Is Lorentz Contraction Indistinguishable from Standard Relativity?

[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?

 

[JesseM]

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

I assumed above that the rod's length in its new inertial rest frame S' after it had stopped accelerating was d' = d*gamma. Then I showed that if we apply the length contraction equation, this implies that its length in the launch frame S must be d'/gamma = (d*gamma)/gamma = d.

 

[cfrogue]

I assumed above that the rod's length in its new inertial rest frame S' after it had stopped accelerating was d' = d*gamma. Then I showed that if we apply the length contraction equation, this implies that its length in the launch frame S must be d'/gamma = (d*gamma)/gamma = d.

Good, you finally made it.

So, where do the length contraction of SR apply?

1) We start out in a rest frame and measure d.
2) d accelerates and stops.
3) After stopping, the length is d*gamma.
4) Now, the rest frame applies LT since we have relative motion and applies length contraction as advertised, (d*gamma)/gamma = d.
5) No real length contraction.

 

[JesseM]

Good, you finally made it.

"Finally"? This is what I have been saying all along, apparently you weren't paying attention?

So, where do the length contraction of SR apply?

1) We start out in a rest frame and measure d.
2) d accelerates and stops.
3) After stopping, the length is d*gamma.
4) Now, the rest frame applies LT since we have relative motion and applies length contraction as advertised, (d*gamma)/gamma = d.
5) No real length contraction.

"Length contraction" does not refer to the length contracting over time in one frame, it refers to a comparison between two frames, with one frame seeing the length "contracted" relative to the length in the rest frame. If we consider the rod after it's finished accelerating, the length in the launch frame S is contracted relative to the length in the rod's rest frame S', this is all that is ever meant by "length contraction" in SR.

 

[cfrogue]

"Finally"? This is what I have been saying all along, apparently you weren't paying attention?

"Length contraction" does not refer to the length contracting over time in one frame, it refers to a comparison between two frames, with one frame seeing the length "contracted" relative to the length in the rest frame. If we consider the rod after it's finished accelerating, the length in the launch frame S is contracted relative to the length in the rod's rest frame S', this is all that is ever meant by "length contraction" in SR.

The logic is simple.

If a rod starts out at d and is accelerated to v, then the actual length of the rod is dλ in the moving frame. Thus, there is no length contraction for the rest frame.

 

[JesseM]

The logic is simple.

If a rod starts out at d and is accelerated to v, then the actual length of the rod is dλ in the moving frame. Thus, there is no length contraction for the rest frame.

You are using the phrase "length contraction" incorrectly. Did you even read my post above? "Length contraction" in SR always refers to a comparison of lengths in two frames, it does not refer to a change in length in a single frame, so the fact that the length doesn't change in the launch frame doesn't mean it's correct to say "there is no length contraction for the rest frame".

 

[Al68]

Good, you finally made it.

So, where do the length contraction of SR apply?

1) We start out in a rest frame and measure d.
2) d accelerates and stops.
3) After stopping, the length is d*gamma.
4) Now, the rest frame applies LT since we have relative motion and applies length contraction as advertised, (d*gamma)/gamma = d.
5) No real length contraction.

The logic is simple.

If a rod starts out at d and is accelerated to v, then the actual length of the rod is dλ in the moving frame. Thus, there is no length contraction for the rest frame.

The logic is simple:

You just said that the length of the rod in the moving frame is dλ while its length in the rest frame is d (length contraction for the rest frame).

Then you say there is no length contraction for the rest frame.

And this after it has been repeatedly pointed out that length contraction is the difference in length between two frames (d vs dλ).

I can only assume this is some kind of bizarre practical joke.

 

[A.T.]

And this after it has been repeatedly pointed out that length contraction is the difference in length between two frames (d vs dλ).

I can only assume this is some kind of bizarre practical joke.

The more common word for this kind of practical joke is 'trolling'. His entire argument is based on using the same term for two different quantities, and wondering about the contradictions that arise from it. This was already pointed out to him hundreds of posts ago. He will continue to ignore the clarification because he doesn't want to understand anything, just to argue like a kid.

 

[cfrogue]

You are using the phrase "length contraction" incorrectly. Did you even read my post above? "Length contraction" in SR always refers to a comparison of lengths in two frames, it does not refer to a change in length in a single frame, so the fact that the length doesn't change in the launch frame doesn't mean it's correct to say "there is no length contraction for the rest frame".

I have this part figured out.

I will say it again.

Two frames are at rest.

They measure a rod length to be d.

The frame O' takes the rod and accelerates to v and stops accelerating. O stays at rest.

The length of the rod has increased in O' to d*λ because of the acceleration.

Now, O applies LT length contraction to the rod.

(d*λ)/λ = d.

Thus, the rod expands from the acceleration and LT contracts it back to d from the POV of O.

What is the problem?

Note, that the "rest" length of the rod is d and the LT moving length is d.

 

[JesseM]

I have this part figured out.

I will say it again.

Two frames are at rest.

They measure a rod length to be d.

The frame O' takes the rod and accelerates to v and stops accelerating. O stays at rest.

The length of the rod has increased in O' to d*λ because of the acceleration.

Now, O applies LT length contraction to the rod.

(d*λ)/λ = d.

Thus, the rod expands from the acceleration and LT contracts it back to d from the POV of O.

What is the problem?

I have no problem with this--it was you who seemed to say there was a problem, that somehow length contraction does not apply in this example.

 

[cfrogue]

I have no problem with this--it was you who seemed to say there was a problem, that somehow length contraction does not apply in this example.

You about have it.

Length contraction does not apply to any problem that talks about an "at rest" rod distance.

Do you know any other kind?