In Peskin and Schroeder page 37, it is written that

  • Using vector and tensor fields, we can write a variety of Lorentz-invariant equations.
  • Criteria for Lorentz invariance: In general, any equation in which each term has the same set of uncontracted Lorentz indices will naturally be invariant under Lorentz transformations.

I would like to explicitly show that the above criteria is valid for Maxwell's equations $\partial^{\mu} F_{\mu \nu} = 0$ or $\partial^{2}A_{\nu}-\partial_{\nu}\partial^{\mu}A_{\mu}=0$.

  • Solution 1: Maxwell's equations follow from the Lagrangian $$\mathcal{L}_{MAXWELL}=-\frac{1}{4}(F_{\mu \nu})^{2} = -\frac{1}{4}(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})^{2}$$ which is a Lorentz scalar, so this means that the equation of motion is Lorentz-invariant as well. That's one way to convince yourself that the above Maxwell's equations are, in fact, Lorentz invariant. Is this correct?

  • Solution 2: I would like to actively transform the electromagnetic field strength tensor $F_{\mu \nu}$ and show that the Maxwell's equations $\partial^{\mu} F_{\mu \nu} = 0$ or $\partial^{2}A_{\nu}-\partial_{\nu}\partial^{\mu}A_{\mu}=0$ remain Lorentz invariant.

I can see that $\partial^{2}$ and $\partial^{\mu}A_{\mu}$ will not Lorentz transform as they are Lorentz scalars.

Under an active Lorentz transformation, $V^{\mu}(x) \rightarrow \Lambda^{\mu}_{\nu}V^{\nu}(\Lambda^{-1}x)$. So, will $A_{\nu}$ and $\partial_{\nu}$ Lorentz transform in the same way?

  • $\begingroup$ Note that transformation matrices involved in transforming the upper and lower and upper indices are the inverses of each other and the derivative operator with the index downstairs is the derivative w.r.t the coordinate with the index upstairs. $\endgroup$ Commented Nov 19, 2015 at 20:51

1 Answer 1


Maxwell's equations can be written in the form $$ \partial_{\mu}F^{\mu\nu} = \frac{4\pi}{c} j^{\nu},\qquad \partial_{\lambda}F_{\mu\nu}+ \textrm{cyclic}(\lambda,\mu,\nu)=0 $$ with $F_{\mu\nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}$. Let us look at the first set: the right hand side is a vector (and this can be proven looking at the equation for the conservation of the charge), therefore the left hand side must be a vector too. Being $\partial_{\mu}$ the covariant components of a dual form, $F^{\mu\nu}$ must be the components of a rank $(0,2)$ tensor. As such, under Lorentz transformations, they will transform as $$ {F^{\mu\nu}}'(x') = \Lambda^\mu_{\phantom{\mu}\gamma}(x)\,\Lambda^\nu_{\phantom{\nu}\rho}(x)F^{\gamma\rho}(x) $$ (where one can equivalently take $\Lambda^{-1}$ according to how such map is defined). Expanding the above in terms of the electric and magnetic field (as derived from the vector potential $A_{\mu}$ such that $F_{\mu\nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}$) one can obtain the explicit transformation laws for the two fields. Once so, you can directly plug them into the Maxwell's equations by brute force and check that they are left invariant.

Another way, as you pointed out yourself, is to notice that the action is left invariant under the effect of some particular transformations; hence, the equations of motion must be as well.

  • $\begingroup$ We have shown here that $F'^{\mu}{\nu}$ would transform in the specific fashion like a tensor should with the help of $\Lambda^{\mu}_{\gamma}$ but this $\Lambda$ can well be of a Galilean transformation. How does this prove invariance under Lorentz transformation? $\endgroup$ Commented Dec 8, 2017 at 8:28
  • $\begingroup$ @NamanAgarwal That is the definition of invariance of a tensor form on a manifold. We assume that the manifold is provided with the Lorentz metric and afterwards we just write down the transformation properties of its tensor fields. Why we start with such hypotheses is another question (answer: because the speed of light is invariant and one can prove the Lorentz metric must be the one describing physical observers). $\endgroup$
    – gented
    Commented Dec 8, 2017 at 11:01
  • $\begingroup$ Is it the same as Minkowski metric? Also so here you suggest that we have defined some manifold called the lorentz manifold and lorentz transformations are the only possible transformations on this manifold? $\endgroup$ Commented Dec 8, 2017 at 11:26
  • $\begingroup$ Yes, this is in fact the case. $\endgroup$
    – gented
    Commented Dec 8, 2017 at 13:18

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