Derivation of Maxwell's equations from field tensor lagrangian

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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3 Answers

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 at 2:22

We vary the action $$\delta \int {Ldt} = \delta \int {\int {\Lambda \left( {A_\nu ,\partial _\mu A_\nu } \right)d^3 xdt = 0} }$$ ${\Lambda \left( {A^\nu ,\partial _\mu A^\nu } \right)}$ is the density of lagrangian of the system.\ So: $$\int {\int {\left( {\frac{{\partial \Lambda }}{{\partial A^\nu }}\delta A_\nu + \frac{{\partial \Lambda }}{{\partial \left( {\partial _\mu A_\nu } \right)}}\delta \left( {\partial _\mu A_\nu } \right)} \right)d^3 xdt = 0} }$$ By integrating by parts we obtain: $$\int {\int {\left( {\frac{{\partial \Lambda }}{{\partial A_\nu }} - \partial _\mu \frac{{\partial \Lambda }}{{\partial \left( {\partial _\mu A_\nu } \right)}}} \right)\delta A_\nu d^3 xdt = 0} }$$ $$\frac{{\partial \Lambda }}{{\partial A_\nu }} - \partial _\mu \frac{{\partial \Lambda }}{{\partial \left( {\partial _\mu A_\nu } \right)}} = 0$$ W e have to determine the density of lagrangian. One terms deals with the interaction of the charges with the electromagnetic field. This term is $J^\nu A_\nu$. One term is the density of energy of the electromagnetic field: this term is the difference of the magnetic field and the electris 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: $$\partial _\mu \frac{{\partial \Lambda }}{{\partial \left( {\partial _\mu A_\nu } \right)}} = \frac{1}{{4\mu _0 }}\partial _\mu \left( {\frac{\partial }{{\partial \left( {\partial _\mu A_\nu } \right)}}F^{\kappa \lambda } F_{\kappa \lambda } } \right) = \frac{1}{{4\mu _0 }}\partial _\mu \left( {\frac{\partial }{{\partial \left( {\partial _\mu A_\nu } \right)}}\left( {\left( {\partial ^\kappa A^\lambda - \partial ^\lambda A^\kappa } \right)\left( {\partial _\kappa A_\lambda - \partial _\lambda A_\kappa } \right)} \right)} \right) =$$ $$= \frac{1}{{4\mu _0 }}\partial _\mu \left( {\frac{\partial }{{\partial \left( {\partial _\mu A_\nu } \right)}}\left( {\partial ^\kappa A^\lambda \partial _\kappa A_\lambda - \partial ^\kappa A^\lambda \partial _\lambda A_\kappa - \partial ^\lambda A^\kappa \partial _\kappa A_\lambda + \partial ^\lambda A^\kappa \partial _\lambda A_\kappa } \right)} \right) =$$ The third and the fourth are the same of first and the second terms. You can do $k \leftrightarrow \lambda$: $$\partial _\mu \frac{{\partial \Lambda }}{{\partial \left( {\partial _\mu A_\nu } \right)}} = \frac{1}{{2\mu _0 }}\partial _\mu \left( {\frac{\partial }{{\partial \left( {\partial _\mu A_\nu } \right)}}\left( {\partial ^\kappa A^\lambda \partial _\kappa A_\lambda - \partial ^\kappa A^\lambda \partial _\lambda A_\kappa } \right)} \right)$$ but: $$\frac{\partial }{{\partial \left( {\partial _\mu A_\nu } \right)}}\left( {\partial ^\kappa A^\lambda \partial _\kappa A_\lambda } \right) = \partial ^\kappa A^\lambda \frac{\partial }{{\partial \left( {\partial _\mu A_\nu } \right)}}\left( {\partial _\kappa A_\lambda } \right) + \partial _\kappa A_\lambda \frac{\partial }{{\partial \left( {\partial _\mu A_\nu } \right)}}\left( {\partial ^\kappa A^\lambda } \right) =$$ $$= \partial ^\kappa A^\lambda \delta _\kappa ^\mu \delta _\lambda ^\nu + g^{\kappa \alpha } g^{\lambda \beta } \partial _\kappa A_\lambda \frac{\partial }{{\partial \left( {\partial _\mu A_\nu } \right)}}\left( {\partial _\alpha A_\beta } \right) = 2\partial ^\mu A^\nu$$ We have: $$\frac{\partial }{{\partial \left( {\partial _\mu A_\nu } \right)}}\left( {\partial ^\kappa A^\lambda \partial _\lambda A_\kappa } \right) = 2\partial ^\nu A^\mu$$ so: $$\partial _\mu \left( {\frac{{\partial \Lambda }}{{\partial \left( {\partial _\mu A_\nu } \right)}}} \right) = \frac{1}{{\mu _0 }}\partial _\mu \left( {\partial ^\mu A^\nu - \partial ^\nu A^\mu } \right) = \frac{1}{{\mu _0 }}\partial _\mu F^{\mu \nu }$$ 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. – Oaoa Jan 25 at 14:59
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