Relativity of simultaneity help

[striphe]

If I have a 1 light second long spacecraft with a clock at each end that are synchronised before take off, then I accelerate the craft to 0.5C. Are the clocks synchronised or is the leading clock 0.5 seconds ahead of the trailing clock?

 

[jtbell]

In the ship's new rest frame, the two clocks remain synchronized. In the ship's initial rest frame (the "Earth frame" if it starts from Earth), the clock at the front of the ship is now 0.5 second behind the clock at the rear.

 

[striphe]

I'm having particular amount of difficulty understanding how this outcome arises as the clocks have been synchronised in the Earth frame. Could please explain.

 

[chinglu1998]

I'm having particular amount of difficulty understanding how this outcome arises as the clocks have been synchronised in the Earth frame. Could please explain.

I will help you.

You apply SR acceleration equations in the context of the accelerated frame.

T(t) = c/a arsinh(a t/c)

This is absolute.



http://www.ias.ac.in/currsci/feb252007/416.pdf
http://www.ejournal.unam.mx/rmf/no521/RMF52110.pdf
http://users.telenet.be/vdmoortel/dirk/Physics/Acceleration.html
http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html
http://en.wikipedia.org/wiki/Twins_paradox
http://arxiv.org/PS_cache/physics/pdf/0411/0411233v1.pdf

 

[Dale]

In the ship's new rest frame, the two clocks remain synchronized. In the ship's initial rest frame (the "Earth frame" if it starts from Earth), the clock at the front of the ship is now 0.5 second behind the clock at the rear.

Doesn't it depend on the details of the acceleration profiles of the front and back clocks (e.g. Born rigid vs. simultaneous acceleration in the launch frame vs. something else)?

 

[jtbell]

Right, I was basically assuming Born rigid motion (I think).

It occurred to me after I posted, that if the two clocks are not attached to a spaceship or to each other, and each has its own rocket attached to it, and the two rockets accelerate identically in the Earth frame up to their final velocity, then at the end, the two clocks are still synchronized in the Earth frame, and the distance between them is not length-contracted in the Earth frame (both unlike the case with a Born-rigid rocket). This puts us in the territory of "Bell's spaceship paradox" which has been discussed here extensively.

 

[Dale]

Yes, I agree. I think the OP needs to completely specify the acceleration profile of both clocks and specify in which frame we are doing the comparison after the acceleration.

 

[striphe]

I think i was going with born rigid as i was talking about one spacecraft rather than two and run with an on-board perspective.

I understand that length contracts on board the spacecraft due to acceleration to an observer who is at the initial velocity of the craft. This would mean that the leading clock would be ahead of the trailing, as the leading was going slower than the trailing during acceleration, due to the clocks converging, due contraction of the vessel, during acceleration of the whole craft.

 

[Dale]

I think i was going with born rigid as i was talking about one spacecraft rather than two and run with an on-board perspective.

Then you want to study Rindler coordinates:
http://en.wikipedia.org/wiki/Rindler_coordinates

In units where c=1 and g=1 where g is the acceleration of the clock at the rear of the rocket we have the metric:
ds² = -x² dt² + dx² + dy² + dz²

In these coordinates the clock at the rear of the rocket (x=1) has proper time equal to coordinate time:
ds² = -dt²

and the clock at the front of the rocket (x=1+L) shows gravitational time dilation:
ds² = -(1+L)² dt²

So if the rear accelerates at g for a time T (and the front clock accelerates Born-rigidly with the rear clock) then the front clock will read (1+L) T.

 

[striphe]

Can you give an example of how this is applied to the situation and what the outcome will be?

 

[Dale]

So let's say that we are using units of years and lightyears such that c = 1 lightyear/year and g = 1 lightyear/year² = 9.5 m/s². To accelerate from rest to 0.5 c requires a time T = arctanh(0.5) = 0.55 year. The length of the ship is L = 1 lightsecond = 3.2E-8 lightyear.

So at the end of the acceleration in the rest frame of the ship the rear clock will read
T = 0.55 year

The front clock will read
(1+L) T = 0.55 year

The difference will be
(1+L) T - T = L T = 1.7E-8 year = 0.55 s

 

[striphe]

Just to confirm. This calculation does not include relativity of simultaneity, The front clock will be ahead of the back clock and the effect will be come less as you lengthen the duration of acceleration to the particular target velocity.

 

[Dale]

Please read the link on Rindler coordinates. You asked for the "on-board" perspective. This is it, including all relativistic effects for this non-inertial frame.

 

[yuiop]

In the ship's new rest frame, the two clocks remain synchronized. In the ship's initial rest frame (the "Earth frame" if it starts from Earth), the clock at the front of the ship is now 0.5 second behind the clock at the rear.

Right, I was basically assuming Born rigid motion (I think).

It occurred to me after I posted, that if the two clocks are not attached to a spaceship or to each other, and each has its own rocket attached to it, and the two rockets accelerate identically in the Earth frame up to their final velocity, then at the end, the two clocks are still synchronized in the Earth frame, and the distance between them is not length-contracted in the Earth frame (both unlike the case with a Born-rigid rocket). This puts us in the territory of "Bell's spaceship paradox" which has been discussed here extensively.

Your analysis of the second case when the two rockets accelerate identically in the Earth frame up to their final velocity. The clocks will still be synchronised in the Earth frame but in the rocket frame the the rockets will be further apart and the clocks will be out of sync. To synchronise the clocks in the new frame, the rear clock will have to be advanced or the front clock retarded.

In the case of Born rigid motion, the clocks will out of sync in the Earth frame AND in the rocket frame, even though the spatial separation of the clocks appears to be the same in the rocket frame. In this case the de-synchronisation of the clocks will be even greater than the case above, because the rear clock has been traveling faster and lost even more time relative to the front clock, when it should be ahead of the front clock (by L*v/c^2 in the Earth frame) to be synchronised in the rocket frame.

Neither acceleration scheme allows the clocks to remain synchronised during or after the acceleration phase. To keep the clocks synchronised would require the front clock to accelerate faster than the rear clock and the length of the rocket would increase in both the Earth frame and the rocket frame (i.e. its proper length would have to increase and the rocket would probably be torn apart). A practical way to keep the clocks synchronised would be to have a separate rocket for each clock. It would be interesting to work out the required acceleration formulas to keep accelerating clocks synchronised.

This old thread contains calculations and diagrams for similar problem https://www.physicsforums.com/showthread.php?p=1614307#post1614307 and this old thread too https://www.physicsforums.com/showthread.php?p=1753166#post1753166 (See diagram in post #76)

 

[striphe]

The difference will be
(1+L) T - T = L T

Based on this, if I was to have the rocket instantaneously accelerate, there would be no effect. Correct?

 

[Dale]

Then it wouldn't be Born rigid acceleration so my derivation wouldn't apply.