Is Lorentz Contraction Indistinguishable from Standard Relativity?

[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?

 

[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 ...