# non-zero divergence of stress energy tensor

are there spacetime metrics where $\nabla_{\nu}T^{\mu\nu} \ne 0$ ?

if the divergence of the energy momentum tensor is non-zero, what does that tell us about the spacetime aside from the fact that energy momentum is not conserved?

• @BenCrowell Yes. Still the fact that you have to use the field equations in your answer worries me. Maybe you should add that these are classical statements. (But you are right, I misread the question.) – marmot Sep 19 '18 at 13:41

General relativity does not allow a stress-energy tensor with a nonvanishing divergence. The Einstein field equations set the Einstein curvature tensor proportional to the stress-energy tensor. The Einstein curvature tensor is constructed in such a way that it is automatically divergence-free.

are there spacetime metrics where $\nabla_{\nu}T^{\mu\nu} \ne 0$ ?

No. If you write down a metric, the way you find the corresponding stress-energy tensor is to use the Einstein field equations, but the field equations will always give a divergence-free stress-energy tensor.

• The only comment I would add is the fact that there are space-times that "almost" have a violation of stress-energy that is created by assuming a situation that cannot be in equilibrium (such as two static black holes next to each other). These kinds of assumptions result in the various "strut" and "string" singularities in solutions to Einstein equations. – Void Sep 19 '18 at 13:38
• how about for an accelerating spacetime? I read somewhere that energy momentum is not conserved. – jboy Sep 19 '18 at 22:50
• @jboy: Energy-momentum is always locally conserved in GR, and the zero divergence of the stress-energy tensor is the expression of that fact. There is not always any way to define a globally conserved energy-momentum. – Ben Crowell Sep 20 '18 at 13:21

Stress-Energy Tensor is defined, at least in QFT, as the Noehter current associated with infinitesimal space-time translations. Let me explain, consider an infinitesimal space-time translations such as

$x'^{\mu} = x^{\mu}-\epsilon^{\mu}$

If we require the field $\phi$ to be invariant under the transformation we've just written down, we can then apply Noether's Theorem. Thus, we say there are four conserved currents, one for each space-time direction. Putting this four currents all together in a big $4\times4$ matrix, we get the Stress-Energy Tensor $T^{\mu\nu}$

$T^{\mu\nu} = \frac{\partial\mathcal{L}}{\partial(\partial_\mu \phi)}\partial_{\nu \phi}-\delta^{\mu\nu}\mathcal{L}$

It can be easly shown that, since the $T^{\mu\nu}$ is made up of conserved currents, it has zero divergence:

$\partial_{\nu} T^{\mu\nu} = 0$

The Einstein-Hilbert stress-energy-momentum (SEM) tensor is always covariantly conserved, provided one of two conditions are satisfied:

• The gravitational equations of motion hold.

• The matter field equations of motion hold. This however does assume that the matter field action is diffeomorphism-invariant.

It is important and interesting to note that only one of these conditions need to be satisfied.

Let's start with the second. If $\psi$ is a matter field, or a collection of matter fields with action $$S_m[\psi,g]=\int d^4x\ \mathcal L_m(\psi,\partial\psi,g,\partial g)$$ that is diffeomorphism-invariant, because of diffeomorphism-invariance, the action will be invariant under an infinitesimal diffeomorphism (duh). An infinitesimal diffeo is always generated by a vector field $X$. The variation of the matter action is then $$\delta_X S_m=\int d^4x\sqrt{-g}\ \left( \frac{\delta S_m}{\delta\psi}\delta_X\psi+\frac{\delta S_m}{\delta g^{\mu\nu}}\delta_Xg^{\mu\nu} \right)=0.$$ If the matter field EoMs hold, then the first functional derivative vanishes, and we obtain $$\delta_X S_m=\int d^4x\sqrt{-g} \ \frac{\delta S_m}{\delta g^{\mu\nu}}\delta_Xg^{\mu\nu}=0.$$

On the other hand, the fact that the variation is an infinitesimal diffeo, it means that $$\delta_X g^{\mu\nu}=-\mathcal L_X g^{\mu\nu}$$ where $\mathcal L_X$ is the Lie derivative (the minus sign is conventional - I define transformations via pushforward, and the Lie derivative is change along a pullback). But we have $$-\mathcal L_X g^{\mu\nu}=\nabla^\mu X^\nu + \nabla^\nu X^\mu.$$ If we name $$T_{\mu\nu}=\frac{-2}{\sqrt{-g}}\frac{\delta S_m}{\delta g^{\mu\nu}},$$ then we have $$\delta_X S_m=\int d^4x\sqrt{-g}\left(-\frac{1}{2}\right)T_{\mu\nu}(\nabla^\mu X^\nu + \nabla^\nu X^\mu)=-\int d^4x\sqrt{-g} T_{\mu\nu}\nabla^\mu X^\nu \\ =\int d^4x \sqrt{-g}\nabla_\mu T^\mu_{\ \nu} X^\nu=0,$$ and since this is true for all $X$, we obtain $$\nabla_\mu T^{\mu\nu}=0.$$ Note that in the last two equalities I have used that $T$ is symmetric and then I have partially integrated (with boundary termsthrown away).

Now, you can also use this procedure but with the gravitational action (without even needing to use the EoMs!) to prove that $$E_{\mu\nu}=\frac{1}{\sqrt{-g}}\frac{\delta S_G}{\delta g^{\mu\nu}}$$ is divergenceless (if the gravitational action is the usual Einstein-Hilbert action, then this tensor if the $G_{\mu\nu}$ Einstein tensor which is known to be divergenceless, however this procedure work for a general gravitational action).

Then, the gravitational EoMs are $$E^{\mu\nu}=\frac{1}{2}T^{\mu\nu},$$ so if the gravitational EoMs hold, then because $E$ is divergenceless, so must be $T$.

The conclusion then is that it is impossible to have the EFE (Einstein's field equations) hold without $\nabla_\mu T^{\mu\nu}=0$. Of course, sometimes you may consider fields in a "background" spacetime, without any backreaction to gravity, in which case, it might be possible for the SEM tensor to not be conserved, but it must be noted that then the EFE does not hold .

• Oh okay. So if i "invent" a metric and then use EFE to get an expression for the stress energy tensor and then find that it has nonzero divergence, then that means EFE doesnt hold for that metric? Or that metric is not a solution to EFE? What are the implications if EFE does not hold? – jboy Sep 21 '18 at 10:36
• @jboy If you invent a metric $g_{\mu\nu}$, and you use the EFE to get a SEM tensor, that means you first construct the Einstein-tensor $G^{\mu\nu}[g]=R^{\mu\nu}[g]-\frac{1}{2}R[g]g^{\mu\nu}$ from the metric, and then you define the SEM tensor as $T^{\mu\nu}=\frac{1}{8\pi G}G^{\mu\nu}$. The problem is that $G^{\mu\nu}$ is always divergenceless thanks to the twice-contracted (differential) Bianchi indentity, so $\nabla_\mu G^{\mu\nu}=0$. But then because $T^{\mu\nu}$ is proportional to $G^{\mu\nu}$, you also have $\nabla_\mu T^{\mu\nu}=0$. – Bence Racskó Sep 21 '18 at 13:55