- #1

etotheipi

- Homework Statement
- Derive the source-free Maxwell equations from the action$$S = - \frac{1}{2} \int F \wedge \star F$$

- Relevant Equations
- Some wedgey things

First, I let ##\omega = \sqrt{-g} dx^0 \wedge \dots \wedge dx^3## be the top-form on ##M##, and making use of the inner-product on the space of forms I can write$$\begin{align*}

F \wedge \star F = g(F,F) \omega &= \frac{1}{4} F_{\mu \nu} F_{\rho \sigma}

\begin{vmatrix}

g(dx^{\mu}, dx^{\rho}) & g(dx^{\mu}, dx^{\sigma})\\

g(dx^{\nu}, dx^{\rho}) & g(dx^{\nu}, dx^{\sigma})

\end{vmatrix} \omega \\

&= \frac{1}{4} F_{\mu \nu} F_{\rho \sigma} (\delta^{\mu \rho} \delta^{\nu \sigma} - \delta^{\nu \rho} \delta^{\mu \sigma}) \omega\\

&= \frac{1}{4}\left( F^{\rho \sigma} F_{\rho \sigma} - F^{\sigma \rho} F_{\rho \sigma} \right) \omega = \frac{1}{2} F^{\sigma \rho} F_{\rho \sigma} \omega

\end{align*}$$due to the anti-symmetry of ##F^{\rho \sigma}##. Replacing ##\omega## gives ##F \wedge \star F = \frac{1}{2} F^{\rho \sigma} F_{\rho \sigma} \sqrt{-g} d^4 x##, where I abbreviated ##d^4 x \equiv dx^0 \wedge \dots \wedge dx^3##. Now we can vary the action$$\delta S = - \frac{1}{4} \int d^4 x \left(\delta(\sqrt{-g}) F^{\rho \sigma} F_{\rho \sigma} + \sqrt{-g} F^{\rho \sigma} \delta(F_{\rho \sigma}) + \sqrt{-g} \delta(F^{\rho \sigma}) F_{\rho \sigma}\right)$$Because of following property of diagonal matrices ##A##,$$\frac{1}{\mathrm{det} A} \delta(\mathrm{det} A) = \mathrm{tr}(A^{-1} \delta A)$$we can write down the variation of the metric determinant$$\delta(\sqrt{-g}) = - \frac{1}{2} \sqrt{-g} g_{\mu \nu} \delta(g^{\mu \nu})$$But I don't know how to re-write the variations of the ##F_{\rho \sigma}## and ##F^{\rho \sigma}##. Wondered if someone could give a few pointers? Thanks

(Also, whilst I'm here, is there a faster/better way to work out ##F \wedge \star F##?)

F \wedge \star F = g(F,F) \omega &= \frac{1}{4} F_{\mu \nu} F_{\rho \sigma}

\begin{vmatrix}

g(dx^{\mu}, dx^{\rho}) & g(dx^{\mu}, dx^{\sigma})\\

g(dx^{\nu}, dx^{\rho}) & g(dx^{\nu}, dx^{\sigma})

\end{vmatrix} \omega \\

&= \frac{1}{4} F_{\mu \nu} F_{\rho \sigma} (\delta^{\mu \rho} \delta^{\nu \sigma} - \delta^{\nu \rho} \delta^{\mu \sigma}) \omega\\

&= \frac{1}{4}\left( F^{\rho \sigma} F_{\rho \sigma} - F^{\sigma \rho} F_{\rho \sigma} \right) \omega = \frac{1}{2} F^{\sigma \rho} F_{\rho \sigma} \omega

\end{align*}$$due to the anti-symmetry of ##F^{\rho \sigma}##. Replacing ##\omega## gives ##F \wedge \star F = \frac{1}{2} F^{\rho \sigma} F_{\rho \sigma} \sqrt{-g} d^4 x##, where I abbreviated ##d^4 x \equiv dx^0 \wedge \dots \wedge dx^3##. Now we can vary the action$$\delta S = - \frac{1}{4} \int d^4 x \left(\delta(\sqrt{-g}) F^{\rho \sigma} F_{\rho \sigma} + \sqrt{-g} F^{\rho \sigma} \delta(F_{\rho \sigma}) + \sqrt{-g} \delta(F^{\rho \sigma}) F_{\rho \sigma}\right)$$Because of following property of diagonal matrices ##A##,$$\frac{1}{\mathrm{det} A} \delta(\mathrm{det} A) = \mathrm{tr}(A^{-1} \delta A)$$we can write down the variation of the metric determinant$$\delta(\sqrt{-g}) = - \frac{1}{2} \sqrt{-g} g_{\mu \nu} \delta(g^{\mu \nu})$$But I don't know how to re-write the variations of the ##F_{\rho \sigma}## and ##F^{\rho \sigma}##. Wondered if someone could give a few pointers? Thanks

(Also, whilst I'm here, is there a faster/better way to work out ##F \wedge \star F##?)