Is my understanding of the equivalence principle correct?

[stevendaryl]

Yes. That is how its defined in the GR literature. e.g. Principle of Equivalence, F. Rohrlich, Ann. Phys. 22, 169-191, (1963), page 173

I think that doesn't really make any sense. And the fact that something appears in the literature doesn't by itself make it sensible. If you look at how the equivalence principle is actually used in physics, it doesn't have anything to do with curvature. The way it's actually used (this is the way Einstein used it originally) is to predict (approximately) how clocks and other physical things behave on Earth by comparison with a rocket accelerating in flat spacetime. That use does not depend on curvature vanishing (it doesn't vanish on the Earth). What it depends on is that the specific experiment under consideration must be insensitive to small changes in the gravitational field.

If you conduct an experiment confined to a small volume of space for a small duration of time, and the variation of the gravitational field within that volume is negligible, then the results will be the same as if one performed the same experiment confined to a small volume of an accelerating rocket, provided that the length $L$ is small enough that $\dfrac{g\ L}{c^2}$ is negligible. Curvature doesn't really enter into it. Whether the variation of gravity within a volume is negligible depends on the size of the volume, but curvature doesn't depend on the size of the volume.

This is what allowed Einstein to make the prediction of gravitational time dilation and gravitational redshift BEFORE he had developed the full theory of GR. In particular, it was before he had the field equations, and before he associated "gravity-free space" with "zero curvature".

If you consider a rotating coordinate system, there are "fictitious gravitational forces" such as the Coriolis force and the Centrifugal force. I think it would be weird to call those forces "uniform", even though the curvature tensor is perfectly zero. The real criterion for the usability of the equivalence principle is whether the variation of the "gravitational forces" are negligible within the volume under consideration.

 

[stevendaryl]

The term "uniform", as used in the relativity litertature, means that there are no tidal forces (i.e. no spacetime curvature). It doesn't mean what you just said.

Well, my point is that, whether or not that is common terminology, it is BAD terminology. It gives a several misleading impressions: (1) It leads people to think that the fictitious gravitational force experienced by an accelerating rocket is uniform, with the usual meaning of "uniform"---it's not. (2) It leads people to think that the equivalence principle is only useful in calculations involving flat spacetime--it's not. The equivalence principle can be used perfectly well in curved spacetime, if one understands that one must only apply it to experiments taking place in a sufficiently confined region of spacetime that the variation of the "gravitational field" (that is, the connection coefficients) within that region is negligible.

 

[ApplePion]

"The term "uniform", as used in the relativity litertature, means that there are no tidal forces (i.e. no spacetime curvature). It doesn't mean what you just said."

Usually the presence or absense of tidal forces is a good indication of whether the Riemann Tensor vanishes. But not always.

In the example I gave you where g00 was of a form (ignoring unimportant constants) g00 = x^2 , there are tidal forces. The gravitational field is approximately the first derivative of g00, and so it is proportional to x. So we have a case where the gravitational field in the x direction varies with x. So there are tidal forces. But if you plug the g00 = (x^2) into the formula for the Riemann tensor, you will see the Riemann Tensor vanishes. (Do it, if you do not believe me.) So we have an example if a situation with a non-vanishing tidal force wherethe spacetime is flat.

 

[stevendaryl]

Another point about why it's very misleading to identify "uniform gravitational field" with "zero curvature": According to GR, zero curvature means flat spacetime. It means no gravitational sources anywhere. It means that the gravitational field can be eliminated everywhere through a coordinate transformation. That's obviously the case with an accelerating rocket--you transform to an inertial coordinate system, and the field disappears. So saying that an accelerating rocket is equivalent to zero curvature spacetime is a complete triviality: of course, an accelerating rocket in flat spacetime is equivalent to flat spacetime. You really haven't said anything by making this equivalence.

The interesting thing about the equivalence is to compare the behavior of clocks aboard a rocket to clocks at different heights near the surface of the Earth. But for that comparison, curvature doesn't matter---you're comparing a zero-curvature case with a nonzero curvature case.

 

[PeterDonis]

In the example I gave you where g00 was of a form (ignoring unimportant constants) g00 = x^2 , there are tidal forces.

You are defining "tidal forces" in a non-standard way. The standard definition of "tidal forces" requires initially parallel geodesics--freely falling worldlines--to converge or diverge. Try computing the geodesic equation for your modified metric; you will find that geodesics that are initially parallel remain parallel, indicating zero tidal forces in the standard sense. In other words, "tidal forces" as they are standardly defined are invariants; you can't make them appear or disappear by changing coordinate charts.

 

[PeterDonis]

His most conceptually elegant example of such a system is ball of fluid. Imagine it orbiting the Earth (for simplicity, eliminate the moon, sun, etc.). It will not be a perfect sphere due to tidal effects. The interesting fact is that its non-sphericity (ratio of axes in different directions) remains constant as the ball shrinks, as long as you assume it is a perfect fluid with no surface tension (its density remaining constant).

Does "non-sphericity" refer to the equilibrium state of the ball in which tidal effects are exactly balanced by internal forces within the fluid? (I ask because a ball of dust, with no internal forces, has a non-sphericity that changes with time; but it seems like Ohanian is talking about a steady state.)

 

[stevendaryl]

Usually the presence or absense of tidal forces is a good indication of whether the Riemann Tensor vanishes. But not always.

Umm. I would say that the Riemann Tensor is the precise definition of tidal forces. Tidal forces are about geodesic deviation, which is exactly what the Riemann tensor describes.

In the example I gave you where g00 was of a form (ignoring unimportant constants) g00 = x^2 , there are tidal forces.

I would say that in that case the "gravitational field" is nonuniform, but I wouldn't say that there are tidal forces in that case.

 

[PAllen]

Does "non-sphericity" refer to the equilibrium state of the ball in which tidal effects are exactly balanced by internal forces within the fluid? (I ask because a ball of dust, with no internal forces, has a non-sphericity that changes with time; but it seems like Ohanian is talking about a steady state.)

Yes, he is talking about equilibrium - a geoid surface shaped only by balance of tidal force, self gravitation, and pressure (with density constant and viscosity constant, and zero surface tension). Under these specific assumptions, the tidal forces go down (assuming r is radius of ball) by r^2, the self gravity by r, leaving tidal bulge size going down by r. But then the ratio of tidal bulge to radius (asphericity) remains constant.

 

[atyy]

Thus, mathematically, the basis for strong EP is local flatness, but Ohanian argues (without disputing the math) that the physics of certain systems makes the mathematical statement unphysical. I find this argument interesting, and not so easy to brush off.

Basically, the second derivative is local in the sense that it is well-defined at a point - that's calculus.

However, because notionally, the second derivative compares two pairs of points, whereas the first derivative only one pair, the second derivative is said to be more nonlocal than the first derivative. Admittedly, this is not as rigourous a definition as the first. But maybe one can formulate it based on Fermi normal coordinates.

So it depends on what one means by local.

The tiny instruments that detect local curvature use the first definition of local. OTOH, all experiments that show that SR is a tremendously good approximation in some regimes, show that the second definition show that the second definition is "physically meaningful" in some sense.

Because our current "fundamental" laws of physics (standard model of particle physics) can be coupled unambiguously to the spacetime metric by "comma to semicolon", the equivalence principle does hold at quite a basic level. However, the "derived" laws of physics which may involve second derivatives cannot be unambiguously coupled to the spacetime metric, so for those laws, the EP could be said to fail.

I'm not sure I got that completely right, but basically there are two notions of "local" in play, and there are two sorts of physical laws - those specified using first derivatives only, and those using second derivatives. It so happens out "fundamental" laws use only first derivatives at the moment.

BTW, I do like Ohanian's books very much:)