Writing the Euler-Lagrange equation for variation of an action with respect to metric using only the Lagrangian

Question

Given Lagrangian which dependent on collection of fields ${\phi^a},a=1,...,N$ and on a tensor metric $g^{\mu\nu}$ such that the action in $d$ dimension which describes the system is $$S=\int d^dx \sqrt{-g}\mathscr{L}[\phi^a,g^{\mu\nu}].\tag{1}$$

I know That variation of such general action with respect to the metric gives rise to the following Euler-Lagrange equation:

$$ \frac{\delta S}{\delta g^{\alpha\beta}}=0 \iff \frac{\partial\mathscr{L}}{\partial g^{\alpha\beta}}-\frac{1}{2}g_{\alpha\beta}\mathscr{L}=T_{\alpha\beta}\tag{2}$$

Where $T_{\alpha\beta}$ is the stress-energy tensor.

On the other hand, there is a general expression for the stress-energy tensor given a Lagrangian: $$T_{\alpha\beta}=\sum_a{\frac{\partial\mathscr{L}}{\partial{(\partial_{\alpha}\phi^a)}}\partial_{\beta}\phi^a}-g_{\alpha\beta}\mathscr{L}\tag{3}$$

combining both equations above, we get the Euler-Lagrange equation expressed only through the Lagrangian and its derivatives:

$$\frac{\partial\mathscr{L}}{\partial g^{\alpha\beta}}-\sum_a{\frac{\partial\mathscr{L}}{\partial{(\partial_{\alpha}\phi^a)}}\partial_{\beta}\phi^a}+\frac{1}{2}g_{\alpha\beta}\mathscr{L}=0.\tag{4}$$

Is this a correct formula? because I didn't see this formula in any place.

Answer 1

Formula (3) defines the canonical energy-momentum tensor of the field, which is defined only in flat Minkowski spacetime and is the Noether current associated to translation symmetry.

By contrast, formula (2) defines the Einstein-Hilbert energy momentum tensor, which is also defined in a general spacetime background, but is not a Noether current, at least not in the sense the canonical tensor is.

It is well-known that the canonical EM tensor often requires "improvement", whereas it is generally accepted that the Einstein-Hilbert EM tensor is the "correct" EM tensor of a given field.

Formula (4) would be valid if the EHEM tensor (Einstein-Hilbert energy momentum tensor) and the CEM tensor (canonical energy-momentum tensor) turned out to be always equivalent to one another. This is not true in general however.

This paper by Forger and Römer examines the relationship between these two tensors in great detail.

I will give a simpler analysis here. Suppose that the Lagrangian is $\mathcal L(\phi,\partial\phi,g)$ (the factor of $\sqrt{|\mathfrak g|}$ is considered part of the Lagrangian) and the field theory has some concept of diffeomorphism symmetry and the Lagrangian is diffeomorphism invariant. The Lie derivative of the $\phi=(\phi^i)$ field is written generally as $$ \delta_X\phi^i=X^\mu\phi^i_{,\mu}+Q^{i,\mu}{}_\nu(\phi,\partial\phi)X^\nu_{,\mu} $$for any vector field $X^\mu$. (For fields which are not in the representation of the diffeomorphism group, see the paper by Forger and Römer).

Under a diffeomorphism, the Lagrangian is a scalar density, thus $$ \delta_X\mathcal L=\partial_\mu(X^\mu\mathcal L)=\frac{1}{2}\mathfrak{T}^{\mu\nu}\delta_X g_{\mu\nu}+E_i\delta_X\phi^i+\partial_\mu(P^\mu_i\delta_X\phi^i), $$ where $$ \mathfrak T^{\mu\nu}:=2\frac{\delta \mathcal L}{\delta g_{\mu\nu}},\quad E_i =\frac{\delta\mathcal L}{\delta\phi^i},\quad P^\mu_i=\frac{\partial\mathcal L}{\partial\phi^i_{,\mu}},$$and finally, $\delta_Xg_{\mu\nu}=2\nabla_{[\mu}X_{\nu]}$ is the metric Lie derivative.

If we expand the Lie derivatives and integrate by parts on the vector field $X$ and rearrange the terms to $0$, we get $$ 0=[E_i\phi^i_{,\nu}-\partial_\mu(E_i Q^{i,\mu}{}_\nu)-\nabla_\mu\mathfrak T^{\mu}{}_\nu]X^\nu+\partial_\mu[\mathfrak T^\mu{}_\nu X^\nu+P^\mu_i\delta_X\phi^i+E_iQ^{i,\mu}{}_\nu X^\nu]. $$

If we integrate this and take $X^\mu$ to be compactly supported, we get that the first bracket $[...]$ must vanish identically. This gives $$ \nabla_\mu\mathfrak T^\mu{}_\nu=E_i\phi^i_{,\nu}-\partial_\mu(E_iQ^{i,\mu}{}_\nu) $$as an off-shell relation. If the field $\phi$ is on-shell, then this reduces to $\nabla_\mu\mathfrak T^\mu{}_\nu=0$. Let the vector density whose total divergence appears in the second bracket $\partial_\mu[...]$ be denoted $S^\mu_X$.

Then we also get that $\partial_\mu S^\mu_X=0$ for all possible choices of $X$ and $\phi$ and $g$. If we expand the Lie derivative in $S^\mu_X$ as well, we get $$S^\mu_X=(\mathfrak T^\mu{}_\nu+P^\mu_i\phi^i_{,\nu}-\mathcal L\delta^\mu_\nu+E_iQ^{i,\mu}{}_\nu)X^\nu+P^\mu_iQ^{i,\kappa}{}_\nu X^\nu_{,\kappa}. $$ It is then fairly easy to show that $\partial_\mu S^\mu_X=0$ for all choices of $X^\mu$ implies that $$ S^\mu_X=\partial_\rho K^{\mu\rho}_X,\quad K^{\mu\rho}_X=P^{[\mu}_i Q^{i,\rho]}{}_\nu X^\nu, $$ so $K^{\mu\rho}_X$ is skew-symmetric in $\mu\rho$, moreover $K^{\mu\rho}_X$ is unique (the higher order analogue of this proof is much more difficult and the uniqueness is lost!).

Now evaluate this entire relation ($S^\mu_X=\partial_\rho K^{\mu\rho}_X$) in flat spacetime with $X^\mu=\tau^\mu$ a constant translation vector! Moreover, evaluate it on-shell for the $\phi$ field, i.e. $E_i=0$. Then we get $$ S^\mu=(T^\mu{}_\nu+P^\mu_i\phi^i_{,\nu}-\mathcal L\delta^\mu_\nu)\tau^\nu=\partial_\rho(P^{[\mu}_i Q^{i,\rho]}{}_\nu)\tau^\nu, $$ and using that $\tau$ is any constant vector field, we get $$ T^\mu{}_\nu=\mathcal L\delta^\mu_\nu-P^\mu_i\phi^i_{,\nu}+\partial_\rho(P^{[\mu}_i Q^{i,\rho]}{}_\nu). $$

Here $T^\mu{}_\nu$ is the (flat spacetime limit of the) Einstein-Hilbert energy momentum tensor, the $\Theta^\mu{}_\nu:=\mathcal L\delta^\mu_\nu-P^\mu_i\phi^i_{,\nu}$ is the canonical energy momentum tensor, and the divergence terms are correction terms.

This is the correct relation at least under the assumptions made on the initial form of the Lagrangian.