# Why can't General Relativity be written in terms of physical variables?

I am aware that the field in General Relativity (the metric, $g_{\mu\nu}$) is not completely physical, as two metrics which are related by a diffeomorphism (~ a change in coordinates) are physically equivalent. This is similar to the fact that the vector potential in electromagnetism ($A_\mu$) is not physical. In electromagnetism, the equations can be written in terms of physical (i.e. gauge invariant) quantities -- the electric and magnetic fields. Why can't Einstein's equations similarly be written in terms of physical variables? Is it just that nobody has been able to do so, or is there some theorem/argument saying that it can't be done?

EDIT: Let me rephrase: Prove/argue that there is no explicit prescription that can be given which would uniquely fix coordinates for arbitrary physical spacetimes. I.e., show that there is no way to fix gauge in the full theory of general relativity (unlike in E&M or linearized GR where gauge can be fixed).

• I'm not so sure that they can't - certainly people are working on it. I have a friend who was involved in a project to express pulsar signal distortion in terms of the Riemann curvature or something like that, I'll see if I can convince him to pipe in here. – David Z Jan 10 '12 at 1:52
• Is the Riemann tensor physical? Its trace (the ricci tensor) appears in Einstein's eqn, so I would think not. Also, is the pulsar stuff in linear perturbation theory? I think that can be made gauge invariant, but I was more interested in the full theory. – Joss L Jan 10 '12 at 2:05
• I'm not sure which notion of "physical" we are using here. If you argue that the metric is not physical, because after a diffeomorohism it's "another metric", then any vector or tensor, like the force or the field strength is not physical either. If you want to get rid of the coordinates, then just write the Einstein equations as $G=c\ T$. And regarding the other issue, the equation $f'(x)=k\ f(x)$ also doesn't fix a single solution. – Nikolaj-K Jan 10 '12 at 8:23
• Is your question "Maxwell's eqns are written in terms of $F_{\mu\nu}$ which doesn't change under gauge transformations, but Einstein's eqns are written in terms of $R_{\mu\nu}$ (etc) which does change under gauge transformations (and you take GR gauge transformations to be diffeos) ? If so it may be useful to take a look at this link – twistor59 Jan 10 '12 at 8:57
• @Nick Firstly, I don't see how the question reduces to just singling out a coordinate system - the Diff(M) gauge group is a group of active diffeomorphisms, so a gauge orbit consists of different metrics, not just different coordinate reps for a fixed metric. Secondly, however you do it, to gauge fix, you need to convince yourself there's no Gribov ambiguity don't you ? On a space as complex as this I wouldn't know how to do this. – twistor59 Jan 15 '12 at 14:10

Wigner always complained about people who used the word « invariant » (this was, of course, in the context of Special Relativity): he said one should say that the principle of relativity requires « covariance,» not invariance. Einstein's own papers on GR tend to carry out Wigner's request: the theory of GR (which is more general than Einstein's theory of gravity or his field equations) is always expressed as the demand for covariance with respect to arbitray coordinate changes. Now another myth, according to me, is that relativity requires that the laws be transformed by the group into other laws « of the same form.» This is just verbiage until you define what you mean by « form,» and even worse, it would then be just linguistics instead of physics. Examining Einstein's practice in this matter, and his occasional explicit pronouncements, GR really says: the laws of physics must take the form of equating a tensor to zero. This works because the tensor has the same covariance properties under change of coordinates as zero does.

Therfore the requirement of covariance has nothing to do with $f^* g$ for $f$ a diffeomorphism and $g$ a tensor field. This can be seen another way: for Einstein, $M$ is not physical, it is $g$ that is physical. Hence $f$ has to be regarded as a change of coordinates which does not actually move the mathematical points of $M$. The formula which mathematicians use for $f^*g$ has to be reinterpreted as coming from $f$ as the transition function between two charts of $M$ around a given point $x$, i.e., qua diffeomeorphism, it is the identity. Let me put this another way: a change of coordinates does not move the points, it just changes the charts. Therefore, a change of coordinates is the trivial identity map when you look at it in the mathematicians' invariant, coordinate free definitions of $M$, $f$, $f^*$, and tensors.

And now let me put this a third way, tying it up with Wigner's point of view in Special Relativity: what GR requires is that $g_{\mu\nu}$ be covariant, and in your set-up, this is the requisit for $f^* g$ to even be well-defined. This is what is needed to define a group action of diff($M$) on the set of metric tensors, and is the strict analogy to Wigner's requiring that one work with a representation of the Lorentz group. Covariance means the group action is defined, not that it is trival.

That is one reason why diff($M$) is not a good analogy to the gauge transformations of EM or Weyl's theory. But there is another: in EM, the relation between the potential and the field is one thing, but the relation between $g_{\mu\nu}$ and the Christoffel symbols (the affine connection) is quite another. Yes, mathematically the two relations have something similar but from the stanpoint of the symmetries involved there is a crucial difference: the metric field is covariant (a tensor) but the Christoffel symbols are not, whereas in EM, the fields transform well under the Lorentz group too. Hence, by the philosophy of GR, the metric tensor field has to be regarded as more physical than the Christoffel symbols even though everyone, and Einstein too, calls the metric the « gravitational potential » and the Christoffel symbols the « gravitational field.» This suggested analogy just shouldn't be taken too seriously, and in fact, Einstein himself constantly oscillates between this terminology and the seemingly contradictory calling the metric tensor field « the gravitational field,» which, at least to me, suggests he didn't think the distinction was that important.

And there is another: in EM we can only measure differences in potentials, of course, so that is why we introduce a choice of gauge and gauge transformations. But (pace extreme positivists), we can in principle measure the metric tensor using light rays and clocks and travelling rods, as explained by Weyl and Einstein. (Of course only because this is an ideal classical world so we can make the masses negligible...). Einstein's equations are irrelevant! Just as, in discussion of what a choice of gauge is and what a gauge transformation is in EM, Maxwell's equations were irrelevant! That is, the definition or concept of gauge and gauge transformation make sense, and one can think about their physicality and desirability, even without considering Maxwell's equations. And following that road, one could first decide, on physical grounds, what the gauge transformations were, and then search for the Law of Nature that was gauge invariant in that sense.

But the Christoffel symbols, although they can obviously be measured in a sense since they can be calculated from the metric, are not physical because they are not covariant. Too much argument about what « physical » means would be philosophical, but all I want to really insist on is that for GR, if something isn't even covariant then it is not objective and « real », so this destroys the analogy with gauges in EM all by itself.

Having now disposed of diff($M$), I briefly state what everyone already knows: for any Riemannian or Lorentzian manifold, there exists a gauge which makes the metric field a tensor, this is explained by Weyl in his book, except he calls it a calibration. So that answers your question for classical GR.

EDIT for the comment by the OP.

The principle of general relativity is that there is no natural way to distinguish between one set of coordinates and another. That is the whole point of GR, its philosophy, if you will. There is no physical criterion to use to say one coordinate system is better than another.

Perhaps you already knew that, so let us consider choices which have no physical motivation or significance but look pretty. E.g., geodetic coordinates. For any $M$ and any given point $p$ you can define local coordinates in a small neighbourhood of $x$ in $M$ which are geodetic in the sense that they nicely describe parallel transport along the coordinate axes. But they have no global significance, they don't do anything for the whole potato, only for the one point $x$, because as soon as you parallel transport something a finite distance away from $x$, what you get depends on the path you took to get there. They have « local » significance, not « global » significance, and the reason there is a difference between local and global is the geometric fact of non-integrability, which is inherent in the curved geometry of $M$. Only if $M$ is flat is the situation « integrable.» In fact, this is the definition of curvature. Curvature is defined as the deviation from integrability of this parallel transport you do in a geodetic coordinate system.

So the answer to your question is: there is no coordinate system in the large with nice properties, unless $M$ is flat.

You see, the question was confused between choosing a gauge and choosing a coordinate system, these are not the same things. If this confusion is straightened out, it gets two different answers: If $M$ is pseudo-Riemannian, yes there exists a choice of gauge which means the metric can be represented by a tensor, not a twisted tensor. But no, there does not exist any prescription for coordinates which have nice properties in the large unless $M$ is flat.

• Nice description. Re the "or prove it can't be done" request in the original question, Torre uses the 3+1 canonical framework to describe the problems with constructing Dirac observables for GR: If $\Gamma$ is the phase space for GR (cotangent bundle of the space of 3-metrics) then a physical metric must lie in the subspace $\bar{\Gamma}$ of $\Gamma$ which is defined by the Hamiltonian and diffeomorphism (momentum) constraints. – twistor59 Jan 17 '12 at 14:14
• He then demonstrates the impossibility constructing a Dirac observable which is an integral of a local function of the phase space variables and their derivatives, local meaning being restricted to a finite number of derivatives. – twistor59 Jan 17 '12 at 14:15
• @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"). – Joss L Jan 17 '12 at 17:01
• @JossL Fair enough, I have enlarged my answer to try to deal with what you clarify here in your helpful comment. – joseph f. johnson Jan 17 '12 at 19:31
• 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. – Joss L Jan 17 '12 at 21:10

Let us reformulate OP's question(v1) as follows.

Can General Relativity in $d$ bulk spacetime dimensions be written in terms of physical/propagating variables only?

The best one can do seems to be the following. For weak gravitational fields, one can write the curved metric

$$g_{\mu\nu}~=~\eta_{\mu\nu}+h_{\mu\nu}$$

as a sum of a flat Minkowski background $\eta_{\mu\nu}$ and a fluctuation part $h_{\mu\nu}$, which is symmetric and therefore contains $\frac{d(d+1)}{2}$ independent components.

Now use light-cone coordinates for the flat metric $\eta_{\mu\nu}$. The fluctuation part $h_{\mu\nu}$ then splits into $2d$ unphysical auxiliary variables (which can be eliminated), and $\frac{d(d-3)}{2}$ physical variables (=the traceless transversal part).

Reference:

Barton Zwiebach, A first course in String Theory, Section 10.6.

• Thanks. I'm aware that linearized perturbations off a background solution (at least Minkowski and FLRW; I'm not sure about general backgrounds) can be written in terms of gauge invariant variables, but I was more interested in the full (nonperturbative) theory. – Joss L Jan 10 '12 at 14:41
• I see that OP changed the question so that the new question(v2) now specifically excludes linearized gravity considerations. I might try to answer the new question(v2) in a future update. – Qmechanic Jan 16 '12 at 21:44
• 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. – Joss L Jan 17 '12 at 0:27

If the notion of being physical is gauge-invariance, then the Ricci scalar in the Einstein-Hilbert action is a "physical" variable, in the same sense that $F_{\mu \nu} F^{\mu \nu}$~$(|E|^2-|B|^2)$ and $F_{\mu \nu} \tilde{F}^{\mu \nu}$~$(E \cdot B)$ are the fundamental gauge invariant quantities in pure Yang-Mills theories. But Einstein field equations are not built from an invariant in the same way that Yang-Mills field equations are not built from its invariants. Nevertheless, these field equations remain unchanged under gauge transformations of the fields, because the extra contribution is a total derivative term in the Lagrangian (unless the manifold has a boundary, in which case a Gibbons-Hawking term has to be added to the Lagrangian to absorb the extra contribution)

Note that $E$ and $B$ fields themselves are not gauge invariant as your question seems to suggest.

I am not sure if Ricci curvature is the only fundamental invariant of Riemannian manifolds. Is Yamabe invariant fundamental? Would be nice if someone could post a list of (fundamental and derived) invariants.

• Sorry for the incorrect remark. My mistake! $E$ and $B$ are indeed invariant, but only because QED is abelian gauge. In non-abelian Yang-Mills, the only lowest order invariants under gauge symmetry are the ones I mentioned. Varying Yang-Mills w.r.t. $A$ gives its field equations (which is not put-together using those invariants). In GR, the gauge symmetry is GCT. Under GCT, $R$ is an invariant. Varying Einstein-Hilbert action w.r.t. metric $g$ gives its field equations. – GuSuku Jan 15 '12 at 4:57
• A better way to draw parallels between the two worlds is to consider matter spinors coupled to GR in vielbein formalism. Here, each term has to be invariant under both gauge symmetries. – GuSuku Jan 15 '12 at 5:17
• You are referring to an isometry, and not invariance under GCT. $R$ is called a scalar precisely because it is invariant under GCT/diffeomorphism. – GuSuku Jan 15 '12 at 11:47
• Some other scalars/invariants: en.wikipedia.org/wiki/… . Looks like Kretschmann scalar corresponds to $F_{\mu \nu}F^{\mu \nu}$ of Yang-Mills and Chern-Pontryagin scalar to $F_{\mu \nu} \tilde{F}^{\mu \nu}$. I dont know what Euler scalar would correspond to. – GuSuku Jan 15 '12 at 12:30
• Curvature invariant $\neq$ diffeomorphism invariant. Under the infinitesimal diffeomorphism generated by the vector field $\xi$, $R$ transforms by $R \rightarrow R + \mathcal{L}_{\xi}R$. If $\xi$ is not a Killing vector then this diffeo is not an isometry (by defn of Killing vector). In any case, this is all irrelevant to my original question, which was about gauge invariants, not curvature invariants. – Joss L Jan 15 '12 at 20:33