I Inertial & non-inertial frames & principle of equivalence

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1. Nov 20, 2017

Frank Castle

By this do you mean that one can use any coordinate system you want (inertial or non-inertial) such that the laws of physics are those of SR for a sufficiently small neighbourhood - one can calculate the geodesic deviation in any of these coordinate systems to determine how small this neighbourhood has to be in order for the approximation to of SR to hold?

2. Nov 20, 2017

Staff: Mentor

Yes.

3. Nov 20, 2017

Frank Castle

Ok great, I think I'm getting it now. So is the point that if one considers larger regions of a given coordinate system the approximation breaks down and one has to take into account the effects of the gravitational field? The equations (for the non-gravitational laws of physics) will look the same as they do in a non-inertial reference frame (i.e. including connection terms), however, the Riemann tensor will be non-zero indicating that the spacetime is curved (this is true in an infinitesimal neighbourhood of a point too, but the point is the tidal effects are too small to be observable for small enough regions). Furthermore, the geodesic deviation for finite patches of the coordinate system will be non-negligible meaning that full GR is required in order to correctly describe physical experiments.

4. Nov 20, 2017

Staff: Mentor

Not larger regions of a given coordinate system. Larger regions of the spacetime. All of this is independent of any choice of coordinates. As I said, it depends on how curved the spacetime is and how accurate your measurements are. Those are independent of coordinates.

5. Nov 20, 2017

Frank Castle

Sorry, by larger region I was assuming this corresponded to covering a larger region of spacetime.

So depending on how accurate one's measurements are and how curved spacetime actually is will determine the size of the region of spacetime around each point in which the laws of SR (approximately) hold, and this will be true for any coordinate system?

If one can always choose a RNC system, and furthermore, because the laws of physics are in tensorial form, one can choose any coordinate system in which the laws of physics are those of SR for sufficiently small neighbourhoods of each point, is it the case that the only real point where GR comes in is determining the geodesics of spacetime such that a RNC can be constucted, and working out the geodesic deviation such that one can determine how small the region around each point has to be in order for curvature to be negligible?

Last edited: Nov 20, 2017
6. Nov 20, 2017

Staff: Mentor

Yes.

It is true independently of coordinates. You seem to have the logic backwards. You don't first choose coordinates and then figure out the size of the region. You first figure out the size of the region, using coordinate-independent facts (the accuracy of your measurements and the curvature of spacetime are both coordinate-independent), and then, if you must, you choose coordinates and calculate what the coordinate-independent facts translate to in those coordinates.

You make it sound like this isn't very much. In fact it's everything. "GR comes in" in determining the actual curved geometry of the spacetime. That is everything. It's not just a small thing added on.

7. Nov 20, 2017

Frank Castle

Sorry, I realise it's a much bigger deal than I make it sound. I was just wondering how this enters the non-gravitational laws of physics - since they are in tensorial form they "look" the same whether or not spacetime is curved, it's just in coordinate form that they differ, i.e. partial derivatives becoming covariant derivatives and the metric becoming non-Minkowski, however, this would be true in a non-inertial frame in flat spacetime too. Can differences be seen, for example, from the EM wave equation, in which a term proportional to curvature appears in curved spacetime?

8. Nov 20, 2017

Staff: Mentor

Both of these statements are independent of coordinates.

9. Nov 20, 2017

Frank Castle

Ah ok. So this is the key point - the fact that the derivatives become covariant derivatives and the metric non-Minkowski is a due to the manifold being curved, which is a coordinate independent statement.

10. Nov 20, 2017

Staff: Mentor

Yes.

11. Nov 20, 2017

Frank Castle

Ok great, I think it’s finally sinking in. Thanks for your time and patience, it’s much appreciated.