16 hours ago, md65536 said:

what I just described doesn't seem to make sense from the tracks frame.

I think I figured out the basic idea in the track frame, for Born rigidity or anything else that approximately maintains the proper length of the train with non-instantaneous acceleration.

In the track frame, the back of the train always accelerates at a higher rate than the front, so that as velocity increases, the train contracts. Then when acceleration stops, the back of the train stops accelerating first, when it reaches velocity v. The rest of the train is moving slower and continues to accelerate (and contract), with the front of the train being the last to stop accelerating, at which point the whole train is moving at v and is length-contracted by gamma. This is consistent with allowing the whole train to stop accelerating simultaneously in a moving frame (v's frame, if I got it right).

Edited December 18, 20223 yr by md65536

50 minutes ago, md65536 said:

In the track frame, the back of the train always accelerates at a higher rate than the front, so that as velocity increases, the train contracts. Then when acceleration stops, the back of the train stops accelerating first, when it reaches velocity v. The rest of the train is moving slower and continues to accelerate (and contract), with the front of the train being the last to stop accelerating, at which point the whole train is moving at v and is length-contracted by gamma.

Right. And you proved that with your comment about the cars stopping simultaneously in the frame moving at the final speed vfinal relative to the tracks.

50 minutes ago, md65536 said:

This is consistent with allowing the whole train to stop accelerating simultaneously in a moving frame (v's frame, if I got it right).

Exactly. And therefore the ending solution in the vfinal frame must be the same as the beginning solution in the track frame, except the n's are reversed (n=0 at the front) and the formula applies to negative values of t (relative to the stopping time). Now write down that solution (it's in the "primed" frame) and transform it into the track frame to see what it looks like! 😋

Edited December 18, 20223 yr by Lorentz Jr

1 hour ago, md65536 said:

I think I figured out the basic idea in the track frame, for Born rigidity or anything else that approximately maintains the proper length of the train with non-instantaneous acceleration.

In the track frame, the back of the train always accelerates at a higher rate than the front, so that as velocity increases, the train contracts. Then when acceleration stops, the back of the train stops accelerating first, when it reaches velocity v. The rest of the train is moving slower and continues to accelerate (and contract), with the front of the train being the last to stop accelerating, at which point the whole train is moving at v and is length-contracted by gamma. This is consistent with allowing the whole train to stop accelerating simultaneously in a moving frame (v's frame, if I got it right).

I think in moving frame the train should stretch again. Therefore forward part of the train stoped earlier in S'.Because forward part of the train (in S) is backward part of the train in S'. Maybe you are correct but then you should correctly remake the model.

12 hours ago, DimaMazin said:

I think in moving frame the train should stretch again. Therefore forward part of the train stoped earlier in S'.Because forward part of the train (in S) is backward part of the train in S'. Maybe you are correct but then you should correctly remake the model.

Yes, I agree. Sorry for the confusion, I've ended up discussing what would happen in both the instantaneous and gradual acceleration cases, and the whole train stopping simultaneously only applies to the gradual case.

In the instantaneous case, yes the train stretches again, because the parts of the train don't stop simultaneously in S', and the "car 0" end (the back end in S) continues moving for a short time while the other end has stopped, in the S' frame.

 

12 hours ago, Lorentz Jr said:

Exactly. And therefore the ending solution in the vfinal frame must be the same as the beginning solution in the track frame, except the n's are reversed (n=0 at the front) and the formula applies to negative values of t (relative to the stopping time). Now write down that solution (it's in the "primed" frame) and transform it into the track frame to see what it looks like! 😋

Well it seems to work, but it's confusing. In this case of gradual acceleration, in the vfinal frame, what is observed is that we start with the train having constant velocity -v and is length contracted. Then to start the acceleration phase, the n=N end of the train begins to decelerate first, and the train begins to stretch very gradually at first. Other cars decelerate in turn until the n=0 car begins decelerating last. The n=0 car has the greatest proper acceleration (which all frames must agree on), but since it started last, it maintains a greater speed (with negative velocity) than the rest of the train in the vfinal frame during the entire acceleration phase. Then finally all cars arrive at the same velocity of 0 simultaneously, the train fully stretched to its full proper length.

This does sound exactly like a time reversal of what is seen in the track frame, with everyone agreeing that the n=0 end has the higher proper acceleration.

But everything I'm describing has a locally constant proper acceleration during the acceleration phase, and satisfies Born rigidity as far as I understand it. Are we talking about the same thing?

Another possible source of confusion is that constant proper acceleration isn't constant coordinate acceleration in any single frame, and I may have mistook one for the other in what you wrote.

Edited December 19, 20223 yr by md65536

6 hours ago, md65536 said:

Yes, I agree. Sorry for the confusion, I've ended up discussing what would happen in both the instantaneous and gradual acceleration cases, and the whole train stopping simultaneously only applies to the gradual case.

In the instantaneous case, yes the train stretches again, because the parts of the train don't stop simultaneously in S', and the "car 0" end (the back end in S) continues moving for a short time while the other end has stopped, in the S' frame.

 Yes, in gradual  negative acceleration the train stretches during long time.

On 12/18/2022 at 4:22 AM, md65536 said:

 

Interesting that the second form is what you posted in the very first post, and the first form I posted later, and they are the same but I didn't realize it. We were talking about the same thing all along.

Our non-simultaneity works exectly like Minkowski's model of light cone.
https://www.youtube.com/watch?v=0hj8ZCQtrpU

 

On 12/13/2022 at 11:03 AM, md65536 said:

 

I figure that the correct velocity should be greater than c for all v<c, otherwise the accelerations won't be simultaneous in any frame. As well, when v approaches 0, this velocity should approach infinity, because for vanishing v, the frame in which the accelerations are simultaneous approaches the track's frame.

I think that is important because we can imagine any non-simultaneity between two points on distance as velocity or as instant pure acceleration to the velocity. When we transform the non-simultaneity in S' then we get the same only future and past are opposite because the velosity is opposite in S'. We can transform other non-simultaneities into other velosities or into their instant pure accelerations when the velosities are bigger than c . Only then for transform in the S' we should use relativistic addition of velosities .

 

On 12/11/2022 at 10:26 AM, Mordred said:

 

 

to be honest though as far as the OP is concerned I'm still trying to fathom what he means by neutral simultaneity. Might just be a translation error  

 

For example body travels at v and has momentum m*gamma*v. Then let's define at what velocity  the body has 1/2 the momentum.

gamma*v/2=v/2(1-v²/c²)^1/2

v/2(1-v²/c²)^1/2=gamma'*u=u/(1-u²/c²)^1/2

u=v/(4-3v²/c²)^1/2

I don't understand how relativity works, but do you think the velosity is another in S'(the frame is frame of the body at v)? And if non-simultaneities of u(in S) and u'(in S') are simultaneous then why we cannot use them as neutral simultaneity?

8 minutes ago, DimaMazin said:

For example body travels at v and has momentum m*gamma*v. Then let's define at what velocity  the body has 1/2 the momentum.

gamma*v/2=v/2(1-v²/c²)^1/2

v/2(1-v²/c²)^1/2=gamma'*u=u/(1-u²/c²)^1/2

u=v/(4-3v²/c²)^1/2

I don't understand how relativity works, but do you think the velosity is another in S'(the frame is frame of the body at v)? And if non-simultaneities of u(in S) and u'(in S') are simultaneous then why we cannot use them as neutral simultaneity?

There is only one frame in your example. What you've calculated is, that if body A moves with velocity v and body B of the same mass moves with velocity u=v/(4-3v²/c²)^1/2, then momentum of the body B equals half of momentum of the body A. All in the same one frame.

1 hour ago, Genady said:

There is only one frame in your example. What you've calculated is, that if body A moves with velocity v and body B of the same mass moves with velocity u=v/(4-3v²/c²)^1/2, then momentum of the body B equals half of momentum of the body A. All in the same one frame.

When body A is moving with velocity v relative to frame S then it is observer in S' and has velocity o in S'. Body O is observer in S. What is velocity of body B in frame S' when it travels with velocity u=v(4-3v²/c²)^1/2 in S frame?

26 minutes ago, DimaMazin said:

When body A is moving with velocity v relative to frame S then it is observer in S' and has velocity o in S'. Body O is observer in S. What is velocity of body B in frame S' when it travels with velocity u=v(4-3v²/c²)^1/2 in S frame?

The origin of S moves with the speed -v in S'. The body B moves with the speed u in S. Use relativistic velocity addition formula: the speed of B in S' = (u-v)/(1-uv/c²).

6 hours ago, Genady said:

The origin of S moves with the speed -v in S'. The body B moves with the speed u in S. Use relativistic velocity addition formula: the speed of B in S' = (u-v)/(1-uv/c²).

Theoretically I should get the same module of the velosity in S', but my brain is not working.

7 hours ago, DimaMazin said:

Theoretically I should get the same module of the velosity in S'

I don't know what you mean here, i.e., what should be the same.

The body A has velocity v in S, and velocity v'=0 in S'.

The body B has velocity u=v/(4-3v²/c²)^1/2 in S, and velocity u'=(u-v)/(1-uv/c²) in S'.

18 hours ago, Genady said:

I don't know what you mean here, i.e., what should be the same.

The body A has velocity v in S, and velocity v'=0 in S'.

The body B has velocity u=v/(4-3v²/c²)^1/2 in S, and velocity u'=(u-v)/(1-uv/c²) in S'.

Thanks. That looks like triangle with sides gamma*v , gamma_u*u , and gamma_u'*u' .

If we have many travelers with velocities between 0 and v and we add their gamma*velocity between them then can we get circular arc?Is then gamma*v a diameter or chord of the arc?

4 hours ago, DimaMazin said:

Thanks. That looks like triangle with sides gamma*v , gamma_u*u , and gamma_u'*u' .

If we have many travelers with velocities between 0 and v and we add their gamma*velocity between them then can we get circular arc?Is then gamma*v a diameter or chord of the arc?

I am sorry, I don't understand your idea of a relation between these numbers and a triangle or an arc.

On 7/19/2023 at 2:02 PM, Genady said:

I am sorry, I don't understand your idea of a relation between these numbers and a triangle or an arc.

I incorrectly started to define. There is simple definition. Instant pure acseleration has double disareement in simultaneity. Therefore neutral simultaneity is negative relative to it.

t=t₀-(gamma-1)x/(gamma×v)

Time between events in the non-simultaneity is the same in 2 frames.

30 minutes ago, DimaMazin said:

I incorrectly started to define. There is simple definition. Instant pure acseleration has double disareement in simultaneity. Therefore neutral simultaneity is negative relative to it.

t=t₀-(gamma-1)x/(gamma×v)

Time between events in the non-simultaneity is the same in 2 frames.

I can't decipher this paragraph. Looks like a bad translation.

On 7/18/2023 at 2:18 PM, Genady said:

and velocity u'=(u-v)/(1-uv/c2) in S'.

Correct formula should be

u'=v-u/gamma

We can use it for velocities of non-simultaneouses ,wich are defined by MD65536.

2 hours ago, DimaMazin said:

Correct formula should be

u'=v-u/gamma

No. Check relativistic velocities addition.

On 8/14/2023 at 1:09 PM, Genady said:

No. Check relativistic velocities addition.

My definition is observing therefore the formula of scientists is disinformation.

Edited August 15, 20232 yr by DimaMazin

7 minutes ago, DimaMazin said:

My definition is observing therefore the formula of scientists is disinformation.

Mathematics is disinformation?

2 hours ago, Genady said:

Mathematics is disinformation?

What experiment proves the mathematics?

4 minutes ago, DimaMazin said:

What experiment proves the mathematics?

ALL experiments do.

1 hour ago, Genady said:

ALL experiments do.

We should consider million experiments simultaneously?

Just now, DimaMazin said:

We should consider million experiments simultaneously?

No, any one of them. I'd start with something simple.