# Acceleration in SR makes it equivalent to GR?

1. Dec 18, 2012

### arindamsinha

Does putting the concept of acceleration in SR make it equivalent to GR?

I see a lot of sources (including many posts in this forum) which seem to say that SR can handle acceleration fine, but not gravity. However, an acceleration IS gravity by the equivalence principle, so what's the difference? (That historicallly EP was developed after SR does not matter, I think).

In this context, I have seen a lot of explanations saying that the Twin Paradox can be resolved within SR framework and does not require GR. The main differentiator though is that one of the twins preferentially 'feels acceleration'. Isn't it ultimately then putting the explanation in GR territory without openly acknowledging it?

2. Dec 18, 2012

### K^2

SR can handle acceleration of world lines. It cannot handle acceleration of frames of reference. If you consider a space station from which a space ship accelerates away, we have no problem describing this with SR from perspective of space station. The station does not accelerate, so the frame is inertial, and there is no problem. But if you try to describe the same situation from perspective of the accelerating ship, you are now in accelerated frame of reference, and you need GR to address it properly.

Gravity is caused by acceleration of the frame of reference. While standing on the ground, according to GR, the pull you are experiencing is due to you and ground accelerating upwards at 9.8m/s². So if you want to describe motion of object relative to ground, you are working in an accelerated reference frame.

3. Dec 18, 2012

### Bill_K

In both Special and General Relativity there is simply no such thing as an accelerated reference frame in the ordinary sense. Any set of accelerated coordinates such as linear acceleration or rotation eventually exceeds the speed of light at large distances. Acceleration must be described as a collection of accelerating world lines, or equivalently as a collection of Lorentz frames, one at each point. This is not a problem, just a different set of concepts. If treated correctly, acceleration is well within the province of Special Relativity.

4. Dec 18, 2012

### K^2

Or you can just deal with the fact that some 4-velocities will have imaginary components.

5. Dec 18, 2012

### stevendaryl

Staff Emeritus
You can think of the logical progression from SR to GR this way (this is not the way Einstein thought of it, but in hindsight it is a way to think about it):

1. Describe SR in inertial, cartesian coordinates.
2. Describe SR in accelerated, curvilinear coordinates. The physical content of SR is not changed in the transition from 1. to 2. If you know how SR works in inertial, cartesian coordinates, then you can derive how it works in accelerated, curvilinear coordinates by just using calculus. There is no new physics. However, using curvilinear coordinates allow us to describe how things look from the point of view of an accelerated observer. What we find when we do this is:
• If you drop an object, it appears to fall away as if there were a universal force (acting the same way on all objects) were acting on it.
• There are apparent velocity-dependent forces, such as the Coriolis force, Centrifugal force.
• Clock rates appear to be position-dependent.
Once again, these effects are not new physical predictions. They are simply the predictions of ordinary SR described in different coordinates.
3. Note that SR in curvilinear coordinates has a lot of the same features as Newtonian gravity:
• There is an apparent force affecting all objects.
• This force is universal, leading to the same (coordinate) acceleration for all objects, regardless of their mass or composition.
4. Descrbe SR in curved spacetime. Once you have figured out how to do SR using curvilinear coordinates, it's not a huge leap to asking how to do SR in curved spacetime. Here an analogy with 2D Euclidean geometry might help. Most of Euclid's geometry is about flat, 2D spaces. So knowing Euclidean geometry, how can we analyze things on a curved surface such as the surface of the Earth? The key insight is this: any small enough region on a curved surface will look approximately flat. (This actually isn't true if the surface is fractal, but let's ignore that complication, since I don't know how to deal with it.) So a recipe for analyzing geometry on the surface of the Earth is:
• Partition the surface into lots of overlapping small regions.
• Use ordinary flat 2D geometry within a region.
• Use curvilinear coordinates to interpolate between neighboring regions.
The same recipe can be used to do SR in curved spacetime: Partition spacetime into lots of little regions of spacetime. Apply SR within each region. Use curvilinear coordinates to interpolate between neighboring regions.

This development actually goes beyond SR, but not a whole lot beyond SR. If you know how some physical theory, such as Maxwell's equations, works in SR, then you can come up with a pretty good guess as to how it works in curved spacetime. It's not uniquely determined what the theory ought to be like in curved spacetime, because there might be effects that couple to the curvature tensor that you can't guess just from knowing the flat spacetime version of the theory. But you can get a good idea of how things work in regions where the curvature is negligible.
5. Speculate that gravity is nothing but a manifestation of curved spacetime. As noted in Development 2., SR in curvilinear coordinates has some of the same features as gravity. But it's "fake" gravity, because it can always be eliminated through choosing inertial cartesian coordinates. On the other hand, doing SR in curved spacetime requires the use of curvilinear coordinates when considering a large enough region of spacetime. So the "fake" gravitational effects cannot be eliminated completely. So one might guess that "real" gravity is just a manifestation of noninertial, curvilinear coordinates that are forced on us because spacetime is curved. This guess is the Equivalence Principle.
6. Describe the dynamics of the curvature tensor. Development 3. is only half-way to General Relativity. It describes physics in a fixed curved "background" spacetime. It tells you how spacetime curvature affects physics (or at least, it gives you an approximate idea), but it doesn't tell you how physics affects spacetime curvature. Once you have an equation describing how matter and energy affects the curvature tensor, then you've got GR.

This is not the historical order of things: Einstein guessed the equivalence between gravity and noninertial coordinates (Development 4) before systematically developing the theory of curved spacetime (Development 3). Developments 3 and 5 were never done separately--Einstein didn't explore SR in curved spacetime before trying to come up with equations describing the curvature.

To do physics on board an accelerating rocket requires only developments 1 and 2. To do physics near the surface of the Earth requires the Equivalence Principle (development 4).

6. Dec 18, 2012

### nitsuj

I would guess the issue of acceleration is used to illustrate the "break in symmetry".

The details of how acceleration breaks the symmetry can be as simple as this twin felt acceleration.

7. Dec 18, 2012

### Staff: Mentor

This is only true locally. Once you go beyond a small local patch of spacetime, there *is* a difference between acceleration and gravity: gravity requires spacetime curvature. SR can only deal with flat spacetime; to deal with curved spacetime requires GR. That's the difference.

The standard twin paradox scenario is set in flat spacetime, so yes, SR can resolve it just fine. There are alternate versions set in curved spacetime which require GR to resolve.

No; observers can feel acceleration in flat spacetime. They just need to turn on their rocket engines.

8. Dec 18, 2012

### Staff: Mentor

The difference is tidal gravity. If you have no tidal gravity then you have a flat spacetime and can use SR. If there is tidal gravity then spacetime is curved and you need GR and the EFE.

9. Dec 18, 2012

### nitsuj

In books I have read that mention the "Weak / Strong Equivalence Principle" mention with a "big BUT" this is only true locally, just like Peter Donis already mentioned.

As you've seen it is a great segue into similar GR principals...locally

Curved spacetime in SR = Ship is inertial fires photon and accelerates while photon is en route. Not for real, but the picture is great!

10. Dec 18, 2012

### Staff: Mentor

This isn't curved spacetime. The worldline of the ship is curved--it has nonzero proper acceleration, for at least a portion of it--but spacetime itself is still flat.

11. Dec 18, 2012

### nitsuj

Never said it was...
"Not for real but the picture is great!" and it is...weak equivalence.

Dalespam already improved the distinction with mentioning tidal forces/

Last edited: Dec 18, 2012
12. Dec 18, 2012

### haushofer

13. Dec 18, 2012

### Staff: Mentor

Well, you did say "curved spacetime in SR". What was that supposed to mean? If you were drawing an analogy between spacetime curvature in *GR* (not SR) and path curvature of an accelerated worldline, I don't think that analogy is valid.

14. Dec 18, 2012

### nitsuj

oh comon Peter, pretty sure it's considered equivalent... at least in principle

The photon makes a nice curved path & visually is great picture of curvature. True it is not because of the spacetime, merely because the scenario uses proper acceleration to get the curved path.

15. Dec 18, 2012

### Staff: Mentor

And I'm pretty sure it's not. Path curvature and spacetime curvature are different things. They are not equivalent, and pretending that they are can easily cause confusion. I've seen it happen on previous threads here on PF.

Of *path* curvature. *Not* of spacetime curvature. (And even path curvature in this case is problematic because the photon is traveling on a geodesic; it's the observer's path that is curved.)

Exactly. If you want to claim that the two cases are nevertheless somehow "equivalent", the burden of proof is on you to demonstrate the equivalence. Just waving your hands and saying you're "pretty sure" is not sufficient.

16. Dec 18, 2012

### nitsuj

burden of proof! so formal

Peter, I'll have you know I didn't merely wave my hands; I also stomped my feet!

I'm going on proper acceleration "force" is same as gravitational "force", that the ship observer couldn't distinguish the difference between the curved path of the photon being because of curved space time or proper acceleration just by looking at the photon path.

17. Dec 18, 2012

### Staff: Mentor

And this is incorrect. By hypothesis, the photon path is a geodesic; that means it has zero proper acceleration. The proper acceleration of any path is an invariant, so any observer can in principle calculate it and get the same answer, including the ship observer.

You have just exhibited the confusion I was talking about.

18. Dec 18, 2012

### grav-universe

Acceleration in SR is equivalent to gravity in a constantly accelerating field for a single accelerating observer, for instance, right. The gravitational field around a mass, however, is not constant, but varies with distance from the center, so SR is only valid locally, neglecting the tidal gradient of the field, while GR relates the positions and accelerations between many static observers. That is to say, static observers at varying distances from the center of a gravitational field will measure different local accelerations, so for a body in freefall, we must not only apply SR or the equivalence principle for the local acceleration of a single static observer, but we must apply full GR in order to relate the different accelerations of a family of static observers at varying distances to each other in order to determine what each measures of the freefalling body at different times or at different places while falling through the field.

Last edited: Dec 18, 2012
19. Dec 18, 2012

### Staff: Mentor

More precisely, proper acceleration in free space is equivalent, locally, to being *at rest* in a static gravitational field of the appropriate strength to require the same proper acceleration to hold you at rest. So, for example, if you are inside your rocket ship and are feeling a 1 g acceleration, you can't tell, locally, whether that's due to your rocket engine firing in free space, or your rocket sitting at rest on the surface of the Earth. But as you point out, if you are allowed to make measurements over a finite region, you can tell the difference by the presence of tidal gravity in the second case, but not in the first.

20. Dec 18, 2012

### Naty1

This is a very interesting discussion. I have tried to compare statements above with explanations I have in my notes.....some successful, some I am still puzzling about.

Dalespams is one of the puzzles: