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I've started reading Peskin and Schroeder on my own time, and I'm a bit confused about how to obtain Maxwell's equations from the (source-free) lagrangian density $L = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$ (where $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$ is the field tensor).

Substituting in for the definition of the field tensor yields $L = -\frac{1}{2}[(\partial_\mu A_\nu)(\partial^\mu A^\nu) - (\partial_\mu A_\nu)(\partial^\nu A^\mu)]$. I know I should be using $A^\mu$ as the dynamical variable in the Euler-Lagrange equations, which become $\frac{\partial L}{\partial A_\mu} - \partial_\mu\frac{\partial L}{\partial(\partial_\mu A_\nu)} = - \partial_\mu\frac{\partial L}{\partial(\partial_\mu A_\nu)}$, but I'm confused about how to proceed from here.

I know I should end up with $\partial_\mu F^{\mu\nu} = 0$, but I don't quite see why. Since $\mu$ and $\nu$ are dummy indices, I should be able to change them: how do the indices in the lagrangian relate to the indices in the derivatives in the Euler-Lagrange equations?

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See in Sean carrol's book. Full of derivation there – user97642 Jan 20 at 1:09
    
Why is it not enough to plug the $F^{\mu\nu} = \partial{[\mu}A{\nu]}$ into Maxwell's equations and show that they hold? – Daniel Mahler Apr 29 at 6:35
up vote 12 down vote accepted

Well, you are almost there. Use the fact that $$ {\partial (\partial_{\mu} A_{\nu}) \over \partial(\partial_{\rho} A_{\sigma})} = \delta_{\mu}^{\rho} \delta_{\nu}^{\sigma}$$ which is valid because $\partial_{\mu} A_{\nu}$ are $d^2$ independent components.

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Dear amc, first, write your Lagrangian density as $$ L = -\frac{1}{4} F_{\mu\nu}F^{\mu\nu} = -\frac{1}{2} (\partial_\mu A_\nu) F^{\mu\nu} $$ Is that fine so far? The $F_{\mu\nu}$ contains two terms that make it antisymmetric in the two indices. However, it's multiplied by another $F^{\mu\nu}$ that is already antisymmetric, so I don't need to antisymmetrize it again. Instead, both terms give me the same thing, so the coefficient $-1/4$ simply changes to $-1/2$.

Now, the field equations force you to compute the derivatives of the Lagrangian with respect to $A_\mu$ and its derivatives. First of all, the derivative of the Lagrangian $L$ with respect to $A_\mu$ components themselves vanishes because the Lagrangian only depends on the partiial derivatives of $A_\mu$. Is that clear so far?

So the equations of motion will be $$0 = -\partial_\mu [\partial L / \partial(\partial_\mu A_\nu)] = \dots $$ Whoops, you already got to this point. But now, look at my form of the Lagrangian above. The derivative of the Lagrangian with respect to $\partial_\mu A_\nu$ is simply $$-\frac{1}{2} F^{\mu\nu}$$ because $\partial_\mu A_\nu$ simply appears as a factor so the equations of motion will simply be $$ 0 = +\frac{1}{2} \partial_\mu F^{\mu\nu} $$ However, I have deliberately made one mistake. I have only differentiated the Lagrangian with respect to $\partial_\mu A_\nu$ included in the first factor of $F_{\mu\nu}$, with the lower indices. However, $\partial_\mu A_\nu$ components also appear in $F^{\mu\nu}$, the second factor in the Lagrangian, one with the upper indices. If you add the corresponding terms from the Leibniz rule, the result is simply that the whole contribution will double. So the right equation of motion, including the natural coefficient, will be $$ 0 = \partial_\mu F^{\mu\nu} $$ The overall normalization is important because this equation may get extra terms, like the current, whose coefficient is obvious, and you don't want to get a relative error of two between the derivative of $F$ and the current $j$.

Cheers Lubos

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+1 Elegant! I like it. – nervxxx May 12 '13 at 2:22
    
Hey I know this is 5 years late but maybe you'll see this: Why is $\frac{\partial}{\partial (\partial_{\mu} \phi)} (\partial_{\mu}A_{\nu}F^{\mu \nu}) = F^{\mu \nu}$. Doesn't the tensor also depend on the partial derivatives? Don't we have to use the product rule then? – user17574 Jan 29 at 15:36
    
Hi @user17574 - doesn't "which" tensor depend on the partial derivatives? Surely the stress-energy tensor does, and so does the Lagrangian. That's why the derivative of it with respect to the partial derivatives is nonzero. The derivative is calculated in the answer. The product rule indeed works and it's why one cancels the factor of $1/2$. Have you tried to read the answer? – Luboš Motl Jan 29 at 17:31

We vary the action $$\delta \int {L\;\mathrm{d}t} = \delta \int {\int {\Lambda \left( {A_\nu ,\partial _\mu A_\nu } \right)\mathrm{d}^3 x\;\mathrm{d}t = 0} } $$ ${\Lambda \left( {A_\nu ,\partial _\mu A_\nu } \right)}$ is the density of lagrangian of the system.

So, $$\int {\int {\left( {\frac{{\partial \Lambda }}{{\partial A_\nu }}\delta A_\nu + \frac{{\partial \Lambda }}{{\partial \left( {\partial _\mu A_\nu } \right)}}\delta \left( {\partial _\mu A_\nu } \right)} \right)\mathrm{d}^3 x\;\mathrm{d}t = 0} } $$ By integrating by parts we obtain: $$\int {\int {\left( {\frac{{\partial \Lambda }}{{\partial A_\nu }} - \partial _\mu \frac{{\partial \Lambda }}{{\partial \left( {\partial _\mu A_\nu } \right)}}} \right)\delta A_\nu \mathrm{d}^3 x\;\mathrm{d}t = 0} } \implies \frac{{\partial \Lambda }}{{\partial A_\nu }} - \partial _\mu \frac{{\partial \Lambda }}{{\partial \left( {\partial _\mu A_\nu } \right)}} = 0$$ We have to determine the density of the lagrangian. One terms deals with the interaction of the charges with the electromagnetic field, $J^\mu A_\mu$. The other term is the density of energy of the electromagnetic field: this term is the difference of the magnetic field and the electric field. So we have: $$\Lambda = J^\mu A_\mu + \frac{1}{{4\mu _0 }}F^{\mu \nu } F_{\mu \nu } $$ We have: $$\frac{{\partial \Lambda }}{{\partial A_\nu }} = J^\nu $$ so: \begin{align}\partial _\mu \frac{{\partial \Lambda }}{{\partial \left( {\partial _\mu A_\nu } \right)}} &= \frac{1}{{4\mu _0 }}\partial _\mu \left( {\frac{\partial }{{\partial \left( {\partial _\mu A_\nu } \right)}}F^{\kappa \lambda } F_{\kappa \lambda } } \right) \\&= \frac{1}{{4\mu _0 }}\partial _\mu \left( {\frac{\partial }{{\partial \left( {\partial _\mu A_\nu } \right)}}\left( {\left( {\partial ^\kappa A^\lambda - \partial ^\lambda A^\kappa } \right)\left( {\partial _\kappa A_\lambda - \partial _\lambda A_\kappa } \right)} \right)} \right) \\&= \frac{1}{{4\mu _0 }}\partial _\mu \left( {\frac{\partial }{{\partial \left( {\partial _\mu A_\nu } \right)}}\left( {\partial ^\kappa A^\lambda \partial _\kappa A_\lambda - \partial ^\kappa A^\lambda \partial _\lambda A_\kappa - \partial ^\lambda A^\kappa \partial _\kappa A_\lambda + \partial ^\lambda A^\kappa \partial _\lambda A_\kappa } \right)} \right)\end{align} The third and the fourth are the same of first and the second terms. You can do $k \leftrightarrow \lambda $: \begin{align}\partial _\mu \frac{{\partial \Lambda }}{{\partial \left( {\partial _\mu A_\nu } \right)}} & = \frac{1}{{2\mu _0 }}\partial _\mu \left( {\frac{\partial }{{\partial \left( {\partial _\mu A_\nu } \right)}}\left( {\partial ^\kappa A^\lambda \partial _\kappa A_\lambda - \partial ^\kappa A^\lambda \partial _\lambda A_\kappa } \right)} \right)\;.\end{align} But \begin{align}\frac{\partial }{{\partial \left( {\partial _\mu A_\nu } \right)}}\left( {\partial ^\kappa A^\lambda \partial _\kappa A_\lambda } \right) &= \partial ^\kappa A^\lambda \frac{\partial }{{\partial \left( {\partial _\mu A_\nu } \right)}}\left( {\partial _\kappa A_\lambda } \right) + \partial _\kappa A_\lambda \frac{\partial }{{\partial \left( {\partial _\mu A_\nu } \right)}}\left( {\partial ^\kappa A^\lambda } \right) \\ &= \partial ^\kappa A^\lambda \delta _\kappa ^\mu \delta _\lambda ^\nu + g^{\kappa \alpha } g^{\lambda \beta } \partial _\kappa A_\lambda \frac{\partial }{{\partial \left( {\partial _\mu A_\nu } \right)}}\left( {\partial _\alpha A_\beta } \right)\\& = 2\partial ^\mu A^\nu \;.\end{align}

We have:

\begin{align}\frac{\partial }{{\partial \left( {\partial _\mu A_\nu } \right)}}\left( {\partial ^\kappa A^\lambda \partial _\lambda A_\kappa } \right) &= 2\partial ^\nu A^\mu \;. \end{align}

So,

\begin{align}\partial _\mu \left( {\frac{{\partial \Lambda }}{{\partial \left( {\partial _\mu A_\nu } \right)}}} \right)& = \frac{1}{{\mu _0 }}\partial _\mu \left( {\partial ^\mu A^\nu - \partial ^\nu A^\mu } \right)\\ & = \frac{1}{{\mu _0 }}\partial _\mu F^{\mu \nu } \;.\end{align} The lagrangian equations provide the non homogeneus maxwell equations:

$$\partial _\mu F^{\mu \nu } = \mu _0 J^\nu \;. $$

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Thanks for this detailed solution. – FraSchelle Jan 25 '13 at 14:59
    
And can't we derive the homogeneous Maxwell equations from the Lagrangian? – omephy Jul 19 '13 at 16:56
    
@ome, did you look at the answer from Lubos? – Alfred Centauri Jul 19 '13 at 21:12
    
Yah, I have just seen it. :P – omephy Jul 19 '13 at 22:50

One method is to vary the Maxwell action (set $J^\mu = 0$ if you want, for the source-free case) $$ S = \int d^4 x {\mathcal{L}} = - \int d^4 x \left(\frac{1}{4} F^{\mu \nu} F_{\mu \nu} + J^\mu A_\mu\right). $$ First note that $$ \begin{align} \delta \left(F^{\mu \nu} F_{\mu \nu}\right) &= 2 F^{\mu \nu} \delta F_{\mu \nu} \\ &= 2 F^{\mu \nu} \left(\partial_\mu \delta A_\nu - \partial_\nu \delta A_\mu\right) \\ &= 4 F^{\mu \nu} \partial_\mu \delta A_\nu \\ &= 4\left[\partial_\mu\left(F^{\mu \nu} \delta A_\nu\right) - \partial_\mu F^{\mu \nu} \delta A_\nu\right], \end{align} $$ where we've used the fact that $F$ is antisymmetric.

Notice also that the $\partial_\mu\left(F^{\mu \nu} \delta A_\nu\right)$ term will vanish upon converting it to a surface integral, using the standard argument that $\delta A_\mu$ vanishes at the integration boundary.

Using the above, the variation of the action is $$ \delta S = - \int d^4x \ \delta A_\nu \left(-\partial_\mu F^{\mu \nu} + J^\nu \right), $$ which, since $\delta A_\nu$ is arbitrary, yields the desired result $$ \partial_\mu F^{\mu \nu} = J^\nu. $$

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