# Acceleration and GR

1. Sep 3, 2014

### ShayanJ

In introductory texts about GR, it is sometimes stated that one of the things that guided Einstein to GR, was the effects of accelerated motion ( don't know whether proper or coordinate acceleration) , things like Ehrenfest paradox.
But after that, they completely forget about acceleration and nothing is said about it.
I mean, what they say at early stages, implies that acceleration should be able to curve space-time and that's very confusing. In fact, for me, its the single wall stopping me from understanding GR completely.
Thanks

2. Sep 3, 2014

### m4r35n357

Look into "Rindler Space". Here is a link to give you a flavour:The Rindler Horizon

One of my favourite relativistic effects: constant acceleration creates a (temporary) event horizon, ie. you can outrun light if you get enough of a head start.

Last edited: Sep 3, 2014
3. Sep 3, 2014

### A.T.

Yes, see:
http://en.wikipedia.org/wiki/Equivalence_principle

Not "curve" in the intrinsic sense. But an accelerated frame can be modeled with curve-linear coordinates. Intrinsic curvature is introduced, if you want to extend the local equivalence to larger area with gravity of different magnitudes and directions.

Locally you have curve-linear coordinates, like in an accelerating frame, but no intrinsic curvature (negligible tidal effects). A cone is intrinsically flat:

On a larger scale, where tidal effects are not negligible, intrinsic curvature is inevitable. See case C:

Here you also see the space-time geometry for the inside and the other side of a planet:

4. Sep 3, 2014

### Staff: Mentor

You are not the first person to be confused here and I blame the textbook authors

Here's one way of working through the maze:
1) Acceleration means you are not following the path that inertia says you ought to be following.
2) Classical physics says that #1 is always due to net force that is pushing you off that inertial path.
3) Classically, when you are standing on the surface of the earth you are not accelerating because two forces (gravity pulling you down, earth pushing you up) are exactly canceling so there is no net force. This is consistent with #1 and #2.
4) Classically, when you are standing on the floor of an accelerating spaceship you are accelerating (along with the floor) because the floor is pushing you and there's a net force because there's no counteracting force. This is also consistent with #1 and #2.
5) Einstein's insight: #3 and #4 are (locally) indistinguishable. Therefore, it should be possible to describe the #3 situation as if it were a #4 situation.
6) So we try describing #3 as if there is only one force at work, the earth pushing on you.

It turns out #6 works remarkably well - we just need an explanation for why the natural inertial path in #1 seems to want to intersect the ground. And that's easy to see if you just imagine that the ground wasn't there but the gravitational field still is (imagine that the earth suddenly shrinks to a point leaving you floating above its surface - $F=Gm_1m_2/r^2$ still works and you're still at the same distance $r$ from the center of the earth). No earth pushing on you, so you're in free fall and experiencing no acceleration as you fall into an elliptical orbit around the center of the earth. When we say that objects "fall" towards the earth, we're actually saying that their inertial no-force-acting-on-them trajectory intersects the surface of the earth, which then applies forces that accelerate them off that inertial trajectory.

Curvature is the reason why the inertial trajectory takes on the shape that it does. Mass creates that curvature, but acceleration does not. Instead, acceleration happens when forces act on you to push you off your inertial trajectories; firing a rocket engine is one way of creating such a force, and objects like the surface of the earth pushing on you because they're in the way of your curvature-determined inertial trajectory is another.

Last edited: Sep 3, 2014
5. Sep 3, 2014

6. Sep 3, 2014

### ShayanJ

Thanks people. Things are getting more clear!

Could you suggest a few?

7. Sep 3, 2014

### ShayanJ

Let me give it a try:

1- When in an accelerated frame, you have inertial forces which, if you consider yourself at rest, cause you to think there is a force which gives objects accelerations independent of their mass and that's indistinguishable from gravity. So Einstein came to think that gravity and effects of accelerated motion should have similar explanations.

2- Effects like Ehrenfest paradox suggested the explanation should somehow involve geometry. But because accelerated motion is something about the observer, not the universe, that involvement should be through coordinates used by the accelerated observer. In fact the accelerated coordinates are somehow, that according to them, the Space-Time has some curvature, even if its actually flat.( Here we can safely use the word "actually" because gravity and acceleration are absolute, unlike uniform rectilinear motion.)

3- Because the explanation of gravity and the effects of accelerated motion are the same, and because gravity is something about the universe not a particular observer, Einstein came to the conclusion that in the presence of energy, Space-Time actually gets curved.

How was that?

8. Sep 3, 2014

### WannabeNewton

The spatial geometry of the rotating disk is curved in the frame of the disk. This does not mean the space-time is curved; it clearly isn't since we're still in flat space-time. Einstein's insights that led him to GR were not always correct intuitions; this is understandable of course because GR is an extremely profound theory that doesn't necessarily lay out a clear road towards its inception. That he managed to come up with such a beautiful and profound theory is amazing in and of itself, even if the path was marred with shaky insights.

The often claimed statement that the problem of the rigidly rotating disk (not Ehrenfest's paradox which is different) helped Einstein in his thinking of gravity as space-time geometry, even if true, is an incorrect insight because the space-time geometry is not curved in the frame of the disk, there is simply a fictitious rotating gravitational field that introduces spatial curvature and gravitational fields are not the gravitational tidal forces that one identifies with space-time curvature.

9. Sep 3, 2014

### WannabeNewton

10. Sep 3, 2014

### ShayanJ

I didn't mean that the space-time is actually curved. If I want to put it in technical terms, in the non-inertial coordinates, the metric isn't Minkowskian but the fact that makes it different from the case of having actual curvature, is that we can make the metric Minkowskian.
So an accelerated observer sees a "removable" curvature(Which is no way curvature of space-time, only a consequence of the "wrong" coordinates) but an observer in free fall in a gravitational field, sees an "essential" curvature.

11. Sep 3, 2014

### Staff: Mentor

The metric (which is a coordinate-independent tensor) is unaffected by the choice of coordinates - it's just that its components when written in this coordinate system are not the $(-1,1,1,1)$ that you more often use with this metric. It's no more or less Minkowskian than if you choose to to use spherical coordinates so that the metric components are $(-1,1,r^2,r^2sin^2\theta)$.

I'd rather say that both observers see exactly the same coordinate-independent curvature of spacetime. The accelerated observer finds non-local calculations more difficult if he chooses to use the non-inertial coordinates in which he is at rest, as the representation of an unaccelerated inertial path is messy in those coordinates. On the other hand, these may be perfectly satisfactory coordinates for local calculations - indeed, that's what we're doing when we teach high-school students that $F_g=mgh$.

12. Sep 3, 2014

### stevendaryl

Staff Emeritus
There is something a little weird about the historical development of General Relativity, and its relationship to noninertial coordinate systems. On the one hand, you can argue as follows:

1. If $x^\mu$ is an inertial coordinate system, then the path of an unaccelerated point mass is given by: $\dfrac{d^2 x^\mu}{d\tau^2} = 0$
2. If instead, you use noninertial coordinates, this gets modified to $\dfrac{d^2 x^\mu}{d\tau^2} + \Gamma^\mu_{\nu \eta} \dfrac{dx^\nu}{d\tau} \dfrac{dx^\eta}{d\tau} = 0$
3. The extra terms look like a kind of force term. But interpreted as a force, it implies a force that accelerates all masses in the same way.
4. Hey! That's just like gravity!
5. Maybe what we're calling the "force of gravity" is actually a term that arises from using a noninertial coordinate system.
6. Working that out requires that spacetime be curved. (Because if spacetime were not curved, then the "gravitational force" could be eliminated everywhere by a suitable choice of coordinates.)

So this way of motivating GR (at this point, I'm not claiming that it's the thought process that Einstein followed, just that it is a possible line of thinking) involves noninertial coordinates in an essential way. However, in the resulting theory, GR, inertial versus noninertial coordinates plays no role whatsoever. The equations are written in a way that gives no preference to inertial coordinates. So the original motivation for GR seems missing in the final product.