Maxwell's action with differential geometry formalism

Question

I'm having problems in showing that the following identity, regarding Maxwell's action, holds true:

$$ S_{Maxwell}=-\frac{1}{2}\int F\wedge\star F=-\frac{1}{4}\int \sqrt{-g}\,F_{\mu\nu}F^{\mu\nu}\mathrm{d}^4x. $$

The way I'm trying to show it is the following:

$$ -\frac{1}{2}\int F\wedge\star F=-\frac{1}{2}\int F_{\alpha\beta}\frac{1}{2}\sqrt{-g}\,\epsilon_{\xi\eta\mu\nu}F^{\xi\eta}(\mathrm{d}x^{\alpha}\wedge\mathrm{d}x^{\beta})\wedge(\mathrm{d}x^{\mu}\wedge\mathrm{d}x^{\nu}) $$ Where I've used the fact that $$(\star\omega)_{\mu\nu}=\frac{1}{2}\sqrt{-g}\,\epsilon_{\alpha\beta\mu\nu}\omega^{\mu\nu}.$$ That's all I'm able to do, I can't seem to figure out how to solve the identity. I get that, since it is a top form, the integrand will only have one component, but that's as far as I can get. Any help would me much appreciated.

Answer 1

Unfortunately there are too many very small errors (signs, factors of $2$, $4!$ etc...) in the question statement and other answers to just leave comments. I will follow the conventions of ([1] Sections 3.1, 7.5, 7.6).

Given $$F = \frac{1}{2} F_{\mu_1 \mu_2} dx^{\mu_1} \wedge dx^{\mu_2}$$ and assuming $\varepsilon_{0123} = + 1$ so that ([1], 7.6) $$*(dx^{\mu_1} \wedge dx^{\mu_2}) = \frac{1}{2!} \sqrt{-g} g^{\mu_1 \nu_1} g^{\mu_2 \nu_2} \varepsilon_{\nu_1 \nu_2 \mu_3 \mu_4} dx^{\mu_3} \wedge dx^{\mu_4} $$ we have $$*F = \frac{1}{4} \sqrt{-g} \varepsilon_{\mu_3 \mu_4 \nu_3 \nu_4} F^{\nu_3 \nu_4} dx^{\mu_3} \wedge dx^{\mu_4}$$ Now, using $\varepsilon^{0123} = - 1$ we also have $$dx^{\mu_1} \wedge dx^{\mu_2} \wedge dx^{\mu_3} \wedge dx^{\mu_4} = - \varepsilon^{\mu_1 \mu_2 \mu_3 \mu_4} dx^0 \wedge dx^1 \wedge dx^2 \wedge dx^3 = - \varepsilon^{\mu_1 \mu_2 \mu_3 \mu_4} d^4 x.$$ Finally, recalling the identity ([1], 3.1) $$\varepsilon^{\mu_1 \mu_2 \rho_1 \rho_2} \varepsilon_{\nu_1 \nu_2 \rho_1 \rho_2} = - 2! 2! \delta^{\mu_1 \mu_2}_{\nu_1 \nu_2} $$ where $\delta^{\mu_1 \mu_2}_{\nu_1 \nu_2}$ satisfies $F^{\mu \nu} = F^{\rho \sigma} \delta^{\mu \nu}_{\rho \sigma}$, we have \begin{align} - \frac{1}{2} F \wedge *F &= - \frac{1}{2} (\frac{1}{2} F_{\mu_1 \mu_2} dx^{\mu_1} \wedge dx^{\mu_2}) \wedge *(\frac{1}{2} F_{\mu_3 \mu_4} dx^{\mu_3} \wedge dx^{\mu_4}) \\ &= - \frac{1}{2} (\frac{1}{2} F_{\mu_1 \mu_2} dx^{\mu_1} \wedge dx^{\mu_2}) \wedge (\frac{1}{4} \sqrt{-g} \varepsilon_{\mu_3 \mu_4 \nu_3 \nu_4} F^{\nu_3 \nu_4} dx^{\mu_3} \wedge dx^{\mu_4} ) \\ &= - \frac{1}{16} \sqrt{-g} F_{\mu_1 \mu_2} F^{\nu_3 \nu_4} (\varepsilon_{\mu_3 \mu_4 \nu_3 \nu_4} dx^{\mu_1} \wedge dx^{\mu_2} \wedge dx^{\mu_3} \wedge dx^{\mu_4} ) \\ &= - \frac{1}{16} \sqrt{-g} F_{\mu_1 \mu_2} F^{\nu_3 \nu_4} (- \varepsilon_{\nu_3 \nu_4 \mu_3 \mu_4} \varepsilon^{\mu_1 \mu_2 \mu_3 \mu_4} d^4 x) \\ &= - \frac{1}{16} \sqrt{-g} F_{\mu_1 \mu_2} F^{\nu_3 \nu_4} (+ 2! 2! \delta_{\nu_3 \nu_4}^{\mu_1 \mu_2} d^4 x) \\ &= - \frac{1}{4} d^4 x \sqrt{-g} F_{\mu_1 \mu_2} F^{\mu_1 \mu_2} \end{align}

Reference:

1. Freedman and van Proeyen, "Supergravity", 1st Ed.

Answer 2

I think what you need (or what I like to use actually) is turning the exterior product of these Einstein-summed one-forms into another Levi-Civita symbol and non-Einstein summed four-form.

$$\frac{1}{4!}dx^{\alpha}\wedge dx^{\beta} \wedge dx^{\mu} \wedge dx^{\nu} = \epsilon^{\alpha\beta\mu\nu} dx^{0}\wedge dx^{1} \wedge dx^{2} \wedge dx^{3}$$

Putting that into your expression, we shall have:

$$ -\frac{1}{2}\int F\wedge \ast F = -\frac{4!}{4 }\int F_{\alpha\beta}\epsilon_{\xi\eta\mu\nu}F^{\xi\eta}\epsilon^{\alpha\beta\mu\nu}\sqrt{-g} \, dx^{0}\wedge dx^{1} \wedge dx^{2} \wedge dx^{3}$$

In the last expression we recognise the Riemannian volume form induced from the metric, denote it $\eta$ following the Straumann's book, or $\sqrt{-g}\,d^{4}x$ if you wish.

Now we partially contract the Levi-Civita symbols. I will use the results given in https://en.wikipedia.org/wiki/Levi-Civita_symbol.

As it turns out, the contractions follow a general relation. In the article it is given in terms of the Levi-Civita tensor, or equivalently the Riemannian volume form expressed as a totally antisymmetric tensor, but it is very much applicable since the square roots of the metric determinants cancel.

$$ \epsilon^{\mu_{1}\cdots\mu_{p}\alpha_{1}\cdots\alpha_{n-p}}\epsilon_{\mu_{1}\cdots\mu_{p}\beta_{1}\cdots \beta_{n-p}}=(-1)^{q}p!\delta^{\alpha_{1}\cdots\alpha_{n-p}}_{\beta_{1}\cdots\beta_{n-p}}$$

where $q$ is the number of negative signs in the signature of the metric. Here $q=1$. The

$$ \delta^{\alpha_{1}\cdots\alpha_{n-p}}_{\beta_{1}\cdots\beta_{n-p}} = (n-p)!\delta_{\beta_{1}}^{[\alpha_{1}}\cdots\delta^{\alpha_{n-p}]}_{\beta_{n-p}}$$

is called the generalized Kronecker delta.

In case of our contractions, we just have two indices:

$$ \epsilon_{\xi\eta\mu\nu}\epsilon^{\alpha\beta\mu\nu} =-1\cdot2!\delta_{\xi\eta}^{\alpha\beta} = - 2\cdot(2!)\delta^{[\alpha}_{\xi}\delta^{\beta]}_{\eta} = -4 \frac{1}{2}\big{(} \delta^{\alpha}_{\xi}\delta_{\eta}^{\beta} - \delta^{\beta}_{\xi}\delta_{\eta}^{\alpha} \big{)}$$

Contract that further with the Maxwell tensor:

$$ -2\big{(} \delta^{\alpha}_{\xi}\delta_{\eta}^{\beta} - \delta^{\beta}_{\xi}\delta_{\eta}^{\alpha} \big{)} F_{\alpha\beta}F^{\xi\eta} = -2 \big{(} F_{\alpha\beta}F^{\alpha\beta} - F_{\alpha\beta}F^{\beta\alpha}\big{)} = -4 F_{\alpha\beta}F^{\alpha\beta}$$

EDIT

I have just noticed I am a bit late. Leaving the answer anyway, hope you easily proceed after receiving them both.

Answer 3

You are almost there. First of all, as you have stated correctly, we have that

$$\ast F=\frac{1}{2!}F^{\alpha\beta}\sqrt{-g}\epsilon_{\alpha\beta\gamma\delta}\mathrm{d}x^{\gamma}\wedge\mathrm{d}x^{\delta}.$$

As a second ingredient, we need the fact that

$$-\epsilon^{\alpha\beta\gamma\delta}\mathrm{d}^{4}x=\mathrm{d}x^{\alpha}\wedge\mathrm{d}x^{\beta}\wedge\mathrm{d}x^{\gamma}\wedge\mathrm{d}x^{\delta}.$$

This is all you need:

$$F\wedge\ast F=\frac{1}{2!}F_{\mu\nu}F^{\alpha\beta}\sqrt{-g}\epsilon_{\alpha\beta\gamma\delta}\mathrm{d}x^{\mu}\wedge\mathrm{d}x^{\nu}\wedge\mathrm{d}x^{\gamma}\wedge\mathrm{d}x^{\delta}=-\frac{1}{2!}F_{\mu\nu}F^{\alpha\beta}\sqrt{-g}\epsilon_{\alpha\beta\gamma\delta}\epsilon^{\mu\nu\gamma\delta}\mathrm{d}^{4}x.$$

Now we are done, using the relation

$$\epsilon_{\alpha\beta\gamma\delta}\epsilon^{\mu\nu\gamma\delta}=\epsilon_{\alpha\beta\gamma\delta}\epsilon^{\gamma\delta\mu\nu}=-2!(\delta_{\alpha}^{\mu}\delta_{\beta}^{\nu}-\delta_{\alpha}^{\nu}\delta_{\beta}^{\mu})$$

and hence

$$F\wedge\ast F=2F_{\mu\nu}F^{\mu\nu}\sqrt{-g}\mathrm{d}^{4}x.$$

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EDIT: From the comments: Mathematically, the above is true. However, in physics, the functions $F_{\mu\nu}$ are by definition the coordinate functions of the "Faraday-tensor", which is a covariant rank $2$-tensor field, i.e. $$F=F_{\mu\nu}\,\mathrm{d}x^{\mu}\otimes\mathrm{d}x^{\nu}.$$ By definition, a differential form of rank $k$ is a covariant alternating rank $k$-tensor field and hence, we can think of $F$ also as a $2$-form and this is also what we need, since we are working with differential forms on the left-hand side above. Hence, we have have to write

$$F=\color{red}{\frac{1}{2}}F_{\mu\nu}\mathrm{d}x^{\mu}\wedge\mathrm{d}x^{\nu}$$

and accordingly

$$\ast F=\frac{1}{2!\color{red}{2}}F^{\alpha\beta}\sqrt{-g}\epsilon_{\alpha\beta\gamma\delta}\mathrm{d}x^{\gamma}\wedge\mathrm{d}x^{\delta}.$$

In the end, we hence have that

$$F\wedge\ast F=\color{red}{\frac{1}{2}}F_{\mu\nu}F^{\mu\nu}\sqrt{-g}\mathrm{d}^{4}x,$$

which is what we want.