Time in gravitational reference frame?

[phinds]

No, it isn't. It's tidal gravity. But tidal gravity is not what makes things fall into the floor locally on the surface of a planet (or in an accelerating rocket).

Thank you Peter. I KEEP getting that wrong.

 

[MikeGomez]

I don’t have a problem with the statement that gravity is spacetime curvature, because I don’t feel the need to qualify that with “tidal” gravity. We had the discussion in earlier posts that no uniform gravitational field exists and I assumed that was true because nobody challenged it, and that indicates to me that all gravity is tidal.

I guess there is a terminology issue that I have been missing regarding spacetime curvature. If spacetime curvature is present globally then it seems to me it must exist locally. By my own logic, spacetime curvature would be a broad term with many aspects, some of which have a global effect on separated spacetime events, and some of which have an effect locally in a area where the separation is small enough that the EP holds. When you say “… spacetime curvature, at least of the form that is present around an ordinary spherical planet, disappears in this limit” I thought that the one form of spacetime curvature which disappears in the limit is due to the spherical shape of the planet, and not not fundamentally different from any other form of spacetime curvature (or gravity) which remains. Anyway, I guess that’s wrong. Could you please discuss a little about this non-tidal gravity which makes things fall to the floor locally.

Thanks

 

[PeterDonis]

I don’t have a problem with the statement that gravity is spacetime curvature, because I don’t feel the need to qualify that with “tidal” gravity.

I'm not sure this is a good policy on your part. See below.

We had the discussion in earlier posts that no uniform gravitational field exists and I assumed that was true because nobody challenged it, and that indicates to me that all gravity is tidal.

The "gravity" that makes a rock fall locally on the surface of a planet is not tidal gravity, as I have already said. But the term "gravity" is very commonly used to refer to this phenomenon, so the statement "all gravity is tidal" is, at the very least, liable to cause confusion.

It is true, AFAIK, that "no uniform gravitational field exists" in the sense that there is no known solution to the Einstein Field Equation that can correctly be described as a "uniform gravitational field". But going from that to "all gravity is tidal" is a logic error; the only correct logical deduction you can make from it is "all gravitational fields are non-uniform". (We're also assuming here that the term "gravitational field" has an unambiguous meaning, which is not really true; that term also has multiple meanings in the literature.) As I noted just now, there are phenomena taking place in non-uniform gravitational fields which are referred to as "gravity" but are not "tidal gravity".

If spacetime curvature is present globally then it seems to me it must exist locally.

The term "locally" is yet another term with multiple meanings. In one sense of the term, spacetime curvature is indeed local, since it is described by a tensor, the Riemann tensor, and tensors are "local" objects--they are defined in the tangent space at each point of spacetime.

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.

some of which have an effect locally in a area where the separation is small enough that the EP holds.

No effects of spacetime curvature are detectable in a patch of spacetime small enough that the EP holds. That is the definition of what "small enough" means in terms of the EP--if any effects of spacetime curvature are detectable, it means you aren't focusing on a small enough patch of spacetime for the EP to apply.

spacetime curvature, at least of the form that is present around an ordinary spherical planet, disappears in this limit”

We should probably drop that particular example, since I've already said several times that AFAIK it doesn't correspond to any valid solution of the EFE. Saying that "well, you just take a spherical planet and take the limit as its radius goes to infinity" doesn't mean that limit corresponds to anything valid.

Could you please discuss a little about this non-tidal gravity which makes things fall to the floor locally.

See above.

One other note: you appear to me to be getting hung up on terminology instead of focusing on the physics. This is one reason why discussions of this sort are best done with math, not ordinary language. The math underlying everything I've said is completely unambiguous, and raises no awkward questions about what "gravity" means or what "local" means or what "spacetime curvature" means.

 

[MikeGomez]

The term "locally" is yet another term with multiple meanings. In one sense of the term, spacetime curvature is indeed local, since it is described by a tensor, the Riemann tensor, and tensors are "local" objects--they are defined in the tangent space at each point of spacetime.

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.

Is this the vanishing of the Christoffel symbols (or making them small enough to be 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 what would be a good name for this type of gravity? Having to say “the gravity that makes you fall into the floor” is a mouthful, and “apple falling gravity” or “apple gravity” sounds dumb.

One other note: you appear to me to be getting hung up on terminology instead of focusing on the physics. This is one reason why discussions of this sort are best done with math, not ordinary language. The math underlying everything I've said is completely unambiguous, and raises no awkward questions about what "gravity" means or what "local" means or what "spacetime curvature" means.

Very well then. A few terminology issues remain (for me) but that can be postponed for another time. What I am still hung up on is Ibis’s comments in post #12, which seem to indicate that the physics (and not just terminology) are different for the two cases of falling to the floor on the planet and falling to the floor in the rocket. If that is what he is really saying, I don't think the math will support his argument.

 

[Nugatory]

Is this the vanishing of the Christoffel symbols (or making them small enough to be negligible)?

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.

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

"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).

Very well then. A few terminology issues remain (for me) but that can be postponed for another time. What I am still hung up on is Ibis’s comments in post #12, which seem to indicate that the physics (and not just terminology) are different for the two cases of falling to the floor on the planet and falling to the floor in the rocket. If that is what he is really saying, I don't think the math will support his argument.

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. The equivalence principle just says that we can make that difference arbitrarily small.

 

[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.