Proof than an equation is Lorentz invariant

[spaghetti3451]

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?

 

[dextercioby]

This is the manifestation of the famous issue covariant vs. invariant. The Maxwell equations are readily shown to be Lorentz covariant, if one uses Lorentz tensors, QFT books do not resort to geometric definitions of tensors (vectors/covectors), so that you need to pay attention to each definition involving correct index placement and distinguishing between a Lambda matrix and its inverse.

 

[bcrowell]

So, will $A_{\nu}$ and $\partial_{\nu}$ Lorentz transform in the same way?

Yes. That's why we notate $\partial_\nu$ that way.

 

[spaghetti3451]

All right, so, under the active Lorentz transformation $V^{\mu}(x) \rightarrow \Lambda^{\mu}_{\nu}V^{\nu}(\Lambda^{-1}x),$

$(\partial^{2}A_{\nu})(x)-(\partial_{\nu}\partial^{\mu}A_{\mu})(x)=0$ becomes

$(\Lambda^{\nu}_{\mu}\partial^{2}A_{\nu})(\Lambda^{-1}x)-(\Lambda^{\nu}_{\mu}\partial_{\nu}\partial^{\mu}A_{\mu})(\Lambda^{-1}x)=0$ so that

$\Lambda^{\nu}_{\mu}(\partial^{2}A_{\nu}-\partial_{\nu}\partial^{\mu}A_{\mu})(\Lambda^{-1}x)=0$.

Now, do I just peel off $\Lambda^{\nu}_{\mu}$ from the above transformed equation and prove Lorentz invariance?

 

[spaghetti3451]

Bummp!

 

[vanhees71]

Sure, since $\hat{\Lambda}$ is an invertible matrix, the equation $\hat{\Lambda} z=0$ necessarily implies $z=0$, where $z$ is an arbitrary four-vector.

 

[spaghetti3451]

So, there's no steps missing in my proof, then?

 

[vanhees71]

No, it's correct :-))!

 

[spaghetti3451]

Thank you so much!