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I'm confused by this question: Why can't General Relativity be written in terms of physical variables?

I can't quite see how you can make any change to the metric without either: (a) changing the physics; or (b) changing which point in your manifold represents which physical point in spacetime. Either of these would, it seems to me, disqualify the change as being a gauge transformation, at least in the usual sense.

So is there some cunning way of changing the metric that avoids either (a) or (b), or is there some sense in which (b) can still be thought of as a gauge transformation?

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Harry Johnston
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very smart people got this part really wrong. As you intelligently point out, EM is also invariant under diffeomorphisms, that alone does not make diff(M) a gauge group of EM. So we really need a better argument for saying that diff(M) is the gauge group of GR.

This has been pointed out before in the past, but i don't think there is a intelligent reply to these issues.

If we were really into finding a Lie group that would preserve GR physics, we might need to consider spacetimes which are allowed to have different curvature but test particles have the same apparent trajectories. I briefly speculate about such things in this question, but i don't think no one has take the time on these problems, me included.

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lurscher
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    EM is only invariant under diffeomorphisms if you introduce local fields that describe the external geometry, and these fields are nondynamical, the metric tensor doesn't have propagating modes in this conception. – Ron Maimon Jan 17 '12 at 04:37
  • @Ron, can you give more details? i don't understand what local fields you mean that need to be introduced, but certainly the diffeomorphic invariance of EM is not dependant on any additional degrees of freedom other than a vector potential defined up to a scalar gradient. – lurscher Jan 17 '12 at 04:45
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    I see, you are talking about dF=0 and dF=0. Sorry. I was talking about more general field equation which include non-form fields, like EM with a source. What I said is not accurate, but still you need a volume form to define "F" on a general coordinate system, so there is some geometrical information. If you just do a general transformation of a radiative electromagnetic field, it doesn't transform into a radiative electromagnetic field in a background with no metric tensor, but in a background with the transformed metric tensor. General covariance means "g is dynamical". – Ron Maimon Jan 17 '12 at 05:48
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Weyl, the inventor of gauge field theory, introduced the following notion, see his book pp. 121--123. If you have a manifold $M$, which has an affine connection given by the Christoffel symbols $\Gamma$, but is not necessarily pseudo-Riemannian, then you can define a notion which is subtly different from a metric tensor, which is a conformal structure instead. Then if you fix a gauge you can compare absolute values of tangent vectors at different points, but this notion of size might not be gauge invariant. Naively writing down the $g_{\mu\nu}$ you get does not produce a tensor, it produces a « tensor of weight one » which twists by a scalar factor as you change the gauge. I.e., it's not a tensor strictly speaking, but you can think of it as a twisted tensor (to me this is analogous to the sheaf of twisted differential operators which act on a line bundle but are not differential operators since they don't act on the functions on the manifold.) But if $M$ is pseudo-Riemannian (i.e., if the connection comes from a pseudo-Riemannian metric) you can choose a gauge for which $g_{\mu\nu}$ is a tensor.

joseph f. johnson
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A diffeomorphism (the gauge transformations in GR) does move around the points in spacetime, but it also changes all of the fields, including the metric, leaving the physics unchanged. An example would be if you had some time function on spacetime $t$ and you did a "time evolution" diffeomorphism which mapped each surface of constant $t$ into the surface of constant $t$ that is 10 seconds to the future of the original surface. As long as you similarly map ("pullback/pushforward") the metric and all of the matter fields forward in time, there should be no way to physically distinguish the new configuration from the old one. Mathematically, though, you have changed the metric and the matter fields on the spacetime manifold, since the new fields on the $t=0$ slice are different than the old fields on the same slice, since the new fields at $t=0$ have the values that the old fields have at $t=-10s$.

Joss L
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  • Yes, I understand that a diffeomorphism doesn't change the physics, but I don't think that "doesn't change the physics" automatically means "is a gauge transformation"! – Harry Johnston Jan 16 '12 at 04:15
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    In conventional electromagnetism, you could equally well perform a rotation, a translation, or a Lorentz boost, and you wouldn't have changed the physics, but I wouldn't consider that a gauge transformation. – Harry Johnston Jan 16 '12 at 04:16
  • Mathematically we can define an infinitesimal gauge transformation as a field variation which changes the Lagrangian by a total divergence (plus some conditions...). In E&M there is a fixed background metric, so a rotation would not qualify as a gauge transformation, but adding $\partial_\mu \xi$ to $A_\mu$ would qualify, since this changes the Lagrangian by 0. In GR, adding $\mathcal{L}\xi g{\mu\nu}=2\nabla_{(\mu}\xi_{\nu)}$ to the metric changes the Lagrangian by a total divergence (of $\xi_\mu$ times the Lagrangian). The difference is that there are no fixed background fields in GR. – Joss L Jan 16 '12 at 04:48
  • Yeah, I agree "doesn't change the physics" is not a complete definition of gauge. – Joss L Jan 16 '12 at 04:56
  • What is a "total divergence"? – Harry Johnston Jan 16 '12 at 20:42
  • $\nabla_\mu v^\mu$ – Joss L Jan 17 '12 at 00:12
  • That doesn't help, since I have no idea what either symbol means. – Harry Johnston Jan 17 '12 at 01:44
  • 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... – Joss L Jan 17 '12 at 02:14
  • ...So if the Lagrangian changes by a total divergence, the action (which is the integral of the Lagrangian) is unchanged. – Joss L Jan 17 '12 at 02:17
  • OK, but that only works if spacetime is empty. Surely a "proper" gauge transformation would work for the interacting case too? Certainly a conventional EM gauge transformation works (i.e., doesn't affect the physics) regardless of the presence of charges and any other forces that may be acting on them. – Harry Johnston Jan 17 '12 at 03:25
  • There is a crucial distinction between letting a diffeomorphism act on the manifold and also letting it act on the matter field...failing to observe this distinction means this answer is wrong. In fact diff(M) is not the gauge group of the metric or of GR. The metric is a tensor and so it makes sense to discuss how a diffeomorphism qua change of coordinates acts on it, but it is « covariant », i.e. although the formulas for its components in the first coordinates are no longer valid for its components in the changed coordinates, it's still the same tensor even if the map was not an isometry. – joseph f. johnson Jan 17 '12 at 04:28
  • And this has absolutely nothing to do with the Einstein field equations...There are in fact gauge transformations of a non-Riemannian manifold with an arbitray affine connections, and these do not preserve the metric tensor, but they are not coordinate transformations, either, i.e. they are not elements of diff(M). They would be the appropriate candidates for a gauge symmetry...I.e. the analogies between GR and EM are not completely valid, the analogy which talks of the tensor as the gravitational potential and the Christoffel symbols as the gravitational field are not strict analogies with EM – joseph f. johnson Jan 17 '12 at 04:29
  • @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). – Joss L Jan 17 '12 at 05:17
  • @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). – Joss L Jan 17 '12 at 05:29
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    This is not what I think of as a gauge transformation, which always has to do with some kind of bundle over the manifold, which makes « twisted tensors » the truly natural object, instead of tensors. So it has nothing to do with the field equations and nothing to do with the simple functoriality you are talking about. – joseph f. johnson Jan 17 '12 at 06:25
  • 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. – Joss L Jan 17 '12 at 06:36
  • Yes they do, but this is a controversial topic and they are, as they are entitled to do, making up their own meaning for « gauge » if they want to. It is not Weyl's original meaning, it is not a strict analogy with gauges in EM, and if you followed the links to this issue where it is discussed more explicitly, they will have taken you to the great Streater's valuable page, « Lost Causes in Theoretical Physics,» http://www.mth.kcl.ac.uk/~streater/lostcauses.html#XXII where he discusses why diff($M$) should not be regarded as a gauge group. – joseph f. johnson Jan 17 '12 at 16:51
  • @JossL: I think the upshot here is that while there are ways in which a diffeomorphism can usefully be thought of as a gauge transformation, it is not a gauge transformation in the sense most important to your original question - it doesn't mean that the variables being transformed are not physical. (Joseph's answer to your question explains this very well.) – Harry Johnston Jan 17 '12 at 22:24
  • Are you saying that the metric is physical? If so, I think that's a pretty radical assertion. – Joss L Jan 19 '12 at 00:37
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One sense in which a change of coordinates is a gauge transformation is described here.

Harry Johnston
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  • Yes it is a sense, but not a very tight analogy, and there may be other senses of gauge transformation which are different and useful for different purposes. – joseph f. johnson Jan 17 '12 at 04:42