I Variation of matter action under diffeomorphism (Carroll)

[chartery]

Queries on Carroll's derivation of matter action $S_M$ under a diffeomorphism:

(Book B.23+4 Notes 5.35+6)

$\frac{\delta S_{M}}{\delta g_{\mu\nu}} \delta g_{\mu\nu} = \frac{\delta S_{M}} {\delta g_{\mu\nu}} \left( 2 \nabla _{(\mu}V_{\nu)} \right) =\left( 2 \right) \frac{\delta S_{M}}{\delta g_{\mu\nu}}\nabla _{\mu}V_{\nu}$

He explains dropping the symmetrisation by symmetry of the fraction, but would the double contraction not do so irrespective of metric symmetry?Also (Book B.25 Notes 5.37)

$0 = \int d^{n}x \frac{\delta S_{M}}{\delta g_{\mu\nu}}\nabla _{\mu}V_{\nu} = -\int d^{n}x \sqrt{-g} V_{\nu}\nabla _{\mu}\left( \frac{1}{\sqrt{-g}}\frac{\delta S_{M}}{\delta g_{\mu\nu}} \right)$

Could someone explain the steps to get right hand side from left (of second equality)?

Please, thanks.

 

[ergospherical]

For the first, $A^{\mu \nu} B_{(\mu \nu )} = \frac{1}{2} A^{\mu \nu} (B_{\mu \nu} + B_{\nu \mu}) = \frac{1}{2}(A^{\mu \nu} + A^{\nu \mu}) B_{\mu \nu} = A^{(\mu \nu)} B_{\mu \nu}$
And if $A^{\mu \nu}$ is symmetric then $A^{(\mu \nu)} = A^{\mu \nu}$.

Your second equality is integration by parts (note $\nabla \sqrt{-g} = 0$).

 

[chartery]

For the first, $A^{\mu \nu} B_{(\mu \nu )} = \frac{1}{2} A^{\mu \nu} (B_{\mu \nu} + B_{\nu \mu}) = \frac{1}{2}(A^{\mu \nu} + A^{\nu \mu}) B_{\mu \nu} = A^{(\mu \nu)} B_{\mu \nu}$
And if $A^{\mu \nu}$ is symmetric then $A^{(\mu \nu)} = A^{\mu \nu}$.

@ergospherical, thanks very much.

I wasn't doubting that logic, just wondering whether my application of index rules was shaky. It seemed to me that the combination of symmetry $\nabla _{(\mu}V_{\nu)}$ with the dual contraction meant there was no need to rely on symmetry of the $\frac{\delta S_{M}}{\delta g_{\mu\nu}}$ term?
Or even, if $A^{(\mu \nu)} = A^{\mu \nu}$, why not $\nabla _{(\mu}V_{\nu)} = \nabla _{\mu}V_{\nu}$ ?

EDIT: Oops, the last sentence highlighted my confusion (between notation and actual symmetry), so please ignore above queries and instead:

Is $\frac{\delta S_{M}}{\delta g_{\mu\nu}}$ effectively contravariant in the two indices, to conform to the summation convention? Is it as straightforward as $\left( \delta g_{\mu\nu} \right)^{-1} = \delta g^{\mu\nu}$ ?

Your second equality is integration by parts (note $\nabla \sqrt{-g} = 0$).

For the absent antiderivative:

Is $\frac{\delta S_{M}}{\delta g_{\mu\nu}}$ a fraction with $\delta S_{M} =0$,
or is it a single symbol, zeroed with boundary conditions,
or some other reason ?

 

[ergospherical]

Is $\frac{\delta S_{M}}{\delta g_{\mu\nu}}$ effectively contravariant in the two indices, to conform to the summation convention? Is it as straightforward as $\left( \delta g_{\mu\nu} \right)^{-1} = \delta g^{\mu\nu}$ ?

Downstairs indices in the denominator count as upstairs indices, yes.

For the second sentence you have to be careful. Remember $\delta^{\mu}_{\nu} = g^{\mu \rho} g_{\rho \nu}$, and taking the variation $0 = g^{\mu \rho} \delta g_{\rho \nu} + \delta g^{\mu \rho} g_{\rho \nu}$ and hence $\delta g^{\mu \nu} = -g^{\mu \rho} g^{\nu \sigma} \delta g_{\rho \sigma}$.

For the absent antiderivative:

Is $\frac{\delta S_{M}}{\delta g_{\mu\nu}}$ a fraction with $\delta S_{M} =0$,
or is it a single symbol, zeroed with boundary conditions,
or some other reason ?

You can turn it into
\begin{align*}
\int d^4 x \sqrt{-g} \nabla_{\mu} \left(\frac{1}{\sqrt{-g}} \frac{\delta S_M}{\delta g_{\mu \nu}} V_{\nu} \right)
\end{align*}
And for any vector $X^{\mu}$ which vanishes sufficiently fast towards infinity then
\begin{align*}
\int d^4 x \sqrt{-g} \nabla_{\mu} X^{\mu} = \int d^4 x \partial_{\mu} (\sqrt{-g} X^{\mu}) = \int_{\partial} dS_{\mu} X^{\mu} \sqrt{-g} = 0
\end{align*}
by Stokes, where we used the divergence formula ${X^{\mu}}_{;\mu} = (-g)^{-1/2} ((-g)^{1/2} X^{\mu})_{,\mu}$ in the first equality.

 

[chartery]

@ergospherical, again many thanks.

Self-taught, I realise I don't have enough knowledge of variational (and tensor) manipulations, but there does seem to be a lot of casually presupposed ability packed suddenly into that one bald equality in his notes !