I'm trying to understand the following equation, used in the derivation of the equations of motion. Let $S = \frac{-1}{2} \int F \wedge \ast F$ and $F = dA$. Let $\delta$ denote variation. Then

$$\delta S = - \int d \delta A \wedge \ast F.$$

(In the full derivation we then int. by parts and get $d\ast F = 0$ for the EOM with no currents.)

In tensor notation, $F = F_{\mu\nu}$ and (I think) $(\ast F) = (\ast F)_{\gamma \delta} = \sqrt{|g|} \epsilon^{\alpha\beta}_{\,\,\,\,\,\, \gamma \delta} F_{\alpha \beta}$ (where $|g|$ is the determinant of the metric). The wedge product is antisymmetric, so (I think) $F\wedge \ast F = \epsilon^{\mu\nu \gamma\delta} F_{\mu\nu} (\ast F)_{\gamma \delta}$. After these considerations, I do not see how we end up with the above result; I'd expect two terms from the product rule for variations, for instance, and where did the factor of $\frac12$ go?


Actually, once being in differential form formalism, the best is to stay in it and thereby benefit from it. In this sense let's evaluate the variation of the EM-action using F=dA:

$$\delta S =-\frac{1}{2}\delta\int_V dA \wedge\star dA = -\frac{1}{2}\int_V(d\delta A\wedge \star dA + dA \wedge \star d \delta A)=-\frac{1}{2}\int_V 2 d\delta A\wedge \star dA$$

Here we used a hodge operator rule for forms $\lambda, \omega \in \Lambda^k$ for $\lambda = d\delta A$ and $\omega=dA$:

$$\lambda \wedge \star \omega = (-1)^q <\lambda,\omega>e = (-1)^q <\omega,\lambda>e =\omega\wedge \star \lambda$$

with $$e: = e_1 \wedge e_2 \wedge \ldots \wedge e_n \in \Lambda^n$$ where $e_i$ form an orthogonal base in $\Lambda$ and the index $q$ is related with the signature of the metric of the used space. The braket pair $<a,b>$ represents the scalar product on the space $\Lambda^k$.

Second step (product rule):

$$ d(\delta A\wedge\star dA) = d\delta A\wedge\star dA -\delta A\wedge d\star dA$$

It follows:

$$ -d\delta A\wedge\star d A = -d(\delta A\wedge\star dA)-\delta A\wedge d\star dA$$

Therefore we get for the varied action integral:

$$\delta S = \int_V d(\delta A\wedge\star dA) + \int_V \delta A\wedge d\star dA $$

We transform the first volume integral into an integral over the volume's surface according to Stokes theorem and realize that the variations $\delta A$ vanish on the surface: $$0=\delta S = \int_{\partial V}\delta A\wedge\star dA + \int_V \delta A\wedge d\star dA = \int_V \delta A\wedge d\star dA$$

As the integral $$\int_V \delta A\wedge d\star dA=0$$ for all variations $\delta A$ we can conclude that:

$$ d\star dA=0 $$ which corresponds to the second part of the free Maxwell equations.

  • $\begingroup$ Thank you. Is the scalar product $< \lambda, \omega >$ symmetric and bilinear for general $\lambda,\omega$? And how is $q$ defined in $(-1)^q$? $\endgroup$ – Dwagg Oct 6 '18 at 16:31
  • $\begingroup$ 1) If $\lambda = \frac{1}{k!} \lambda_{i_1, i_2,\ldots,i_k} dx^{i_1}\wedge dx^{i_2}\ldots dx^{i_k}$ and $\omega$ in the same way, then $<\lambda,\omega>=\frac{1}{n!} \lambda_{i_1, i_2,\ldots,i_k} \omega^{i_1, i_2,\ldots,i_k}$ which is symmetric and bilinear. 2) a metric with the signature $(p,q)$ is assumed. If $g_{ik}=diag(1,-1,-1,-1): q=3$. $\endgroup$ – Frederic Thomas Oct 6 '18 at 22:36

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