# A few questions about GR

1. ### alexbib

62
1. The time distortion due to relative velocity does not cause any "real" loss if time in itself,loss of time only happens with acceleration, right? How can I predict the amount of time that has been lost relative to an inertial point of view(if I know the value for the other variables)?

2. Is the gravitational field created by an object relative? ie: does the m in the formula vary with 1/(1-sqrt(v^2/c^2))?

3. c is the limit of the velocity an object can have. Is there a limit to the acceleration an object can have (I am speaking only of instantaneous acceleration here)?

2. ### Ambitwistor

837
No. But you may need to accelerate in order to compare clocks.

http://math.ucr.edu/home/baez/physics/Relativity/SR/clock.html

What variables? And what do you mean by "amount of time lost relative to an inertial point of view"?

In general, to find out how much time elapses between two events according to some observer, integrate the proper time along that observer's worldline.

The source of the gravitational field is not just mass, it is stress-energy. One component of the stress-energy tensor is mass-energy density, which varies like $$\gamma^2 = 1/(1-v^2/c^2)$$.

Not in general relativity.

3. ### lethe

657
i m not sure what it means for time to be "lost". time runs slower in one frame relative to another, if the frame is in motion. time a frame with a stronger gravitational field runs slower relative to an inertial frame as well, since, by the equivalence principle, that frame is accelerating.

yes, the motion of the particle effects the gravitational field, although it is not so simple as just multiplication by gamma. there are gravito-magnetic effects.

no limit to the acceleration a particle can have, in its rest frame, as far as i know.

4. ### alexbib

62
" m not sure what it means for time to be lost" I mean that if you speed up your watch to .95c and then turn it around and make it comeback to its original position it will be behind a clock that has not moved by a certain amount of time. When you compare them, the rate at which they count time should be equal, but the clock that has moved will be behind the other.

5. ### Ambitwistor

837
There's no such thing as a "clock that has not moved"; motion is relative. A clock that doesn't move in one frame, will move in any other frame.

Other than that, you're describing the usual SR twin paradox, and you determine the aging of the clocks by the method I described.

If the clock that made the trip comes back at relative rest to the original clock, it will be ticking at a rate equal to the original clock when you compare them (but the total amount of time elapsed on it will be less; they were ticking at different rates during the trip).

6. ### alexbib

62
false. By hasn't moved, I meant that has not accelerated. I use this clock as reference frame. When the other clock comes back from the trip, it will be lagging behind (not in counting rate, but in the hour shown). This has even been proven experimentally. Once they are back at the same place, and are at relative rest, the clock that has accelerated is late. At this moment, from any reference frame, clock that has accelerated IS late.

Edit: sorry, I misread your last post, we are in fact agreeing on this. But obviously, the fact that less time has passed on the clock that made the trip depends on acceleration, not relative speed. What depends on relative speed at any moment is the rate at which they count the time from each other's prespective.

Last edited: Nov 20, 2003
7. ### Ambitwistor

837
If you work out the formula for the elapsed time for the travelling twin, it depends directly on the relative velocity, not the acceleration. If you work out the total elapsed time according to either clock, you do so by integrating the gamma factor (which depends only on relative speed) over the worldline of the clock.

(Acceleration of course can affect relative speed and so can indirectly influence the relative clock rate by means of a changing relative velocity, but the duration of acceleration can be made to have an arbitrarily small effect on the elapsed proper time --- no effect, in the case of instantaneous acceleration.)

Last edited: Nov 20, 2003
8. ### alexbib

62
but if you have two clocks that move with relative velocity v, wihtout accelerating (or rather, let's say that they are in orbit around a planet, so they undergo the same acceleration), and they come back to the same initial position, after completing a turn in opposite directions, their clocks will agree (the question here is: do they agree, in the hour they display, acording to the observers travelling with the clocks, or do theym only agree to a "neutral" observer that we use as a reference frame?), even though each one sees the other running slow.

If they only agree according to a neutral observer, then you are right and the fact that they disagree in the situation where one of them accelerates is only due to the fact that the observers share the reference frame of the inertial watch

9. ### Ambitwistor

837
If you set up a symmetric situation, then you will get symmetric results. If you set up an asymmetric situation, such as one accelerating and the other not (although there ways to set up an asymmetric situation other than by accelerating), then they will get different results -- but, for the reasons I gave, the elapsed time even in this case does not fundamentally involve the acceleration: the period of acceleration can be made to contribute an arbitrarily small amount to the elapsed proper time.

10. ### alexbib

62
yeah, I get what you mean. but we agree that in the situation where one of the clocks is accelerating and then coming back at rest(relative to the other clock) at its initial position, once they are at the same position and at relative rest, their reference frames become one and the same. So they both agree that the inertial clock has counted more time? How can you explain that, from the point of view of the clock that has accelerated, the other clock must have actually been running faster, even though it had a relative velocity? By your method, from the point of view of the clock that has accelerated, it is the inertial clock that should be late, while in reality (and in experiement), they both agree that the accelerating clock is late.

11. ### Ambitwistor

837
Yes.

At all times, each observer finds that the other's clock is running more slowly than his own, by the same factor (let's assume instantaneous turnaround to keep things simple). This does not imply that their clocks should read the same elapsed time at the end.

There are many different ways of resolving the twin paradox. Take your pick:

The one I usually use is the one the FAQ calls "the spacetime diagram explanation". You sound like you might prefer reading "the Doppler shift explanation" or "the time gap objection".

If by "late" you mean "less proper time has elapsed", then no. According to me, more proper time should elapse for the inertial observer.

12. ### alexbib

62
So you're saying that from the point of view of the accelerating clock, at every moment, it sees the other clock running slow, but in the end it sees it displaying more elapsed time? And even if it is so, the only difference between clock A and B is that B has accelerated, and it displays less elapsed time. Then the acceleration should be the cause of the "missing" time on it. Because, as we have seen with symmetrical situations, relative velocity alone does not cause a difference in elapsed time.

Thanks for the link, I'll take a look!

13. ### Ambitwistor

837
Well, you have to be a little careful: the observers can visually see the other clock running slower or faster (the Doppler shift explanation), but that's due only partly to the actual time dilation; it's also partly due to optical effects arising from the finite propagation speed of light. If you correct for that, using an Einstein rods-and-clocks synchronized grid of observers, then each observer finds that the other clock is really running slower (by a factor gamma).

In this example, the acceleration breaks the symmetry, because the travelling twin exists in two inertial frames. However, since the acceleration period itself makes an arbitrarily small contribution to the elapsed proper time of the travelling twin, it's not really accurate to say that acceleration causes time differences; see also the original clock postulate FAQ I cited.

14. ### Janus

2,367
Staff Emeritus
If you break the trip down by stages, this is what the accelerating clock measures:

1. Accelerating away.
Inertial clock changes from in sync to running slow.

2. Coasting away.
Inertial clock running slow.

3. decelerating to a stop relative to the inertial clock.
Inertial clock running fast.

4. Accelerating towards inertial clock.
Inertial clock running fast.

5. Coasting towards inertial clock.
Inertial clock runs slow.

6. Decelerating to a stop next to inertial clock.
Inertial clock goes from running slow to being in sync.

After all these combined effects are taken into account, the accelerating clock will have found that less time has past for it.

Meanwhile, the inertial clock will just see the accelerating clock run slow the whole time just due to relative velocity.

(One point. during phases 3 and 4, the accelerating clock will see two effects operating on the inertial clock: the tendancy to run fast due to the acceleration, and the tendancy to run slow due to relative velocity. Which one dominates at any given time depends on the distance form the inertial clock, the relative velocity and the acceleration. The total measured elasped time for the inertial clock will always end up as more for these periods.)

15. ### alexbib

62
Thx, good explanation!

"the tendancy to run fast due to the acceleration"

how does the acceleration cause this?

16. ### Ambitwistor

837
Acceleration itself does not determine the rates of clocks. Velocity does. Acceleration only influences the rates of clocks insofar as it changes the velocity.

17. ### Ambitwistor

837
It may not be clear from your statements that what somebody "sees" when they look at the other clock (due to time dilation combined with optical effects) is different from what the relative rates of the clocks actually are (as defined by an Einstein synchronization procedure). (I will refer to the former as "seeing" and the latter as "measuring".) What one of the clocks measures is always that the other clock is running more slowly by a factor $$\mbox{\gamma=1/\sqrt{1-(v/c)^2}}$$, for either clock and during any leg of the journey. There is never any extra time dilation factor arising from the acceleration.

18. ### alexbib

62
No, I think Janus is right, the inertial clock actually running faster has to be due to acceleration. Go read the GR explanation from the link you provided.

19. ### Ambitwistor

837
It's all well and good to work in a non-inertial frame, if you want. But if you actually compute the time dilation factor between the inertial and non-inertial clocks at any instant of time, you will find that it can be expressed solely in terms of their relative velocity (computed in a hypersurface of simultaneity), even in the GR explanation.

The "gravitational field" in the non-inertial frame is precisely arranged to have this property (i.e., the acceleration depends on the relative velocity in just such a way to make the time dilation depend only on relative velocity); it has to, in order to agree with the observations in an inertial frame.

The reason why the analysis goes differently in an accelerating frame is not because the time dilation factor depends directly on the proper acceleration or anything like that (if it doesn't in one frame, it doesn't in any frame) -- it's just because the acceleration changes the surfaces of simultaneity.

It is an empirical fact that SR time dilation depends only on velocity, not acceleration. That was the point of the clock postulate FAQ on the same site.

20. ### alexbib

62
I get your point. As you say, it depends only on the velocity, and the best way to calculate it is to do it with time distortion relative to the inertial frame.
However, I also think it is important to understand why you still get more elapsed time on the inertial clock from the non-inertial clock's point of view, which is (in my opinion) explained pretty well on the link you posted.