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asked Are all static solutions of Einstein's equations spherically symmetric?
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comment What is a gauge transformation of the metric in GR?
Are you saying that the metric is physical? If so, I think that's a pretty radical assertion.
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17
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Jan
17
comment Why can't General Relativity be written in terms of physical variables?
Thanks Joseph, your edit addresses the question I meant to ask. Are "geodetic coords" the same as "Riemannian normal coords"? If so, I see why such coords can not be used to cover a potato globally (geodesics intersect). It seems very reasonable to guess that there is no other prescription that will work globally (or "almost globally" if that means anything) for all potatoes (I certainly can't think of any). But is there any sort of proof of this nonexistence? If not, I'm willing to leave it at that, but I posed this question since I was having trouble arguing this nonexistence convincingly.
Jan
17
comment Why can't General Relativity be written in terms of physical variables?
@Joseph: Thanks for your detailed analysis, but I think this whole issue of how "gauge" should be defined is beyond my pay grade, although I will try to read the sources you provided. The question I was really hoping to have answered is "Why can't an explicit prescription be given to uniquely fix coordinates on an arbitrary spacetime (or on the surface of an arbitrary potato)?" (I didn't know I was being controversial in suggesting this could be called "making a gauge choice").
Jan
17
comment What is a gauge transformation of the metric in GR?
I don't know what twisted tensors are, but what I've wrote down is certainly what is called a gauge transformation in the GR textbooks I've read, and what most relativists refer to as gauge transformations.
Jan
17
comment What is a gauge transformation of the metric in GR?
@Joseph: I don't totally understand your comments, but I'll just note that when I said "changes all of the fields, including the metric" in my answer, I had in mind the following idea: Let $M$ be the spacetime manifold and let $\chi$ be the collection of dynamical fields (including the metric and any matter fields). Let $\phi:M\rightarrow M$ be a diffeomorphism, and let $\phi^*$ be the map that pulls back tensor fields under $\phi$ (defined in appendix C of Wald). Then we call the field transformation $(M,\chi)\rightarrow(M,\phi^*\chi)$ a "diffeomorphism" (an abuse of language, I know).
Jan
17
comment What is a gauge transformation of the metric in GR?
@Harry: As long as any additional matter fields (charges, stressed solids, fluids, whatever) are put in in such a way that the total Lagrangian remains diffeomorphism covariant ("no fixed background"), then the diffeomorphisms are still gauge transformations (as long as you transform the matter fields as well).
Jan
17
comment What is a gauge transformation of the metric in GR?
...So if the Lagrangian changes by a total divergence, the action (which is the integral of the Lagrangian) is unchanged.
Jan
17
comment What is a gauge transformation of the metric in GR?
Oh, sorry. $v^\mu$ is a spacetime vector field (4 components, t,x,y,z) and $\nabla_\mu$ is the derivative operator associated with the metric. In flat spacetime $\nabla_\mu = \partial/\partial x^\mu$ so using the convention of summing over repeated indices, $\nabla_\mu v^\mu = \partial v^t/\partial t + \partial v^x/\partial x +...$, which is the generalization of the usual 3d divergence $\vec\nabla \cdot \vec v$. On curved spacetime its a bit more complicated, but Stokes' theorem still holds which says the integral of a total divergence over a region is zero if boundary terms can be ignored...
Jan
17
comment Why can't General Relativity be written in terms of physical variables?
I would argue that I didn't "change" the question, since the original question (which is the first paragraph, unchanged) said "Einstein's equations", not "linearized Einstein's equations", and made no reference to the linearized theory. I added the second paragraph to make it more explicit. I certainly look forward to your future update, if you decide to do one. BTW, I'm pretty sure the question is essentially equivalent to "why can't you uniquely fix coordinates on an arbitrary surface of a potato?". It's intuitively obvious you can't, but I'm not sure how to argue it.
Jan
17
comment What is a gauge transformation of the metric in GR?
$\nabla_\mu v^\mu$