I Time difference caused by length contraction

[externo]

TL;DR Summary

Threat 3 : When a cabin accelerates rigidly it contracts in the starting reference frame. What is the calculation that gives the difference in the passage of time between the top and the bottom of the cabin?

The goal is to calculate the difference in the passage of time between two ends of a cabin which is accelerating upwards due to length contraction. To help in the calculations we can consult:
https://arxiv.org/pdf/1807.05338.pdf
There is an old problem called the 4/3 problem which has been solved by taking into account the length contraction experienced by the electron during its acceleration. When we ignore this contraction there is a difference of 4/3 between the electromagnetic mass of the electron and its electromagnetic energy, but when we take it into account the two are equal.
On page 10 we find, for the acceleration of a rigid body, the formula:
g₀/g₁ = 1 + hg₀/c²
In our situation I believe that g₀ is the acceleration of the bottom of the cabin and g₁ at the top of the cabin.
With this formula one should be able to do the calculations.

 

[Dale]

What calculation specifically do you want to do? In the frame of the rocket there is no length contraction and no relative velocity. In an inertial frame there is length contraction that steadily increases over time. What specific calculation do you think relates that to the time dilation?

 

[externo]

In the starting reference frame there is length contraction, it is therefore the difference in the proper times between the front and the back in the starting reference frame that I am looking for.

 

[externo]

You can see here the situation.

https://en.wikipedia.org/wiki/Bell's_spaceship_paradox#/media/File:BornRigidHorizon.svg

in the starting frame, time does not flow in the same rate in the front and in the back

 

[Dale]

In the starting reference frame there is length contraction, it is therefore the difference in the proper times between the front and the back in the starting reference frame that I am looking for.

How do you intend on relating that difference in proper times due to length contraction? I mean, I can easily calculate the difference in proper times using their velocity, but length contraction doesn’t enter in at all.

 

[externo]

In a rigid acceleration the front and the rear do not go at the same speed, which causes a difference in the passage of time. You can see it very well in this diagram.

https://en.wikipedia.org/wiki/Bell's_spaceship_paradox#/media/File:BornRigidHorizon.svg

 

[Dale]

in a rigid acceleration the front and the rear do not go at the same speed, which causes a difference in the passage of time.

Yes, but how do you want to relate that to length contraction? In other words, a difference in speed doesn’t imply length contraction. I could have two objects going in different directions or even circling around each other at different speeds. They would exhibit different proper times, but not length contraction.

So how do you want to involve length contraction in the calculation? The calculation is naturally a velocity calculation not a length contraction calculation.

 

[topsquark]

Summary: Threat 3 : When a cabin accelerates rigidly it contracts in the starting reference frame. What is the calculation that gives the difference in the passage of time between the top and the bottom of the cabin?

The goal is to calculate the difference in the passage of time between two ends of a cabin which is accelerating upwards due to length contraction. To help in the calculations we can consult:
https://arxiv.org/pdf/1807.05338.pdf
There is an old problem called the 4/3 problem which has been solved by taking into account the length contraction experienced by the electron during its acceleration. When we ignore this contraction there is a difference of 4/3 between the electromagnetic mass of the electron and its electromagnetic energy, but when we take it into account the two are equal.
On page 10 we find, for the acceleration of a rigid body, the formula:
g₀/g₁ = 1 + hg₀/c²
In our situation I believe that g₀ is the acceleration of the bottom of the cabin and g₁ at the top of the cabin.
With this formula one should be able to do the calculations.

First, this article mostly focuses on electrons, which have extra properties over "ordinary" matter you find in most SR problems. So most of this paper is useless to you. I'd recommend finding a good textbook on the subject.

Second, the equation you are talking about is for a system of point particles, not a cabin accelerating upward. In order for you apply this you are really doing a problem of two point particles rigidly connected by a massless rod accelerating upward.

-Dan

 

[Ibix]

In the starting reference frame there is length contraction, it is therefore the difference in the proper times between the front and the back in the starting reference frame that I am looking for.

That's easy, then. You just look up the elapsed time along a hyperbola.

Note that the difference in proper times between front and back clocks as measured in the starting rest frame will not get you the time dilation as measured in the rocket frame unless you also take into account the changing simultaneity criterion. That's why it's a mistake to think of this as "caused by length contraction".

 

[externo]

Yes, but how do you want to relate that to length contraction? In other words, a difference in speed doesn’t imply length contraction. I could have two objects going in different directions or even circling around each other at different speeds. They would exhibit different proper times, but not length contraction.

So how do you want to involve length contraction in the calculation? The calculation is naturally a velocity calculation not a length contraction calculation.

On the rigid acceleration diagram the left world line is the back of the rocket and the right world line is the front of the rocket. The two ends do not have the same velocity because the rocket is contracting. The length contraction brings the two ends closer together, so their velocity and proper time are different.

 

[Dale]

On the rigid acceleration diagram the left world line is the back of the rocket and the right world line is the front of the rocket. The two ends do not have the same velocity because the rocket is contracting. The length contraction brings the two ends closer together, so their velocity and proper time are different.

So to be clear there are three velocities here: the velocity of the front $v_f$, the velocity of the back $v_b$, and the closing velocity $v_d=v_b-v_f$. Of those, only the closing velocity has anything directly to do with length contraction.

So what exactly do you want to calculate?

 

[PeterDonis]

it is therefore the difference in the proper times between the front and the back in the starting reference frame that I am looking for.

Just to make sure I'm clear about what you want: at a given instant of coordinate time $t$ in the starting inertial frame, the $x$ coordinates of the front and the back will be $x_f$ and $x_b$. You are asking for the difference in proper time elapsed along the front and back worldlines from the start at $t = 0$ to $t$, correct?

Assuming the above is correct, what makes you think length contraction will appear anywhere in the resulting formulas?

 

[externo]

Note that the difference in proper times between front and back clocks as measured in the starting rest frame will not get you the time dilation as measured in the rocket frame unless you also take into account the changing simultaneity criterion. That's why it's a mistake to think of this as "caused by length contraction".

But does the changing simultaneity have anything to do with 1 + gh/c² ? Is it the changing simultaneity which gives 1+ gh/c² or is it the length contraction? The change in simultaneity does not depend on the acceleration but on the velocity, whereas the time lag between the two ends from length contraction does indeed depend on the acceleration and could be 1 + gh/c².

So to be clear there are three velocities here: the velocity of the front, the velocity of the back, and the closing velocity. Of those, only the closing velocity has anything directly to do with length contraction. So what exactly do you want to calculate?

The closing velocity is only the difference between the velocity of the back and the velocity of the front. So all three velocity have something directly to do with length contraction.

Just to make sure I'm clear about what you want: at a given instant of coordinate time $t$ in the starting inertial frame, the $x$ coordinates of the front and the back will be $x_f$ and $x_b$. You are asking for the difference in proper time elapsed along the front and back worldlines from the start at $t = 0$ to $t$, correct?

Assuming the above is correct, what makes you think length contraction will appear anywhere in the resulting formulas?

Initially the proper times are t_f and t_b, with t_f = t_b.
After an acceleration, the proper time of the front and the back will no longer be identical. We will have t_f > t_b.
I'm looking for t_f - t_b.
The length contraction will not appear in the result, but the length contraction is the reason why t_f > t_b.

 

[Ibix]

But does the changing simultaneity have anything to do with 1 + gh/c² ? Is it the changing simultaneity which gives 1+ gh/c² or is it the length contraction? The change in simultaneity does not depend on the acceleration but on the velocity, whereas the time lag between the two ends from length contraction does indeed depend on the acceleration and could be 1 + gh/c².

The changing simultaneity affects the times at which you are comparing the clocks, so of course it affects the ratio of clock rates and elapsed times.

 

[externo]

I want calculations. I don't think changing simultaneity gives the 1 + gh/c² difference.

 

[Ibix]

I want calculations. I don't think changing simultaneity gives the 1 + gh/c² difference.

Why not do them yourself? All you need to do is look up the interval along a timelike hyperbola if you can't derive it for yourself. Then you consider the rates of change of interval along two such hyperbolae assuming Einstein and Rindler simultaneity respectively.

 

[externo]

The change in simultaneity cannot exceed the proper length of the rocket. There is a relation that says :
(proper length)² - (contracted length)² = (time difference between the two ends)²

 

[Dale]

The closing velocity is only the difference between the velocity of the back and the velocity of the front. So all three velocity have something directly to do with length contraction.

Ok, so which velocity do you want to use and what do you want to do with it?

I want calculations. I don't think changing simultaneity gives the 1 + gh/c² difference.

So what calculations do you want?

 

[Dale]

Initially the proper times are t_f and t_b, with t_f = t_b.
After an acceleration, the proper time of the front and the back will no longer be identical. We will have t_f > t_b.
I'm looking for t_f - t_b.
The length contraction will not appear in the result, but the length contraction is the reason why t_f > t_b.

We can calculate $t_f-t_b$ but:

1) it will not give the time dilation in the accelerated frame
2) there will be no length contraction involved in the calculation
3) there will be no length contraction involved in the result

 

[externo]

1)Ok
2)Yes, the length contraction is the reason for the difference in the velocities between the back and the front, but we don't need to calculate it to find the result.
3)Ok

 

[Dale]

2)Yes, the length contraction is the reason for the difference in the speeds between the back and the front, but we don't need to calculate it to find the result.

You claim that, but at no point will length contraction actually show up in the calculations, neither as inputs nor as outputs, nor as intermediate steps. So to claim that is the reason is very suspect.

If I say that "x is due to y" then somewhere in the derivation of x I would expect y to appear. To claim that "x is due to y" when y never shows up anywhere is unprecedented and makes no sense

 

[externo]

I believe I was wrong. What I'm looking for is not Tf - Tb, but the difference in the frequency of proper time between the back and the front of the rocket (or the bottom and the top of a cabin) at a given moment during acceleration.
I think this relation : g₀/g₁ = 1 + hg₀/c² mus t be used.

You claim that, but at no point will length contraction actually show up in the calculations, neither as inputs nor as outputs, nor as intermediate steps. So to claim that is the reason is very suspect.

If you want it's not due to length contraction, we'll see that later, but I think Ibix understood very well that it came from there.
https://www.physicsforums.com/threa...ed-by-length-contraction.1046087/post-6807039

 

[Dale]

What I'm looking for is not Tf - Tb, but the difference in the frequency of time between the back and the front of the rocket (or the bottom and the top of a cabin) at a given moment during acceleration.

So this is time dilation, not length contraction, but yes this can be calculated. Are you sure you want the difference rather than the ratio?

 

[externo]

So this is time dilation, not length contraction, but yes this can be calculated. Are you sure you want the difference rather than the ratio?

The ratio is the best

 

[externo]

You claim that, but at no point will length contraction actually show up in the calculations, neither as inputs nor as outputs, nor as intermediate steps. So to claim that is the reason is very suspect.

So this is time dilation, not length contraction, but yes this can be calculated.

Question: Where does the velocities and accelerations difference between the rear and the front come from?

 

[PeterDonis]

There is a relation that says :
(proper length)² - (contracted length)² = (time difference between the two ends)²

Where are you getting such a relation from? Please show your work.

 

[PeterDonis]

at a given moment

At a given moment according to which simultaneity convention? That of the starting inertial frame, or that of the momentarily comoving inertial frame (the latter is also the simultaneity convention of the accelerated Rindler frame)?

 

[PeterDonis]

I think this relation : g0/g1 = 1 + hg0/c² mus t be used.

Where are you getting this from?

 

[externo]

I think this relation : g0/g1 = 1 + hg0/c² mus t be used.

Where are you getting this from?

From the paper quoted in my introductory message.

 

[PeterDonis]

From the paper quoted in my introductory message.

What section? Page? Equation number?

 

[externo]

(proper length)² - (contracted length)² = (time difference between the two ends)²

Where are you getting such a relation from? Please show your work.

I might open another thread about it.
------------------------------------------------------------------

I think this relation : g₀/g₁ = 1 + hg₀/c² mus t be used.

What section? Page? Equation number?

3.1 Relativistic description of accelerated rigid body
Page 8, 9, 10 equation 20.

 

[externo]

At a given moment according to which simultaneity convention? That of the starting inertial frame, or that of the momentarily comoving inertial frame (the latter is also the simultaneity convention of the accelerated Rindler frame)?

That of the starting inertial frame was my thought, but both would be best, as a matter of fact. So we can compare them.

 

[PeterDonis]

3.1 Relativistic description of accelerated rigid body
Page 8, 9, 10 equation 20.

At a given moment according to which simultaneity convention?

That of the starting inertial frame.
Both, as a matter of fact.

The equation you reference is valid in the momentarily comoving inertial frame of the accelerated particles. More precisely, it is valid at the instant of coordinate time in that frame in which it is comoving, if, in that frame, we assign the coordinate $x = 0$ to the "back" particle (the one with proper acceleration $g_0$) and the coordinate $x = h$ to the "front" particle (the one with proper acceleration $g_i$).

The starting inertial reference frame is only momentarily comoving with the accelerated particles at time $t = 0$ in that frame. So the given equation is only valid in that frame at the instant $t = 0$.

We could also treat the equation as an equation in the accelerated (non-inertial) frame in which the particles are always at rest. This amounts to using Kottler-Moller coordinates as described on this page:

https://en.wikipedia.org/wiki/Rindler_coordinates

In these coordinates, as in the momentarily comoving inertial frame, the "back" particle is at $x = 0$ and the "front" particle is at $x = h$. The constant $\alpha$ in the metric as given in the Wikipedia article is the same as what the paper you reference calls $g_0$, the proper acceleration of the particle at $x = 0$.

In these coordinates, yes, the ratio of time dilation factors, i.e., of "rates of time flow", is the inverse of the ratio of proper accelerations: the rate of time flow of the "front" particle as compared to the "back" particle is larger by the same ratio that its proper acceleration is smaller. Note, however, that this property does not generalize to curved spacetime; for example, it is not true in the gravitational field of the Earth.

The above also shows that none of this has anything to do with "length contraction", since, in either the momentarily comoving inertial frame or the accelerated frame, the distance between the particles is constant; it is always $h$.

 

[Ibix]

But does the changing simultaneity have anything to do with 1 + gh/c² ? Is it the changing simultaneity which gives 1+ gh/c² or is it the length contraction? The change in simultaneity does not depend on the acceleration but on the velocity, whereas the time lag between the two ends from length contraction does indeed depend on the acceleration and could be 1 + gh/c².

Here is why simultaneity matters. I have sketched a Minkowski diagram showing the worldlines of the front (blue) and rear (red) of a rocket, and added a pair of fine grey lines showing two consecutive clock ticks in the inertial frame in which the rocket was at rest at some time. In this frame, the ratio of clock rates is the ratio of proper times along the two worldlines between the grey lines (in the shaded area).

Now, here's the same diagram but showing two simultaneity planes in the rocket frame. Again, the clock rate in this frame is the ratio of the proper times in the shaded areas.

Hopefully it's obvious the red and blue lengths in the shaded areas aren't the same in the two diagrams. Apart from anything else, the value in the rocket frame will be independent of time, but the value in the inertial frame will not be. So what simultaneity criterion you use matters.

(Note that the above diagrams are an approximation - the clock rate is actually the limit of the ratio as the simultaneity planes get close to one another.)

 

[externo]

Hopefully it's obvious these aren't the same. Apart from anything else, the value in the rocket frame will be independent of time, but the value in the inertial frame will not be.

Thanks.

In these coordinates, yes, the ratio of time dilation factors, i.e., of "rates of time flow", is the inverse of the ratio of proper accelerations: the rate of time flow of the "front" particle as compared to the "back" particle is larger by the same ratio that its proper acceleration is smaller.

The above also shows that none of this has anything to do with "length contraction", since, in either the momentarily comoving inertial frame or the accelerated frame, the distance between the particles is constant; it is always h.

In the momentarily comoving inertial frame, the distance between the particles is not constant, as you can see in the diagrams above, otherwise they would keep the same proper time. If the rocket's length is constant in the rocket's frame it cannot be constant also in the starting frame. It must contract.

 

[externo]

https://en.wikipedia.org/wiki/Rindler_coordinates
"Note that Rindler observers with smaller constant x coordinate are accelerating harder to keep up. This may seem surprising because in Newtonian physics, observers who maintain constant relative distance must share the same acceleration. But in relativistic physics, we see that the trailing endpoint of a rod which is accelerated by some external force (parallel to its symmetry axis) must accelerate a bit harder than the leading endpoint, or else it must ultimately break. This is a manifestation of Lorentz contraction. As the rod accelerates its velocity increases and its length decreases. Since it is getting shorter, the back end must accelerate harder than the front. Another way to look at it is: the back end must achieve the same change in velocity in a shorter period of time. This leads to a differential equation showing that, at some distance, the acceleration of the trailing end diverges, resulting in the Rindler horizon."

 

[PeterDonis]

In the momentarily comoving inertial frame, the distance between the particles is not constant

If you keep using the same frame when it is no longer momentarily comoving, yes, that's true, since once the frame is no longer comoving the particles are moving in it, and at different speeds. But that's not what I was describing.

Each momentarily comoving frame is only momentarily (i.e., for one instant of its time) comoving with the accelerating particles. And at the instant when each momentarily comoving frame is comoving with the particles, the distance between the particles in that frame is $h$.

If the rocket's length is constant in the rocket's frame

Here "the rocket's frame" can only be the non-inertial frame described by Rindler coordinates (in one of several versions, which are described in the Wikipedia article I linked to). In this frame, yes, the rocket's length is constant.

it cannot be constant also in the starting frame. It must contract.

Yes, that's correct. But the claim you have repeatedly made is stronger than this: not just that the rocket contracts in the starting frame, but that its length contraction in the starting frame is the cause of the time dilation. That is the claim that others in this thread have objected to, and which you have not defended; you just continue to assert it without any supporting argument, and without addressing any of the arguments against it that others have posted.

 

[PeterDonis]

https://en.wikipedia.org/wiki/Rindler_coordinates
"Note that Rindler observers with smaller constant x coordinate are accelerating harder to keep up. This may seem surprising because in Newtonian physics, observers who maintain constant relative distance must share the same acceleration. But in relativistic physics, we see that the trailing endpoint of a rod which is accelerated by some external force (parallel to its symmetry axis) must accelerate a bit harder than the leading endpoint, or else it must ultimately break. This is a manifestation of Lorentz contraction. As the rod accelerates its velocity increases and its length decreases. Since it is getting shorter, the back end must accelerate harder than the front. Another way to look at it is: the back end must achieve the same change in velocity in a shorter period of time. This leads to a differential equation showing that, at some distance, the acceleration of the trailing end diverges, resulting in the Rindler horizon."

Everyone else in the thread is quite familiar with all this. Is there a reason why you posted it?

 

[Ibix]

It must contract.

The problem is that length contraction is a phenomenon in the original inertial rest frame, with its definition of simultaneity. But the equivalence principle derivation does not care about that inertial frame - it is talking about the simultaneity of the rocket's Rindler frame. Length contraction does not come into that.

Basically you seem very confused about which frame you want to work in. You are interested in the equivalence principle - fine, you should use the Rindler frame. But you want to invoke length contraction explanations that require you not to use the Rindler frame. You can work in either frame but not both, and the length contraction will always disappear from the maths as you attempt to derive the gravitational time dilation, because the latter is only well defined in the frame where the rocket length is constant.

 

[externo]

The effect of length contraction does not disappear in the frame of reference of the rocket, look at that : https://en.wikipedia.org/wiki/Bell's_spaceship_paradox#/media/File:BornRigidHorizon.svg
If there is no length contraction during acceleration, as here, do you think there will be the same time difference between the back and front of the rocket in the rocket's frame :
https://en.wikipedia.org/wiki/Bell's_spaceship_paradox#/media/File:BornBellHorizon.svg

 

[externo]

Yes, that's correct. But the claim you have repeatedly made is stronger than this: not just that the rocket contracts in the starting frame, but that its length contraction in the starting frame is the cause of the time dilation. That is the claim that others in this thread have objected to, and which you have not defended; you just continue to assert it without any supporting argument, and without addressing any of the arI must thinkguments against it that others have posted.

I must think about this.

 

[Ibix]

The effect of length contraction does not disappear in the frame of reference of the rocket

Of course it does. Otherwise the rocket's length would be changing in its own frame.

If there is no length contraction during acceleration, as here, do you think there will be the same time difference between the back and front of the rocket?

Now you are just confusing Born rigid and Bell acceleration. They are different. In the Bell case the separation of the rockets changes in their own frame and so does the relative clock rate. In the Born rigid case (hint: rigid) the length of the rocket does not change in its rest frame and the time dilation between front and rear is constant.

 

[PeterDonis]

The effect of length contraction does not disappear in the frame of reference of the rocket, look at that : https://en.wikipedia.org/wiki/Bell's_spaceship_paradox#/media/File:BornRigidHorizon.svg
If there is no length contraction during acceleration, as here, do you think there will be the same time difference between the back and front of the rocket in the rocket's frame :
https://en.wikipedia.org/wiki/Bell's_spaceship_paradox#/media/File:BornBellHorizon.svg

The Bell spaceship paradox scenario is irrelevant to the question under discussion in this thread because that scenario does not represent an accelerating rocket whose proper length remains constant. A "rocket" accelerating according to the Bell spaceship paradox scenario (front and back both with the same proper acceleration) would stretch until it came apart, just as the string in the Bell spaceship paradox breaks.

 

[externo]

Basically you seem very confused about which frame you want to work in. You are interested in the equivalence principle - fine, you should use the Rindler frame. But you want to invoke length contraction explanations that require you not to use the Rindler frame. You can work in either frame, but the length contraction will always disappear from the maths as you attempt to derive the gravitational time dilation, because the latter is only well defined in the frame where the rocket length is constant.

Here is an argument: The contraction is the cause that the rear and the front do not have the same proper acceleration, and the proper acceleration is felt in the reference frame of the rocket. So how can you say that the contraction has no effect in the reference frame of the rocket when it is at the origin of the difference in accelerations between the two ends and so at the origin of the difference in gravitational potentials between the two ends ?

 

[PeterDonis]

The contraction is the cause that the rear and the front do not have the same proper acceleration

No, you have this backwards. The fact that the rear and the front do not have the same proper acceleration is the cause of the length contraction in the starting frame, not the other way around. Length contraction itself is not the cause of anything in this scenario; it's an effect, not a cause.

how can you say that the contraction has no effect in the reference frame of the rocket

Apart from the above, there is no length contraction in the (non-inertial) rest frame of the rocket, so obviously length contraction in this frame can't cause anything since it doesn't exist. This has already been pointed out to you.

 

[externo]

No, you have this backwards. The fact that the rear and the front do not have the same proper acceleration is the cause of the length contraction in the starting frame. Length contraction itself is not the cause of anything in this scenario; it's an effect, not a cause.

And why don't they have the same proper acceleration ?
Remember that the purpose of Born's rigid acceleration is to keep the proper length invariant in the frame of the rocket. So the acceleration comes from the rigidity condition. So if you want it is not length contraction, it is the Born rigidity which is the reason.

Apart from the above, there is no length contraction in the (non-inertial) rest frame of the rocket, so obviously length contraction in this frame can't cause anything since it doesn't exist. This has already been pointed out to you.

The contraction is not noticeable in the rocket due to the change in simultaneity, but the difference in acceleration is still felt. If you prefer, let's say that the rigidity condition is the primary cause.

 

[PeterDonis]

why don't they have the same proper acceleration ?

You yourself give the answer:

the purpose of Born's rigid acceleration is to keep the proper length invariant in the frame of the rocket. So the acceleration comes from the rigidity condition.

if you want it is not length contraction, it is the Born rigidity which is the reason.

Born rigidity as a specification leads to the Rindler congruence of worldlines, yes. And once you have the Rindler congruence of worldlines, then you can easily show that the time dilation is present.

The contraction is not noticeable in the rocket

Not just not "noticeable", it does not exist, period. The proper length of the rocket remains constant.

due to the change in simultaneity

No. The reason the rocket's proper length is constant is that physically, proper length requires you to look at spacelike curves in a spacelike surface that is orthogonal to the worldlines you are interested in. That is an invariant specification and does not depend on any choice of coordinates. And it makes the "length contraction" in the starting inertial frame irrelevant.

If you prefer, let's say that the rigidity condition is the primary cause.

It's not a matter of what I or anyone prefers. The physics of the situation, as I have described it above, is perfectly objective and doesn't depend on anyone's preferences, or choice of frames, or anything else subjective.

 

[externo]

Not just not "noticeable", it does not exist, period. The proper length of the rocket remains constant.

How do you know the contraction does not exist ? How do you know that the length of the rocket remains constant in reality? You are violating forum rules here with this statement. You are attempting to argue the superiority or veracity of BU. If you think BU is the true reality, say so straight up and state it in your rules, and don't pretend to be impartial.
I am trying myself to avoid this with the help of impartial reasoning. I am simply saying that even if the contraction does not take place, the difference in accelerations, which causes the contraction, is felt in the rocket. So we cannot say that the effect of contraction is not felt in the rocket.

 

[Dale]

How do you know the contraction does not exist ? How do you know that the length of the rocket remains constant in reality? You are violating forum rules here with this statement. You are attempting to argue the superiority or veracity of BU.

You are misrepresenting his statement. He was speaking of the rocket’s frame. Both LET and BU agree that the length is constant in the rocket’s frame. That you think this is a BU argument shows that you do not understand LET. In LET the rocket’s length is indeed constant in the rocket’s frame.

 

[externo]

But in LET what is the rocket's frame? if the front and the back are not synchronized, how can we speak of a reference frame?
In BU there is a different rocket's length for each reference frame, but in LET there is only one rocket's length in an absolute present.