Lorentz Contraction: Exploring Standard Relativity & Bell's Paradox

[atyy]

I guess cfrogue's confusion comes from something like:

1) In SR it is not moving objects that Lorentz contract, but space itself.
2) The distance between the ships is moving, so it should Lorentz contract, since space itself contracts.
3) Yet this moving distance remains the same, thus we have obtained a contradiction.

Well, something must be wrong above. I myself never use this space itself contracts or that time itself dilates, so I don't really have an intuition where the error occurs, but maybe cfrogue can confirm that I have summarized his confusion correctly, and someone else can write the correct version of the above statements 1-3.

 

[matheinste]

Fine, I will argue my point from SR here.

Any logic that assumes the ships will have a greater distance after acceleration stops from the accelerating frame will contradict the launch frame that sees the distance as constant and then contracted by the normal LT calculations. You must have logic and equations to support this position.

I want to see all this.

Why, so you can ignore them.

Matheinste.

 

[Al68]

More directly, if you support the proposition that the accelerating ships will retain their > d positions after the acceleration stops, you will find that the launch frame will completely disagree with your assessment since the ships maintained their distance.

So, where is your math to reconcile this?

Supporting that proposition is equivalent to supporting the proposition that the ships retain their velocity relative to earth, since their "> d positions" are due only to relative velocity.

Maybe this will be more clear:

1) The distance between the ships remained constant in the launch frame.
2) The distance between the ships increased in the new co-moving ship frame.
3) The distance between the ships after acceleration is shorter in the launch frame than it is in the co-moving frame.

The math is simple:

Distance between the ships in launch frame equals distance between the ships in co-moving frame times sqrt(1-v^2/c^2). It's that simple.

The distance between the ships will be different in the launch frame than in a co-moving frame as long as there is relative velocity between them.

 

[JesseM]

If cfrogue wants some more detailed math, here it is:

Say that in the launch frame, the left ship is at position x=0 at time t=0, and the right ship is at position x=d at time t=0. At t=0 they both begin to accelerate with constant coordinate acceleration a in the launch frame, and they both stop accelerating at time t=t1 in the launch frame. Then prior to t1, v(t) for both ships will be given by v(t) = a*t, and x(t) for the left ship will be given by x(t) = (1/2)*a*t^2 while x(t) for the right ship will be given by x(t) = (1/2)*a*t^2 + d. So at time t1 when they stop acceleration, both ships will have a velocity in the launch frame of a*t1, and the position of the left ship will be (1/2)*a*t1^2 while the position of the right ship will be (1/2)*a*t1^2 + d. After that they both move at constant velocity v1 = a*t1, so after t1 x(t) for the left ship will be given by x(t) = v1*t - v1*t1 + (1/2)*a*t1^2 while x(t) for the right ship will be given by x(t) = v1*t - v1*t1 + (1/2)*a*t1^2 + d.

Now we can use the formula I derived in post 134 of the twins thread, which tells us that if we pick the event E of the left ship stopping acceleration, and pick another event E' on the worldine of the right ship which is simultaneous with E in the ship's own inertial rest frame, in the launch frame where the right ship is moving at speed v1 and was at a distance of d from E when it happened, the coordinate time between E and E' must be (d*v1/c^2)*gamma^2, so if E occurred at time t1, E' occurred at time t = t1 + (d*v1/c^2)*gamma^2. Plugging this time into the x(t) for the right ship, E' must have occurred at a position of x = v1*[t1 + (d*v1/c^2)*gamma^2] - v1*t1 + (1/2)*a*t1^2 + d = (d*v1^2/c^2)*gamma^2 + (1/2)*a*t1^2 + d.

So, coordinates of E in the launch frame:
x = (1/2)*a*t1^2, t = t1.
Since v1 = a*t1, the coordinates of E can be rewritten as:
x = (1/2)*v1*t1, t = t1

Coordinates of E' in the launch frame:
x = (d*v1^2/c^2)*gamma^2 + (1/2)*a*t1^2 + d, t = t1 + (d*v1/c^2)*gamma^2.

Since v1 = a*t1, the coordinates of E' can be rewritten as:
x = (d*v1^2/c^2)*gamma^2 + (1/2)*v1*t1 + d, t = t1 + (d*v1/c^2)*gamma^2

And remember, E was an event on the worldline of the left ship--the event of it stopping its acceleration--and E' was an event on the worldline of the right ship--the event that happened simultaneously with E in the right ship's inertial rest frame once it finished accelerating. So, now we can figure out the x' and t' coordinates of E and E' in the rest frame of the ships after they stop accelerating, using the Lorentz transformation, and the difference in x' coordinates of the two events will be the distance between the ships in their final inertial rest frame once the left ship stops accelerating.

The coordinates of E in the ship's final inertial frame are:

x' = gamma*((1/2)*v1*t1 - v1*t1) = gamma*(-1/2)*(v1*t1)

t' = gamma*(t1 - (1/2)*t1*v1^2/c^2)

The coordinates of E' in the ship's final inertial frame are:

x' = gamma*([(d*v1^2/c^2)*gamma^2 + (1/2)*v1*t1 + d] - v1*[t1 + (d*v1/c^2)*gamma^2])
= gamma*(d + (-1/2)*v1*t1)

t' = gamma*([t1 + (d*v1/c^2)*gamma^2] - v1/c^2*[(d*v1^2/c^2)*gamma^2 + (1/2)*v1*t1 + d])
= gamma*(t1 + gamma^2*(d*v1/c^2)*[1 - v1^2/c^2] + (-1/2)*t1*v1^2/c^2 - d*v1/c^2)
= gamma*(t1 + [1/(1 - v1^2/c^2)]*(d*v1/c^2)*[1 - v1^2/c^2] - (1/2)*t1*v1^2/c^2 - d*v1/c^2)
= gamma*(t1 - (1/2)*t1*v1^2/c^2)

So, you can see here from direct calculation that both E and E' do have the same t' coordinate in their final rest frame, regardless of whether you trusted my derivation in post 134 of the twins thread. You can also see that the difference in x' coordinate between E and E' is gamma*d, so this must be the distance between the two ships in their final rest frame, which is greater than the distance between them in the launch frame by a factor of gamma.

 

[cfrogue]

What's paradoxical about that? The distance in the non-inertial frame began to grow larger than d from the moment they began accelerating. Anyway, you don't even need to consider a non-inertial frame here; just using the Lorentz transformation and their x(t) functions in the launch frame, you can easily show that the distance between them in other inertial frames can be larger than d. Different inertial frames always disagree on the distance between a pair of objects, just like they disagree on the time between a pair of events, that's just a feature of how the Lorentz transformation works.

That was not the issue.

Does the distance remain after the acceleration.

If so, then the accelerating frame will believe the distance has grown and the launch frame after the acceleration will believe the distance has contracted as d/gamma.

But, the launch frame and prior accelerating frame will disagree about this d.

 

[cfrogue]

I guess cfrogue's confusion comes from something like:

1) In SR it is not moving objects that Lorentz contract, but space itself.
2) The distance between the ships is moving, so it should Lorentz contract, since space itself contracts.
3) Yet this moving distance remains the same, thus we have obtained a contradiction.

Well, something must be wrong above. I myself never use this space itself contracts or that time itself dilates, so I don't really have an intuition where the error occurs, but maybe cfrogue can confirm that I have summarized his confusion correctly, and someone else can write the correct version of the above statements 1-3.

Actually, I am concerned about the distance between the ships after the accelerations stops from both POV's.

 

[cfrogue]

That quote is from p. 4, and if you look at the context it's clear that S is what we have been calling the "launch frame". Do you disagree?

I agree with you

 

[cfrogue]

Supporting that proposition is equivalent to supporting the proposition that the ships retain their velocity relative to earth, since their "> d positions" are due only to relative velocity.

Maybe this will be more clear:

1) The distance between the ships remained constant in the launch frame.
2) The distance between the ships increased in the new co-moving ship frame.
3) The distance between the ships after acceleration is shorter in the launch frame than it is in the co-moving frame.

The math is simple:

Distance between the ships in launch frame equals distance between the ships in co-moving frame times sqrt(1-v^2/c^2). It's that simple.

The distance between the ships will be different in the launch frame than in a co-moving frame as long as there is relative velocity between them.

The co-moving frame is gone after the acceleration discontinues.

Besides, one does not need a co-moving frame after acceleration is done since the two ships will be in the same frame at that point anyway.

Now, the question is using only ships1 and ship2, is the distance between them > d once the acceleration discontinues from their perspective?

 

[cfrogue]

If cfrogue wants some more detailed math, here it is:

Say that in the launch frame, the left ship is at position x=0 at time t=0, and the right ship is at position x=d at time t=0. At t=0 they both begin to accelerate with constant coordinate acceleration a in the launch frame, and they both stop accelerating at time t=t1 in the launch frame. Then prior to t1, v(t) for both ships will be given by v(t) = a*t, and x(t) for the left ship will be given by x(t) = (1/2)*a*t^2 while x(t) for the right ship will be given by x(t) = (1/2)*a*t^2 + d. So at time t1 when they stop acceleration, both ships will have a velocity in the launch frame of a*t1, and the position of the left ship will be (1/2)*a*t1^2 while the position of the right ship will be (1/2)*a*t1^2 + d. After that they both move at constant velocity v1 = a*t1, so after t1 x(t) for the left ship will be given by x(t) = v1*t - v1*t1 + (1/2)*a*t1^2 while x(t) for the right ship will be given by x(t) = v1*t - v1*t1 + (1/2)*a*t1^2 + d.

I am confused. The SR acceleration equations give the following for the launch frame.
v(t) = a t / sqrt[ 1 + (a t)^2 ]
x(t) = 1/a [ sqrt( 1 + (a t)^2 ) - 1 ]

 

[JesseM]

That was not the issue.

Does the distance remain after the acceleration.

In every inertial frame, as well as the non-inertial frame I defined, the distance at the moment the last ship stops accelerating (since in many frames they do not stop accelerating simultaneously) will also be the distance at a later time when they are traveling inertially.

If so, then the accelerating frame will believe the distance has grown and the launch frame after the acceleration will believe the distance has contracted as d/gamma.

Huh? Why do you think answering yes to "does the distance remain after the acceleration" would imply the launch frame thinks the distance contracts? Again, in any of these frames, the distance at the moment the last ship stops accelerating is the same as the distance later when both ships are moving inertially, so since the launch frame sees the distance as d throughout the acceleration, it will continue to see the distance as d once they are both moving inertially. If you define d' as the distance between the ships in their new inertial rest frame after they stop accelerating (which is the same as the final distance between them in the non-inertial frame I defined), then it is true that the d in the launch frame is equal to d'/gamma, but no one will measure a distance of d/gamma (again assuming d is defined as the constant distance between them in the launch frame while they were accelerating).

But, the launch frame and prior accelerating frame will disagree about this d.

Please don't mix your uses of the symbol d! If d refers specifically to the distance in the launch frame, then you shouldn't refer to the distance in the non-inertial frame as "this d". Anyway, it is certainly true that the final distance d in the launch frame is different than the final distance d' in the non-inertial frame, but if this is all you're saying, then again I don't see why you consider this a paradox.

That quote is from p. 4, and if you look at the context it's clear that S is what we have been calling the "launch frame". Do you disagree?

I agree with you

So do you also agree that nowhere in the paper do they use any variables or do any calculations that involve a non-inertial frame?

I am confused. The SR acceleration equations give the following for the launch frame.
v(t) = a t / sqrt[ 1 + (a t)^2 ]
x(t) = 1/a [ sqrt( 1 + (a t)^2 ) - 1 ]

That's for constant proper acceleration, I was assuming constant coordinate acceleration, which makes the math simpler. The Bell spaceship paradox doesn't specify anything about the detailed nature of the acceleration beyond the fact that both ships have identical coordinate acceleration at any given moment in the launch frame, so assuming constant coordinate acceleration (which implies increasing proper acceleration) is fine.

 

[cfrogue]

Huh? Why do you think answering yes to "does the distance remain after the acceleration" would imply the launch frame thinks the distance contracts? Again, in any of these frames, the distance at the moment the last ship stops accelerating is the same as the distance later when both ships are moving inertially, so since the launch frame sees the distance as d throughout the acceleration, it will continue to see the distance as d once they are both moving inertially. If you define d' as the distance between the ships in their new inertial rest frame after they stop accelerating (which is the same as the final distance between them in the non-inertial frame I defined), then it is true that the d in the launch frame is equal to d'/gamma, but no one will measure a distance of d/gamma (again assuming d is defined as the constant distance between them in the launch frame while they were accelerating).

How can you do this?

The launch frame does not know a d'.
How does the launch frame conclude d'?

Please don't mix your uses of the symbol d! If d refers specifically to the distance in the launch frame, then you shouldn't refer to the distance in the non-inertial frame as "this d". Anyway, it is certainly true that the final distance d in the launch frame is different than the final distance d' in the non-inertial frame, but if this is all you're saying, then again I don't see why you consider this a paradox.

Because, the launch frame will not longer know the correct value to perform d/gamma.
This is simple.

So do you also agree that nowhere in the paper do they use any variables or do any calculations that involve a non-inertial frame?

I agree with you.

That's for constant proper acceleration, I was assuming constant coordinate acceleration, which makes the math simpler. The Bell spaceship paradox doesn't specify anything about the detailed nature of the acceleration beyond the fact that both ships have identical coordinate acceleration at any given moment in the launch frame, so assuming constant coordinate acceleration (which implies increasing proper acceleration) is fine.

I do not understand what you mean with this.

 

[JesseM]

How can you do this?

The launch frame does not know a d'.
How does the launch frame conclude d'?

d' is not measured in the launch frame! What I wrote was "If you define d' as the distance between the ships in their new inertial rest frame after they stop accelerating"

Because, the launch frame will not longer know the correct value to perform d/gamma.
This is simple.

Frames are just coordinate systems and don't "know" anything about other frames, but intelligent beings such as ourselves can certainly relate distances in one frame to distances in another, that's what the length contraction equation $$l' = l/\sqrt{1 - v^2/c^2}$$ does after all.

That's for constant proper acceleration, I was assuming constant coordinate acceleration, which makes the math simpler. The Bell spaceship paradox doesn't specify anything about the detailed nature of the acceleration beyond the fact that both ships have identical coordinate acceleration at any given moment in the launch frame, so assuming constant coordinate acceleration (which implies increasing proper acceleration) is fine.

I do not understand what you mean with this.

Do you understand the difference between coordinate acceleration and proper acceleration? Coordinate acceleration is just the rate the coordinate velocity is changing (derivative of v(t) in that frame), and coordinate velocity is just the rate the coordinate position is changing with coordinate time (derivative of x(t) in that frame). Proper acceleration at any given moment is defined as an object's coordinate acceleration in the object's instantaneous inertial rest frame at that moment, not in any other frame. If an object has constant proper acceleration, its coordinate acceleration in the launch frame is decreasing, as you can see from the http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html equation v = at / sqrt[1 + (at/c)^2]. But there is nothing in the statement of the Bell spaceship problem that says they must have constant proper acceleration, it only says their coordinate acceleration must be identical.

 

[cfrogue]

d' is not measured in the launch frame! What I wrote was "If you define d' as the distance between the ships in their new inertial rest frame after they stop accelerating"

Here, instead of all this background noise, do you believe this extra distance holds between the ships after the acceleration discontinues yes or no.

 

[JesseM]

Here, instead of all this background noise, do you believe this extra distance holds between the ships after the acceleration discontinues yes or no.

In any frame where the distance increases as they accelerate (i.e. not the launch frame), the distance will stay increased after the acceleration discontinues, yes. I already told you that in post 290:

Again, in any of these frames, the distance at the moment the last ship stops accelerating is the same as the distance later when both ships are moving inertially, so since the launch frame sees the distance as d throughout the acceleration, it will continue to see the distance as d once they are both moving inertially.

 

[Dmitry67]

That what Max Tegmark was talking about the 'mere labels': when we say

F=ma

We think that Force F equal to (m)*** multiplied by (a)cceleration.
The same formula

Ы=Я*Ж

wont tell you anything
But in TOE if won't be important.

 

[Al68]

Does the distance remain after the acceleration.

If so, then the accelerating frame will believe the distance has grown and the launch frame after the acceleration will believe the distance has contracted as d/gamma.

But, the launch frame and prior accelerating frame will disagree about this d.

Of course the distance will be different for different frames. That's what length contraction is.

(1) Constant distance between the ships in the launch frame is stipulated in the scenario.
(2) Also stipulated is that the velocity of the ships will increase relative to the launch frame.
(3) The ratio (gamma) between "ship frame distance" and "launch frame distance" increases with velocity, ie "Ship frame distance" > "launch frame distance" for any velocity > 0.

1+2+3= Distance between the ships in the ships' final inertial frame is greater than the distance between the ships in the launch frame.

Now, the question is using only ships1 and ship2, is the distance between them > d once the acceleration discontinues from their perspective?

Absolutely, assuming that d is the distance between the ships in the launch frame.

For example, if the ships are 600 km apart in the launch frame, accelerate to 0.8 c while maintaining the 600 km distance in the launch frame, then the distance between the ships in their inertial frame after they cut their engines will be 1000 km.

 

[cfrogue]

In any frame where the distance increases as they accelerate (i.e. not the launch frame), the distance will stay increased after the acceleration discontinues, yes. I already told you that in post 290:

Then the launch frame cannot correctly calculate LY once the acceleeration is finished since the launch frame still has the original d as the distance between the ships.

 

[cfrogue]

Of course the distance will be different for different frames. That's what length contraction is.

(1) Constant distance between the ships in the launch frame is stipulated in the scenario.
(2) Also stipulated is that the velocity of the ships will increase relative to the launch frame.
(3) The ratio (gamma) between "ship frame distance" and "launch frame distance" increases with velocity, ie "Ship frame distance" > "launch frame distance" for any velocity > 0.

1+2+3= Distance between the ships in the ships' final inertial frame is greater than the distance between the ships in the launch frame.
Absolutely, assuming that d is the distance between the ships in the launch frame.

Then how does the launch frame correctly apply LT after the acceleration.

Should the launch frame calculate a new d and abandon the old d?

 

[A.T.]

Then how does the launch frame correctly apply LT after the acceleration.

In the same way as during the acceleration, based on the current speed of the rockets.

Should the launch frame calculate a new d and abandon the old d?

Isn't d constant?

assuming that d is the distance between the ships in the launch frame

 

[cfrogue]

In the same way as during the acceleration, based on the current speed of the rockets.

No, because the distance between the ships from the launch frame does not agree with the ships assessment.

 

[A.T.]

No, because the distance between the ships from the launch frame does not agree with the ships assessment.

Neither did it during the acceleration. So?

Maybe you should explain what you mean by '...the launch frame correctly apply LT...'. Apply LT to calculate what?

 

[cfrogue]

Neither did it during the acceleration. So?

Maybe you should explain what you mean by '...the launch frame correctly apply LT...'. Apply LT to calculate what?

It is OK what happens during acceleration, but after, the original d cannot be used for LT calculations since it will be wrong.

But, I do not see this as a problem.

This would imply that the launch frame would apply a new d based on the acceleration equations rather than the old d prior to acceleration.

I do not see this as a problem, just a necessary calculation.

As I see it, this thread is solved.

 

[A.T.]

It is OK what happens during acceleration, but after, the original d cannot be used for LT calculations since it will be wrong.

Isn't d the distance between the ships in the launch frame? It is constant:

d before acceleration = d during acceleration = d after acceleration

But, I do not see this as a problem.

Me neither

 

[JesseM]

Then the launch frame cannot correctly calculate LY once the acceleeration is finished since the launch frame still has the original d as the distance between the ships.

You mean LT? Frames don't "calculate" anything about other frames, only people do that. A person at rest in the launch frame can certainly apply LT to distances/coordinates in the launch frame to find distances/coordinates in some other frame. The standard length contraction equation tells you that an object moving inertially which has length d in your frame, and which is moving at speed v in your frame, will have length d' = d*gamma in its own rest frame. That applies to the two rockets with the string between them once they are both moving inertially--why wouldn't it?

 

[cfrogue]

You mean LT? Frames don't "calculate" anything about other frames, only people do that. A person at rest in the launch frame can certainly apply LT to distances/coordinates in the launch frame to find distances/coordinates in some other frame. The standard length contraction equation tells you that an object moving inertially which has length d in your frame, and which is moving at speed v in your frame, will have length d' = d*gamma in its own rest frame. That applies to the two rockets with the string between them once they are both moving inertially--why wouldn't it?

You mean LT? Frames don't "calculate" anything about other frames, only people do that.
LOL

Yes, I understand the above. But is seems the logic of this thread dictates that the original distance d is not correct after the acceleration.

Thus, when the "launch frame calculates" the length contraction of the ships after the acceleration is terminated, it will be using the wrong d, no?

 

[Dale]

No. At least not unless you do the math wrong.

 

[cfrogue]

No. At least not unless you do the math wrong.

OK, so you are of the view that the ships expand past d while they accelerate and then snap back to d after the acceleration.

That was the way I saw the math, but that was termed magic on this thread.

 

[Dale]

OK, so you are of the view that the ships expand past d while they accelerate and then snap back to d after the acceleration.

No.

That was the way I saw the math, but that was termed magic on this thread.

I wouldn't call it magic, I would call it doing the math wrong.

 

[JesseM]

Yes, I understand the above. But is seems the logic of this thread dictates that the original distance d is not correct after the acceleration.

Why? You never explain any of the "logic" you are talking about.

Thus, when the "launch frame calculates" the length contraction of the ships after the acceleration is terminated, it will be using the wrong d, no?

Why? A person in one frame who sees an object moving at speed v with length d in his frame should understand that the object's length in its own rest frame is greater than d by a factor of gamma, not contracted to a length smaller than d in its own rest frame. After all, the length contraction equation says that if an object has length L in its own rest frame, then its length L' in a frame where it's moving at speed v is given by L' = L / gamma, so if you know L' (in this case d) and you want to find L, just multiply both sides by gamma to get the equation L = L' * gamma.

If you disagree, perhaps you could give the actual equation you think the launch frame would be using that would involve "the wrong d".

 

[cfrogue]

Why? You never explain any of the "logic" you are talking about."

Sure I did. I said the increasing relative v was relative to the instantaneous at rest frame and once the acceleration stopped, then this instantaneous at rest frame goes away. It is this instantaneous at rest frame where the increasing distance between the ships is calculated.
So I felt, once it is gone, then this increased distance is gone.

Why? A person in one frame who sees an object moving at speed v with length d in his frame should understand that the object's length in its own rest frame is greater than d by a factor of gamma, not contracted to a length smaller than d in its own rest frame. After all, the length contraction equation says that if an object has length L in its own rest frame, then its length L' in a frame where it's moving at speed v is given by L' = L / gamma, so if you know L' (in this case d) and you want to find L, just multiply both sides by gamma to get the equation L = L' * gamma.

If you disagree, perhaps you could give the actual equation you think the launch frame would be using that would involve "the wrong d".

Do you recall the latest paper and solution has the distance between the ships increasing?
Now, the question came up as to whether this increased distance remained after the acceleration ended.

I think you took a position on this d', do you recall what it was?

 

[A.T.]

...then this instantaneous at rest frame goes away...

A frame of reference 'goes away'? I think you have some misconceptions about what a frame of reference is.

Now, the question came up as to whether this increased distance remained after the acceleration ended.

I think that was answered several times already, for example:

In every inertial frame, as well as the non-inertial frame I defined, the distance at the moment the last ship stops accelerating (since in many frames they do not stop accelerating simultaneously) will also be the distance at a later time when they are traveling inertially.

How many reiterations do you need to acknowledge something?

 

[cfrogue]

A frame of reference 'goes away'? I think you have some misconceptions about what a frame of reference is.

Maybe so.

Can you explain how the instantaneous at rest frame remains once the acceleration stops?

Also, can you tell me the v relative to the instantaneous at rest frame once the acceleration discontinues?

 

[A.T.]

Can you explain how the instantaneous at rest frame remains once the acceleration stops?

Also, can you tell me the v relative to the instantaneous at rest frame once the acceleration discontinues?

What does 'remain' mean in respect to a frame of reference? What is the 'instantaneous at rest frame'? What is v?

 

[cfrogue]

What does 'remain' mean in respect to a frame of reference? What is the 'instantaneous at rest frame'? What is v?

It is all in this paper from the first couple of pages of this thread.
http://arxiv.org/PS_cache/arxiv/pdf/0906/0906.1919v2.pdf

 

[A.T.]

Also, can you tell me the v relative to the instantaneous at rest frame once the acceleration discontinues?

What does 'remain' mean in respect to a frame of reference? What is the 'instantaneous at rest frame'? What is v?

It is all in this paper from the first couple of pages of this thread.
http://arxiv.org/PS_cache/arxiv/pdf/0906/0906.1919v2.pdf

The paper doesn't say what 'frame remains' or 'frame goes away' means. You make this stuff up.

I assume by 'instantaneous at rest frame' you mean the instantaneous rest frame of one of the rockets?

But what is v? The velocity of the rocket in it's own instantaneous rest frame is 0, per definition.

So what is the point of your question?