General-Relativistic Covariance

Abstract

This is an essay about general covariance, and what it says (or doesn’t say) about spacetime structure. After outlining a version of the dynamical approach to spacetime theories, and how it struggles to deal with generally covariant theories, I argue that we should think about the symmetry structure of spacetime rather differently in generally-covariant theories compared to non-generally-covariant theories: namely, as a form of internal rather than external symmetry structure.

Covariance, Unvariance, and Transformations

[40, Appendix A] seek to show that “minimally coupled dynamical equations in GR manifest local Poincaré symmetry, when written in normal coordinates at any $p \in M$.” Here, I critically review their proof.

Read et al. begin by assuming that any minimally coupled dynamical equation in GR is of the form

$$\begin{aligned} O_1 + O_2 + \cdots + O_m = 0 \end{aligned}$$

(18)

where each $O_i$ is either:

• a tensor;

• a partial derivative of a tensor; or

• a partial derivative of a connection coefficient.

The reason to exclude (undifferentiated) connection coefficients—according to Read et al.—is that we are assuming the equation (18) is written in normal coordinates, in which (at the point p under consideration) connection coefficients vanish. However, this already risks introducing confusion. There is, I claim, a significant difference between an equation involving a partial derivative, and an equation involving a covariant derivative whose connection coefficients happen to be zero, even though the formal expression of these two equations will be the same. Specifically, two such equations will involve different transformation rules, and hence will have different invariance properties.

To see this, consider the two equations

$$\begin{aligned} \nabla _\mu J^\nu = 0 \end{aligned}$$

(19)

and

$$\begin{aligned} \partial _\mu J^\nu = 0 \end{aligned}$$

(20)

and suppose that we are in a flat space to which our coordinates are adapted, such that $\nabla _\mu J^\nu = \partial _\mu J^\nu$ (since $\Gamma ^\rho _{\mu \nu } = 0$). It follows that the two equations pick out exactly the same class of solutions. However, if we apply a coordinate transformation

$$\begin{aligned} x^\mu \mapsto \widetilde{x}^\mu \end{aligned}$$

(21)

then we transform $\nabla _\mu J^\nu$ as a rank-(1, 1) tensor, but transform $\partial _\mu J^\nu$ as the partial derivative of a components of a vector. This means that our two equations are transformed into

$$\begin{aligned} \frac{\partial \widetilde{x}^\alpha }{\partial x^\mu } \frac{\partial x^\nu }{\partial \widetilde{x}^\beta } \widetilde{\nabla }_\alpha \widetilde{J}^\beta = 0 \end{aligned}$$

(22)

and

$$\begin{aligned} \frac{\partial \widetilde{x}^\alpha }{\partial x^\mu } \widetilde{\partial }_\alpha \left( \frac{\partial x^\nu }{\partial \widetilde{x}^\beta } \widetilde{J}^\beta \right) = 0 \end{aligned}$$

(23)

respectively. The first of these is equivalent to $\widetilde{\nabla }_\alpha \widetilde{J}^\beta = 0$ (in the sense of having the same solutions), and so equation (19) is invariant under the coordinate transformation; but the second is not equivalent to $\widetilde{\partial }_\alpha \widetilde{J}^\beta = 0$, and so equation (20) is not invariant under the coordinate transformation. Another way of seeing what’s going on here is to observe that equation (19) is more fully expressed as

$$\begin{aligned} \partial _\mu J^\nu + \Gamma ^\nu _{\mu \rho } = 0 \end{aligned}$$

(24)

and although $\Gamma ^\nu _{\mu \rho }$ takes the value zero, its transformation rule means that the coordinate transformation (21) will (in general) transform it away from zero—in just such a way, of course, as to cancel out the extra terms arising from the transformation of $\partial _\mu J^\nu$. Equation (20), on the other hand, has no connection coefficients figuring at all (whether zero-valued or not); so such coefficients cannot step out from the shadows to guarantee invariance, in the way they do for equation (19).

What this means is that we should include connection coefficients on the list of possible ingredients for our equation: although such coefficients might be zero in the coordinate system we start with, if we are assessing which coordinate transformations preserve the form of the equations, we need to check that they will preserve the vanishing of those connection coefficients! Fortunately, adding them to the list of ingredients doesn’t make a significant difference to Read et al.’s next observation: that for affine coordinate transformations, all the ingredients transform tensorially. Recall that an affine coordinate transformation is a transformation of the form

(25)

where and $a^\mu$ are constant. Note that

(26)

and

(27)

where is the inverse to (i.e. is the matrix such that ). As is well-known, connection coefficients transform tensorially under affine coordinate transformations (since the non-tensorial part of the transformation rule features a second partial derivative). In the interests of space, I do not reproduce Read et al.’s proof that partial derivatives of tensors or partial derivatives of connection coefficients transform tensorially under affine transformations.

However, they then proceed to say

We have found that each of the $O_i$ featuring in any minimally coupled dynamical equation in GR, written in normal coordinates at a point $p \in M$, is covariant—i.e., transforms tensorially—under affine coordinate transformations. However, we have yet to show that all such equations are invariant—i.e. take the same form—under affine coordinate transformations. In fact, this is in general not the case.

Prima facie, this claim is somewhat surprising. For consider again the expression (18). In order for the left-hand-side to be well-formed, each $O_i$ must have the same index structure: i.e., they must have the same free covariant and contravariant indices (where a “free index” is one that has not been contracted with another index). But if two terms have the same index structure, then when they are transformed tensorially, they will pick up the same partial derivative terms; owing to the linearity of tensor calculus these terms can then be uniformly multiplied away, as we did in observing that equation (22) is equivalent to $\widetilde{\nabla }_\alpha \widetilde{J}^\beta = 0$. And note that the presence of bound indices (those which have been contracted) doesn’t make any difference: if we have an expression of the form

(28)

then applying the tensorial transformation rule yields

(29)

and so such indices “cancel out”.

Read et al. argue that this doesn’t hold, in general, “due to the potential contraction of indices in some terms with respect to the metric” (p. 11). As an example, they give (in my notation)

$$\begin{aligned} \partial _\nu F^{\mu \nu } = J^\mu \end{aligned}$$

(30)

They argue that the affine transformation (25) transforms this (again, using my notation) into

(31)

and that this latter equation is only equivalent to (30) if ; i.e., if is a Lorentz transformation (and hence, (25) a Poincaré transformation).

Now, this last assertion is correct in the sense that the right-hand equality in (31) is, indeed, only equivalent to (30) if is a Lorentz transformation. But the left-hand equality also only holds if is a Lorentz transformation: that’s the only way to use $\eta$ to raise or lower indices and convert into .⁴⁹ And if we look at the leftmost term in equation (31), we observe that—just as our general discussion of contracted indices would lead us to expect—we have a matrix term and its inverse ; cancelling these out, we see that (30) transforms into

(32)

and (32) is equivalent to (30). Thus, the supposed counterexample is invariant under arbitrary affine transformations (not just Poincaré transformations).

Indeed, it seems to me that we have good grounds to expect equations formed from minimal coupling to be invariant under arbitrary coordinate transformations (not just affine transformations)—i.e. to be generally covariant. The reason why the above proof was limited to affine transformations is that, in general, partial derivatives and connection coefficients will transform non-tensorially, and hence we will get “extra” terms showing up in the transformed equation—terms which will prevent the transformed equation from being equivalent to the original. But if partial derivatives and connection coefficients show up together, then it may be that the extra terms from the one cancel out the extra terms from the other, and we do get invariance under (non-affine) coordinate transformations.

The condition under which such cancellations happen is, of course, that the partial derivatives and connection coefficients in the equation are such as to form a covariant derivative—indeed, the whole point of covariant differentiation is that the result of applying a covariant derivative to a tensor is, itself, another tensor (as reflected in the transformation (22)). But now consider Read et al.’s definition of minimal coupling (pp. 2–3):

Minimally coupled dynamical equations for matter fields in GR are constructed from dynamical equations for matter fields featuring coupling to a fixed Minkowski metric field $\eta _{ab}$ and no curvature terms, by replacing all instances of $\eta _{ab}$ with a generic Lorentzian metric field $g_{ab}$, and replacing all instances of the torsion-free derivative operator compatible with $\eta _{ab}$ with the torsion-free derivative operator compatible with $g_{ab}$.

Thus, on the face of it, one would expect a minimally coupled dynamical equation to consist of tensors and covariant derivatives of tensors (with respect to the torsion-free derivative operator compatible with $g_{ab}$)—that is, tensors and tensors. And clearly, if all the terms $O_i$ in (18) are tensors (with the same index structure), then (18) will be invariant under arbitrary coordinate transformations.

The above, I claim, is the standard argument for the claim that generally covariant equations—including those obtained through minimal coupling—to be invariant under arbitrary (smooth) coordinate transformations. However, there is a different analysis one can give.⁵⁰ Suppose that our dynamical equation features the metric; very schematically, we give it the form

$$\begin{aligned} \cdots g_{\mu \nu } \cdots = 0 \end{aligned}$$

(33)

Now, if we have written this equation in normal coordinates, then $g_{\mu \nu } = \eta _{\mu \nu }$, where $\eta _{\mu \nu }$ denotes the matrix of coefficients (6). What is it for this expression to “retain the same form” under a coordinate transformation, from $x^\mu$ to $\widetilde{x}^\mu$? First answer: it is for it to be (or to be equivalent to) an equation with the same syntactic structure, albeit with tildes over everything; schematically, the form is preserved if (33) is transformed into something (equivalent to)

$$\begin{aligned} \widetilde{\cdots } \widetilde{g}_{\mu \nu } \widetilde{\cdots } = 0 \end{aligned}$$

(34)

(A non-schematic example is given by the comparison of (30) with (32).) This first answer is the answer that the argument above assumed, and so this is the sense in which its conclusion holds.

Second answer: it is for it to have the same syntactic structure, and for certain simplifying identities to continue to hold. In the case at hand, this will mean that the form is preserved if (33) is transformed into something equivalent to (34), and—in addition—$\widetilde{g}_{\mu \nu } = \eta _{\mu \nu }$. In defence of this answer, one can argue that part of what makes the normal-coordinate form the “simplest” form is that writing the equation out in components would be considerably simpler if the metric diagonalises than if it does not; and it is in this sense that a transformation away from normal coordinates makes the equation into one with a less simple form, and so ipso facto one with a different form.⁵¹

Evidently, if we require this second (stronger) sense of invariance, then any equation featuring the metric will be invariant only under coordinate transformations which are Poincaré in form, i.e. where

$$\begin{aligned} \frac{\partial \widetilde{x}^\alpha }{\partial x^\mu } \frac{\partial \widetilde{x}^\beta }{\partial x^\nu } \eta _{\alpha \beta } = \eta _{\mu \nu } \,, \end{aligned}$$

(35)

since it is only these equations which preserve the metric’s being diagonal.

Thus, the question becomes which of these two senses is more appropriate. The problem with the latter sense is that it risks overgenerating, and will lead us to underestimate the size of a theory’s symmetry group. In the context of electromagnetic theory on curved spacetime, for example, there will exist at any point Riemann normal coordinates with respect to which $J^\mu = (\rho , 0, 0, 0)$. Adopting those coordinates will simplify the dynamical equations, just as the adoption of Riemann normal coordinates does; and such equations will be preserved only under spatial translations and rotations. This suggests that on the stronger sense of invariance just canvassed, the equations of electromagnetism turn out to be invariant only under the Newton group of transformations. This seems to me a reason to prefer the former, weaker, notion of invariance.