Two watches, one thrown in the air.

[pervect]

I should have made myself more clear, this was in response to CarlB's solution, using special relativity.

OK, was this his post #20 that you were criticizing?

 

[pervect]

A general comment:

Much of the confusion about dealing with accelerations in special relativity is addressed in the sci.physics.faq

Does a clock's acceleration affect its timing rate?

I'll take the liberty of quoting the first few paragraphs

It's often said that special relativity is based on two postulates: that all inertial frames are of equal validity, and that light travels at the same speed in all inertial frames. But in real world scenarios, objects almost never travel at constant velocity, and so we might never find an inertial frame in which such an object is at rest. To allow us to make predictions about how accelerating objects behave, we need to introduce a third postulate.

This is often called the "clock postulate", but it applies to much more than just clocks, and in fact it underpins much of advanced relativity, both special and general, as well as the notion of covariance (that is, writing the equations of physics in a frame-independent way).

The clock postulate can be stated in the following way. First, we take the rate that our frame's clocks count out their time, and compare that to the rate that a moving clock counts out its time. Before the clock postulate was ever thought of, all that was known was that when the moving clock has a constant velocity v (measured relative to the speed of light c), this ratio of rates is the Lorentz factor

gamma = 1/sqrt(1-v2)

The clock postulate generalises this to say that even when the moving clock accelerates, the ratio of the rate of our clocks compared to its rate is still the above quantity. That is, it only depends on v, and does not depend on any derivatives of v, such as acceleration. So this says that an accelerating clock will count out its time in such a way that at anyone moment, its timing has slowed by a factor (gamma) that only depends on its current speed; its acceleration has no effect at all.

In other words, the accelerated clock's rate is identical to the clock rate in a "momentarily comoving inertial frame" (MCIF), which we can imagine is holding an inertial clock that for a brief moment slows to a stop alongside of the accelerated clock, so that their relative velocity is momentarily zero. At that moment they are ticking at the same rate. A moment later, the accelerated clock has a new MCIF, again one that is moving momentarily to match its speed, and there is a new inertial clock that briefly slows to a stop alongside of the accelerated clock. And again, the rates of accelerated clock and the new inertial one will momentarily be the same.

 

[pervect]

I've trimmed a lot of text, it may be necessary to read the original post to get the full context.

I'm not suggesting this is a proper view of the universe as seen by T_a, but it all seems to be consistent with a special relativistic view of the problem and, to me at least, casts doubt on the validity of any quasi-inertial approach to treating accelerating reference frames. I've always assumed the general theory could account for these intervals of acceleration properly, but I don’t see it happening from a calculation like the one you did.

Feel free to tell me what's wrong with my description of the scenario. That' why I'm throwing it out there.

The view of an accelerated observer is actually even weirder than you described.

An accelerated observer has a local coordinate system, but this coordinate system does not cover all of time and space. It has an event horizon, known as the "Rindler horizon".

Peeking at a clock hidden behind an event horizon is going to lead to some major contradictions - I'd suggest that you refrain from doing this, even though it was in your thought experiment.

For someone accelerating at 1G, the rindler horizon will be about 1 light year behind him. Signals from the region of space-time behind this horizon will never reach the accelerating observer as long as he continues to accelerate.

The best discussion I'm aware of is in MTW's "Gravitation", in the section on accelerated observers, though I assume other GR textbooks talk about the issue.

One way of discussing the dificulties is to look at the metric of space-time according to an accelerated observer with an acceleration of 'g', the "rindler metric"

ds^2 = -(1 + g*x)^2 dt^2 + dx^2 + dy^2 + dz^2

(note that geometric units are used, so it is assumed that the speed of light is unity).

You can see that g_00 on this metric becimes zero when g*x becomes -1.

Compare this to the Schwarzschild metric of a black hole previously posted

ds^2 = (-1 + r/rs) dt^2 + dr^2/(1-r/r_s) + theta, phi terms

and note that g_00 = 0 at the event horizon of the schwarzschild coordinate system when r=r_s

So we see some similarites between the Rindler metric and the Schwarzschild metric right away, the Rindler horizon is often used to help illustrate some of the properties of the event horizon of a black hole, and vica-versa.

A feature of the Rindler metric (of an accelerating observer) not shared by that of the black hole/Schwarzschild metric is that |g_00| becomes very large in the Rindler metric for large positive x. This corresponds to clocks ahead of the spaceship running really, really fast - the larger the distance the clock is in front of the ship, the faster it runs.

But guess what - all of this weirdness is tied to using the coordiante system of an accelerated observer, one which corresponds to the above metric.

When you use the coordinate system of an inertial observer, the metric coefficients are always

ds^2 = -dt^2 + dx^2 + dy^2 + dz^2

And when you use this metric, all clocks run normally - there is no gravitational field, and there is no gravitational time dilation effects.

(see my previous comments about the sci.physics.faq and accelerating clocks).

And you can always use this equation to find the Lorentz interval between any two events by integrating ds over the path, as long as you are in a non-accelerated coordinate system.

The thread

https://www.physicsforums.com/showthread.php?t=77334

may also be of some interest with regards to the viewpoint of an accellerated observer, it discusses some similar issues.

And of course there is the sci.physics.faq on the relativistic rocket (I've previously posted the link).

 

[OlderDan]

For someone accelerating at 1G, the rindler horizon will be about 1 light year behind him. Signals from the region of space-time behind this horizon will never reach the accelerating observer as long as he continues to accelerate.

Thank you for replying to my question. It will take some work on my part if I am going to comprehend the implications of the metrics you have talked about. I'd like to follow up qualitatively on this one paragraph. In particular, if the rindler horizon is behind the accelerating observer, what is going on in front of him? Is there a limiting horizon in both directions? If while T_a is at rest relative to T_i they both see the same time on the star clock at a distance of 20 light years, as T_a accelerates the distant clock must run off nearly 20 years of time, and travel 19+ light years so that when he stops accelerating the star clock is correctly unsynchronized relative to the Earth clock, and less than a light year away. Can he witness this evolution, or is this like some magicians curtain that he cannot see behind until he stops accelerating, with an altered universe revealed when the curtain is lifted?

 

[pervect]

The short answer to your question is "yes", he can witness the evolution in fast motion. Now for the long answer.

The horizon is behind the accelerated observer. Light signals can go both to and fron an event in front of the accelerated observer, so there is no horizon in front of him. The accelerated observer, while always having a velocity slower than light, can still outrun light given a "head-start", which is why light from regions behind him does not reach him.

The rate of time-flow in a general metric is just sqrt(|g_00|). This is easy to see. If you set all the spatial components to zero

dtau^2 = -g_00 dt^2

And taking the square root, we get dtau = sqrt(|g_00|) dt, as the relationship between proper time and coordinate time.

If you look at the Rindler metric, I gave, sqrt(|g_00|) is just 1+g*x. (To see the metric I gave derived, you'll need to look at a GR book, such as "Gravitation", the version I quoted was on pg 173 of that book).

The mathematical consequence of this is that the accelerated observer does in fact see that things age incredibly rapidly in front of him in his coordinate system - gravitational time acceleration, if you would.

For a 1 g acceleration, which is apprxomiately 1 light year/year^2, events 1 light year ahead of the accelerated observer age at twice the normal rate, events 2 light years ahead of him age at 3x the normal rate, etc.

Note that he doesn't literally _see_ them happen in fast motion, this "fast-motion" effect happens to the observer with the "psychic vision" you mentioned, not what he physically sees with a telescope.

What he sees through a telescope is pretty interesting too, but I don't think I need to get into that, this is already long enough.

 

[OlderDan]

The short answer to your question is "yes", he can witness the evolution in fast motion. Now for the long answer.

Well that's awsome. I appreciate your taking the time to respond. What's even more intriguing than the time evolution is the rapid change in the distance from the accelerating observer to the distant star. It seems the star must appear to be moving substantially faster than c, but I guess that's OK as long as the light from the star is staying ahead of it. I guess I'll finally have to do some reading on this GR stuff.

 

[CarlB]

But guess what - all of this weirdness is tied to using the coordiante system of an accelerated observer, one which corresponds to the above metric.

When you use the coordinate system of an inertial observer, the metric coefficients are always

ds^2 = -dt^2 + dx^2 + dy^2 + dz^2

And when you use this metric, all clocks run normally - there is no gravitational field, and there is no gravitational time dilation effects.

I assume you mean this in the general sense, that is, for the watch thrown in the air, when the calculation is made in the rest frame of that watch, there is no gravitational time dilation, even though the watch is clearly in the "gravitational field" of the Earth.

I agree with this, but it is such a subtle point that I wasn't willing to say it out loud. The problem is that everyone thinks that gravitational fields cause time dilation and to argue otherwise is very difficult.

What I believe we agree on here, is that it truly is only velocity that causes time dilation. "Gravitational time dilation" is not included in Einstein's relativity, except when you make certain (very natural) assumptions about the metric you will work in.

Another way of putting this is that time dilation can only be a relative thing.

Carl

 

[pervect]

I assume you mean this in the general sense, that is, for the watch thrown in the air, when the calculation is made in the rest frame of that watch, there is no gravitational time dilation, even though the watch is clearly in the "gravitational field" of the Earth.

Actually I was restricting this comment to the problem where the watch was thrown upwards in an accelerating rocket.

When you view this situation from an inertial frame, there is no gravity anywhere. Hence there is no need to apply "gravitational time dilation", and to attempt to do so would be wrong. Viewied from any inertial frame of reference, all there is is flat-space time, a non-accelerating watch, and an accelerating rocket - there is no gravity, or pseudo-gravity, that could cause gravitational time dilation. The rocket is accelerating, but from the point of view of the inertial frame this is not due to any sort of gravity (which does not exist in the inertial frame), it's due to the rocket's engines.

I don't believe it's possible to have a coordinate system with a diagonal Lorentzian metric for a watch being thrown upwards on the Earth, so this remark doesn't apply to throwing a watch upwards on the Earth - you can't make the metric uniformly equal to ds^2 = dt^2 - dx^2 -dy^2 - dz^2 on the Earth over any extended region. So you can't sidestep gravitational time dilation on Earth the same way you can on a rocket.

What I believe we agree on here, is that it truly is only velocity that causes time dilation. "Gravitational time dilation" is not included in Einstein's relativity, except when you make certain (very natural) assumptions about the metric you will work in.

Another way of putting this is that time dilation can only be a relative thing.

Carl

We may be getting tied up in semantics here. I would say that there is in general such a thing as gravitational time dilation, but that it's due to the metric coefficients, and only occurs when the metric coefficients vary as a function of position.

I'd also agree that the equations which determine the rate at which a clock ticks (the clock's proper time) involve only the position of the clock, the rate of change of that position, and the metric coefficients.. The equations for proper time specifically do not include the second derivative of the clocks position with respect to time, i.e. they do not include the accleration of the clock. This is a point that the sci.physics.faq also makes.