting phase before and after.

Where did I use the word "force"? AG 

" During the turnaround the motion is NOT force free."

There is a force but not necessarily due to gravity. The problem can be modeled as if the traveler is beyond the gravity of Earth, and fires rockets to return.  And Yes, the turnaround time can be arbitrarily short, but that involves frame jumping which I want to avoid since I am unsure how a clock behaves under that circumstance. AG

Brent

[Alan Grayson]

On Sunday, September 7, 2025 at 7:49:01 PM UTC-6 Brent Meeker wrote:

On 9/7/2025 5:44 PM, Alan Grayson wrote:

On Sunday, September 7, 2025 at 2:38:40 PM UTC-6 Brent Meeker wrote:

A complicated explanation of the triplet paradox.  Length contraction is consistent, but it's not necessary to understand the effect.  AG will reject it because he doesn't "believe in" handing off clock readings.

Brent

No, that's not it. Rather, I am uncomfortable with de-facto frame-jumping because I am unsure what happens to time when this is included in a solution. And if the twins are at rest and juxtaposed as the scenario begins -- which, BTW, is how the TP is habitually DEFINED -- the traveling twin MUST accelerate to begin his journey. But in the final analysis it's "your way or the highway", meaning that alternate solutions are unacceptable for you. 

Not at all.  You think it depends on acceleration.  Fine, then here's an alternate version with acceleration.  The twins each accelerates exactly the same level for exactly the same duration. 

 

 No. This is wrong and is the cause of the so-called paradox. AG

 

But Red is still younger than Blue for exactly the same reason; his path is longer in space and therefore shorter in spacetime.

But why do the paths have different lengths? Answer; one twin is accelerating, the other not. AG

[John Clark]

On Mon, Sep 8, 2025 at 2:19 PM Alan Grayson <agrays...@gmail.com> wrote:

> I'm not sure the impossibility of absolute simultaneity solves the problem,

Watch the video! If you follow what he does step-by-step you will see that he is right. It's not difficult. 

I wish I was taught the Twin's Paradox this way! 

I'll definitely watch it, very soon, but a-priori the impossibility of absolute simultaneity can't solve the paradox because it's not its cause. Can you succinctly state the cause of the paradox?

Yes, it's impossible for all observers to agree that two events are simultaneous unless they happen at the same place. But I'm sure that is a little too succinct to convince you, that's why you need to watch the video, it's only a little longer. And neither acceleration nor General Relativity is the key to resolving the twin "paradox", as far as this matter is concerned you'd do best to forget about them. 

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

fai

[Alan Grayson]

The twins ARE at the same place when the thought experiment begins, and since they're juxtaposed when the thought experiment begins, their clocks are synchronized without anything to do with absolute simultaneity! And second, using a spacetime diagram, the paths are unequal, and that's because only one path represents the accelerating twin who is traveling. If you don't believe me, look at (ds)^2 for each twin to confirm my claim. Take notice of the second order differentials. AG 

[Alan Grayson]

CORRECTION: The spacetime path lengths are INVARIANT, but along the path of the stationary twin, the second order differentials are zero since that twin is NOT accelerating, but those second order differentials for the traveling twin are NON-ZERO since he's accelerating. Now look at the proper times along both paths and you'll see that it's greater for stationary twin because the path lengths are invariant. Hence, stationary twin ages more than traveling twin! Note also that this analysis uses SR. If the video is using a simultaneity argument, then the video is wrong. AG

[John Clark]

On Tue, Sep 9, 2025 at 1:38 PM Alan Grayson <agrays...@gmail.com> wrote:

> I'm not sure the impossibility of absolute simultaneity solves the problem,

Watch the video! If you follow what he does step-by-step you will see that he is right. It's not difficult. 

I wish I was taught the Twin's Paradox this way! 

 

I'll definitely watch it, very soon, but a-priori the impossibility of absolute simultaneity can't solve the paradox because it's not its cause. Can you succinctly state the cause of the paradox?

Yes, it's impossible for all observers to agree that two events are simultaneous unless they happen at the same place. But I'm sure that is a little too succinct to convince you, that's why you need to watch the video, it's only a little longer. And neither acceleration nor General Relativity is the key to resolving the twin "paradox", as far as this matter is concerned you'd do best to forget about them. 

The twins ARE at the same place when the thought experiment begins, and since they're juxtaposed when the thought experiment begins, their clocks are synchronized without anything to do with absolute simultaneity! And second, using a spacetime diagram, the paths are unequal, and that's because only one path represents the accelerating twin who is traveling. If you don't believe me, look at (ds)^2 for each twin to confirm my claim. Take notice of the second order differentials. AG 

CORRECTION: The spacetime path lengths are INVARIANT, but along the path of the stationary twin, the second order differentials are zero since that twin is NOT accelerating, but those second order differentials for the traveling twin are NON-ZERO since he's accelerating. Now look at the proper times along both paths and you'll see that it's greater for stationary twin because the path lengths are invariant. Hence, stationary twin ages more than traveling twin! Note also that this analysis uses SR. If the video is using a simultaneity argument, then the video is wrong. AG

In the time it took you to write all that you could have just watched the video.  

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

ffff

[Alan Grayson]

Why don't you budget your time as you see fit, and I'll do the same, if you'll be so kind as to permit me that freedom. Your aggression is not appreciated. Oh, FYI, I have been watching it, but not finished. So far I'm not impressed. His model seems more complicated than just recognizing that the assumption of symmetry, which causes the paradox, is wrong. Oh, I see I made a mistake to refer to second order differentials in (ds)^2, so I'm not sure how acceleration fits into the solution using a spacetime diagram. I'll have to go back and see how Brent uses it. AG

[Alan Grayson]

On Saturday, September 6, 2025 at 5:56:36 PM UTC-6 Brent Meeker wrote:

No. You're over complicating the problem.  It's as simple as the fact that two different thru spacetime are different lengths.  Because the spatial coordinate distance, X, appears with a minus sign relative to the coordinate time, T, the proper time, S (which is what a clock measures).  So the rocket, which takes the longer spatial path, experiences less proper time lapse.

If the spacetime distance between two points in invariant, how can you say one path has a different length than another? BTW, considering acceleration does NOT necessarily mean invoking GR, and the invariance is a result of SR. AG 

This result has nothing to do with accelerations experience by the traveler. 

I disagree since it is what indicates lack of symmetry, which causes the paradox. Moreover, referring to acceleration does not necessarily invoke GR. AG

 

This a common specious  "explanation" trying to connect it to general relativity.  This is most easily seen by the triplet version of the paradox.  In this version the one triplet stays home, one travelers away from Earth, and one who has been far away returns to Earth.  When the outbound triplet passes the inbound triplet he hands off the time to the inbound one, so together they measure the same path as the turnaround twin.



Notice that I have also avoided any acceleration at the beginning and end, so no triplet ever accelerates.

Brent

[Alan Grayson]

On Tuesday, September 9, 2025 at 2:19:19 PM UTC-6 Alan Grayson wrote:

On Saturday, September 6, 2025 at 5:56:36 PM UTC-6 Brent Meeker wrote:

No. You're over complicating the problem.  It's as simple as the fact that two different thru spacetime are different lengths.  Because the spatial coordinate distance, X, appears with a minus sign relative to the coordinate time, T, the proper time, S (which is what a clock measures).  So the rocket, which takes the longer spatial path, experiences less proper time lapse.

If the spacetime distance between two points IS invariant, how can you say one path has a different length than another? BTW, considering acceleration does NOT necessarily mean invoking GR, and the invariance is a result of SR. AG 

CORRECTION: Invariance refers to the fixed length from the pov of different frames under a LT, not that all paths connected by two points in spacetime have the same length. AG 

[Brent Meeker]

But notice that the acceleration is entirely incidental, as illustrated by the case in which Red and Blue each accelerates the same amount.  IT'S JUST GEOMETRY.  ONE PATH IS LONGER THAN THE OTHER.

Brent

[Brent Meeker]

On 9/9/2025 12:59 AM, Alan Grayson wrote:

On Sunday, September 7, 2025 at 7:49:01 PM UTC-6 Brent Meeker wrote:

...

Later I posted why my GR model doesn't work. There's no obvious way for the twins to compare clock and determine their relative ages. It might depend on the paths taken, and I don't see how to do a calculation for any particular path for the traveling twin. Nonetheless, your denial of acceleration is mistaken. In your diagram with two spacetime paths, the proper times differ because along one path all the spatial derivatives are zero, unlike along the other path of the traveling twin. 

Spatial derivatives of what??  Red and Blue each accelerate the same level for the same duration.

This is your de-facto admission that differences in accelerations is the key to solving the paradox. 

There are no differences in accelerations.  Look at the damn diagram!

Your solution is ostensibly simpler because you fail to state exactly why the proper times are different along the two paths. AG

The proper times are different because one path is longer than the other (in Mikowski 4-space metric).  And I explicitly note the times.

and clocks in gravitational fields slow down, and this applies solely to the traveling twin. Notice, I never used or applied the concept of force, 

Above you seem to think the equivalence principle means acceleration implies gravity

Brent

When you're accelerating, it seems as if you're in a local gravitational field; that is, you cannot distinguish your acceleration from local gravity field. If that's not what the EP is, what's your take? AG

My take is that it's true but completely irrelevant to the twins paradox.  Even in the usual story the acceleration level and duration for the traveling twin can be made arbitrarily small compared to the geometric effect simply by choosing a more distant turnaround point.

Brent

[Brent Meeker]

On 9/9/2025 1:19 PM, Alan Grayson wrote:

On Saturday, September 6, 2025 at 5:56:36 PM UTC-6 Brent Meeker wrote:

No. You're over complicating the problem.  It's as simple as the fact that two different thru spacetime are different lengths.  Because the spatial coordinate distance, X, appears with a minus sign relative to the coordinate time, T, the proper time, S (which is what a clock measures).  So the rocket, which takes the longer spatial path, experiences less proper time lapse.

If the spacetime distance between two points in invariant, how can you say one path has a different length than another? 

That's like saying there's a fixed distance between L.A. and NYC so how can there be different path lengths between them.

BTW, considering acceleration does NOT necessarily mean invoking GR, and the invariance is a result of SR. AG 

This result has nothing to do with accelerations experience by the traveler. 

I disagree since it is what indicates lack of symmetry, which causes the paradox. 

See (1) my triplet example in which nobody accelerates and (2) my example in which Red and Blue accelerate exactly the same.

Brent

Moreover, referring to acceleration does not necessarily invoke GR. AG

 

This a common specious  "explanation" trying to connect it to general relativity.  This is most easily seen by the triplet version of the paradox.  In this version the one triplet stays home, one travelers away from Earth, and one who has been far away returns to Earth.  When the outbound triplet passes the inbound triplet he hands off the time to the inbound one, so together they measure the same path as the turnaround twin.



Notice that I have also avoided any acceleration at the beginning and end, so no triplet ever accelerates.

Brent

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[Alan Grayson]

In the original statement of the "paradox', the traveling twin must accelerate to return so the clocks can be compared. Please explain how this can happen without acceleration. You seem to defying basic physics if this is your claim. I don't deny that the original problem can be restated in a way which avoids acceleration, and IMO this is what you've done. AG 

Brent

[Alan Grayson]

On Tuesday, September 9, 2025 at 5:39:58 PM UTC-6 Brent Meeker wrote:

On 9/9/2025 12:59 AM, Alan Grayson wrote:

On Sunday, September 7, 2025 at 7:49:01 PM UTC-6 Brent Meeker wrote:

...

Later I posted why my GR model doesn't work. There's no obvious way for the twins to compare clock and determine their relative ages. It might depend on the paths taken, and I don't see how to do a calculation for any particular path for the traveling twin. Nonetheless, your denial of acceleration is mistaken. In your diagram with two spacetime paths, the proper times differ because along one path all the spatial derivatives are zero, unlike along the other path of the traveling twin. 

Spatial derivatives of what??  Red and Blue each accelerate the same level for the same duration.

If you had followed my posts carefully, you'd realize that my statement about spatial derivatives was wrong. I confused (dx)^2 with the 2nd order derivative; same for y and x coordinates. Also, I clearly stated that my use of GR didn't get the result I thought it would and I stated that as well. But I am not saying GR cannot be used to solve the problem. AG  

This is your de-facto admission that differences in accelerations is the key to solving the paradox. 

There are no differences in accelerations.  Look at the damn diagram!

Your diagram has no relation to the TP as originally stated. You claim both spaceships have the same acceleration. How this relates to the stationary twin who never changes his location you provide no clue! AG 

Your solution is ostensibly simpler because you fail to state exactly why the proper times are different along the two paths. AG

The proper times are different because one path is longer than the other (in Mikowski 4-space metric).  And I explicitly note the times.

and clocks in gravitational fields slow down, and this applies solely to the traveling twin. Notice, I never used or applied the concept of force, 

Above you seem to think the equivalence principle means acceleration implies gravity

Brent

When you're accelerating, it seems as if you're in a local gravitational field; that is, you cannot distinguish your acceleration from local gravity field. If that's not what the EP is, what's your take? AG

My take is that it's true but completely irrelevant to the twins paradox. 

You asked for my interpretation of the EP, so I gave it and you agree. It is irrelevant because my attempt to use GR didn't work, and I stated as such. AG

 

Even in the usual story the acceleration level and duration for the traveling twin can be made arbitrarily small compared to the geometric effect simply by choosing a more distant turnaround point.

If you acknowledge that the paradox is CAUSED by the assumption of symmetry, and that there is no symmetry due to spatial acceleration, you have a solution. But you claim there is no spatial acceleration, or maybe the twin at rest is really accelerating at the same rate as the traveling twin, then your solution makes no sense and violates basic physics. AG

Brent

[Brent Meeker]

I've shown two different ways without acceleration and I've also shown the paradox with equal accelerations by both twins.  Why can't you just accept that it's geometry; that one path is longer than the other.

You seem to defying basic physics if this is your claim. I don't deny that the original problem can be restated in a way which avoids acceleration, and IMO this is what you've done. 

But I've done more than that.  I've done it while maintaining exactly the same paradox.

Brent

AG 

Brent

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[Brent Meeker]

On 9/9/2025 10:23 PM, Alan Grayson wrote:

On Tuesday, September 9, 2025 at 5:39:58 PM UTC-6 Brent Meeker wrote:

...

This is your de-facto admission that differences in accelerations is the key to solving the paradox. 

There are no differences in accelerations.  Look at the damn diagram!

Your diagram has no relation to the TP as originally stated. 

You say no relation, but it produces exactly the same paradox for exactly the same reason.

You claim both spaceships have the same acceleration. How this relates to the stationary twin who never changes his location you provide no clue! 

From the parting of ways to reuniting, Blue doesn't change his location.  Do I have to draw in Earth for you too.  Look at the damn diagram!

Brent

[Alan Grayson]

On Tuesday, September 9, 2025 at 11:25:37 PM UTC-6 Brent Meeker wrote:

On 9/9/2025 9:57 PM, Alan Grayson wrote:

On Tuesday, September 9, 2025 at 5:18:14 PM UTC-6 Brent Meeker wrote:

On 9/8/2025 8:45 PM, Alan Grayson wrote:

On Monday, September 8, 2025 at 9:35:09 PM UTC-6 Brent Meeker wrote:

On 9/8/2025 11:19 AM, Alan Grayson wrote:

On Monday, September 8, 2025 at 5:06:36 AM UTC-6 John Clark wrote:

On Mon, Sep 8, 2025 at 7:00 AM Alan Grayson <agrays...@gmail.com> wrote:

> I'm not sure the impossibility of absolute simultaneity solves the problem,

Watch the video! If you follow what he does step-by-step you will see that he is right. It's not difficult. 

I'll definitely watch it, very soon, but a-priori the impossibility of absolute simultaneity can't solve the paradox because it's not its cause. Can you succinctly state the cause of the paradox? It's the application of time dilation in SR, under the mistaken assumption that the twins take symmetric paths; that their situations are symmetric. This results in the situation that when they meet and compare clock readings, each concludes the other is younger. 

No that's wrong.  The stay at home twin has a clock that indicates a longer interval than the traveling twins clock.   They agree that the traveling twin is younger.

Brent

Can't you understand English? I was stating the paradox and its cause. With an accurate analysis, the traveling twin is younger. Also, FWIW, for the traveling twin to return for the clock comparison, some acceleration is necessary, although it can be minimized if the comparison is done by fly-by. a AG 

But notice that the acceleration is entirely incidental, as illustrated by the case in which Red and Blue each accelerates the same amount.  IT'S JUST GEOMETRY.  ONE PATH IS LONGER THAN THE OTHER.

In the original statement of the "paradox', the traveling twin must accelerate to return so the clocks can be compared. Please explain how this can happen without acceleration. 

I've shown two different ways without acceleration and I've also shown the paradox with equal accelerations by both twins.  Why can't you just accept that it's geometry; that one path is longer than the other.

If both twins are accelerating, then you've redefined the TP. If you have two paths in spacetime, starting at the same point and ending at the same point, or at a  different point, how can you tell which is longer? AG 

[Alan Grayson]

It's OK to claim one twin is accelerating when he never accelerates, but don't also claim this is easily understood. AG 

Brent

[Brent Meeker]

On 9/9/2025 10:36 PM, Alan Grayson wrote:

On Tuesday, September 9, 2025 at 11:25:37 PM UTC-6 Brent Meeker wrote:

On 9/9/2025 9:57 PM, Alan Grayson wrote:

On Tuesday, September 9, 2025 at 5:18:14 PM UTC-6 Brent Meeker wrote:

On 9/8/2025 8:45 PM, Alan Grayson wrote:

On Monday, September 8, 2025 at 9:35:09 PM UTC-6 Brent Meeker wrote:

On 9/8/2025 11:19 AM, Alan Grayson wrote:

On Monday, September 8, 2025 at 5:06:36 AM UTC-6 John Clark wrote:

On Mon, Sep 8, 2025 at 7:00 AM Alan Grayson <agrays...@gmail.com> wrote:

> I'm not sure the impossibility of absolute simultaneity solves the problem,

Watch the video! If you follow what he does step-by-step you will see that he is right. It's not difficult. 

I'll definitely watch it, very soon, but a-priori the impossibility of absolute simultaneity can't solve the paradox because it's not its cause. Can you succinctly state the cause of the paradox? It's the application of time dilation in SR, under the mistaken assumption that the twins take symmetric paths; that their situations are symmetric. This results in the situation that when they meet and compare clock readings, each concludes the other is younger. 

No that's wrong.  The stay at home twin has a clock that indicates a longer interval than the traveling twins clock.   They agree that the traveling twin is younger.

Brent

Can't you understand English? I was stating the paradox and its cause. With an accurate analysis, the traveling twin is younger. Also, FWIW, for the traveling twin to return for the clock comparison, some acceleration is necessary, although it can be minimized if the comparison is done by fly-by. a AG 

But notice that the acceleration is entirely incidental, as illustrated by the case in which Red and Blue each accelerates the same amount.  IT'S JUST GEOMETRY.  ONE PATH IS LONGER THAN THE OTHER.

In the original statement of the "paradox', the traveling twin must accelerate to return so the clocks can be compared. Please explain how this can happen without acceleration. 

I've shown two different ways without acceleration and I've also shown the paradox with equal accelerations by both twins.  Why can't you just accept that it's geometry; that one path is longer than the other.

If both twins are accelerating, then you've redefined the TP. 

No.  But Blue does not accelerate while Red travels out and back.

If you have two paths in spacetime, starting at the same point and ending at the same point, or at a  different point, how can you tell which is longer? AG 

These are paths in spacetime.  They start and end at the same event, a point in 4-space.  The obvious way to tell which is longer in proper time is to carry an ideal clock along the two paths and compare the measured intervals.  You could also measure the space distance X along the paths and and compute proper time S=\sqrt{T^2 - X^2}  where T is the coordinate time difference (in the same reference frame you measured distance).

Brent

You seem to defying basic physics if this is your claim. I don't deny that the original problem can be restated in a way which avoids acceleration, and IMO this is what you've done. 

But I've done more than that.  I've done it while maintaining exactly the same paradox.

Brent

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[Alan Grayson]

On Wednesday, September 10, 2025 at 12:08:39 AM UTC-6 Brent Meeker wrote:

On 9/9/2025 10:36 PM, Alan Grayson wrote:

On Tuesday, September 9, 2025 at 11:25:37 PM UTC-6 Brent Meeker wrote:

On 9/9/2025 9:57 PM, Alan Grayson wrote:

On Tuesday, September 9, 2025 at 5:18:14 PM UTC-6 Brent Meeker wrote:

On 9/8/2025 8:45 PM, Alan Grayson wrote:

On Monday, September 8, 2025 at 9:35:09 PM UTC-6 Brent Meeker wrote:

On 9/8/2025 11:19 AM, Alan Grayson wrote:

On Monday, September 8, 2025 at 5:06:36 AM UTC-6 John Clark wrote:

On Mon, Sep 8, 2025 at 7:00 AM Alan Grayson <agrays...@gmail.com> wrote:

> I'm not sure the impossibility of absolute simultaneity solves the problem,

Watch the video! If you follow what he does step-by-step you will see that he is right. It's not difficult. 

I'll definitely watch it, very soon, but a-priori the impossibility of absolute simultaneity can't solve the paradox because it's not its cause. Can you succinctly state the cause of the paradox? It's the application of time dilation in SR, under the mistaken assumption that the twins take symmetric paths; that their situations are symmetric. This results in the situation that when they meet and compare clock readings, each concludes the other is younger. 

No that's wrong.  The stay at home twin has a clock that indicates a longer interval than the traveling twins clock.   They agree that the traveling twin is younger.

Brent

Can't you understand English? I was stating the paradox and its cause. With an accurate analysis, the traveling twin is younger. Also, FWIW, for the traveling twin to return for the clock comparison, some acceleration is necessary, although it can be minimized if the comparison is done by fly-by. a AG 

But notice that the acceleration is entirely incidental, as illustrated by the case in which Red and Blue each accelerates the same amount.  IT'S JUST GEOMETRY.  ONE PATH IS LONGER THAN THE OTHER.

In the original statement of the "paradox', the traveling twin must accelerate to return so the clocks can be compared. Please explain how this can happen without acceleration. 

I've shown two different ways without acceleration and I've also shown the paradox with equal accelerations by both twins.  Why can't you just accept that it's geometry; that one path is longer than the other.

If both twins are accelerating, then you've redefined the TP. 

No.  But Blue does not accelerate while Red travels out and back.

But earlier you claimed both Red and Blue have the same acceleration, and that's what your diagram shows. AG

If you have two paths in spacetime, starting at the same point and ending at the same point, or at a  different point, how can you tell which is longer? AG 

These are paths in spacetime.  They start and end at the same event, a point in 4-space. 

If the paradox is resolved, then the clocks should read different values when finally compared. So the end point events are NOT the same. AG

 

The obvious way to tell which is longer in proper time is to carry an ideal clock along the two paths and compare the measured intervals.  You could also measure the space distance X along the paths and and compute proper time S=\sqrt{T^2 - X^2}  where T is the coordinate time difference (in the same reference frame you measured distance).

Brent

You seem to defying basic physics if this is your claim. I don't deny that the original problem can be restated in a way which avoids acceleration, and IMO this is what you've done. 

But I've done more than that.  I've done it while maintaining exactly the same paradox.

So you admit you're defying the laws of physics? AG 

Brent

[Brent Meeker]

On 9/9/2025 11:52 PM, Alan Grayson wrote:

On Wednesday, September 10, 2025 at 12:08:39 AM UTC-6 Brent Meeker wrote:

On 9/9/2025 10:36 PM, Alan Grayson wrote:

On Tuesday, September 9, 2025 at 11:25:37 PM UTC-6 Brent Meeker wrote:

...

If both twins are accelerating, then you've redefined the TP. 

No.  But Blue does not accelerate while Red travels out and back.

But earlier you claimed both Red and Blue have the same acceleration, and that's what your diagram shows. AG

But not  "while Red travels out and back".  Which IS what my diagram shows.  You can imagine Blue as decelerating, landing on Earth, and then accelerating to join up again as Red comes back by.  Next time read the whole sentence, not just up until a question occurs to you.

If you have two paths in spacetime, starting at the same point and ending at the same point, or at a  different point, how can you tell which is longer? AG 

These are paths in spacetime.  They start and end at the same event, a point in 4-space. 

If the paradox is resolved, then the clocks should read different values when finally compared. So the end point events are NOT the same. AG

The clock readings don't define events any more that odometer readings define distances.

 

The obvious way to tell which is longer in proper time is to carry an ideal clock along the two paths and compare the measured intervals.  You could also measure the space distance X along the paths and and compute proper time S=\sqrt{T^2 - X^2}  where T is the coordinate time difference (in the same reference frame you measured distance).

Brent

You seem to defying basic physics if this is your claim. I don't deny that the original problem can be restated in a way which avoids acceleration, and IMO this is what you've done. 

But I've done more than that.  I've done it while maintaining exactly the same paradox.

So you admit you're defying the laws of physics? AG 

Why?  Because something "seemed to defy basic physics" to you?  Can you explain how basic physics is defied?

Brent

[Alan Grayson]

On Wednesday, September 10, 2025 at 1:55:07 PM UTC-6 Brent Meeker wrote:

On 9/9/2025 11:52 PM, Alan Grayson wrote:

On Wednesday, September 10, 2025 at 12:08:39 AM UTC-6 Brent Meeker wrote:

On 9/9/2025 10:36 PM, Alan Grayson wrote:

On Tuesday, September 9, 2025 at 11:25:37 PM UTC-6 Brent Meeker wrote:

...

If both twins are accelerating, then you've redefined the TP. 

No.  But Blue does not accelerate while Red travels out and back.

But earlier you claimed both Red and Blue have the same acceleration, and that's what your diagram shows. AG

But not  "while Red travels out and back".  Which IS what my diagram shows.  You can imagine Blue as decelerating, landing on Earth, and then accelerating to join up again as Red comes back by.  Next time read the whole sentence, not just up until a question occurs to you.

 Thank you for replying, but please cease adding crap to it. I read it, and looked at your plot. It sure seemed as if both are accelerating. AG 

If you have two paths in spacetime, starting at the same point and ending at the same point, or at a  different point, how can you tell which is longer? AG 

These are paths in spacetime.  They start and end at the same event, a point in 4-space. 

If the paradox is resolved, then the clocks should read different values when finally compared. So the end point events are NOT the same. AG

The clock readings don't define events any more that odometer readings define distances.

That's what I thought. But when traveling twin returns, it's to a different event because the time label on the coordinate differs from the  event. So the return event is not the same as starting event when the twins are juxtaposed. AG 

The obvious way to tell which is longer in proper time is to carry an ideal clock along the two paths and compare the measured intervals.  You could also measure the space distance X along the paths and and compute proper time S=\sqrt{T^2 - X^2}  where T is the coordinate time difference (in the same reference frame you measured distance).

Brent

You seem to defying basic physics if this is your claim. I don't deny that the original problem can be restated in a way which avoids acceleration, and IMO this is what you've done. 

But I've done more than that.  I've done it while maintaining exactly the same paradox.

So you admit you're defying the laws of physics? AG 

Why?  Because something "seemed to defy basic physics" to you?  Can you explain how basic physics is defied?

I now see you as a magician. In your special model of the problem, your traveling twin somehow turns around without acceleration. Or maybe you mean one of the triplets. In physics as I understand it, change in direction is acceleration and that's how the traveling twin returns. AG 

Brent

[Brent Meeker]

On 9/10/2025 1:30 PM, Alan Grayson wrote:

On Wednesday, September 10, 2025 at 1:55:07 PM UTC-6 Brent Meeker wrote:

On 9/9/2025 11:52 PM, Alan Grayson wrote:

On Wednesday, September 10, 2025 at 12:08:39 AM UTC-6 Brent Meeker wrote:

On 9/9/2025 10:36 PM, Alan Grayson wrote:

On Tuesday, September 9, 2025 at 11:25:37 PM UTC-6 Brent Meeker wrote:

...

If both twins are accelerating, then you've redefined the TP. 

No.  But Blue does not accelerate while Red travels out and back.

But earlier you claimed both Red and Blue have the same acceleration, and that's what your diagram shows. AG

But not  "while Red travels out and back".  Which IS what my diagram shows.  You can imagine Blue as decelerating, landing on Earth, and then accelerating to join up again as Red comes back by.  Next time read the whole sentence, not just up until a question occurs to you.

 Thank you for replying, but please cease adding crap to it. I read it, and looked at your plot. It sure seemed as if both are accelerating. AG 

If you have two paths in spacetime, starting at the same point and ending at the same point, or at a  different point, how can you tell which is longer? AG 

These are paths in spacetime.  They start and end at the same event, a point in 4-space. 

If the paradox is resolved, then the clocks should read different values when finally compared. So the end point events are NOT the same. AG

The clock readings don't define events any more that odometer readings define distances.

That's what I thought. But when traveling twin returns, it's to a different event because the time label on the coordinate differs from the  event. So the return event is not the same as starting event when the twins are juxtaposed. AG 

You don't understand what an event is.  It's a point in spacetime.  It's independent of what coordinate labeling exists, just as a point on Earth is independent of what map you use.  It's a physical thing.  Have you never read a book on relativity?

The obvious way to tell which is longer in proper time is to carry an ideal clock along the two paths and compare the measured intervals.  You could also measure the space distance X along the paths and and compute proper time S=\sqrt{T^2 - X^2}  where T is the coordinate time difference (in the same reference frame you measured distance).

Brent

You seem to defying basic physics if this is your claim. I don't deny that the original problem can be restated in a way which avoids acceleration, and IMO this is what you've done. 

But I've done more than that.  I've done it while maintaining exactly the same paradox.

So you admit you're defying the laws of physics? AG 

Why?  Because something "seemed to defy basic physics" to you?  Can you explain how basic physics is defied?

I now see you as a magician. In your special model of the problem, your traveling twin somehow turns around without acceleration. Or maybe you mean one of the triplets. In physics as I understand it, change in direction is acceleration and that's how the traveling twin returns. AG 

The point of the slingshot turnaround is that is shows the effect on the clock has nothing to do with acceleration.  The whole trip is in free-fall.  An accelerometer would read zero the whole time.

Brent

[Alan Grayson]

On Wednesday, September 10, 2025 at 7:19:01 PM UTC-6 Brent Meeker wrote:

On 9/10/2025 1:30 PM, Alan Grayson wrote:

On Wednesday, September 10, 2025 at 1:55:07 PM UTC-6 Brent Meeker wrote:

On 9/9/2025 11:52 PM, Alan Grayson wrote:

On Wednesday, September 10, 2025 at 12:08:39 AM UTC-6 Brent Meeker wrote:

On 9/9/2025 10:36 PM, Alan Grayson wrote:

On Tuesday, September 9, 2025 at 11:25:37 PM UTC-6 Brent Meeker wrote:

...

If both twins are accelerating, then you've redefined the TP. 

No.  But Blue does not accelerate while Red travels out and back.

But earlier you claimed both Red and Blue have the same acceleration, and that's what your diagram shows. AG

But not  "while Red travels out and back".  Which IS what my diagram shows.  You can imagine Blue as decelerating, landing on Earth, and then accelerating to join up again as Red comes back by.  Next time read the whole sentence, not just up until a question occurs to you.

 Thank you for replying, but please cease adding crap to it. I read it, and looked at your plot. It sure seemed as if both are accelerating. AG 

If you have two paths in spacetime, starting at the same point and ending at the same point, or at a  different point, how can you tell which is longer? AG 

These are paths in spacetime.  They start and end at the same event, a point in 4-space. 

If the paradox is resolved, then the clocks should read different values when finally compared. So the end point events are NOT the same. AG

The clock readings don't define events any more that odometer readings define distances.

That's what I thought. But when traveling twin returns, it's to a different event because the time label on the coordinate differs from the  event. So the return event is not the same as starting event when the twins are juxtaposed. AG 

You don't understand what an event is.  It's a point in spacetime.  It's independent of what coordinate labeling exists, just as a point on Earth is independent of what map you use.  It's a physical thing.  Have you never read a book on relativity?

Every point in spacetime has a label, called cooordinates. So every event happens at some label. If not that, then what?  I see this in those spacetime diagrams. If their clocks disagree at the reunion, how do you define "the event"? AG

The obvious way to tell which is longer in proper time is to carry an ideal clock along the two paths and compare the measured intervals.  You could also measure the space distance X along the paths and and compute proper time S=\sqrt{T^2 - X^2}  where T is the coordinate time difference (in the same reference frame you measured distance).

Brent

You seem to defying basic physics if this is your claim. I don't deny that the original problem can be restated in a way which avoids acceleration, and IMO this is what you've done. 

But I've done more than that.  I've done it while maintaining exactly the same paradox.

So you admit you're defying the laws of physics? AG 

Why?  Because something "seemed to defy basic physics" to you?  Can you explain how basic physics is defied?

I now see you as a magician. In your special model of the problem, your traveling twin somehow turns around without acceleration. Or maybe you mean one of the triplets. In physics as I understand it, change in direction is acceleration and that's how the traveling twin returns. AG 

The point of the slingshot turnaround is that is shows the effect on the clock has nothing to do with acceleration.  The whole trip is in free-fall.  An accelerometer would read zero the whole time.

How does the traveling twin separate from his twin? In free fall? I think you've changed the TP model to get the result you desire. AG 

Brent

[Brent Meeker]

Wow, what a clever solution.  If the clocks read differently when right beside one another there's no paradox.  Somebody go wake up Albert.

Brent

On 9/9/2025 11:52 PM, Alan Grayson wrote:

[Brent Meeker]

On 9/10/2025 6:41 PM, Alan Grayson wrote:

Every point in spacetime has a label, called cooordinates. So every event happens at some label. 

But the label is arbitrary.  You draw a different map and give it a different label.  But it's the same event.  It doesn't become two different events because two different brought different clocks to it.

If not that, then what?  I see this in those spacetime diagrams. If their clocks disagree at the reunion, how do you define "the event"? AG

It's the event of Red and Blue meeting again.

Brent