Time difference caused by length contraction

[externo]

TL;DR

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?