komponisto comments on [SEQ RERUN] Mach's Principle: Anti-Epiphenomenal Physics - Less Wrong

2 Post author: MinibearRex 16 May 2012 06:59PM

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Comment author: pragmatist 17 May 2012 11:00:07PM *  3 points [-]

My understanding (which you seem to confirm) is that "inertial motion" refers to geodesic paths. But those are precisely the paths which (the theory says) describe all motion! In other words, there's no such thing as "non-inertial motion". (Remember that GR is a theory of gravity alone -- as far as it's concerned, the other forces of nature don't exist, so everything is always "in freefall under gravity" at all times.)

Some worldlines satisfy the geodesic equation, others don't. The ones which do are geodesics. It's not true that GR cannot incorporate other forces of nature. It can, as long as these forces are amenable to a field-theoretic formulation. See here, for instance.

How so? Your link agrees with my definition.

You're right! I'm sorry, I read your definition wrong the first time.

I thought that in special relativity, spacetime was represented by a specific Lorentzian manifold: Minkowski space itself. (Or, perhaps more precisely, an affine space over Minkowski space.) In other words, the manifold is required to be flat (have zero curvature). Whereas in general relativity, the curvature is determined by the equations of motion.

That's right. The point I was making is that representing spacetime as a Lorentzian manifold (even a Lorentzian manifold with arbitrary curvature) is insufficient (and unnecessary) to get "no privileged reference frame". What that requires is that the laws in the theory are formulated in such a way that they do not presume anything about the reference frame in which they hold. Incidentally, both special relativity and Newtonian mechanics can be re-formulated in this way, it's just that they usually are not. That a space-time theory is generally covariant does not express a constraint on its content, only on its formulation. On the other hand, whether or not the manifold is Lorentzian is a matter of content. See here.

It seems to me that you're mixing up the local and global structures of spacetime. There is no fact of the matter about whether spacetime is flat or curved locally, because many of the permissible coordinate changes turn straight lines into curves and vice-versa. However, there is a fact of the matter about the global curvature of the manifold.

I'm not sure what you mean by whether or not a spacetime is curved locally. If the Riemann curvature tensor at a point vanishes in one frame of reference, then it must vanish in every frame of reference. So one cannot go from non-zero to zero curvature tensor locally by a change of coordinates.

It is true that one can change coordinates so that the metric at a point is Minkowski. This is what people usually mean when they talk about using a locally flat coordinate frame. But the metric reducing to Minkowski at a point does not mean that the curvature at that point vanishes. Curvature has to do with the second derivatives of the metric.

Comment author: komponisto 18 May 2012 12:51:56AM 0 points [-]

Some worldlines satisfy the geodesic equation, others don't. The ones which do are geodesics. It's not true that GR cannot incorporate other forces of nature

This is a matter of terminology, but I would maintain that a physical theory is defined by its equations of motion, and that GR is defined by the Einstein equation(s); and since the other forces do not appear in the Einstein equation(s), they are not part of GR. (There is no question of being able to incorporate them; the theory either does or does not incorporate them.) If you're working with Maxwell's equations in curved spacetime, you're not working in GR; you're working in a hybrid of GR and Maxwell's theory.

Thus defined, GR itself does not (as far as I know) allow non-geodesic paths to be worldlines. (An Einstein-Maxwell hybrid theory, on the other hand, might; but I would be tempted to suspect in that case that it isn't formulated "properly".)

The point I was making is that representing spacetime as a Lorentzian manifold (even a Lorentzian manifold with arbitrary curvature) is insufficient (and unnecessary) to get "no privileged reference frame". What that requires is that the laws in the theory are formulated in such a way that they do not presume anything about the reference frame in which they hold.

But that is the same thing, only expressed in old-fashioned physicist's language instead of modern mathematician's language. If you translate "the laws in the theory are formulated in such a way that they do not presume anything about the reference frame in which they hold" into modern mathematician's language, what you get is "the equations of motion must be expressed in terms of objects which are well-defined on a manifold (of the appropriate type)".

It is true that one can change coordinates so that the metric at a point is Minkowski. This is what people usually mean when they talk about using a locally flat coordinate frame. But the metric reducing to Minkowski at a point does not mean that the curvature at that point vanishes. Curvature has to do with the second derivatives of the metric.

I could hardly have said it better myself. The ability to use a locally flat coordinate system at a point (regardless of the value of the Riemann tensor at that point) is all that

you could never tell the difference between firing your rocket to accelerate through flat spacetime, and firing your rocket to stay in the same place in curved spacetime.

means.

Comment author: pragmatist 18 May 2012 05:06:29AM 0 points [-]

I could hardly have said it better myself. The ability to use a locally flat coordinate system at a point (regardless of the value of the Riemann tensor at that point) is all that

you could never tell the difference between firing your rocket to accelerate through flat spacetime, and firing your rocket to stay in the same place in curved spacetime.

means.

I don't see how it could mean that. Rockets are extended objects, as are people. While we can always find coordinates that make the metric Minkowski at a single point, it is not true that we can always find coordinates that make the metric Minkowski over a finite region, no matter how small.

Comment author: komponisto 18 May 2012 05:49:06AM *  0 points [-]

While we can always find coordinates that make the metric Minkowski at a single point, it is not true that we can always find coordinates that make the metric Minkowski over a finite region, no matter how small.

Indeed; this is why the concept of an "inertial frame" does not exist in general relativity, except in the infinitesimal limit.

But as long as we're going to permit ourselves to speak about the motion of an entire rocket or person, rather than the motion of its parts (thus in effect modeling the object as a point-particle), we can equally well describe the same rocket or person as being at rest.

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