DeltaT of clocks at endpoints of a rocket post acceleration

[Jeronimus]

Following scenario:

A rocket measuring 10 lightseconds in length accelerates near instantaneously to v=0.4c. Pre-acceleration two clocks placed at the endpoints of the rocket were synced.

What will be the time difference between the two clocks post acceleration?

I tried to figure it out, using two x-t diagrams, drawing the worldlines of the rocket's endpoints (green lines) as i would imagine them.
Considering the innacuracy of the drawing, my rough estimate is that the time difference will be slightly over 4 seconds. Did i get it right?

Here are the x-t diagrams

edit: Also, is the time difference between the two clocks post acceleration decided only by the final velocity or does the "type" of acceleration (faster/slower) matter?

 

[PeterDonis]

A rocket measuring 10 lightseconds in length accelerates near instantaneously to v=0.4c. Pre-acceleration two clocks placed at the endpoints of the rocket were synced.

What will be the time difference between the two clocks post acceleration?

For finite acceleration (which is a more realistic assumption for several reasons--see below), it depends on the specific value you assume for the acceleration, which in itself will be different for the front and rear ends of the rocket.

For the "instantaneous" case you are assuming, you have two problems, apart from the obvious point that no actual object can have infinite acceleration, which is what "instantaneously" implies.

The first problem is that there is no way to define a "rest frame" for the rocket through this entire process. You can define a rest frame for it before any acceleration occurs; and you can define a different rest frame for it starting a sufficient time after the acceleration has occurred. But in between, there is no well-defined rest frame for the rocket, because the two ends of the rocket are moving relative to each other. Modeling the scenario using finite acceleration allows you to fix this problem.

The second problem is that the length of the rocket, in the frame in which it is at rest, is different after the acceleration than before (and in fact there is a period when it is undefined because the rocket's "rest frame" is undefined, as above). Modeling the scenario using finite acceleration allows you to fix this problem too: you can define motions for the ends of the rocket such that the distance between them, in the rest frame of the rocket, is constant.

Leaving all of those objections aside, you have one other issue: how is it determined when, according to their own clocks, the front and rear ends accelerate? The simplest assumption would appear to be that they both accelerate when their clocks read a certain predetermined value (say t = 0), but the spacetime diagrams you have drawn don't appear to show that. This assumption also affects the answer.

 

[Jeronimus]

Leaving all of those objections aside, you have one other issue: how is it determined when, according to their own clocks, the front and rear ends accelerate? The simplest assumption would appear to be that they both accelerate when their clocks read a certain predetermined value (say t = 0), but the spacetime diagrams you have drawn don't appear to show that. This assumption also affects the answer.

The spacetime diagrams show "exactly" when the back of the rocket and the front of the rocket accelerate. Assuming the clocks at the back and end of the rocket were in sync pre-acceleration, the back of the rocket would accelerate when the clock displays t = -1s roughly. The front of the rocket would accelerate at t = 1s roughly.
The rocket clocks would be in sync with the violet clocks pre-acceleration. The rocket is in the rest-frame of the violet clocks pre-acceleration.

The rocket keeps its size in whichever rest-frame it is temporarily even during the acceleration. It certainly has the same size post acceleration in its rest frame.
The cyan colored clocks on the right x-t diagram are 10 lightseconds apart from each other. Exactly the size of the rocket. The worldlines of the rocket will eventually overlap with the cyan colored clocks' worldlines according to logic (my logic maybe).

I do believe that my drawings accurately reflect the scenario, but prove me wrong. Someone certainly would know how to calculate the exact time difference between two such clocks.
And while i am not certain at the moment, i do believe that the time difference is independent on if you would accelerate instantaneously or choose a less brutal acceleration to reach 0.4c.

 

[PeterDonis]

Assuming the clocks at the back and end of the rocket were in sync pre-acceleration, the back of the rocket would accelerate when the clock displays t = -1s roughly. The front of the rocket would accelerate at t = 1s roughly.

Ah, ok.

The rocket keeps its size in whichever rest-frame it is temporarily even during the acceleration.

No, it can't, because during the acceleration the rocket as a whole is not at rest in any frame: the two ends are moving relative to each other. This is obvious from the diagrams you drew.

It certainly has the same size post acceleration in its rest frame.

Once both ends have accelerated, yes, that will be the case given the way you have chosen the times for each end to accelerate. But that is only true for those specific, exact timings.

 

[Ibix]

The spacetime diagrams show exactly when the back of the rocket and the front of the rocket accelerate. Assuming the clocks at the back and end of the rocket were in sync pre-acceleration, the back of the rocket would accelerate when the clock displays t = -1s roughly. The front of the rocket would accelerate at t = 1s roughly.

OK - but that isn't instantaneous acceleration except in one particular frame (and then I'm making assumptions about what the middle parts of the rocket are doing). Different parts of the rocket may be instantaneously accelerating, but the whole rocket is likely doing something rather more complex.

The rocket keeps its size in whichever rest-frame it is temporarily even during the acceleration. It certainly has the same size post acceleration in its rest frame.

There isn't a well-defined rest frame while the rocket is accelerating, as Peter already told you. Some parts of the rocket are in motion and others aren't.

For what it's worth, I agree your analysis of the time difference in the right frame of two clocks initially synchronised in the left frame and following the worldlines you have drawn. However, this doesn't represent a plausible physical model of a 10 light second long rocket, since the front of the rocket accelerating is not causally connected to the back of the rocket accelerating - the events are space-like separated. Edit: Unless you've got rocket motors all along the length of your rocket, and they're set to fire synchronously in some frame, or something of that sort.

 

[PeterDonis]

my rough estimate is that the time difference will be slightly over 4 seconds

Given your two diagrams, it's easy to see this. In your first diagram, the rocket's front end accelerates two seconds after the back end. In your second diagram, the rocket's front end accelerates two seconds before the back end. The total time difference will be the sum of these two differences.

 

[Jeronimus]

No, it can't, because during the acceleration the rocket as a whole is not at rest in any frame: the two ends are moving relative to each other. This is obvious from the diagrams you drew.

Yes, i was a bit careless when i wrote "The rocket keeps its size in whichever rest-frame it is temporarily even during the acceleration" and was about to delete it, but you beat me to it. That would make for another interesting question later.

 

[Jeronimus]

since the front of the rocket accelerating is not causally connected to the back of the rocket accelerating - the events are space-like separated. Edit: Unless you've got rocket motors all along the length of your rocket, and they're set to fire synchronously in some frame, or something of that sort.

The proper way to do it, would be to place motors along the rocket with clocks, programmed to trigger the motors according to the times we get in the left diagram. The back motor would trigger at roughly -1s. A motor at the middle at 0s and a motor at the front at roughly 1s.
This would guarantee the least amount of stress to the rocket, given we could actually build such precise motors to trigger that fast, which i doubt we can.

Even better, would be if we could generate some in-homogeneous forcefield which matches exactly the acceleration pattern required, but that is sci-fi...

 

[SlowThinker]

Isn't it enough to simply re-run Einstein clock synchronization on the running rocket?
Surely the people on the rocket cannot experience time shifts as they walk back and forth (other than what SR predicts).
I seem to get $\Delta t=\gamma v x=\frac{0.4\times10}{\sqrt{1-0.4^2}}=4.36...$

 

[jartsa]

Constant proper acceleration to some speed takes proper time t when proper acceleration is a. In our case a * t must be about 0.42 c, because a * t is a thing called rapidity, and as velocity is 0.4 c when rapidity is about 0.42 c, we know that constant proper acceleration a must be 0.42 c / t .

In the accelerating rocket there seems to exist an uniform gravity field, where potential is a * h. Acceleration times height. Gravitational time dilation factor is proportional to potential, if we set c=1, then the proportionality constant is 1.

So let us set set c=1, and t=1 s.

a = 0.42 / 1 s
h = 10 s
time dilation factor = 0.42 1/s * 10 s = 4.2

So when clock at rear advances 1 second, a clock at front advances 4.2 seconds, according to observer at rear.

Post acceleration clock difference is 3.2 seconds.

(I assumed that according to the rear observer the bow stays all the time 10 light seconds away. Besides of making the calculation easy, it is a way to end up with a normal length post acceleration rocket)

 

[Ibix]

Isn't it enough to simply re-run Einstein clock synchronization on the running rocket?

Not in general. For the purposes of this experiment, the back of the rocket switches from a "stationary" Einstein frame to a "moving" Einstein frame. In the left hand diagram, the line of "now" used by the back of the rocket swings from horizontal to sloped up to the right at the point when the back of the rocket accelerates. That defines a wedge to the right. The de-synchronisation depends on the length (well, the interval) of the worldline of the front of the rocket in that wedge, which obviously depends on the acceleration profile of the front of the rocket.

Edit: There are significant problems with this line of thinking in general, but it serves here.

Surely the people on the rocket cannot experience time shifts as they walk back and forth (other than what SR predicts).

Indeed. But no law of nature requires the clocks on the rocket to be in sync, unless they were inertial and already in sync.

I seem to get $\Delta t=\gamma v x=\frac{0.4\times10}{\sqrt{1-0.4^2}}=4.36...$

I haven't checked any maths, but I believe PeterDonis and Jeronimus are correct with exactly 4s.

 

[Ibix]

@jartsa - aren't you assuming a rocket under steady acceleration there? I suspect that doesn't work for an impulsive change as Jeronimus has modeled here.

 

[jartsa]

@jartsa - aren't you assuming a rocket under steady acceleration there? I suspect that doesn't work for an impulsive change as Jeronimus has modeled here.

From zero to 0.4c in one second is not impulsive enough?

Well, if the time is one millisecond, then the time dilation factor becomes 4000, which means that when rear clock advances one millisecond, front clock advances 4000 milliseconds according to rear observer, and post acceleration clock difference is 3999 milliseconds. Strange, I expected it to still be 3 seconds.

... Oh, speed is not proper acceleration times proper time , as I thought. But a thing called rapidity is proper acceleration times proper time. Now that I have corrected my calculation I'm getting 3.2 seconds clock difference when acceleration occurs in one second and also when acceleration occurs in one millisecond.

 

[Jeronimus]

ü

From zero to 0.4c in one second is not impulsive enough?

Well, if the time is one millisecond, then the time dilation factor becomes 4000, which means that when rear clock advances one millisecond, front clock advances 4000 milliseconds according to rear observer, and post acceleration clock difference is 3999 milliseconds. Strange, I expected it to still be 3 seconds.

... Oh, speed is not proper acceleration times proper time , as I thought. But a thing called rapidity is proper acceleration times proper time. Now that I have corrected my calculation I'm getting 3.2 seconds clock difference when acceleration occurs in one second and also when acceleration occurs in one millisecond.

According to the x-t diagrams, the difference in times should be slightly over 4 seconds. Not exactly 4 seconds. I expect a value between 4 and 4.1, less than the 4.36s value Slowthinker got. In theory i could calculate the exact value by getting the exact times/places for when the worldlines of the cyan and violet colored clocks cross but that would require me to figure out the math.

edit: It is the second question which i find more interesting to answer once we have an agreement on the time difference. Does the type of acceleration matter for the time difference we get, or is it decided only by the final velocity?
I think the latter is the case but i could not fully wrap my mind around it yet.

 

[Ibix]

I make it 4.17s (actually $10(5-\sqrt {21})$s). The reasoning is this:

In the original rest frame let the back of the rocket start accelerating at the origin of coordinates. The front of the rocket follows $x=L$ before accelerating and $x=vt+L/\gamma$ afterwards. The front-of-rocket acceleration event is where these two lines meet, which is $x=L$ and $t=(L/v)(1-1/\gamma)$. In the post-acceleration frame, this is $x'=L$ and $t'=-(L/v)(1-1/\gamma)$.

For the back of the rocket the elapsed time between $t=0$ and $t'=0$ is zero. For the front of the rocket, though, it's the time from $t=0$ to the acceleration event at $t=(L/v)(1-1/\gamma)$ plus the time from the acceleration event at $t'=-(L/v)(1-1/\gamma)$ to $t'=0$. That is, the extra elapsed time is $2(L/v)(1-1/\gamma)$.

Edit: That is, the times Jeronimus marked as "about 2s" in the left and right diagrams in the OP are actually both $(L/v)(1-1/\gamma)$ of proper time.

Put in the L=10 and v=0.4 and out drops my number.

Regarding your second question, it's easy to see that the result depends on the acceleration profile. All I did was calculate the interval along the "front" worldline from where it crossed the $t=0$ surface to where it crossed the $t'=0$ surface (cheating by using the fact that the line is a "rest" worldline in one frame or other). A different acceleration profile is a different line. A different line can have a different length.

 

[Jeronimus]

I make it 4.17s (actually $10(5-\sqrt {21})$s). The reasoning is this:

In the original rest frame let the back of the rocket start accelerating at the origin of coordinates. The front of the rocket follows $x=L$ before accelerating and $x=vt+L/\gamma$ afterwards. The front-of-rocket acceleration event is where these two lines meet, which is $x=L$ and $t=(L/v)(1-1/\gamma)$. In the post-acceleration frame, this is $x'=L$ and $t'=-(L/v)(1-1/\gamma)$.

For the back of the rocket the elapsed time between $t=0$ and $t'=0$ is zero. For the front of the rocket, though, it's the time from $t=0$ to the acceleration event at $t=(L/v)(1-1/\gamma)$ plus the time from the acceleration event at $t'=-(L/v)(1-1/\gamma)$ to $t'=0$. That is, the extra elapsed time is $2(L/v)(1-1/\gamma)$.

Edit: That is, the times Jeronimus marked as "about 2s" in the left and right diagrams in the OP are actually both $(L/v)(1-1/\gamma)$ of proper time.

Put in the L=10 and v=0.4 and out drops my number.

Regarding your second question, it's easy to see that the result depends on the acceleration profile. All I did was calculate the interval along the "front" worldline from where it crossed the $t=0$ surface to where it crossed the $t'=0$ surface (cheating by using the fact that the line is a "rest" worldline in one frame or other). A different acceleration profile is a different line. A different line can have a different length.

4.17s seems too high still, going by the observation of the x-t diagram which are computer generated.

I don't think you can use the same formulas you use to determine the length of a rocket post-acceleration for this scenario. L/γ is not the case during the acceleration phase. But as always, i could be wrong.

 

[Ibix]

All I'm doing is writing down the equations of the lines you drew. If the back of the rocket accelerates at time t=0 then it must follow x=vt. If the rocket is to have length L in its final rest frame then the front of the rocket must (eventually) be (a) traveling at the same speed as the back and (b) be $L/\gamma$ infront of it. That makes the equation of the line $x=vt+L/\gamma$.

During the acceleration phase the length does vary, yes. But that's because sometimes there's a sloped worldline at one end simultaneous with a vertical one at the other. Doesn't change the equation of the lines.

Try it with v=0.8. My formula says the total time difference there is exactly 10s, 5 on each graph.

Edit: I find your graphs very difficult to read since there's so much going on in them, with the light cones and the numbers and the worldlines. In this case, you really just want regular graph paper and the worldlines drawn with the thinnest line setting you can have.

 

[Jeronimus]

All I'm doing is writing down the equations of the lines you drew. If the back of the rocket accelerates at time t=0 then it must follow x=vt. If the rocket is to have length L in its final rest frame then the front of the rocket must (eventually) be (a) traveling at the same speed as the back and (b) be $L/\gamma$ infront of it. That makes the equation of the line $x=vt+L/\gamma$.

During the acceleration phase the length does vary, yes. But that's because sometimes there's a sloped worldline at one end simultaneous with a vertical one at the other. Doesn't change the equation of the lines.

Try it with v=0.8. My formula says the total time difference there is exactly 10s, 5 on each graph.

Well, let me try a different approach.

The Lorentz Transformation formula solving for x' is

x' = γ (x - vt)

The known for the acceleration event at the back of the rocket are:

x = -5 , x' = -5 , v = 0.4c , γ = 1.091... unknow t=?

solving for t we get

t = (x - x'/γ) /v = -1.04356...s t2 = 1.04356...s (the front of the rocket acceleration event)

now that we have t, we can solve for t' using the Lorentz transformation formula solving for t'

t' = γ(t - vx/c²) = 1.04356...s and t2' = - 1.04356...s (x2'=5 t2'=-1.04356...s is the spacetime position of the acceleration even in the right diagram - the violet clock counter displays _positive_ 1.04356...s however at this spacetime position)

deltaT = t1 - t2 ~ 2.08712s
adding the clock counter of the violet clock at the front side, we get

2.08712s + 1.04356...s = 3.13...s

so we have two events which are simultaneous in the right diagram and are on the endpoint worldlines of the rocket

x'=-5ls, t' = 1.04356...s - the clock count of the violet clock at the back of the rocket is showing -1.04356...s
x2' = 5ls, t2' = 1.04356...s - the clock count of the violet clock at the front of the rocket is showing 3.13...s here

edit: I made a little mistake there. The violet clocks are not the clocks placed inside the rocket. They are only in sync with the rocket clocks at the front at end pre-acceleration. Post acceleration, the violet clocks go out of sync(and move away from the clocks placed at the endpoints of the rocket) relative to the clocks placed inside the rocket. The clock counts at 1.04356...s and 3.13...s for the clocks inside the rocket are accurate though, except those are not the violet clocks. their clock counts therefore differ by 4.1742...s

Looks like you were right after all :D

 

[jartsa]

From zero to 0.4c in one second is not impulsive enough?

Well, if the time is one millisecond, then the time dilation factor becomes 4000, which means that when rear clock advances one millisecond, front clock advances 4000 milliseconds according to rear observer, and post acceleration clock difference is 3999 milliseconds. Strange, I expected it to still be 3 seconds.

... Oh, speed is not proper acceleration times proper time , as I thought. But a thing called rapidity is proper acceleration times proper time. Now that I have corrected my calculation I'm getting 3.2 seconds clock difference when acceleration occurs in one second and also when acceleration occurs in one millisecond.

Sorry guys, subtracting two numbers was just too confusing for me.

Constant proper acceleration to 4.0 c in one second according to rear observer:
Gravitational time dilation factor is 4.2.
Upper clock proceeds 4.2 seconds, rear clock proceeds 1 seconds. 4.2 s - 1 s= 3.2 sConstant proper acceleration to 4.0 c in one millisecond according to rear observer:
Gravitational time dilation factor is 4200.
Upper clock proceeds 4200 milliseconds, rear clock 1 milliseconds. 4200 milliseconds - 1 milliseconds = 4199 milliseconds.

 

[PeroK]

Following scenario:

A rocket measuring 10 lightseconds in length accelerates near instantaneously to v=0.4c. Pre-acceleration two clocks placed at the endpoints of the rocket were synced.

What will be the time difference between the two clocks post acceleration?

edit: Also, is the time difference between the two clocks post acceleration decided only by the final velocity or does the "type" of acceleration (faster/slower) matter?

One way to tackle the problem is to consider the acceleration creates a "gravitational" time dilation for clocks within the rocket given by:

$\tau_1 = (1 + \frac{aL}{c^2}) \tau_0$

Where $\tau_0, \tau_1$ are the proper times at the rear and front of the rocket with rest length $L$. This assumes the rocket moves rigidly in its own frame, but length contracting in the original inertial reference frame, with a constant accleration $a$. We have:

$\tau_0 = \frac{c}{a} \tanh^{-1}(\frac{v}{c})$

Where $v$ is the final velocity of the ship. Hence:

$\tau_1 - \tau_0 = \frac{L}{c} \tanh^{-1}(\frac{v}{c})$

I think, however, that this formula won't work for near instantaneous acceleration, as there is an asymmetry in the acceleration at the front and the end of the ship: the front accelerates more slowly (in coordinate time) and keeps accelerating for longer. I make it that as long as $\frac{\gamma Lv}{c^2}$ is small, the the formula is a good approximation.

Edit: I think that formula is valid for any acceleration. We don't have to worry about converting to coordinate time.

 

[Ibix]

I make it that as long as $\frac{\gamma Lv}{c^2}$ is small, the the formula is a good approximation.

But in this case that quantity $20/\sqrt{21}$, which is slightly larger than 4 which is not small in this context. That's why jartsa gets a different answer to Jeronimus and me.

 

[PeroK]

But in this case that quantity $20/\sqrt{21}$, which is slightly larger than 4 which is not small in this context. That's why jartsa gets a different answer to Jeronimus and me.

I've changed my mind. I think that formula in post #20 is correct. We don't have to worry about coordinate time. We can just work within the accelerated coordinate system.

I've edited it.

 

[PeroK]

As an addendum to post #20. You can calculate the difference in the clocks in the IRF in two ways:

1) From post 20, we have that the front clock is $\frac{L}{c} tanh^{-1}(\frac{v}{c})$ ahead in the ship's frame, after the acceleration, when the ship is moving with constant speed $v$. In the original IRF we would have the normal $-\frac{Lv}{c^2}$ difference in the two clocks, so in the original IRF we would have that the front clock is ahead by:

$\frac{L}{c} tanh^{-1}(\frac{v}{c}) -\frac{Lv}{c^2}$

2) To calculate this solely in the original IRF we have:

$t_0 = \frac{\gamma v}{a}$ is the time the rear of the ship stops accelerating and its clock reads $\tau_0$

$t_L = \frac{\gamma v}{a} + \frac{\gamma vL}{c^2}$ is the time the front of the ship stops accelerating and its clock reads $(1 + \frac{aL}{c^2}) \tau_0$

In the time difference $t_L - t_0$ the rear clock advances by the time-dilated $\frac{vL}{c^2}$. And, with $\tau_0 = \frac{c}{a} tanh^{-1}(\frac{v}{c})$ we have the difference in the clock readings of:

$(1 + \frac{aL}{c^2}) \tau_0 -\tau_0 - \frac{vL}{c^2} = \frac{L}{c} tanh^{-1}(\frac{v}{c}) -\frac{Lv}{c^2}$

Note that I used this method to get the correct length contraction of the ship at the end of the acceleration, so it feels quite sound to me now.

 

[0mega]

In the time difference $t_L - t_0$ the rear clock advances by the time-dilated $\frac{vL}{c^2}$. And, with $\tau_0 = \frac{c}{a} tanh^{-1}(\frac{v}{c})$ we have the difference in the clock readings of:

$(1 + \frac{aL}{c^2}) \tau_0 -\tau_0 - \frac{vL}{c^2} = \frac{L}{c} tanh^{-1}(\frac{v}{c}) -\frac{Lv}{c^2}$

Note that I used this method to get the correct length contraction of the ship at the end of the acceleration, so it feels quite sound to me now.

http://cyberleninka.ru/article/n/o-...ostkogo-sterzhnya-2-rassinhronizatsiya-chasov
Formula (29)

 

[Jeronimus]

In the time difference $t_L - t_0$ the rear clock advances by the time-dilated $\frac{vL}{c^2}$. And, with $\tau_0 = \frac{c}{a} tanh^{-1}(\frac{v}{c})$ we have the difference in the clock readings of:

$(1 + \frac{aL}{c^2}) \tau_0 -\tau_0 - \frac{vL}{c^2} = \frac{L}{c} tanh^{-1}(\frac{v}{c}) -\frac{Lv}{c^2}$

I am getting

10s * tanh^-1(0.4) - 0.4s = 0.2364893...s with my calculator.edit: Never mind. I see now. You calculated the difference in clock counts for the original/initial IFR, not the IFR the rocket is at rest in post acceleration.