Does (covariant) divergence-freeness of the stress-energy tensor ${T^{\mu\nu}}_{;\nu}=0$ follow from the Bianchi identity?

Question

I'm working through Chap. $30$ of Dirac's "GTR" where he develops the "comprehensive action principle". He makes a very slick and mathematically elegant argument to show that the stress-energy tensor satisfies ${T^{\mu\nu}}_{;\,\nu}=0$.

$\quad\quad$However, since we already have the Einstein field equation: $$R^{μν}-\frac{1}{2} g^{μν}R = -8πT^{μν} \quad\quad(*)$$ doesn't ${T^{\mu\nu}}_{;\nu}=0$ follow immediately from this corollary of the Bianchi identity:

$$\left( R^{μν}-\frac{1}{2} g^{μν}R \right)_{;\,\nu} = 0 \quad ?$$ (See Dirac "GTR" Eq. $(14.3)$.)

$\quad\quad$The field equation $(*)$ follows from the action principle $\delta(I + I')=0$, where $I$ is the (Hilbert) action for the gravitational field, and $I'$ is the action for any other matter-energy field(s). So we already have $(*)$ in our pocket. Why then don't we automatically have ${T^{\mu\nu}}_{;\,\nu}=0$ from the Bianchi relation?

Answer 1

The law of local conservation of $T$ is consequence of the law of motion of the matter part of the total action matter + curvature $$I[g]+ J[\phi, g]$$ These equations of motion for the matter arise out of $J[g,\phi]$ only, since the variation procedure is performed with respect to the matter fields which do not take place in the curvature part of the action.

Thus, the conservation equation is invariant if we replace $I[g]$ for another geometric action. Using another action $I[g]$ prevents from obtaining your identity (*). This latter arises from the above total action when taking the variational derivative in $g$.

In summary, the conservation equation of $T$ is general, whereas your identity () is true for a special choice of $I[g]$. In this sense, it is false that () is the reason of the conservation equation.

On the other hand, your identity (*) must be compatible with the conservation equation whatever is the explicit form of $J[\phi,g]$. This constraint is automatically satisfied as a consequence of the Bianchi identity, which, in turn, is independent of the explicit form of $J[\phi,g]$.

Answer 2

I haven't read Dirac's book specifically, but usually authors prove the Einstein equations are consistent in part by showing that the divergence on both sides is 0. The construction of stress-energy ensures that the right-hand side will be, which is probably the "slick and elegant" part that you referenced without reproducing here, but that's something to be proven directly. As you noted, the Biachi identity gives it on the left-hand side.

(Otherwise, what would be to stop you from choosing some arbitrary definition of stress-energy, putting it on the RHS of the Einstein equations, and then "proving" that it too is divergence-free by the same argument that you've used.)

Answer 3

${G^{\alpha\beta}}_{;\beta} = 0$ is an identity, but ${T^{\alpha\beta}}_{;\beta} = 0$ is valid only `on-shell'.