Lorentz Contraction: Exploring Standard Relativity & Bell's Paradox

[cfrogue]

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?

You can read the paper and I did not make anything up.
I do not have any more questions. I am satisfied with some of the conclusions in this thread.

So, I have already posted I thought everything is done here.

 

[A.T.]

I do not have any more questions.

So you were actually asking what the velocity of something in it own instantaneous rest frame is? Sometimes if you formulate the question more exactly you see the answer yourself.

 

[JesseM]

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.

Every object has an instantaneous at rest frame. For an object moving inertially, it's just the same as its regular inertial rest frame. Remember the definition I gave for the non-inertial frame I gave before in post 263?

The simplest way to define a non-inertial rest frame for an object which accelerates is to have the frame's time coordinate match up with the object's proper time along its worldline, and then say that at any given event on the object's worldline, the non-inertial frame's definition of simultaneity and distance at the time of that event should match up with the object's instantaneous inertial rest frame at that event on its worldline. In this case, since the distance to the back ship increases in the instantaneous inertial rest frame of the front ship, if we define the front ship's non-inertial rest frame in this way, the distance will increase in the non-inertial frame. Then once the two ships stop accelerating, the non-inertial frame's definition of distance will match up with that of their inertial rest frame.

There is absolutely no logical reason why this definition would cease to function once the ships are moving inertially!

Of course you can also forget non-inertial frames and just consider an inertial frame different from the launch frame, like the inertial frame where the ships come to rest once they stop accelerating. In this frame the distance also increases, and then stops increasing (but doesn't decrease) once the ships stop accelerating.

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.

Of course it does, you haven't given any kind of argument as to why it wouldn't, you just keep asserting you think it wouldn't for no coherent reason.

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

Yes, I said that the d' after they stop accelerating will be gamma*d, where d is the distance between the ships in the launch frame. I also said that in the non-inertial frame defined above, the final d' once both ships have stopped accelerating is equal to the d' at the moment the last ship finishes accelerating. What's the problem?

 

[cfrogue]

Every object has an instantaneous at rest frame. For an object moving inertially, it's just the same as its regular inertial rest frame. Remember the definition I gave for the non-inertial frame I gave before in post 263?


There is absolutely no logical reason why this definition would cease to function once the ships are moving inertially!

OK, this means, S must apply a different definition for d after the acceleration.

Of course you can also forget non-inertial frames and just consider an inertial frame different from the launch frame, like the inertial frame where the ships come to rest once they stop accelerating. In this frame the distance also increases, and then stops increasing (but doesn't decrease) once the ships stop accelerating.

I spelled out two possible outcomes.

I said either the distance remains between the two ships as d' > d or the distance between the ships snaps back to the original d.

If it is the case that a new d' > d is the actual distance after the acceleration, then it is the case that the launch frame will have an incorrect d. Now, no one on this thread addressed this issue but me.

Of course it does, you haven't given any kind of argument as to why it wouldn't, you just keep asserting you think it wouldn't for no coherent reason.

I guess this means I am incoherent.

Yes, I said that the d' after they stop accelerating will be gamma*d, where d is the distance between the ships in the launch frame. I also said that in the non-inertial frame defined above, the final d' once both ships have stopped accelerating is equal to the d' at the moment the last ship finishes accelerating. What's the problem?

I do not have a problem with this.

But, you are discounting the model that the d' remains > d after the acceleration is over.

How do you decide this?

 

[Dale]

If it is the case that a new d' > d is the actual distance after the acceleration, then it is the case that the launch frame will have an incorrect d.

Please post a rigorous derivation of this. It should be pretty easy for us to see where you are making the mistake in your thought process.

 

[cfrogue]

Please post a rigorous derivation of this. It should be pretty easy for us to see where you are making the mistake in your thought process.

Who is us?

The paper defines a frame S'

a system S′ that moves with constant velocity v with respect to the system S.

This frame increases v as the acceleration proceeds.

That is, as the velocity in S increases, the distance between the spaceships in their rest system S′ increases.

Equation 4 describes the distance associated with S'.

More logic

Consequently, the rest frame length in the instantaneous rest system S′ must increase in accordance with Eq. (4), and get longer than the original rest frame length in system S.

Now, since S' is the relative v that increases over acceleration to S, then one can claim that S' remains after the acceleration and thus S will disagree with the distance of the ships as compared with S', ie the ships inertial frame after acceleration.
Therefore, each frame will judge a different distance between the ships.

Now, how do you resolve this.

http://arxiv.org/PS_cache/arxiv/pdf/0906/0906.1919v2.pdf

 

[JesseM]

OK, this means, S must apply a different definition for d after the acceleration.

What is S? The launch frame, or the non-inertial frame, or the inertial frame where the ships are at rest after finishing their acceleration? Either way, I don't see why "S must apply a different definition for d after the acceleration." If you want anyone to understand what you're thinking you need to E-X-P-L-A-I-N the logic, not just make assertions.

I spelled out two possible outcomes.

I said either the distance remains between the two ships as d' > d or the distance between the ships snaps back to the original d.

...in what frame? Please, don't make any statements about distances without specifying what frame you are talking about.

If it is the case that a new d' > d is the actual distance after the acceleration, then it is the case that the launch frame will have an incorrect d. Now, no one on this thread addressed this issue but me.

Perhaps because no one knows what you're talking about, since you never explain your reasoning in a way that's intelligible to anyone but yourself.

I guess this means I am incoherent.

I guess.

I do not have a problem with this.

But, you are discounting the model that the d' remains > d after the acceleration is over.

Do I need a reason to discount a "model" where a distance suddenly changes for no reason whatsoever? We assume objects move continuously in any inertial frame due to their velocity and acceleration in that frame (with acceleration always due to some force, and objects with no forces on them obeying Newton's first law and therefore continuing at constant velocity), a sudden change in distance would amount to an uncaused teleportation.

 

[cfrogue]

What is S? The launch frame, or the non-inertial frame, or the inertial frame where the ships are at rest after finishing their acceleration? Either way, I don't see why "S must apply a different definition for d after the acceleration." If you want anyone to understand what you're thinking you need to E-X-P-L-A-I-N the logic, not just make assertions.

I am not going to read the rest of your post until you read the paper.

S has a definite context in the paper.

 

[JesseM]

I am not going to read the rest of your post until you read the paper.

S has a definite context in the paper.

I read that paper earlier, you hadn't mentioned it in any of your recent posts to me though so I didn't know you were still referring to it. In that paper S simply refers to the launch frame. So, as I expected, "S must apply a different definition for d after the acceleration" makes absolutely no sense whatsoever. If you aren't willing to explain your reasoning (using complete paragraphs, not cryptic one-sentence responses) this discussion is probably not going to go anywhere productive.

 

[0mega]

Moderator sent me a message:
"Making references to other forums are not considered as valid references. Please re-read the PF Rules that you had agreed to".
Ok. If it violates the rules, I am ready to you to explain the incomprehensible. I repeat that this problem has long been solved Котофеич, peregoudov, Пью Чай Ли, В. Войтик.
Given: Two material points, acceleration from a constant proper acceleration from rest in the positive direction of X axis so that the distance between them is relatively laboratory frame (T, X) is always constant and equally. The radius-vector of their relative position is also oriented along Х.
Search: position of the front point on the frame back point (t, x) to the time t in this frame of reference.
Solution. It is essentially based on the transformation of Moller from laboratory frame in the frame associated with the back point.
(1) $$T=\frac{1+Wx}{W}shWt$$
$$X=\frac{(1+Wx)chWt-1}{W}$$
The calculations, which can be found
http://arxiv.org/abs/physics/9810017
or
http://www.sciteclibrary.ru/rus/catalog/pages/9478.html
argue that the length of rigid rod own size x, the back point, which moves with the acceleration of $$W$$ on laboratory frame depends on the time laboratory frame so:
(2) $$L=\frac{\sqrt{(1+Wx)^2+W^2T^2}-\sqrt{1+W^2T^2}}{W}$$
This same equation is suitable for the measurement of any rigid or woozy rod.
We now substitute (1) in (2) and calculate the $$x$$. Obtain, omitting the details that
(3) $$x=\frac{\sqrt{1+W^2 L^2 sh^2Wt}-1+WL chWt}{W}$$
The initial time point was coordinate the front
$$x=L$$, and eventually coordinate $$x$$ is growing.

Sorry for my English

 

[0mega]

I will continue a little further. Speed anterior point on the back gives the derivative (3), i.e.
$$u=\frac{dx}{dt}$$
(4) $$u=\frac{W^2L^2shWtchWt}{\sqrt{1+W^2L^2sh^2Wt}}+WL shWt$$
By the time the $$t$$ to back clock on the front will take time
(5) $$\tau=\int{\sqrt{(1+Wx)^2-u^2}}dt$$
If we substitute here (3) and (4), then integration gives
(6) $$\tau=t+\frac{arsh(WLshWt)}{W}$$
In the problem of Bell's question is at what speed between the points of burst thread, if its length is changed 2 times. Why not answer this question so. The length between the points varies according to the law (3).
Substitute here
$$x = 2L$$ and moments $$t$$ when the string broke. Obtain
(7)$$chWt =\frac{4+5WL}{2 +4 WL}$$
here for not too large $$L$$
$$t =\frac{arch2}{W} =\frac{1,32}{W}$$
Hence speed
(8)$$v=thWt=\frac{\sqrt{(2+WL)(6+9WL)}}{4+5WL}$$
For the real threads, we find that
$$v = 0,866...$$

 

[0mega]

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

Yes. The correct math is based on the transformation Moller

 

[0mega]

The correct formula for the contraction of Lorentz in the symbols of Franklin is
$$d=\frac{\sqrt{1+2a'(1-v^2)d'+a'^2(1-v^2)d'^2}-1}{a'\sqrt{1-v^2}}$$
She also has long been known

 

[Dale]

Who is us?

Any of the PF mentors or science advisors that post here.

The paper defines a frame S'

a system S′ that moves with constant velocity v with respect to the system S.

This frame increases v as the acceleration proceeds.

That is, as the velocity in S increases, the distance between the spaceships in their rest system S′ increases.

Equation 4 describes the distance associated with S'.

More logic

Consequently, the rest frame length in the instantaneous rest system S′ must increase in accordance with Eq. (4), and get longer than the original rest frame length in system S.

Now, since S' is the relative v that increases over acceleration to S, then one can claim that S' remains after the acceleration and thus S will disagree with the distance of the ships as compared with S', ie the ships inertial frame after acceleration.
Therefore, each frame will judge a different distance between the ships.

Now, how do you resolve this.

http://arxiv.org/PS_cache/arxiv/pdf/0906/0906.1919v2.pdf

Yes, S and S' disagree on the distance. In S the distance is fixed at d as a part of the specification of the problem. In S' the distance is found by equation 4:
$$d' = \gamma d$$
or, more explicitly by substituting in equation 1
$$d' = d \frac{1}{\sqrt{1-v^2}}$$

As you can plainly see, d' depends on the velocity v (in units where c=1) and not on the acceleration a. So, when you stop accelerating you do not change v and therefore you do not change d'.

More explicitly, let the acceleration stop at a time $t_f$ (in the unprimed frame) where the ship has obtained a velocity $v_f$ (in the unprimed frame). Since the acceleration has stopped, the velocity at any time $t \geq t_f$ is also $v_f$. Since d' depends only on v then $d'_f$ also remains constant after the acceleration has stopped and equal to $$d'_f = d \frac{1}{\sqrt{1-v_f^2}}$$

 

[0mega]

You are looking at 2 spaceships, which are not hard core. Before solving the task of Bell's we have to consider rigid rod. If the rod is rigid in non-inertial reference frame associated with the back point, then in the laboratory frame of reference he will be nonrigid. The concept of rigidity is relative. The front point of a rigid rod moves at a slower speed than the back point.
The exact formula for the velocity of the point of a rigid rod is
$$V_x=\frac{V}{1+2W(1-V^2)x+W^2(1-V^2)x^2}$$
$$V$$ is velocity origin frame.
This is formula (3.2) in
http://www.sciteclibrary.ru/rus/catalog/pages/9478.html
And you and Franklin ignore relativity concept of rigidity.
After that it's clear, that if in the laboratory frame the front spaceship moves as the back, then in the reference frame associated to the back of the rocket will move front spaceship.
Lorentz transformations for the accelerated motion are not applicable. They apply only to instantaneously comoving frame of reference.
Such cases.

People are silent ...

 

[cfrogue]

Any of the PF mentors or science advisors that post here.Yes, S and S' disagree on the distance. In S the distance is fixed at d as a part of the specification of the problem. In S' the distance is found by equation 4:
$$d' = \gamma d$$
or, more explicitly by substituting in equation 1
$$d' = d \frac{1}{\sqrt{1-v^2}}$$

As you can plainly see, d' depends on the velocity v (in units where c=1) and not on the acceleration a. So, when you stop accelerating you do not change v and therefore you do not change d'.

More explicitly, let the acceleration stop at a time $t_f$ (in the unprimed frame) where the ship has obtained a velocity $v_f$ (in the unprimed frame). Since the acceleration has stopped, the velocity at any time $t \geq t_f$ is also $v_f$. Since d' depends only on v then $d'_f$ also remains constant after the acceleration has stopped and equal to $$d'_f = d \frac{1}{\sqrt{1-v_f^2}}$$

Yes, therefore, the ship's frame and the launch frame will disagree on the distance after acceleration.

The launch frame still has it at d < d'.

Therefore, after acceleration, a launch frame must apply the above equations of yours and the paper's and calculate this new d' itself and then apply this further for LT if that is the goal of the frame.

Do you agree?

 

[Dale]

Yes, therefore, the ship's frame and the launch frame will disagree on the distance after acceleration.

The launch frame still has it at d < d'.

Therefore, after acceleration, a launch frame must apply the above equations of yours and the paper's and calculate this new d'

Yes, to all of this.

and then apply this further for LT if that is the goal of the frame.

Do you agree?

I don't understand what you mean by this last part, especially the phrase "goal of the frame". But if you mean that the distance d' can be used in S' and that you can do Lorentz transforms from S' to other frames, then yes also.

 

[cfrogue]

Yes, to all of this.I don't understand what you mean by this last part, especially the phrase "goal of the frame". But if you mean that the distance d' can be used in S' and that you can do Lorentz transforms from S' to other frames, then yes also.

Oh, I just meant, S must know this new distance before performing any new calculations using LT after the acceleration is over.

This is a minor issue.

 

[Al68]

Therefore, each frame will judge a different distance between the ships.

Now, how do you resolve this.

What needs to be "resolved"? You've already resolved it: The distance between the ships is different in the launch frame than in the ships' frame.

Everyone else already agrees with that, so what is left to resolve?

 

[cfrogue]

What needs to be "resolved"? You've already resolved it: The distance between the ships is different in the launch frame than in the ships' frame.

Everyone else already agrees with that, so what is left to resolve?

Yes, if you read through this thread, everyone else did not conclude this.

I will leave you to your facilities.

 

[matheinste]

cfrogue,

So you are effectively saying that you believe that, in SR, it is not the case that everyone believes measurements made of a spatial distance or length in one reference frame will be different from measurements made of that same spatial distance or length made in a frame moving relative to that first frame.

If this is so, what proportion of respondents do you think do not believe it.

Matheinste

 

[cfrogue]

cfrogue,

So you are effectively saying that you believe that, in SR, it is not the case that everyone believes measurements made of a spatial distance or length in one reference frame will be different from measurements made of that same spatial distance or length made in a frame moving relative to that first frame.

If this is so, what proportion of respondents do you think do not believe it.

Matheinste

Maybe you should read the paper and the effects of acceleration.

 

[Al68]

What needs to be "resolved"? You've already resolved it: The distance between the ships is different in the launch frame than in the ships' frame.

Everyone else already agrees with that, so what is left to resolve?

Yes, if you read through this thread, everyone else did not conclude this.

I'm too lazy to reread the whole thread, but it's irrelevant anyway. I still don't see what there is to resolve. Where's the discrepancy?

 

[matheinste]

Maybe you should read the paper and the effects of acceleration.

Unfortunately that paper has nothing to say about the proportion of people who you think do not believe that distance measurement is frame dependent. It was written long before I asked the question.

Matheinste.

 

[JesseM]

What needs to be "resolved"? You've already resolved it: The distance between the ships is different in the launch frame than in the ships' frame.

Everyone else already agrees with that, so what is left to resolve?

Yes, if you read through this thread, everyone else did not conclude this.

I will leave you to your facilities.

Point out any person on this thread (besides you) that did not conclude the final distance in the launch frame is different than the final distance in the final rest frame of the ships. I suspect the problem lies with your own reading comprehension.

 

[0mega]

Let me explain one more thing. When rigid accelerated motion of the rod all its points move differently. Front point - at a slower speed than the back.
Therefore, the region adjacent to the front end of the rod is reduced less than the back. This leads to differ from the Lorentz formula. Right formula is my post 328.
Ask for what you do not understand.

 

[cfrogue]

Point out any person on this thread (besides you) that did not conclude the final distance in the launch frame is different than the final distance in the final rest frame of the ships. I suspect the problem lies with your own reading comprehension.

Having a bad day?

No, I gave two possibilities and was looking to eliminate one of them that is all.
One possibility I gave was that the distance remains between the two ships.

Now time for the real exercise.

Recall the logic of "rest" distance when considering relative motion? One says there is an "at rest distance" d for a rod length.

Well, one of the frames must accelerate to change that, no?

As such, the accelerated frame's rod lengths will expand to a new d' and will remain after the acceleration, no?

So, now what does the stationary frame calculate for length contraction after the acceleration since now we know this "at rest rod length" is no longer true. It is a new d'.

 

[cfrogue]

Let me explain one more thing. When rigid accelerated motion of the rod all its points move differently. Front point - at a slower speed than the back.
Therefore, the region adjacent to the front end of the rod is reduced less than the back. This leads to differ from the Lorentz formula. Right formula is my post 328.
Ask for what you do not understand.

How does your logic square with the paper presented in this thread?

 

[JesseM]

No, I gave two possibilities and was looking to eliminate one of them that is all.
One possibility I gave was that the distance remains between the two ships.

And yet you never gave the slightest reason to consider the idea of a sudden distance change in the launch frame to be taken seriously as a possibility, which started to get annoying as you repeated the question over and over while refusing to answer questions about the logic behind it. The premise of the question was completely ludicrous, for reasons I explained in post 322 (which you never replied to after I told you I had read the paper):

Do I need a reason to discount a "model" where a distance suddenly changes for no reason whatsoever? We assume objects move continuously in any inertial frame due to their velocity and acceleration in that frame (with acceleration always due to some force, and objects with no forces on them obeying Newton's first law and therefore continuing at constant velocity), a sudden change in distance would amount to an uncaused teleportation.

Now time for the real exercise.

Recall the logic of "rest" distance when considering relative motion? One says there is an "at rest distance" d for a rod length.

What do you mean "at rest distance"? The length of the rod in its instantaneous inertial rest frame (or equivalently in the type of non-inertial frame I mentioned), or its equilibrium length, or something else?

Well, one of the frames must accelerate to change that, no?

If you're talking about the length of the rod in its inertial rest frame, then by definition this is defined solely in terms of the rod's rest frame, it doesn't matter how other frames are moving or what distances they measure.

As such, the accelerated frame's rod lengths will expand to a new d' and will remain after the acceleration, no?

In the Bell's spaceship scenario the rod's length does expand and stay expanded in the non-inertial frame, but I'm not sure what this has to do with "at rest distance" or why you say "as such".

So, now what does the stationary frame calculate for length contraction after the acceleration since now we know this "at rest rod length" is no longer true. It is a new d'.

By "stationary frame" you mean the launch frame? The rod's length in its rest frame d' has changed after the acceleration, but the rod's velocity v relative to the launch frame has changed too, so d'/gamma = d'*sqrt(1 - v^2/c^2) can stay constant.

 

[0mega]

How does your logic square with the paper presented in this thread?

You mean paper of Franklin?
Equation (5) in section 2 strictly speaking incorrect, even though the correct output - a cable is break.The distance between the 2 ships, depending on the time in Bell's paradox is the formula (3) post 325.
In section (3) all right.
In conclusion, I do not agree with the last sentence. Value will be the same as (21)

 

[cfrogue]

And yet you never gave the slightest reason to consider the idea of a sudden distance change in the launch frame to be taken seriously as a possibility, which started to get annoying as you repeated the question over and over while refusing to answer questions about the logic behind it. The premise of the question was completely ludicrous, for reasons I explained in post 322 (which you never replied to after I told you I had read the paper):


What do you mean "at rest distance"? The length of the rod in its instantaneous inertial rest frame (or equivalently in the type of non-inertial frame I mentioned), or its equilibrium length, or something else?

If you're talking about the length of the rod in its inertial rest frame, then by definition this is defined solely in terms of the rod's rest frame, it doesn't matter how other frames are moving or what distances they measure.

In the Bell's spaceship scenario the rod's length does expand and stay expanded in the non-inertial frame, but I'm not sure what this has to do with "at rest distance" or why you say "as such".

By "stationary frame" you mean the launch frame? The rod's length in its rest frame d' has changed after the acceleration, but the rod's velocity v relative to the launch frame has changed too, so d'/gamma = d'*sqrt(1 - v^2/c^2) can stay constant.

Assume the two frames are at rest and O' has a rod of rest length d and both frames agree.

Now, O' accelerates to v. What is the length of the rod now in O' after the acceleration?

 

[JesseM]

Assume the two frames are at rest and O' has a rod of rest length d and both frames agree.

Now, O' accelerates to v. What is the length of the rod now in O' after the acceleration?

It depends on the type of acceleration. If it's Born rigid acceleration, the rest length never changes. But if it's the type of acceleration in the Bell spaceship scenario, where the distance between the ends of the rod stays constant in the inertial "launch frame" where the rod began to accelerate, then in the non-inertial frame where one of the ends of the rod remains at rest (which I presume is what you meant by O'), if this non-inertial frame is defined in the way I defined it in post 263, then the rod's length will have increased in O'.

It's not like I'm telling you anything new here, we've been over this point many times before. Do you have anything new to add or ask or do you just want to repeat the same questions over and over again?

 

[cfrogue]

It depends on the type of acceleration. If it's Born rigid acceleration, the rest length never changes. But if it's the type of acceleration in the Bell spaceship scenario, where the distance between the ends of the rod stays constant in the inertial "launch frame" where the rod began to accelerate, then in the non-inertial frame where one of the ends of the rod remains at rest (which I presume is what you meant by O'), if this non-inertial frame is defined in the way I defined it in post 263, then the rod's length will have increased in O'.

It's not like I'm telling you anything new here, we've been over this point many times before. Do you have anything new to add or ask or do you just want to repeat the same questions over and over again?

http://arxiv.org/PS_cache/arxiv/pdf/0906/0906.1919v2.pdf

Equation 4 says the distance in the accelerated frame is d' = λd.
It says,

We see that this length is greater than the length between the spaceships before
their motion.

So, if the rod was a length d at rest then after acceleration it is of length λd.

Now, please apply LT and length contraction to this expanded rod which was accelerated to v from the "at rest" frame and let me know its length from the rest frame/launch frame.

Please make sure to use the "correct" length of λd.

If I understand it correctly, one takes rod/λ to determine the rod length in the moving frame.

 

[JesseM]

http://arxiv.org/PS_cache/arxiv/pdf/0906/0906.1919v2.pdf

Equation 4 says the distance in the accelerated frame is d' = λd.

Equation 4 does not refer to an accelerated frame, it refers to a new inertial frame S' obtained by doing a Lorentz transformation on the launch frame S (the Lorentz transformation only relates the coordinates of inertial frames). However, since I defined the non-inertial frame to have distances that always agree with the instantaneous inertial rest frame, it doesn't really matter, the distance in the non-inertial frame after the acceleration ended would also be d' = gamma*d.

Now, please apply LT and length contraction to this expanded rod which was accelerated to v from the "at rest" frame and let me know its length from the rest frame/launch frame.

Please make sure to use the "correct" length of λd.

Sure, the length contraction equation says that if an object moving inertially has length L_rest in its own rest frame, then in a frame where the object is moving at speed v, it will have length L_moving = L_rest/gamma. So, if your rod has a length of L_rest = d' = gamma*d in its own inertial rest frame S' after it has finished accelerating, that means that in the launch frame S where it's moving with speed v, the length contraction equation says its length in S will be L_moving = L_rest/gamma = d'/gamma = (gamma*d)/gamma = d.

 

[cfrogue]

Equation 4 does not refer to an accelerated frame, it refers to a new inertial frame S' obtained by doing a Lorentz transformation on the launch frame S (the Lorentz transformation only relates the coordinates of inertial frames). However, since I defined the non-inertial frame to have distances that always agree with the instantaneous inertial rest frame, it doesn't really matter, the distance in the non-inertial frame after the acceleration ended would also be d' = gamma*d.

Sure, the length contraction equation says that if an object moving inertially has length L_rest in its own rest frame, then in a frame where the object is moving at speed v, it will have length L_moving = L_rest/gamma. So, if your rod has a length of L_rest = d' = gamma*d in its own inertial rest frame S' after it has finished accelerating, that means that in the launch frame S where it's moving with speed v, the length contraction equation says its length in S will be L_moving = L_rest/gamma = d'/gamma = (gamma*d)/gamma = d.

What is the length of the rod after acceleration is done?