Special relativity and acceleration

1. Jul 16, 2009

Zman

Is there a relationship in special relativity between acceleration and time dilation?
Or even acceleration and length contraction?

2. Jul 16, 2009

sylas

The primary relation is between relative velocities and dilation/contraction. Acceleration leads to changes in dilation/contraction factors, but only because it alters velocity.

It gets a bit subtle when you have two accelerating observers maintaining a constant separation along the direction of accelerated motion, according to their own accelerating frame of reference. In that case, there is a time dilation between the observers, which is analogous to the dilation observed with altitude in a gravitational field. You can still derive it from the underlying velocity based dilation. In SR (which does not deal with gravity) it all comes down to velocity.

Cheers -- sylas

3. Jul 16, 2009

Zman

The reason that I ask the question is that Einstein’s equivalence principle says (basically) that one can’t differentiate between inertial and gravitational acceleration.
If this is the case then an inertial acceleration of g should also experience the same time dilation as a body at a particular gravitational potential where the acceleration due to gravity is g.
For the gravitational body, its clock rate reference is zero gravity. For the inertial body, its clock rate reference would be an observer that was not accelerating but could have any velocity. The time dilation due to the velocity is not the issue here. Given the equivalence principle there presumably would be a time dilation contribution due to the acceleration otherwise the equivalent principle wouldn’t be entirely equivalent.

4. Jul 16, 2009

sylas

It does: and this is what I describe above in the previous post.

Note that you cannot simply compare an accelerated observer with an unaccelerated observer, because in that case you also get an increasing velocity difference, and that dominates any time dilation.

However, if you have a long spaceship experiencing a constant acceleration at all points, it turns out that the front has slightly less acceleration than the back, and there is a time dilation between the front and the back of the ship... but no change in the distance between them, as measured by anyone on the ship. THIS is what turns out to be exactly analogous to the time difference of two observers at different altitudes in a gravitational field.

Cheers -- sylas

5. Jul 16, 2009

George Jones

Staff Emeritus
Also, common statements about time dilation involve somewhat inconsistent interpretations. Special relativistic time dilation for moving clocks refers to the difference in elapsed coordinate times, not a visual effect that is actually seen through a telescope. In terms of visual effects, moving clocks can run fast or slow. Gravitational time dilation is a visual effect that can be observed through a telescope.

6. Jul 16, 2009

Zman

Why can’t we separate the velocity contribution to time dilation and the acceleration contribution to time dilation? At a particular point in time there will be a given velocity and a given acceleration.

Trying to work this one out, I keep getting more acceleration at the front (less at the rear). How did you work it out?

Cheers Zman

7. Jul 16, 2009

sylas

You can probably do that; but I would prefer to simply integrate proper time over the world line, without trying to decompose it. The point is that the case where there is a continually increasing relative velocity is not going to be a good match with a gravitational time dilation example.

I wrote it from memory, which is not totally reliable. Here's how I rethought it to answer your question... consider a particle with constant proper acceleration a, using the parametric equations, with u as proper time, and x and t as co-ordinates in a suitably chosen inertial frame.
\begin{align*} t &= \frac{c}{a} \sinh (ua/c) \\ x &= \frac{c^2}{a} \cosh (ua/c) \end{align*}​

Let this represent the front of the ship. Now imagine a photon sent backwards at a proper time u-d, and another received forward at time u+d. Their point of crossing defines a rear of the ship, which is a constant distance cd from the front. For a given u, let this rear be at (t', x') in the inertial frame.

Then
\begin{align*} x'+t'c & = \frac{c^2}{a} (\cosh ((u-d)a/c) + \sinh ((u-d)a/c)) \\ &= \frac{c^2}{a} e^{(u-d)a/c} \\ x'-t'c & = \frac{c^2}{a} (\cosh ((u+d)a/c) - \sinh ((u+d)a/c)) \\ &= \frac{c^2}{a} e^{-(u+d)a/c} \\ x' &= \frac{c^2}{a} e^{-da/c} \cosh(ua/c) \\ t' &= \frac{c}{a} e^{-da/c} \sinh(ua/c) \end{align*}​

Hence, the rear of the ship, identified in this way, has an acceleration a', and a proper time u', so that
\begin{align*} a' & = a e^{da/c} \\ u' &= u e^{-da/c} \end{align*}​

I'm taking d and a as positive, and so u' is running slow, just like a clock inside a gravitational well runs slow, and the acceleration at the rear is greater than at the front.

See also: Born Rigidity, Acceleration, and Inertia at www.mathpages.com

Cheers -- sylas

8. Jul 17, 2009

Al68

The reverse of that argument is that according to the equivalence principle, since time dilation occurs for observers "stationary" in an accelerated reference frame (like a rocket) then time dilation should also occur for observers stationary in a gravitational field. This is exactly how gravitational time dilation was predicted by Einstein.

9. Jul 19, 2009

Austin0

What do you think about the tests that seem to indicate that the curvilinear component of the integrated world line due to acceleration, doesn't have an actual ,real world, time dilation correspondence????

Do you think that an accelerated frame , after attaining a new inertial velocity, would then have to resynchronize its clocks to compensate for this asymetric dilation of the rear clocks??
Thanks

10. Jul 19, 2009

Austin0

I thought that this was not the case. That tests like Gravity Probe B , rotational acceleration tests and particle accelerator tests indicated that all observed dilation was due to the sum of instantaneous velocities with no actual dilation attributable to the acceleration itself. DO you have links to other findings??? Thanks

11. Jul 19, 2009

DrGreg

What you say applies to the case when an inertial observer measures an accelerating object. In those conditions, dilation depends only on velocity and not on acceleration.

But it doesn't work the other way round, when an accelerating observer makes the measurement. In that case there is dilation depending on acceleration, even if the object being measured is a fixed distance from the observer in the observer's accelerating frame. The explanation for this is that a fixed distance in an accelerating frame becomes a Lorentz-contracting distance in an inertial frame, so there is movement and dilation.

12. Jul 19, 2009

sylas

In the context of the system I describe, there are two observers, one at the front and one at the rear of a spaceship. Both observers experience a constant acceleration, but the one at the front has a slightly smaller acceleration. The distance between the observers remains constant, as determined by either observer.

There can be no synchronization, and no re-synchronization. This is a steady state example which carries on indefinitely. As long as the acceleration remains constant and the distance remains fixed, the clock at the rear of the ship falls steadily behind the one at the front, because of the time dilation effect calculated.

This is a standard result, and can be calculated from special relativity. I gave a quick outline of the maths above. The conclusion has the status of a mathematical theorem. It's not just my guess about what goes on; it is the necessary implication of relativity for constant acceleration in a space ship.

Cheers -- sylas

13. Jul 20, 2009

Al68

Sure, time dilation for inertial observers is due to velocity, not directly due to acceleration. That's why I specified "observers "stationary" in an accelerated reference frame", since these accelerated observers will have relative velocity between them as measured in an inertial frame. The time dilation calculated from the inertial frame due to relative velocity will equal the "gravitational" time dilation measured by the accelerated observers due to their accelerated frame.

Another way to look at it is that gravitational time dilation is also solely attributable to relative velocity as measured in an inertial frame.

14. Jul 20, 2009

Austin0

Originally Posted by Austin0
Do you think that an accelerated frame , after attaining a new inertial velocity, would then have to resynchronize its clocks to compensate for this asymetric dilation of the rear clocks??

.
If you will check what I said you will see that I was specifically refering to after the period of acceleration [of whatever duration] , when the system was once again in inertial motion.
If one way light speed tests were conducted between the front of the system and the back,,, and the back to the front , do you think that

A --- The tests would result in c as usual???

B --- The tests would result in asymetric and incorrect results due to the desynchronization due to the greater dilation that had occured in the rear clock.
The clocks would have to be resynchronized with light???

I was not questioning either the math or your command of the math, both are beyond me. ANy reservations I have are regarding the physical assumptions behind and the implications deriving from the conclusion. Thanks Stephen

15. Jul 20, 2009

Austin0

Are you refering here to the infinitesimal difference in acceleration or instantaneous velocity due to length contraction???

Measured how??? Does this mean you would choose B in the post above???

Intriguing concept I will have to give more thought to .

16. Jul 20, 2009

sylas

I was describing a continuous never ending constant acceleration. During this acceleration, the clocks run at different rates.

If acceleration stops, then the clocks will run at the same rate again, and so you can sychnronize them if you like. They'll certainly be out of sync after any period of acceleration, given that they were running at different speeds.

Of course, in most cases with which we are familiar, the spaceship is small or the acceleration is weak or the duration of acceleration is short, so that the dilation effect between front and back is very very small.

The speed of light is measured as c, in all cases, by all observers, accelerating or not.

Each clock is assumed to be perfectly correct in measuring the passage of time, in all conditions. They move out of sync because they are correct; because time dilation is a real effect that can be measured by correct clocks.

In a state where acceleration has stopped and the entire ship is moving at one constant velocity, you can sychnronize the clocks because they'll be running at the same speed. You can synchronize them however you like.

No problem. I understand that. I'm just explaining the nature of the assumptions I am making... namely, that relativity is correct. There's no other special assumption needed; the result is necessary consequence of the maths of relativity.

Technically, there's is an assumption of a "rigid" spaceship, which is a very natural assumption that you might not even think of. It means that the spaceship size remains always about the same for passengers on board. You don't have any continuous deformation or compression of the ship. You might not even think of this as an assumption, although it is needed in the derivations.

Cheers -- sylas

17. Jul 20, 2009

Austin0

Given any kind of real world acceleration wouldn't the linear distance difference, in total path length between the front and back of the system over the full course of acceleration, be negligable in terms of relative velocity ???? Or relative acceleration???
If the cumulative overall difference is slight wouldnt the instantaneous or slight interval difference be vanishingly small??
I am assuming that real world acceleration would mean that as system length and mass increased, that time of acceleration/path length would increase also. That if you consider a very long system where the contraction difference would be greater it would also take longer to achieve comparable velocities. ??? Thanks

18. Jul 21, 2009

Austin0

Would you agree that, by definition and convention, any set of clocks that measures the speed of light as c in both directions is synchronized ???

Would you agree that any set of clocks that are not synchronized within the terms of that convention, could not possibly measure the speed of light as c in both directions????

SO if you believe #2 above [which is what I believe] how do you justify #1 above.

In what sense can they be determined or even considered as out of synch if they return correct results for light tests???

In what possible way could the assumed dilation be empirically confirmed ,you think it is not perceived by outside observers in inertial frames and does not effect the functioning of clocks in some observable way within the system itself????

#3a What do you mean by correct in this context????

I have no question that time dilation is a real effect on real world clocks. But there is also no question that there is uncertainty and lack of concensus regarding the :

A physics involved. Is there physics involved?? Is it just a coordinate effect, a purely relative perception without any physical implications?? We see the same effect due to gravitational potential in which case we do assume an underlying physicality to the phenomenon.

B relationship to acceleration. The twins question. While it doesnt seem to actually produce dilation , it in some way is considered to turn relative[reciprocal] dilation, due to velocity, into a nonreciprocal phenomenon .
Kind of a catalytic effect. Whether or not this is correct it is certainly without explanation or reason to be found within the conceptual or mathematical structure of SR.

QUOTE] I'm just explaining the nature of the assumptions I am making... namely, that relativity is correct. There's no other special assumption needed; the result is necessary consequence of the maths of relativity.[/QUOTE]

I am certainly not questioning the assumption that SR is correct. But where in the Lorentz maths does it become inevitable that acceleration causes time dilation???

Without additional assumptions regarding the physics of acceleration.

Without the assumption that the perceived contraction relative to some inertial frame has actual physical meaning and implications. [which may be true but are not known or understood at this time]

Without the assumption of relativistic differentials of velocity between the front and the back.

This whole question seems to assume a conception of acceleration that is divorced from its basic meaning of a change of velocity over time. To disregard the D in D/t.
It seems to say that over the course of an acceleration, the total distance traveled by the rear [ R] dx relative to the total distance traveled by the front - [F] dx could be a relativistic velocity Fdx - Rdx/ t = [relativistically significant] v where Rdx = Fdx +gamma (length of system)
Does this seem realistic to you????
Thanks Stephen

19. Jul 21, 2009

sylas

No. As I have said, all clocks measure the speed of light as c. This applies for ALL clocks, whether they are synchronized or not.

No. All clocks measure the speed of light as c.

Both statements are false. This is fundamental.

In the sense that they run at different rates. Note that measurement of time AND length depends on the frame. Hence there is no contradiction with different observers measuring the same speed for light, even though they measure times and distances with different values. It is light speed that is the same for all frames; but not times or distances.

Time dilation is measured directly using clocks. There are many experiments doing this. My favourite is the family that measured a gravitational time dilation by carrying a small van with three atomic clocks up Mt Rainer for a holiday weekend. Dad took the kinds for an exciting and educational holiday, while Mum stayed home watching over atomic clocks left in the kitchen. It's described in [post=2177891]msg #10[/post] of thread "Gravitational Time Dilation - Confused".

A clock is correct if it lets you measure the passage of time.

There is only uncertainty and lack of consensus with students who don't actually know enough physics yet. The physics is completely unambiguous and any student who can pass an introductory course in relativity should get precisely the same answers. If they don't, then they they are wrong. Relativity is a consistent theory that is thoroughly tested and gives only one possible answer to these questions.

You've come to the right place to learn more about it... but make no mistake... you do need to learn more about it.

I gave the maths before. You can't simply use a Lorentz transformation; that only applies for mapping between non-accelerating frames. But with a bit of calculus applied as well, the result falls out.

This is a bit more advanced than just using the Lorentz transformation itself, but from your initial questions in this post, I think you are best to get thoroughly familiar with inertial frames, and measurement of light speed for inertial observers, before worrying about the accelerating case.

Cheers -- sylas

20. Jul 21, 2009

Al68

The latter.
No, of course not. The way I read that post, there is no proper acceleration when the measurement is taken.
The concept isn't new, this was the basis for Einstein's prediction of gravitational time dilation to begin with. I just worded it in a weird way for this thread.

Obviously, we can predict the elapsed time on each of two accelerated clocks between two defined events by using SR time dilation due to velocity relative to an inertial frame. We can then derive equations that can in turn be used in the accelerated frame to predict the same thing for the same clocks. Then we call it gravitational time dilation in the accelerated frame. That's essentially what Einstein did around 1908? Maybe there is a link to the paper online I could find.