Proof than an equation is Lorentz invariant

1. Nov 19, 2015

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?

2. Nov 19, 2015

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.

3. Nov 19, 2015

bcrowell

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

4. Nov 21, 2015

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?

5. Nov 25, 2015

spaghetti3451

Bummp!

6. Nov 28, 2015

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.

7. Nov 29, 2015

spaghetti3451

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

8. Nov 29, 2015

vanhees71

No, it's correct :-))!

9. Nov 29, 2015

spaghetti3451

Thank you so much!