# Metric for uniform constant acceleration?

1. Apr 7, 2015

### exmarine

What does the metric look like for constant uniform acceleration, say in the x-direction?

ds^2 = g_tt (cdt)^2 + g_xx (dx)^2

g_tt = ?
g_xx = ?

2. Apr 7, 2015

3. Apr 7, 2015

### Staff: Mentor

Strictly speaking, you aren't asking for the metric for uniform acceleration - you already know that, it's the flat-space metric given in Minkowski coordinates by $ds^=-dt^2+dx^2+dy^2+dz^2$. You are asking what coordinates are most convenient to use when you're working with uniform acceleration and what the components of the metric look like using that coordinate system.

As A.T. says, Rindler coordinates are the answer.

But notice that the Rindler transforms $x'=x \cosh(ax)$, $t'=x \sinh(at)$ are more complicated than but not fundamentally different than, for example, the coordinate transforms that you'd use to go from Cartesian to polar coordinates. Neither transformation changes the physics, the spacetime, or the metric; we're just using coordinates that make the problem at hand easier.

4. Apr 7, 2015

### exmarine

Well I obviously don’t understand then. I thought the Pound-Rebka experiment showed that proper clocks run at slightly different rates at the top and bottom of the tower. Then by the equivalence principle, it seems like my pocket watch should run at different rates, depending on how hard I am being accelerated, say in the x-direction? The different rates at the top and bottom of the tower were calculated with the Schwarzschild metric. So wouldn’t the different rates for my pocket watch also be calculated with some metric that was a function of my acceleration rate? The flat Minkowski metric is not a function of any acceleration is it?

Thanks for any help with this.

5. Apr 7, 2015

### DrGreg

There's a terminological issue here. There's a distinction to be made between "metric" and "components of a metric", in the same way that there's a distinction between "vector" and "components of a vector". When you change coordinates, the components of the vector change, and the components of the metric change, but the vector itself, and the metric itself, do not change. So you should read Nugatory's post #3 with this in mind. Strictly speaking, the "Rindler metric" and the "Minkowski metric" are the same thing. What you are really interested in are the Rindler components of the metric.

Having got that technicality out of the way, the answer you are looking for is $$ds^2 = -g^2 x^2 dt^2 + dx^2 + dy^2 + dz^2$$in units where c = 1, where the constant g is the proper acceleration at x = 1/g.

For a clock at rest in Rindler coordinates, this gives$$d\tau = g \, x \, dt$$which shows that the ratio between Rindler coordinate time t and proper time τ varies with position.

6. Apr 7, 2015

### Staff: Mentor

No, that's not what the EP says. The EP says that two clocks that are both accelerating in the $x$ direction, but with one slightly in front of the other, should run at slightly different rates (the one in front will run slightly faster). The difference is not a function of acceleration; it's a function of spatial separation, just as the difference in the Pound-Rebka experiment is a function of the height difference.

7. Apr 8, 2015

### A.T.

Yes, because of their different position along the line of proper acceleration of their common rest frame. Not because they themselves might experience different proper accelerations.

No, that would contradict the clock hypothesis:
http://en.wikipedia.org/wiki/Clock_hypothesis

8. Apr 8, 2015

### exmarine

Well I am not qualified to argue with you guys but this sure is puzzling. I thought the equivalence principle was that it is impossible for any experiment to tell the difference between acceleration and gravity, that gravity and ACCELERATION were equivalent locally.

Let me see if I can rephrase my original question. I understand that a clock rate here on the surface of the earth is equal to the square root of one minus the ratio of r_s over r, etc. Then if I am in distant space and fire the rocket motor at about 1 g acceleration, will my onboard clock run at the same rate as one here on the surface of the earth? If so, then how do you calculate that based on the acceleration? If not, then I don’t know what to think.

9. Apr 8, 2015

### A.T.

Gravitational time dilation is not a local effect.

That is not a clock rate. That is the clock rate ratio compared to a clock at infinity.

No. Gravitational time dilation is related to gravitational potential difference, not to proper acceleration.

10. Apr 8, 2015

### Staff: Mentor

You have to be careful how you phrase this since it is easy to get confused. What the EP says is that, for example, accelerating at 1 g in flat spacetime (i.e., in empty space far away from all other objects) is locally indistinguishable from being at rest on the surface of the Earth and feeling an acceleration of 1 g (which is the proper acceleration required to remain at rest at that position). That's not really the same as saying "gravity and acceleration are equivalent", because the acceleration you feel when you're at rest on the surface of the Earth isn't "gravity"; it's the force that's keeping you at rest.
It depends on how you define "local". Gravitational time dilation can be predicted using an analysis in a single local inertial frame; that's how Einstein's original thought experiment did it. More precisely, gravitational redshift can be predicted using a single LIF, and gravitational time dilation then follows as an interpretation of why the redshift is there. Since the analysis can be done in a single LIF, it qualifies as "local" according to the EP. (In particular, the proper acceleration can be assumed to be the same for both clocks.)

However, making the prediction requires the two clocks to be spatially separated; as the separation goes to zero, the time dilation (meaning the difference in clock rates) goes to zero. So in that sense, it is not "local". But the sense of "local" relevant to the EP is the first sense, described above (analysis done in a single LIF).

11. Apr 8, 2015

### Staff: Mentor

There is no unique answer to this question, because there is no unique way to compare the two clock rates. To be able to compare the rates of two clocks in an invariant manner, they must be either at the same spatial location, or they must be at rest relative to each other in some stationary situation (for example, two clocks at rest relative to the Earth at different altitudes), or there must be some periodicity in their relationship that can be used as a common standard for judging the clock rate (for example, a satellite in orbit can compare its rate with a clock on the ground by measuring the proper time elapsed between two successive passages overhead). The scenario you propose does not meet any of these conditions, so no invariant comparison of clock rates is possible.

What will be true, however, is that, provided your rocket is small enough, if you put one clock at the front and one at the rear, their rates will differ by the same amount as two clocks at rest on Earth that are separated vertically by the same distance. For example, if your spaceship is 22.5 meters long, the difference in rate between a clock at the front and a clock at the rear will be the same as the rate difference observed in the Pound-Rebka experiment on Earth (since 22.5 meters was the height difference tesetd in that experiment).

12. Apr 9, 2015

### exmarine

Thanks for the careful responses.

13. Apr 9, 2015

### bcrowell

Staff Emeritus
14. Apr 10, 2015

### A.T.

I was wondering about a non-concentric spherical cavity inside of a uniform density sphere, which does have uniform Newtonian gravity. But I guess there is no known analytical EFE solution for it.

15. Apr 10, 2015

### bcrowell

Staff Emeritus
I don't understand why you would say that the system you describe would have uniform newtonian gravity. I think the field would vary in both magnitude and direction.

16. Apr 10, 2015

### A.T.

I meant only inside the cavity:
http://home.comcast.net/~peter.m.brown/gr/grav_cavity.htm [Broken]

Last edited by a moderator: May 7, 2017
17. Apr 10, 2015

### bcrowell

Staff Emeritus
I see. That's a surprising result in Newtonian gravity. Cool!

But it's easy to come up with mass distributions that produce a uniform field in Newtonian gravity. A much simpler example is a uniform sheet of mass.

None of this has much to do with the problem of finding a spacetime and set of coordinates in GR that represent a uniform field. If we could find such a spacetime, we could automatically derive the necessary stress-energy tensor using the Einstein field equations. The trouble is that no such spacetime exists. See ch. 5, problem 6 of my GR book, http://www.lightandmatter.com/genrel/ .

Last edited by a moderator: May 7, 2017
18. Apr 10, 2015

### ShayanJ

19. Apr 10, 2015

### bcrowell

Staff Emeritus
Yep, thanks!