In General Relativity Einstein's equation implies that stress-energy tensor on its RHS is conserved (has vanishing divergence), due to the Bianchi identity. Considering variational principles leading to Einstein's equation leads to conclusion that this stress tensor is equal to the variational derivative of full action with respect to the metric tensor. However, on several occasions I heard people stating that quite generally one can define stress tensor for a field theory in this way and it is automatically conserved. In flat spacetime and without any coupling to gravity! I wonder if this is true. I don't see a reason why it should be.


2 Answers 2


Actually, the metric variational definition for the stress-energy tensor (due to Hilbert, as remarked by Qmechanic) is an universal improvement procedure for the canonical stress-energy tensor (and hence not always concides with the latter), in a sense which will be made precise below. Such a procedure is necessary because the canonical stress-energy tensor, although always conserved, often fails to satisfy other physical requirements like gauge invariance (since it is an observable quantity), symmetry (needed if we want it to be a source for the gravitational field) and tracelessness (for locally scale invariant theories). For example, all three requirements fail for pure electrodynamics in four space-time dimensions.

Even if you are dealing with a field theory in Minkowski space-time, it is inevitably coupled to gravity simply because of the fact that the Lagrangian depends on the space-time metric (here taking the particular value of the Minkowski metric). The particular dynamics of the metric is irrelevant - all we need is that there are no other "external" fields besides the metric and that the field action functional is diffeomorphism invariant.

Let $L=L(\phi,g)$ be a local field Lagrangian in the space-time $(M,g)$, and $$S_K[\phi,g]=\int_K L(\phi,g)\sqrt{|\det g|}\mathrm{d}x\ ,\quad K\subset M\text{ any bounded region}$$ the corresponding (family of) action functional(s indexed by $K$ as above). We allow $L$ to have finite but otherwise arbitrary order dependence on $\phi$ and $g$, and no explicit space-time dependence since we want it not to depend on any other fields. The infinitesimal variation of $S_K$ with respect to a vector field $X$ on $M$ (i.e. an infinitesimal diffeomorphism) is then given by $$\delta_X S_K[\phi,g]=\int_K\left(\frac{\delta L(\phi,g)}{\delta g_{\mu\nu}}\delta_X g_{\mu\nu}+\frac{\delta L(\phi,g)}{\delta \phi^j}\delta_X \phi^j+\nabla_\mu(T^{\mu\nu}X_\nu)\right)\sqrt{|\det g|}\mathrm{d}x\ ,\quad X_\rho=g_{\rho\sigma}X^\sigma\ ,$$ where $\frac{\delta L(\phi,g)}{\delta g_{\mu\nu}}$ and $\frac{\delta L(\phi,g)}{\delta \phi^j}$ are respectively the Euler-Lagrange (i.e. variational) derivatives of $L(\phi,g)$ with respect to $g$ and $\phi$, $\nabla$ is the Levi-Civita covariant derivative associated to $g$, $T^{\mu\nu}$ is the (canonical or improved) stress-energy tensor, $$\delta_X g_{\mu\nu}=\nabla_\mu X_\nu+\nabla_\nu X_\mu$$ is the Lie derivative of $g$ along $X$ and the infinitesimal field variation $\delta_X\phi^j$ depends on the particular way we lift $X$ to a projectable vector field on the total space of the fiber bundle over $M$ where the fields $\phi^j$ live (for instance, if they are all scalar fields, we simply have $\delta_X\phi^j=-X\phi^j=-X^\mu\nabla_\mu\phi^j$).

There is an implicit but crucial requirement on the admissible improvements for $T^{\mu\nu}$ - namely, the improved Noether current $j^\mu(L,X)=T^{\mu\nu}X_\nu$ associated with the would-be symmetry $X$ of the action functional should not only be linear in $X$ but depend only on the point values of $X$ (we call this property ultralocality) - therefore, we wrote it already as a tensor contraction. This requirement also affects to a certain extent the definition of $\delta_X\phi^j$, but the details of this are not important in what follows. Why do we insist on this requirement? As we shall see below, ultralocality singles out a unique improvement prescription for $T^{\mu\nu}$ which in addition satisfies all physical desiderata. This idea applies more generally to any local symmetry - for instance, it may be used to improve the canonical Noether current associated with local gauge symmetries.

Diffeomorphism invariance of the action functional means we require that $\delta_X S_K[\phi,g]=0$ for all $X,\phi,g,K$. If, in addition, the fields $\phi^j$ satisfy the Euler-Lagrange equations of motion, we have that $$2\frac{\delta L(\phi,g)}{\delta g_{\mu\nu}}\nabla_\mu X_\nu+\nabla_\mu(T^{\mu\nu}X_\nu)=\left(2\frac{\delta L(\phi,g)}{\delta g_{\mu\nu}}+T^{\mu\nu}\right)\nabla_\mu X_\nu+X_\nu\nabla_\mu T^{\mu\nu}=0\ .$$ The first identity seems trivial but in fact follows from ultralocality of the improved Noether current, as explained above. Since $X$ is arbitrary and therefore we may specify $X_\nu$ and $\nabla_\mu X_\nu$ independently at each point of $M$, we obtain in a single stroke:

  1. The desired variational formula for the improved stress-energy tensor $$T^{\mu\nu}=-2\frac{\delta L(\phi,g)}{\delta g_{\mu\nu}}$$ and therefore the symmetry $T^{\mu\nu}=T^{\nu\mu}$;

  2. The covariant conservation law $\nabla_\mu T^{\mu\nu}=0$;

  3. If the metric happens to obey a dynamics determined by a Lagrangian $L_G(g)$, then $T^{\mu\nu}$ automatically becomes the source to the metric equations of motion. This also guarantees compliance with the second Noether theorem, as it should - the canonical Noether current associated to the total (i.e. metric + field) Lagrangian and to $X$ still vanishes on shell if the total action functional is also diffeomorphism invariant.

Although it is not trivial to show, $T^{\mu\nu}$ also happens to be traceless if the field theory exhibits local scale invariance.

If the fields $\phi^j$ are all scalar and $L(\phi,g)$ is a Lagrangian of first order in $\phi$ with a Klein-Gordon-like kinetic part and not depending on derivatives of $g$, then $T^{\mu\nu}$ coincides with the canonical stress-energy tensor. This is no longer the case for spinor fields, whose Lagrangian usually also depends on the first derivatives of the metric through the spin connection, for scalar fields with non-minimal curvature coupling, or for the electromagnetic field.

The above understanding of the metric variational definition of the stress-energy tensor in full generality came surprisingly late - it was thoroughly developed by M. Forger and H. Römer ("Currents and the Energy-Momentum Tensor in Classical Field Theory: a Fresh Look at an Old Problem". Ann.Phys. 309 (2004) 306-389, arXiv:hep-th/0307199), whose work we warmly recommend for (many) more details and examples.

  • $\begingroup$ At this point I have some trouble with working out formulas you cited, perhaps due to my limited experience with variational calculus. However, this seems to be precisely what I was looking for. I will look in the article and accept your answer when everything becomes clear. $\endgroup$
    – Blazej
    Commented May 19, 2016 at 17:01
  • 1
    $\begingroup$ I don't understand your first formula for variation of the action. Two first terms are clear, and we need also a third term coming from variation of metric determinant. However, I'm getting the following: $\int \frac{1}{2}L g^{\mu \nu} \delta _{X}g_{\mu \nu} \sqrt{|g|}dx$ (this formula for variation of the determinant can be found for example in Wald). $\endgroup$
    – Blazej
    Commented May 19, 2016 at 17:45
  • $\begingroup$ The term coming from the variation of the metric determinant is built into the definition of $T^{\mu\nu}$, which I didn't write a priori - recall that the canonical stress-energy tensor has a term of the form $Lg^{\mu\nu}$, which comes precisely from that variation (more precisely, a term of the form $L\delta^\mu_\rho$ since the canonical stress-energy tensor is actually of type (1,1) - we use a contraction with $g^{\rho\nu}$ to raise the covariant index). $\endgroup$ Commented May 19, 2016 at 17:49
  • $\begingroup$ Sure, but where does the other term in definition of canonical stress tensor come from? That is, assuming $T_{\mu \nu}$ you used is canonical one, which is not completely clear since you wrote it might also be improved tensor. Why should this formula be ambiguous? I worry there might be something fundamental which I don't understand here. $\endgroup$
    – Blazej
    Commented May 19, 2016 at 17:54
  • $\begingroup$ The other term of the canonical stress-energy tensor comes from applying the Leibniz rule to the terms proportional to derivatives of $\delta_X\phi$ and $\delta_X g$, which happen to coincide with the variations of the corresponding derivatives of $\phi$ and $g$ - this is a fundamental fact which is used all the time in calculus of variations. You get a total divergence plus the remaining terms which comprise the Euler-Lagrange derivative of $L$ with respect to $\phi$ and $g$. $\endgroup$ Commented May 19, 2016 at 18:02

Well, you cannot take any ol' matter theory in flat Minkowski space and stick in a curved metric tensor $g_{\mu\nu}$ in the matter action as you like, if that's what you're implying. The caveat is that the resulting matter action should be a general relativistic diffeomorphism-invariant functional of the form $$S_{\rm m}[\Phi, g]~=~ \int \! d^4x ~\sqrt{|g|}~ L(\Phi,\nabla\Phi, g ).$$ Then the Hilbert stress-energy-momentum (SEM) tensor is conserved, cf. e.g. my Phys.SE answers here and here.


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