# GR vs. SR while accelerating away

1. May 30, 2005

### gonzo

A quick question for those fast with the GR and SR math.

Assume you get in a spaceship and start accelerating away from Earth, and during the trip you and the people left behind compare clock speeds periodically (not elapsed time, but rather tick rates).

At what combination of acceleration and speeds would the GR effects from acceleration (which tend to make the clocks on the Earth look faster to you) be more significant than the SR effects from speed (which would tend to make the clocks on Earth look slower to you)? Assuming I'm wording this problem clearly.

Is this only dependent on acceleration (i.e., for a given high enough acceleration, the GR effects will always dominate), or for most accelerations, will there be a speed you eventually reach where the SR effects catch up and pass the GR effects? I would generally assume the later given a constant acceleration, since the SR effects grow, but don't the GR effects also grow with distance?

Anyway, I would appreciate a precise math relationship between the two if someone could whip one up for me (with a qualitative explanation also of course).

Thanks much.

2. May 30, 2005

### gonzo

I started trying to do the math a bit myself, but GR isn't my strong suit, so I got a bit confused.

I started by looking at the formula for gravitational time dilation, and then used the equivalence principle to assume it was the same as being in a gravitational field that produced that force.

I don't know how to insert math symbols, so I'll just "say" the formula I ended up with ... I got dilation by a factor of the square root of one minus (accleration times distance divided by c^2). I plugged in a random accleration to start with, taking 100 m/s^2, and it seemed then that you didn't have to be very far from something before the dilation factor became imaginary (like 10^15 meters). So I must be missing something key here.

Assuming you can treat the acceleration you feel on a ship the as being in a gravitational field, then clocks far away from you from from this effect (discounting relative speed for now) should appear to speed up. Don't that go faster the farther away they are since they are farther away from the gravitational field? Or do I have it backwards somehow?

I am certainly missing some key element here, so any enlightenment would be helpful. Thanks.

3. May 30, 2005

### Mortimer

I'm pretty sure you cannot use the equivalence principle in the way you do. The time dilation in a gravity field should be "equivalenced" with a rotating disc where the edge has a velocity (and acceleration) relative to the center, hence its slower time.
In your example, at any infinitesimal instant the accelerating ship has a defined velocity relative to earth which is the sole element that determines the time dilation as observed from either frame. The acceleration as such does not influence that. So only SR counts, not GR (at least for the spaceship; GR does count on the surface of the earth).

Last edited: May 30, 2005
4. May 30, 2005

### pervect

Staff Emeritus
With the approach you are using, you are going to find that the clocks on earth start running backwards ::. So it's not a very good approach. I'll try to explain why. Basically, you are trying to compare your clocks to the clocks of someone who is beyond an event horizon. This is very similar to trying to compare your clock to the clock of someone who has fallen into a black hole. Two-way signal transmission is not possible, so there isn't really any good way to compare clocks.

To give some specifics, ifyou maintain an acceleration of 1g, there will be an event horizon approximately 1 light year behind you from which you will never receive signals until you stop accelerating.

This is known as the "Rindler horizon". A short way into this journey (somewher around a year or so), the Earth will fall behind this Rindler horizon.

For the usual reasons, you will continue to "see" a faint image of the Earth, but you will never receive signals emitted from the Earth later than a certain time, the time at which it fell behind the horizon.

If you want to do the math, plot the position of the rocket using the relativistic rocket equation at

http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken]

It is simplest to assume that you are acclerating at exactly 1 light year / year^2 (which is very close to 1g), and to use units of light yeras for distance and years for time.

The equation for the position of the rocket in the Earth's reference frame is

x = cosh(tau)-1; t=sinh(tau) (1)

here tau is the "proper time", the time elapsed on the rocket's clock

The equation for light emitted at an earth time T is

x = (t-T), t>T (2)

substiting (1) into (2) we get

cosh(tau)-1 = sinh(tau)-T

cosh(tau)-sinh(tau)=1-T

Because cosh(tau) > sinh(tau), there is no value of tau that solves this equation for T>1.

This will be clearer if you take the time to actually plot the position of the rocket on a graph, and also plot the light signals.

Another way of putting this:

A continuously accelerating observer will outrun a lightbeam, when he has a large enough head start.

Last edited by a moderator: May 2, 2017
5. May 30, 2005

### JesseM

If you consider this problem from the point of view of the inertial observer rather than the accelerating observer, there won't be any additional time dilation effects due to acceleration at all--over any small interval of time, if the accelerating observer's velocity during that interval is v, then his clock is slowed down by a factor of $$\sqrt{1 - v^2/c^2}$$ during that interval.

6. May 30, 2005

### gonzo

I was mainly interested in the accelerated POV. That's intereesting about the horizon effect, that makes me feel better about the imaginary times I was getting at large distances for high constant acceleration.

But what about before you reach this point. I understand if you take a small enough time slice that discounts the accleration the earth clocks will appear to be running slow, but how small is small enough if you are accelerating at 1g? You have a year or so before you are past the Rindler horizon, so during this time what are the conditions for looking back at Earth clocks and seeing them run slow? Are there any conditions in this when you could look back and see them running fast?

The other thing that comes to mind is that the people on Earth also have a 1g accleration from gravity. So if you are undergoing a constant 1g accerlation, then you both have equal GR effects ... is this true?

I just realized I'm not even sure what constant acceleration means in this context given space and time changes. I guess it would have to be defined by the force the passengers feel the whole time? (as opposed to actual speed changes like you do in classical physics ... I assume anyway).

Anyway, I as mentioned I would still like to know if there are any conditions at all where you would see the earth clocks moving faster at any time because of your acceleration. Or will this never happen? Thanks.

7. May 30, 2005

### gonzo

Looking at that link you gave me pervect, it seems for a 1g acceleration you will almost always see the earth clocks as going slightly faster, with their relative speed increasing the farther out you get.

But the equations were too messy for me to get a feel for when this effect would kick in (since it seems to increase with distance, I assume if you are close enough the clocks will seem to be going slower instead? Or does this never happen?)

I'm also confused why if you are accelerating at 1g you would not have the same situation on Earth due to gravity of 1g so both of you would see the other's clocks as going faster in the same way.

And lastly, I assuming if you lower your acceleration this effect is reduced. Where are some random borders for where the lower accleration vs. distance makes the clocks looks slower vs. fast? I'm trying to get a few qualitative markers for the situation.

8. May 30, 2005

### Janus

Staff Emeritus
Because Gravitational time dilation is not related to the local acceleration due to gravity, but due to the difference in gravitational potenial. The gravitational field of the Earth falls off with distance, which limits the difference in gravitational potential between the surface of the Earth and the Ship due to this field. The equivalent gravitational "field" due to the acceleration of the Ship as seen by the occupant of the ship does not fall off with distance but extends for inifinite distance at a constant strength both ahead of and to the rear of the ship.

9. May 30, 2005

### pervect

Staff Emeritus
Hmmm, well - that's not the right answer. How did you arive at that conclusion?

10. May 31, 2005

### gonzo

They have a table there of your time and earth time. Granted, the table starts at 1 year of flight, but for all entries on the table ellapsed earth time is greater than ellapsed ship time, implying that the earth clocks are running faster.

However, as i pointed out I'm not sure. Can you please give me a qualitative right answer then? Or at least a few good comparison points?

11. May 31, 2005

### gonzo

I just looked at the clock postulate page which I hadn't noticed before, which says the opposite ... that acceleration doesn't matter for clock rates, only instantaneous speed. This doesn't seem to make sense to me since GR says there will be a "one way" time effect in a gravitational field. So if two observers are at rest with regard to each other in a gravity field they will both agree on one of them having faster clocks. If they then get some speed relative to each other, they will both note a slowing of clocks for each other based on this speed, but will still have the one way clock difference from GR. At very small speeds, the GR effect would seem to be bigger, and so one would still have faster clocks, but at some point the speed effect would grow larger (assuming the same distance the whole time, just looking at a time slice) and eventually overwhelm the GR effect, and then both would see the other's clock as going slower as in a typical SR scenario (although I would assume the GR component would still make it slightly uneven). Or am I off on this much too?

12. May 31, 2005

### pervect

Staff Emeritus
The table tells you what happens from the Earth's point of view. From the Earth's point of view, the rocket clock's run slow. This is the easiest point of view to describe. But I don't think it was the one you were asking about.

I had the impression you were interested in the rocket's POV.

The rocket's viewpoint is harder to describe. The first point is that the coordinate system used by the accelerating observer on the rocketship is strictly limited in size by the laws of physics. The best brief description of things from the rocket's POV goes something like this.

You can more or less think of a "gravitational field" permeating the entire universe. Objects behind the rocket, like the Earth are lower in the gravity well. This means that clocks behind the rocketship, like the one on Earth, run even slower than the relativistic time dilation formulas would indicate.

Eventually, the Earth clock falls so deep into a gravity well that it falls behind an event horizon. This happens when the distance to the Earth, multiplied by the gravitational acceleration, reaches the speed of light. Recall (or re-read) my remarks about how an accelerating observer can outrun a lightbeam.

When the Earth falls behind an event horizon, it is no longer meaningful to talk about "how fast it's clock is running from the rocket's POV". To determine "how fast a clock is running" when the clocks are separate requires a coordinate system - the purppose of the coordinate system is to say that "this point in space-time in coordinate system #1 (t1,x1) is the same as this other point of space-time in coordinate system #2 (t2,x2). It is not an absolute statement, it is a coordinate dependent statement which requires that the coordinate systems both be defined.

There is no problem with the Earth coordinate system, but as I mentioned, the rocket's coordinate system does not and can not cover the Earth after it falls behind the horiozon, the "Rindler Horizon". Thus there is no meaningful way to determine how fast the Earth's clocks are running. The gravitational time dilation equations willl give you an imaginary elapsed time, for instance, if you plug the numbers in.

On the return trip, the Rindler horizon is on the other side of the rocketship, so it's not a problem anymore. On the reutrn trip, from the rocke'ts POV, the Earth clocks run very fast. This explains from the rocket POV why the Earth clocks read more elapsed time.

BTW, MTW's "Gravitation" has a very good treatment of the accelerated observer. It's possible other textbooks do, too, but I'm not aware of which other books might specifically cover the topic. This would be a good source for more reading with the supporting math if you get really interested, or if you want a reference for why I am saying what I am saying.

13. May 31, 2005

### gonzo

Thanks for that reply. You are right in that I was more interested in the POV from the accelerating observer.

I guess part of my problem has been how to apply the equivalence principle correctly. For example, the idea that on an accelerating rocket ship your "equivalent gravity well" is infinite and doesn't drop off like that from a large mass. I also hadn't thought about the driectional aspect either.

I understand about the event horizon, which is really interesting. But assuming we only talk about events before a ship reaches this point, then what you are saying is that whether you look at it from GR or SR, a ship flying away from the Earth will always see the Earth clocks as running slower? So in this case the effects of velocity and accleration work together? That was the main thing I was initially getting at.

However, you raised some other interesting issues for me about the return trip, since you say in this case the GR effect would tend to make the Earth clocks run faster from the ship POV. But the speed will the make the Earth clocks appear to run slower. So in this situation, where is the typical trade off for when one effect is greater than the other?

Also, does it matter which direction you are actually moving? I mean, if you accelerate away and then want to slow down and stop, you will need to accelerate an equal amount in the opposite direction even though you will still be moving away. This situation seems even more confusing to me.

I just starting reading "Gravity from the ground up" to get some more GR basics, and hope to eventually move on to harder texts than that afterwards. My math is still a bit rusty for the harder stuff, though I'm trying to refresh that as well.

Is "Gravitation" a textbook? Would I need a solid grip on tensors to be able to follow it (one of my week points right now)?

14. May 31, 2005

### pervect

Staff Emeritus
"Gravitation" is a textbook. You would not be able to follow the majority of the book without tensors, but tensors play only a minor role in the chapter on accelerated observers. As I check this issue, they are not, unfortunatley, totally absent though.

You would definitely need 4-vectors to deal with the book. Displacements and velocities are always represented by 4-vectors, as "geometric objects". There is some introduction to this in the book, but a reasonably good understanding of 4-vectors would be a pre-requisite.

You would need to be able to deal with the tensor notation to the extent that $$x^{a}$$ is a 4-vector, and that $$x^a{}_{a}$$ is the norm of the 4-vector. And you'd need to be able to deal with the notation that $$e_{0}$$ was a basis vector of a coordinate system (typically a unit vector in the time direction).

If you can find a simpler book that covers the material, go for the simpler book. However, most of the references I've seen to the problem of the accelerated obserer refer to MTW's treatment, however. I don't know if introductory books like "Space-time Physics" treat the problem.

Yes, the effects work in the same direction initially.

I think you are basically asking for the time dilation formula. You can think of the "gravitational time dilation" as being

$$\sqrt{1+2*a*d/c^2}$$

here a is the acceleration of the rocket ship, and d is the distance to the clock.

when d is positive, the point is "ahead" of the rocket, and the clocks at that point run fast, indicated by a time dilation factor greater than 1.

when d is negative, the point is "behind" the rocket, and the clocks run slow, indicated by a time dilation factor that goes to zero when |2*a*d| = c^2. This is the point where the horizon is.

I'll leave you to work out as to what effects are more important when, except to note that the end result is that the Earth clock has more elapsed time than the rocket clock for a round trip. The "bookeeping" of when the time dilation occurs doesn't really have much physical significance, it's a result of the assumptions made to define the coordinate systems. The comparison of the two clocks at the end of the trip, however, is a real physical event that does not depend on the coordinate systems used.

When you turn around, there will be a huge change in the "bookeeping" of simultaneity, though nothing much physically happens. Basically, one is switching from one coordinate system with one notion of simultaneity to another coordiante system with a different notion - much like the simpler SR case with no accleration.

If you look at the doppler shift of the light that you receive from earth, for instance, it doesn't vary much at the instant of turnaround.

Last edited: May 31, 2005
15. May 31, 2005

### gonzo

Thanks, that was all very helpful. Can you recommend a book with a good intro to tensors?

I've never worked with 4-vectors, but have had quite a bit of 3-vector analysis, so I wouln't expect there would be a major conceptual jump to have to make, besides not being able to picture things well in my head. New notation is always annoying to learn, with the only good way I know being to just do a lot of problems with it to get used to it. So maybe a good tensor textbook with problems and an answer key would be the way to go.

16. May 31, 2005

### pervect

Staff Emeritus
"Gravitation" has an introduction to tensors - but not a whole lot of exercises. If you don't mind a challenge, try picking it up at the library, or ordering it via interlibrary loan.

The chapter on accelerated motion is only about 20 pages long or so (quite photocopyable) - the introductory material about tensors, 4-vectors, geometric objects, and the explanation of the notation used by the book etc. is a lot longer though. The book is pretty good about explaining it's notational system.