Divergence of the stress-energy tensor

I am currently going through The large scale of space-time by Hawking and Ellis and I am having trouble understanding their proof that the divergence of the stress-energy tensor is 0. The part of the book in question in chapter 4 ("General relativity"), section 3.3 ("Lagrangian formulation").

To put things in context, they work in Lagrangian formalism and define the stress-energy tensor as the derivative (up to a factor 2) of the action with respect to the metric -which is quite usual I think.

Now, they claim:

$$\frac{1}{4!}\int_{\mathcal{D}}\!\mathcal{L}_X(L\eta) = \sum_i\int_\mathcal{D}\!\left(\frac{\partial L}{\partial{\psi_{(i)}}^{a\ldots b}_{c\ldots d}} - \left(\frac{\partial L}{\partial{\psi_{(i)}}^{a\ldots b}_{c\ldots d\ ;\ e}} \right)_{;\ e}\right)\mathcal{L}_{X}{\psi_{(i)}}^{a\ldots b}_{c\ldots d}\,\mathrm{d}v + \frac{1}{2}\int_{\mathcal{D}}\!T^{ab}\mathcal{L}_{X}g_{ab}\,\mathrm{d}v$$

$L$ is the lagrangian density, $\mathcal{L}$ denotes the Lie derivative and $\eta$ is the $4$-form $4!\sqrt{|\det(g_{\alpha\beta})|} \mathrm{d}x^1 \wedge \mathrm{d}x^2 \wedge \mathrm{d}x^3 \wedge \mathrm{d}x^4$.

Obviously then, you can cancel the first series of integrals because of the Euler-Lagrange equations and by rewriting the last integral, you deduce $T^{ab}_{\ ;\ a} = 0$. The problem is I don't see how they obtain the equality written above.

1 Answer

Notice that Leibniz's rule $\mathcal{L}_X(L\eta)=(\mathcal{L}_XL)\eta +L(\mathcal{L}_X\eta)$ holds for the Lie derivative. The Lie derivative of the lagrangian is $$\mathcal{L}_XL=\frac{\partial L}{\partial \psi} \mathcal{L}_X\psi+\frac{\partial L}{\partial (\partial_e\psi)}\mathcal{L}_X(\partial_e\psi)+ \frac{\partial L}{\partial g_{ab}}\mathcal{L}_Xg_{ab},$$ omitting the indices of the fields and the summation over them, for simplicity of notation. We will also need the Lie derivative of the volume form $\mathcal{L}_X\eta=(\partial\eta/\partial g_{ab})\mathcal{L}_Xg_{ab}$.

Using $\mathcal{L}_X(\partial_e\psi)=\partial_e(\mathcal{L}_X\psi)$ and integration by parts in the second term we get: $$\int_\mathcal{D}\mathcal{L}_X(L\eta)= \int_\mathcal{D}\eta\left[\frac{\partial L}{\partial \psi} \mathcal{L}_X\psi-\partial_e\left( \frac{\partial L}{\partial(\partial_e\psi)}\right) \mathcal{L}_X(\psi)+ \frac{\partial L}{\partial g_{ab}}\mathcal{L}_Xg_{ab}\right]+ \int_\mathcal{D}L\frac{\partial \eta}{\partial g_{ab}}\mathcal{L}_Xg_{ab}.$$

Finally, the integrands in the third and fourth terms can be written together as $$\left( \frac{\partial L}{\partial g_{ab}}\eta + L\frac{\partial \eta}{\partial g_{ab}} \right)\mathcal{L}_Xg_{ab}= \frac{\partial}{\partial g_{ab}}(L\eta)\mathcal{L}_Xg_{ab}= \frac{\delta S}{\delta g_{ab}}\mathcal{L}_Xg_{ab}= \frac{1}{2}T^{ab}\mathcal{L}_Xg_{ab},$$ as desired.

• Thanks for your answer. But I am still having trouble understanding why we can commute the Lie derivative $\mathcal{L}_X$ with $\partial_e$. Also, I am not sure what kind of derivative you are thinking of when writing $\partial_e\psi$ where $\psi$ is a tensor. A covariant derivative, a Lie derivative, or something else? – John Smith Feb 4 '17 at 11:48