Show charge conservation in a curved spacetime

[TSny]

Anyway, guess I better study more.

Don't push yourself too much. You're doing amazingly well!

 

[vanhees71]

Ok, I think I see what's bothering you. You were hoping that the EL equation for the Lagrangian would be a way to prove the identity $$\nabla_{\mu} \nabla^{\mu} \phi = \frac{1}{\sqrt{|g|}} \partial_{\mu} \left( \sqrt{|g|} g^{\mu \nu} \partial_{\nu} \phi \right)$$.

If you want to do this an alternative way is to couple the field to an external scalar current $J$:
$$\mathcal{L}=\frac{1}{2} \sqrt{|g|} g^{\mu \nu} (\partial_{\mu} \phi)(\partial_{\nu} \phi)-\sqrt{|g|} J \phi.$$
Then you get
$$\Box \phi=J$$
as an equation of motion, and the variational principle tells you
$$\partial_{\mu} \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \phi)}=\frac{\partial \mathcal{L}}{\partial \phi}.$$
Now
$$ \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \phi)}=\sqrt{|g|} g^{\mu \nu} \partial_{\nu} \phi$$
and
$$\frac{\partial \mathcal{L}}{\partial \phi}=\sqrt{|g|} J,$$
from which
$$\partial_{\mu} (\sqrt{|g|} g^{\mu \nu} \partial_{\nu} \phi)=\sqrt{|g|} J \; \Rightarrow \; J=\Box \phi =\frac{1}{\sqrt{|g|}} \partial_{\mu} (\sqrt{|g|} g^{\mu \nu} \partial_{\nu} \phi).$$
QED

 

[JD_PM]

Hello! I was rereading this nice thread and wanted to discuss a bit further (if you are willing to do so as well of course! )

Mmm I am actually trying to show Maxwell's equations in vacuum i.e. $g^{\nu \mu}\nabla_{\mu} F_{\nu \sigma} = 0$ and $\nabla_{[\mu}F_{\nu \sigma]}=0$ using the same Riemann tensor property i.e. $\nabla_{\mu} \nabla_{\nu} A_{\sigma} = R_{\sigma \nu \mu \rho} A^{\rho}$, where $F_{\mu \nu} = \nabla_{\mu} A_{\nu} - \nabla_{\nu} A_{\mu}$.

Let's focus on $g^{\nu \mu}\nabla_{\mu} F_{\nu \sigma} = 0$ first.

We get

\begin{align*}
\nabla_{\mu} F_{\nu \sigma} &= \nabla_{\mu} \nabla_{\nu} A_{\sigma} - \nabla_{\mu} \nabla_{\sigma} A_{\nu} \\
&= R_{\sigma \nu \mu \rho} A^{\rho} - R_{\nu \sigma \mu \rho} A^{\rho} \\
&= 2R_{\sigma \nu \mu \rho} A^{\rho}
\end{align*}

Multiplying both sides by $g^{\nu \mu}$ yields

\begin{equation*}
\nabla^{\nu} F_{\nu \sigma} = 2 R_{\sigma \rho} A^{\rho} = 0
\end{equation*}

Where I used the fact that, in vacuum, $R_{\mu \nu} = 0$.

Does this proof look correct to you?

I am stuck on how to show that $\nabla_{[\mu}F_{\nu \sigma]}=0$ by means of $\nabla_{\mu} \nabla_{\nu} A_{\sigma} = R_{\sigma \nu \mu \rho} A^{\rho}$ though

I start expanding it out

\begin{align*}
\nabla_{[\mu } F_{\nu \rho]} &= \frac{1}{3!} \left( \nabla_{\mu} F_{\nu \rho} + \nabla_{\nu} F_{\rho \mu} + \nabla_{\rho} F_{\mu \nu} -\nabla_{\mu} F_{\rho \nu} -\nabla_{\nu} F_{\mu \rho} -\nabla_{\rho} F_{\nu \mu} \right) \\
&= \frac{1}{3} \left( R_{\rho \nu \mu \sigma} + R_{\mu \rho \nu \sigma} + R_{\nu \mu \rho \sigma} - R_{\nu \rho \mu \sigma} - R_{\rho \mu \nu \sigma} - R_{\mu \nu \rho \sigma}\right) A^{\sigma} \\
&= \frac{2}{3} \left( R_{\rho \nu \mu \sigma}+ R_{\mu \rho \nu \sigma} + R_{\nu \mu \rho \sigma}\right) A^{\sigma}
\end{align*}

But I do not see how to show that the last equation vanishes without having to explicitly write the Riemann tensor as a function of Christoffel symbols (which does not seem to be the best idea ).Regards.

JD.

 

[JD_PM]

\begin{align*}
\nabla_{[\mu } F_{\nu \rho]} &= \frac{1}{3!} \left( \nabla_{\mu} F_{\nu \rho} + \nabla_{\nu} F_{\rho \mu} + \nabla_{\rho} F_{\mu \nu} -\nabla_{\mu} F_{\rho \nu} -\nabla_{\nu} F_{\mu \rho} -\nabla_{\rho} F_{\nu \mu} \right) \\
&= \frac{1}{3} \left( R_{\rho \nu \mu \sigma} + R_{\mu \rho \nu \sigma} + R_{\nu \mu \rho \sigma} - R_{\nu \rho \mu \sigma} - R_{\rho \mu \nu \sigma} - R_{\mu \nu \rho \sigma}\right) A^{\sigma} \\
&= \frac{2}{3} \left( R_{\rho \nu \mu \sigma}+ R_{\mu \rho \nu \sigma} + R_{\nu \mu \rho \sigma}\right) A^{\sigma}
\end{align*}

But I do not see how to show that the last equation vanishes without having to explicitly write the Riemann tensor as a function of Christoffel symbols (which does not seem to be the best idea ).

Alright, I got it.

\begin{equation*}
R_{[\rho \nu \mu]\sigma} = 0
\end{equation*}

 

[etotheipi]

Just for fun:$$
\begin{align*}

\mathrm{d} \star F &= \star j \\

0 = \mathrm{d}\mathrm{d}\star F &= \mathrm{d} \star j \implies \int_{\Omega} \mathrm{d} \star j = \oint_{\partial \Omega} \star j = 0

\end{align*}$$

 

[vanhees71]

If you like the abstract calculus with forms, there's a textbook about electromagnetism, which may be interesting for you:

F. W. Hehl, Y. N. Obukhov, Foundations of classical electrodynamics, Springer (2003)

 

[etotheipi]

Thanks, I'll check that one out! Another nice book I found that teaches similar mathematics is 'Gauge Fields, Knots and Gravity' by John Baez, which is very cool