Is the divergence of the energy tensor related to the equations of motion?

Question

Given a Lagrangian $L[g,\phi]$ we can define its energy tensor as $T=\frac {\delta L}{\delta g}$ and ihe equations of motion for the field $\phi$ are $\frac{\delta L}{\delta \phi} =0$. For the wave equation Lagrangian, it is easy to compute $$ \nabla^a T_{ab} = \square \phi \nabla_b \phi. $$ If I remember correctly this holds also for Klein-Gordon. My question is whether a relation like this holds in general. Is $\nabla ^a T_{ab}$ related to $\frac{\delta L}{\delta \phi}$ in some general way?

Answer 1

The conservation of the energy-momentum tensor is indeed related to the equation of motion in the sense that it is only conserved if the equations of motion are satisfied. One way to see this is the following:

We know that the action is invariant under diffeomorphisms (they correspond to gauge transformations in this theory). Now take a one parameter group of such transformations $\phi_t$. One can show that such groups are always induced by a vectorfield $X$ defined on our manifold.

Now under a general infinitesimal transformation, the action transforms as (Assuming that there is only a scalar field $\Psi$ else there are more terms, however, the argument is identical)

$0 = \delta S = \int \frac{\delta S}{\delta \Phi} \delta \Phi + \frac{\delta S}{\delta g_{ab}}\delta g_{ab}$

where $\delta S = 0$ holds if we take the transformation to be a gauge transformation and $S$ is the matter action. Now under such a transformation, we find that for infinitesimal $t$

$\delta T = t\mathcal{L}_{X}(T)$ for general tensors as can be seen by $\delta T = T - (\phi_{-t})_{*}(T)$ and $\mathcal{L}_X (T) = \lim_{t\rightarrow 0} (1/t) ( (\phi_{-t})_{*}(T)) -T)$ such that for small $t$ the equivalence follows to linear order.

Now $\mathcal{L}_{tX}(\Phi) = tX(\Phi) = tX^a \nabla_a \Phi$ and $\mathcal{L}_{X}(g)_{ab} = \nabla_a tX_b + \nabla_b tX_a$ using that $\Phi$ is a scalar field and $g$ the metric.

Plugging this and $\frac{\delta S}{\delta g_{ab}} = (1/2)\sqrt{-g} T^{ab}$ we find

$0 = \delta S = \int \frac{\delta S}{\delta \Phi} tX^a \nabla_a \Phi + \sqrt{-g}T^{ab}\nabla_atX_b = \int \frac{\delta S}{\delta \Phi} tX^a \nabla_a \Phi - \sqrt{-g}(\nabla_aT^{ab})tX_b$

where the last equality follows from partial integration and assumes the variation to be compactly supported on the interior of the manifold.

As such we get

$\delta S = 0 \iff \frac{\delta S}{\delta \Phi} \nabla_a \Phi = \sqrt{-g}(\nabla_aT^{ab})$

This gives you the relation you found. I.e. that the two are indeed linked. Note that we assumed that there is only a scalar field however one can easily extend this argument to other fields such that essentially you get a sum of such terms.