Time in gravitational reference frame?

[MikeGomez]

The two cases are different. If I drop two objects from my two outstretched hands, and measure the distance between where they hit the floor... On the accelerating spaceship that distance will be equal to the distance between my outstretched hands, and on the surface of a planet it will be slightly less.

I understand that. It is a tidal effect. It doesn’t mean that dropping an object from one hand is any different between the two scenarios.

"Gravity" works fine, and we can use "tidal gravity" for the non-local curvature-dependent effects. (Consider that in natural language no one has any trouble with the convention that "milk" comes from cows even though "soy milk" does not).

But milk from cows and milk from soy are pretty different. With gravity, it’s not like that. Perhaps more like the difference in the air between when it is blowing and when it is not, or the difference in the bath water when the water swirling or when it is not, or the difference between the electrons in AC or DC.

No. You can make the curvature tensors vanish even when the Christoffel symbols do not. Heuristically speaking, the Christoffel symbols capture the effects of both curvature of the manifold and the effects of our choice of coordinates; the Riemann tensor is built out of combinations of these symbols in which the coordinate effects cancel out leaving only the coordinate-independent effects of curvature. For example, in a Euclidean plane the Christoffel coefficients are non-zero in polar coordinates and zero in cartesian; but the plane is flat and the Ricci and Riemann tensors are zero everywhere.

Most excellent! Thank you.

 

[MikeGomez]

BTW, Einstein addressed this issue when Hans Reichenbach made this very same mistake. When he (Einstein) says that it is of no importance whatsoever that the gravitational fields in general cannot be transformed away, he is speaking about the difference between tidal gravity and non-tidal gravity. Is he not?

“I now turn to the objections against the relativistic theory of the gravitational field. Here, Herr Reichenbacher first of all forgets the decisive argument, namely, that the numerical equality of inertial and gravitational mass must be traced to an equality of essence. It is well known that the principle of equivalence accomplishes just that. He (like Herr Kottler) raises the objection against the principle of equivalence that gravitational fields for finite space-time domains in general cannot be transformed away. He fails to see that this is of no importance whatsoever. What is important is only that one is justified at any instant and at will (depending upon the choice of a system of reference) to explain the mechanical behavior of a material point either by gravitation or by inertia. More is not needed; to achieve the essential equivalence of inertia and gravitation it is not necessary that the mechanical behavior of two or more masses must be explainable as a mere effect of inertia by the same choice of coordinates. After all, nobody denies, for example, that the theory of special relativity does justice to the nature of uniform motion, even though it cannot transform all acceleration-free bodies together to a state of rest by one and the same choice of coordinates.”
-A. Einstein

 

[PeterDonis]

what would be a good name for this type of gravity?

The term "acceleration due to gravity" is sometimes used for it; also "the force of gravity". Both of these have issues as well, of course.

When he (Einstein) says that it is of no importance whatsoever that the gravitational fields in general cannot be transformed away, he is speaking about the difference between tidal gravity and non-tidal gravity. Is he not?

Yes. Tidal gravity is the part that can't be transformed away; but the EP only refers to the other part, the part that can.

 

[MikeGomez]

However, in another sense of "local", spacetime curvature is not local, because it involves second derivatives of the metric, and in a small enough patch of spacetime around a given point, we can always adopt coordinates (local inertial coordinates) in which the effects of second derivatives of the metric are negligible. But phenomena that are referred to as "gravity" are still detectable in that small patch of spacetime--for example, bricks fall when dropped in an accelerating rocket, and this can be described, within a local inertial frame, as the rocket accelerating upward while the brick remains at rest, just as a brick falling on Earth can be described, in a local inertial frame, as the surface of the Earth accelerating upward while the brick remains at rest. All of this holds even though the effects of second derivatives of the metric are negligible within a local inertial frame, i.e., spacetime curvature is not "local" in this sense.

So then speaking in the proper language, what makes an object fall to the floor is due to the first derivatives of the metric tensor, not spacetime curvature, which are the second derivatives of the metric tensor. It doesn’t matter if the planet is round, infinite plane, or unicorn shaped.

Yes. Tidal gravity is the part that can't be transformed away; but the EP only refers to the other part, the part that can.

And Einstein is pointing out that the part that can’t be transformed away is of no consequence to how the part that can be transformed away is equivalent for the planet and the rocket.

 

[PeterDonis]

So then speaking in the proper language, what makes an object fall to the floor is due to the first derivatives of the metric tensor

No. What makes an object "fall" to the floor is that the floor is accelerating upwards and the object isn't. Or, in more technical language, the object is following a geodesic curve in spacetime, while the floor is following a non-geodesic curve that intersects the geodesic curve. And this can be described in local inertial coordinates, where the first derivatives of the metric tensor are all zero.

spacetime curvature, which are the second derivatives of the metric tensor

More precisely, spacetime curvature is described by the Riemann tensor, which is a particular combination of second derivatives (and first derivatives) of the metric tensor that transforms as a tensor itself.

Einstein is pointing out that the part that can’t be transformed away is of no consequence to how the part that can be transformed away is equivalent for the planet and the rocket.

Yes.

 

[MikeGomez]

And this can be described in local inertial coordinates, where the first derivatives of the metric tensor are all zero.

And is that because the first derivatives of the metric tensor are transformed away?

No. What makes an object "fall" to the floor is that the floor is accelerating upwards and the object isn't. Or, in more technical language, the object is following a geodesic curve in spacetime, while the floor is following a non-geodesic curve that intersects the geodesic curve.

Perhaps that is how it must be viewed in order to use the mathematical tools of GR, but there must be a better way to describe it than that if we are free to choose whatever reference frame we like. We can say the object is falling to the floor, just as well as the floor is accelerating up. The object is deflecting the planet from its freefall path in addition to the planet deflecting the object from its freefall path.

 

[PeterDonis]

is that because the first derivatives of the metric tensor are transformed away?

Yes; the transformation from a generic coordinate chart to a local inertial coordinate chart makes all of the first derivatives of the metric vanish.

The object is deflecting the planet from its freefall path in addition to the planet deflecting the object from its freefall path.

The planet, or at least the small piece of it that we're talking about, is not on a free-fall path to begin with. It's on an accelerated path. And the object doesn't deflect the planet's surface, because it's not free to be deflected; it's being pushed on by the rest of the planet. So the situation is not symmetric between the planet's surface and the object.

 

[MikeGomez]

No. What makes an object "fall" to the floor is that the floor is accelerating upwards and the object isn't. Or, in more technical language, the object is following a geodesic curve in spacetime, while the floor is following a non-geodesic curve that intersects the geodesic curve. And this can be described in local inertial coordinates, where the first derivatives of the metric tensor are all zero.

I’m certain this is correct (and extremely useful for doing physics) but I have doubts as to whether this is the definitive answer for what makes an object fall to the floor. It seems to be giving preferential treatment to a particular frame of reference in a relative situation. The planet is in freefall. Does choosing a piece of the surface of the planet as having an accelerated path have any more physical significance than selecting a frame of reference, or is there a deeper physical significance here that I am missing?

 

[PeterDonis]

It seems to be giving preferential treatment to a particular frame of reference in a relative situation.

That's because the description "fall to the floor" does the same thing: it gives preferential treatment to the frame in which the planet's surface is at rest. In the object's local inertial frame, the planet is rising up to hit it.

If you really want a "definitive" answer that doesn't make any use of particular frames, you would have to describe the worldlines of the object and the piece of the planet's surface and how they intersect at some particular event.

The planet is in freefall.

The center of mass of the planet is in free fall. But a particular piece of the surface of the planet is not. It has nonzero proper acceleration. Put another way, it feels a force--the force of the piece of the planet below it pushing up on it. The object that is falling, by contrast, feels no force; it is in free fall.

Does choosing a piece of the surface of the planet as having an accelerated path have any more physical significance than selecting a frame of reference, or is there a deeper physical significance here that I am missing?

There is a deeper physical significance: as above, the piece of the planet's surface feels a force, while the falling object does not. That is an invariant statement, independent of any choice of coordinates.