Time difference caused by length contraction

[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.
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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.

 

[Dale]

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?

Reference frames still exist in any theory. There are two common meanings for reference frames:

First, reference frames are coordinate systems used to assign position and time numbers to events. That is part of the mathematical analysis. LET and BU use the exact same math for the coordinate transforms.

Second, reference frames are a system of physical clocks and rulers used to assign position and time numbers to events. That is part of the experimental setup. This is an experimental fact independent of any theory.

In both cases reference frames are every bit a part of LET as BU.

Again, the fact that you think they are not part of LET shows a misunderstanding of LET. Note, this comment is not said as someone promoting a stance on the BU vs LET debate (they are both valid interpretations), but as a professional physics educator. You misunderstand the math of how LET works. It does not work the way you claim. You have a vague grasp of the philosophy, but not the concrete math. So please take some instruction on the topic.

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

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.

This is a fallacious argument. Just because two points on a rigid object have different velocities and accelerations does not imply that it is due to length contraction. In Newtonian physics two objects at rest in a rotating reference frame will have different proper accelerations and in an inertial frame will have different velocities. That doesn’t imply length contraction caused it. Relativistic acceleration is a hyperbolic rotation, so it shares this kinematical fact with Newtonian rotational motion.

To show that length contraction is the cause will require an actual derivation, your simple assertions and rhetorical questions are insufficient.

 

[PeterDonis]

You are violating forum rules here with this statement. You are attempting to argue the superiority or veracity of BU.

I am doing no such thing. And at this point your question has been thoroughly answered, and you have failed to respond to any of the valid answers you have been given, but continue to either repeat the same incorrect statements, or raise irrelevant issues. Enough is enough.

This thread is closed.

 

[Dale]

Sorry @PeterDonis I was working on the math before the closure (actually since yesterday once @externo finally specified what they were looking for) and so I wanted to post it anyway. It just took some time to do.

In Rindler (accelerating) coordinates the metric is $$ds^2 = -c^2 d\tau^2 =-\frac{a^2 x^2}{c^2} dt^2 + dx^2 + dy^2 + dz^2$$ where an observer at $x_0=c^2/a$ experiences a proper acceleration of $a$ and has undilated time. Time dilation is given by $\frac{1}{\gamma} = \frac{d\tau}{dt}$, so for a clock at rest in the Rindler frame $dx=dy=dz=0$ and we have $$\frac{1}{\gamma}=\frac{d\tau}{dt}=\sqrt{\frac{a^2 x^2}{c^4} - \frac{dx^2+dy^2+dz^2}{c^2 dt^2}}= \frac{a x}{c^2}$$ For a clock at $x=x_0+h$ compared to a clock at $x=x_0$ this gives a time dilation ratio of $$ 1+\frac{ah}{c^2}$$ where $ah$ is the pseudo-gravitational potential in the Rindler frame. Note that $h=0$ corresponds to the reference clock which is undilated and has $1/\gamma=1$. So $h$ functions as a parameter that identifies specific members of the Rindler congruence, e.g. a clock at $h=0$ or a clock at $h=5 \mathrm{\ m}$.

Note that this is not a first-order approximation, but is the actual time dilation in the accelerating Rindler frame. By the equivalence principle this is the relationship that holds locally in a real gravitational field over a local region sufficiently small that tidal gravity effects are negligible.

Now, we will transform from the accelerating Rindler frame to an inertial Minkowski frame. If $t=0$ is the moment that the rocket is momentarily at rest in the aether frame, then this transforms to the aether frame, so this is the theoretical LET derivation even if we find it difficult to determine the correct $t$ as a practical matter.

In the inertial frame the worldline of the clock at $h$ is given by $$(T,X,Y,Z)=\left(\left(\frac{c}{a}+\frac{h}{c}\right) \sinh\left( \frac{at}{c} \right),\left(\frac{c^2}{a}+h\right) \cosh\left( \frac{at}{c} \right),0,0\right)$$$$ = \left(T, \left( \frac{c^2}{a}+h \right) \sqrt{1+\frac{a^2 c^2 T^2}{\left( c^2 + a h\right)^2}} ,0,0 \right)$$ where $T$ is time in the inertial frame and $t$ is time in the Rindler frame.

The metric in the inertial frame is $ds^2=-c^2 dT^2 + dX^2 + dY^2 + dZ^2$. Calculating the time dilation for this case (here $dX\ne 0$) we get $$\frac{1}{\gamma}=\sqrt{\frac{(c^2+ah)^2}{(c^2+ah)^2+a^2 c^2 T^2}}$$
Now, we have the time dilation in the inertial frame at any time $T$ for the Rindler clock at $h$. This time dilation, as expected, depends on both $T$ and on $h$. As $T$ increases the clock gets slower, regardless of $h$. So, we can take the ratio of the time dilation for a clock at $h=0$ and a clock at some other $h$. When we do that we get $$\left(c+\frac{ah}{c}\right)\sqrt{\frac{c^2+a^2 T^2}{(c^2+ah)^2+a^2 c^2 T^2}}$$
Now, the issue is that this is quantitatively the wrong value. Importantly, at $T=0$ there is no time dilation with this approach and furthermore the "time dilation" is a function of time, which is not correct for gravitational time dilation. In fact, this value is not even approximately correct. If we do just a first order approximation in $h$ we get $$1+\frac{a^3}{c^4+a^2 c^2 T^2}T^2 h + O(h^2)$$ which does not match the correct expression even to first order. Thus this expression is conclusively ruled out by experiment. The primary reason this is wrong is the one identified and explained to you by @Ibix