# Lagrangian for the Maxwell equations

In his book 'Classical Electrodynamics' Kurt Lechner wants to find a Lagrangian $$\mathcal{L}$$ so that the Euler Lagrange equations $$\partial_\mu\frac{\partial\mathcal{L}}{\partial(\partial_\mu A_\nu)}-\frac{\partial\mathcal{L}}{\partial A_\nu}=0$$ give rise to the Maxwell equations $$\partial_\mu F^{\mu\nu}=j^\nu.$$ He explains heuristically why it is of the form $$\mathcal{L}=\mathcal{L}_1+\mathcal{L}_2$$ with $$\mathcal{L}_1\propto F^{\mu\nu}F_{\mu\nu}$$ and $$\mathcal{L}_2\propto A_\mu j^\mu$$ considering gauge and Lorentz invariance. I think I got this part figured out.

But next he shows that the above $$\mathcal{L}$$ really gives rise to the Maxwell equations and here is where I get lost:

He sets $$\mathcal{L}_1 =-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}$$ to get the normalizations right and then he consideres the variation of $$\mathcal{L}_1$$ under an infintesimal variation of $$\partial A$$: $$$$\delta\mathcal{L}_1=-\frac{1}{2}F^{\mu\nu}\delta F_{\mu\nu}=-\frac{1}{2}F^{\mu\nu}\left(\delta\partial_\mu A_\nu - \delta\partial_\nu A_\mu\right)=-F^{\mu\nu}\delta\left(\partial_\mu A_\nu\right).$$$$

Can someone please explain the steps he made in each of the equalities in the last equation?

Schematically, $$\delta(-\frac{1}{4}F^2)=-\frac{1}{4}(2F)\delta{F}$$.

Again, schematically, note that $$\delta f(x)=\tilde{f}(x)-f(x), \hspace{5mm} \partial f(x)=(f(x+dx)-f(x))/dx$$

This means, for example-$$\delta(\partial A)=\frac{1}{dx}\delta(A(x+dx)-A(x))$$

$$=\frac{1}{dx}\bigg((\tilde{A}(x+dx)-\tilde{A}(x))-(A(x+dx)-A(x))\bigg)$$

$$=\frac{1}{dx}\bigg(\delta A(x+dx)-\delta A(x)\bigg)=\partial(\delta A)$$

i.e. $$\delta$$ and $$\partial$$ commute. This step isn't really necessary here, but would be essential if you were to carry out the computation in all it's mathematical glory. Finally, note that $$F^{\mu\nu}\delta(\partial_\mu A_\nu)=F^{\nu\mu}\delta(\partial_\nu A_\mu)=-F^{\mu\nu}\delta(\partial_\nu A_\mu)$$

Where I have first switched the dummy indices, and then used $$F^{\mu\nu}=-F^{\nu\mu}$$.

The result now follows if you subtract the quantity on the RHS on both sides of the last equation.

Let $$-\frac{1}{2}F^{\mu\nu}\left(\delta\partial_\mu A_\nu - \delta\partial_\nu A\mu\right)=-\frac{1}{2}F^{\mu\nu}\delta\partial_\mu A_\nu+\frac{1}{2}F^{\mu\nu}\delta\partial_\nu A\mu$$ Interchange the dummy index $$\mu$$ and $$\nu$$ in the second term of the equation.You will get $$-\frac{1}{2}F^{\mu\nu}\delta\partial_\mu A_\nu+\frac{1}{2}F^{\nu\mu}\delta\partial_\mu A\nu$$ Since $$F^{\mu \nu}=-F^{\nu \mu}$$ ($$F^{\mu \nu}$$ is Antisymmetric) we can write $$-\frac{1}{2}F^{\mu\nu}\delta\partial_\mu A_\nu+\frac{1}{2}F^{\nu\mu}\delta\partial_\mu A\nu=-\frac{1}{2}F^{\mu\nu}\delta\partial_\mu A_\nu-\frac{1}{2}F^{\mu\nu}\delta\partial_\mu A\nu=-F^{\mu\nu}\delta\left(\partial_\mu A_\nu\right)$$