In most treatments of General Relativity, when the the Einstein-Hilbert action over some manifold $\mathcal{M}$ (plus Gibbons-Hawking-York term if $\mathcal{M}$ has a boundary), given by


(with $8\pi G=\kappa$) is varied, it is done so by an implicit choice of chart. Namely, one writes the action (locally) as


From this, one uses the coordinate equations for $\mathcal{R}$ and vary it with the components of the metric tensor in this particular chart to derive the Einstein field equations, which is a very long and tedious process, prone to mistakes from misplaced indices.

However, it is generally possible to derive equations of motion for field theories without referencing a coordinate chart. In fact, such methods are typically very generalizable and very powerful for calculating conserved quantities. Consider, for instance, (Euclidean) Maxwell theory of a $U(1)$ gauge field $A$. The action is simply

$$S=\frac{1}{2e^2}\int_{\mathcal{M}}F\wedge\star\,F+\int_{\mathcal{M}} A\wedge\star\,j,$$

where $F=\mathrm{d}A$ and $j$ is some matter current. Under a general transformation $A\to A+\delta A$, we have

$$\delta S=\frac{1}{e^2}\int_{\mathcal{M}}\mathrm{d}\delta A\wedge\star\,F+\int_{\mathcal{M}}\delta A\wedge\star\,j\\=\int_{\mathcal{M}}\delta A\wedge\left(\frac{1}{e^2}\mathrm{d}\star F+\star\,j\right)+\int_{\partial\mathcal{M}}\delta A\wedge\star\,F,$$

from which the equations of motion are immediately $\mathrm{d}F=0$ and $\mathrm{d}\star F=-e^2\star\,j$, assuming $\delta A$ has no support on $\partial\mathcal{M}$.

Does there exist a similar completely covariant and coordinate-independent derivation of the Einstein field equations from the Einstein-Hilbert action?

  • $\begingroup$ I will note here in relation to my answer that it seems I was wrong about the parallelizability of 3-manifolds. It seems it is only true for compact, orientable 3-manifolds that they are all parallelizable. However I have heard mathematicians say that "most" noncompact 3-manifolds are parallelizable, but I don't know what does this mean exactly. I need to research this a bit more. $\endgroup$ – Bence Racskó Nov 14 '18 at 10:28

First of all, it is a bit unclear what do you want exactly. As in:

  • Do you want an approach that doesn't use coordinates at all, but can, for example, use frames?

  • Do you want a completely "invariant" approach that absolutely doesn't use any local trivializations of any fiber bundles whatsoever? If so, are you willing to work on the total spaces of said bundles?

  • Do you want an approach that is global, regardless of any other considerations?

  • Do you want an approach that doesn't use indices, but everything else is game?

This is not clear to me. Especially, that despite the fact that your issue with Arnold Neumaier's answer was that it uses frames, but the electrodynamics example you have given essentially also uses frames.

Why? Because if $P\rightarrow M$ is a principal $\text{U}(1)$ bundle with principal Ehresmann connection $\omega$, then what you call $A$ is essentially defined as follows. If $(U,\varphi)$ is a local section of $P$ ($U\subseteq M$ is the domain and $\varphi:U\rightarrow P$ is the map), then we define $A^{(U)}=-i\varphi^\ast\omega$, valid in $U$. Since $\omega$ is a pseudotensorial $1$-form of type $\text{Ad}$ on $P$, these pullbacks in overlapping neighborhoods will not fit into one well-defined global object on $M$. Essentially, using $A$ is the same as using $\omega^a_{\ b}$ (the connection forms) in GR. Also note that thus $A$ is well-defined globally if and only if $P$ admits global sections.

With that said, there are several approaches one can take to vary the Einstein-Hilbert action in an "invariant" manner.

The global geometrical approach:

The biggest difficulty is to derive the variation of the volume element $đ\mu=\sqrt{-g}d^4x$. The easiest is of course to expand in local coordinates, but we want to avoid that for now. Aside from working directly on some fiber bundle, I think one needs to use at least frames here.

One reason why you need to use frames here is that a differential form vanishes when you feed linearly dependent vectors into it. So if $đ\mu$ is a volume form, and $X_1,...,X_n$ are vector fields, then $$ đ\mu(X_1,...,X_n)\neq 0\Leftrightarrow X_1,...,X_n\ \text{is a linearly independent set.} $$ But if these $n$ vector fields are linearly independent, then they form a frame, essentially.

The best thing we can do is to use frames, but only use them "passively". Eg. the volume form is not defined in terms of the frame, but the frame is used to derive relations.

We know that if $\Omega$ is any $n$-form and $đ\mu$ is a volume form, then there is a function $\alpha$ such that $$ \Omega=\alpha\ đ\mu, $$ in particular, since the variation of $đ\mu$ is an $n$-form, there is an $\alpha$ such that $$ \delta đ\mu=\alpha đ\mu. $$

If $(e_1,...,e_n)$ is a positively oriented orthonormal frame, then we have $đ\mu(e_1,...,e_n)=1$, so we have $$ \alpha=(\delta đ\mu)(e_1,...,e_n). $$

Although it is not the most rigorous of approaches, it is not immoral or incorrect to obtain this by considering a first order Taylor expansion. Let $g_\epsilon$ be a one-parameter family of metrics, $đ\mu_\epsilon$ the corresponding 1-parameter family of volume forms, and $e_a^\epsilon$ the corresponding 1-parameter family of orthonormal frames.

We have $g_\epsilon(e^\epsilon_a,e^\epsilon_b)=\eta_{ab}$ for all $\epsilon$, so the $\epsilon$-derivative at $0$ is $$0= \delta g(e_a,e_b)+2g(\delta e_a,e_b). $$ Since the $e_b$ are basis vectors, we have $$ g(\delta e_a,\bullet)=-\frac{1}{2}\delta g(e_a,\bullet), $$ so $$ \delta e_a=-\frac{1}{2}\delta g(e_a,\bullet)^\sharp, $$ where $\sharp$ is "raising an index". In index notation this result is $\delta e^\mu_a=\delta g_{\nu}^{\ \mu}e^\nu_a$. Clearly this expression is linear in $e_a$, so let us define $A(e_a)=\delta g(e_a,\bullet)^\sharp$ to be this linear map.

Now, we know that $$ 1=đ\mu_\epsilon(e_1^\epsilon,...,e^\epsilon_n) $$ for all $\epsilon$s. If we differentiate at $\epsilon=0$, we obtain $$ 0=(\deltađ\mu)(e_1...e_n)+đ\mu(\delta e_1,...,e_n)+...+đ\mu(e_1,...,\delta e_n). $$ I personally find the evaluation of the terms other than the first one (which we wish to express) difficult, so instead of directly differentiating at $\epsilon=0$, we expand in a truncated Taylor series. To first order in $\epsilon$, we have $$ e_a^\epsilon=e_a+\epsilon A(e_a)=(1+\epsilon A)e_a, $$ so to first order in $\epsilon$ we also have $$ 1=đ\mu_\epsilon(e^\epsilon_1,...,e^\epsilon_n)=đ\mu_\epsilon(e_1,...,e_n)\det(1+\epsilon A)=đ\mu_\epsilon(e_1...e_n)(1+\epsilon\text{Tr}A)\\ =(1+\epsilon\alpha)đ\mu(e_1...e_n)(1+\epsilon\text{Tr}A)=(1+\epsilon \alpha)(1+\epsilon\text{Tr}A)=1+\epsilon(\alpha+\text{Tr}A)+O(\epsilon^2), $$ so we obtain $$\alpha=-\text{Tr}A=\frac{1}{2}\text{Tr}_g\delta g=-\frac{1}{2}\text{Tr}_g\delta g^\ast, $$ where $\text{Tr}_g$ is the metric trace (eg. $h_{\mu\nu}\mapsto h_{\mu\nu}g^{\mu\nu}$), and $g^\ast$ is the inverse/dual metric. We have used the well known relation that the variations of the direct and the inverse metrics are negatives of one another.

We have thus obtained in a reasonably invariant manner that $$\delta đ\mu=-\frac{1}{2}\text{Tr}_g\delta g^\ast đ\mu=-\frac{1}{2}g_{\mu\nu}\delta g^{\mu\nu}_\ast đ\mu.$$

For the rest of the derivation, I will use abstract index notation, which is global and coordinate-free.

We know that (see Wald - General Relativity for a derivation) if $\nabla^\prime$ and $\nabla$ are two linear connections, with difference tensor $C:\ \nabla^\prime=\nabla+C$, then the corresponding curvature tensors are related as $$ R^{\prime\rho}_{\ \sigma\mu\nu}=R^\rho_{\ \sigma\mu\nu}+2\nabla_{[\mu} C^\rho_{\nu]\sigma}+2C_{[\mu|\lambda|}^{\rho}C^\lambda_{\nu]\sigma}. $$

In particular, let $\nabla^\epsilon$ be the 1-parameter family of Levi-Civita connections induced by the 1-parameter family of metrics, and $C^\epsilon$ be the 1-parameter family of difference tensors between $\nabla^\epsilon$ and the unperturbed LC-connection. Then $$ \delta\nabla=\frac{d}{d\epsilon}\nabla^\epsilon\ |_{\epsilon=0}=\frac{d}{d\epsilon}\left(\nabla+C^\epsilon\right)_{\epsilon=0}=\frac{d}{d\epsilon}C^\epsilon |_{\epsilon=0}\equiv\delta\Gamma, $$ and the $\epsilon$-family of curvature tensors is $$ (R^{\epsilon}){}^\rho_{\ \sigma\mu\nu}=R^\rho_{\ \sigma\mu\nu}+2\nabla_{[\mu}(C^\epsilon)^\rho_{\nu]\sigma}+2(C^\epsilon)^\rho_{[\mu|\lambda|}(C^\epsilon)^\lambda_{\nu]\sigma}, $$ so $$ \delta R^\rho_{\ \sigma\mu\nu}=\nabla_\mu\delta\Gamma^\rho_{\nu\sigma}-\nabla_\nu\delta\Gamma^\rho_{\mu\sigma}. $$

We have thus resolved the calculation of the variations of the connection, curvature tensor and volume form without using local coordinate or frame expansions. You can fill in the rest of the details yourself.

Further reading:

"Besse: Einstein Manifolds".

This is especially recommended if you really really hate indices, as Besse doesn't use abstract index notation, he uses standard mathematician notation. He does not derive the variation of the volume form however, just leaves it to the reader as an excercise. Relevant sections are Chapter 1 - K: First variations of Curvature Tensor Fields and Chapter 4 - C: Total Scalar Curvature: First Order Properties.

Orthonormal frame approach:

This has been covered by Arnold Neumaier's answer, and this is the approach that it closest to your electrodynamics example in spirit. I will only note one thing, that if we wish for the initial value problem in GR to be well-defined, we want spacetime to be globally hyperbolic, which implies that topologically $M=\mathbb R\times\Sigma$. Since $\mathbb R$ is parallelizable, the parallelizability of $M$ hinges upon whether the 3-manifold $\Sigma$ is parallelizable or not.

On physical grounds, we wish space to be orientable, so $\Sigma$ should be orientable. However it is known that every orientable 3-manifold is parallelizable, so if these physically reasonable requirements are met, then 4-dimensional spacetimes are parallelizable.

Therefore, these orthonormal-frame based approaches to GR are actually global!

Further reading:

"Thirring: Course on Mathematical Physics Vol 2 "

"Straumann: General Relativity"

Books by the Loop Quantum Gravity crowd, like Thiemann, Rovelli, Gambini etc. also tend to treat the action principle for GR via orthonormal frames.

Principal bundle approach:

You can regard every tensor field (on $M$) as a certain array of functions on the frame bundle of $M$ that satisfy certain equivariance properties.

For example, in the usual, frame-based approaches, a vector field is something like $V^a(x)$. But these components do not only depend on the manifold point $x$ but also on a frame chosen at $x$. So these $V^a$ are actually functions on the frame bundle: $$ V^a(x,e), $$ where points of the frame bundle $F(M)\rightarrow M$ are denoted as pairs $(x,e)$, where $e$ is a frame at $x$.

These functions define a vector field if and only if for any $\Lambda\in\text{GL}(n,\mathbb R)$, we have the $$ V^a(x,e\Lambda)=(\Lambda^{-1})^a_{\ b}V^b(x,e) $$ equivariance property. Similarly for other tensor fields.

It is important to note that despite the indices and components, these functions are global and completely invariant.

One can develop a tensor calculus on the frame bundle that essentially mirrors the usual index-based local tensor calculus on the base manifold, but is global and totally frame and coordinate independent.

Further reading:

"David Bleecker: Gauge Theory and Variational Principles"

This book treats gauge theories, gravity and gauge theories + gravity in a completely invariant and mathematically precise manner using principal fiber bundles, action principles included.

  • $\begingroup$ very nice exposition! - I noticed a misprint: paralellizable $\endgroup$ – Arnold Neumaier Nov 7 '18 at 8:22
  • $\begingroup$ An absolutely fantastic answer. Thank you very much! $\endgroup$ – Bob Knighton Nov 7 '18 at 11:11

Does there exist a similar completely covariant and coordinate-independent derivation of the Einstein field equations from the Einstein-Hilbert action?

Yes. A covariant, coordinate-free action-based derivation is given for example in Chapter 4.2 of the textbook

W. Thirring, Course on Mathematical Physics, Vol.2, Classical Field Theory, Springer, New York 1978.

The derivation is elementary: Just replace each field by field+variation, work out the terms of first order in the variation, use integration by parts, and read off the equations of motion. Each step is elementary, covariant and coordinate-free, hence the whole derivation is.

  • 2
    $\begingroup$ While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review $\endgroup$ – user7777777 Nov 2 '18 at 16:42
  • $\begingroup$ @user7777777: The reference is to a well-known textbook, and remains valid even when the link (provided just for convenience) dies. $\endgroup$ – Arnold Neumaier Nov 2 '18 at 16:57
  • $\begingroup$ While this derivation doesn't use a chart, it still requires the existence of an orthogonal frame on the base manifold, which is not guaranteed to exist. Thus, this derivation still only makes sense locally. I was hoping for a derivation that can be applied globally, without referring to a basis for the calculations. $\endgroup$ – Bob Knighton Nov 3 '18 at 19:48
  • 1
    $\begingroup$ @BobKnighton I don't think you can calculate the variation of the canonical volume element without coordinates or frames. Aside from that, there are two considerations: 1) You can derive the variation in a "purely covariant" manner, eg. you never need to use something like $\partial_\mu$ or $\gamma^\rho_{\mu\nu}$. If you then declare that you use abstract index notation, your derivation will be formally coordinate free, aside from the pesky volume element. You can also try to do that without using indices whatsoever, but you will run into notational issues with contractions. $\endgroup$ – Bence Racskó Nov 5 '18 at 11:43
  • 1
    $\begingroup$ @BobKnighton 2) As far as I recall, all physically reasonable spacetimes are paralellizable, so using local orthonormal frames is in fact global. 2+1) You can likely use index-based tensor calculus on the frame bundle, which is formally basis-free and global, despite the use of indices. See Gauge Theory and Variational Principle by David Bleecker for such an approach. $\endgroup$ – Bence Racskó Nov 5 '18 at 11:44

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