# Time reversal transformation of electromagnetic four-potential

1. Nov 12, 2012

### Backpacker

Consider the time-reversal Lorentz transformation given by the 4x4 matrix:

$\Lambda_T = \begin{pmatrix} -1 & 0 & 0 & 0\\ 0 &1 & 0 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 &1 \end{pmatrix}.$

In my relativistic quantum mechanics lecture, we discussed how the electromagnetic 4-potential transforms under this particular Lorentz transformation. Without invoking any sort of mathematical argument, the prof argued that the four-potential transforms as
\begin{align*} A_0 (x^0,x^i)\longmapsto & A'_0 (x'^0,x'^i)=A_0 (-x^0,x^i)\\ A_j (x^0,x^i)\longmapsto & A'_j (x'^0,x'^i)=-A_j (-x^0,x^i) \end{align*}
based on the idea that currents reverse under time-reversal.

Is there a good mathematical reasoning for this? It seems to me that since four-vectors transform as $A\mapsto \Lambda A$, the minus sign should be applied to $A_0$.

2. Nov 12, 2012

### dextercioby

A coupling (scalar, not pseudoscalar) $j^{\mu}A_{\mu}$ is invariant, so that if j0 is invariant, then the space components of j and the ones of A have the same sign, namely -.

3. Nov 13, 2012

### Backpacker

Ok, so I'm comfortable with the fact that the spatial components of the current are inverted under time reversal, i.e. ${j'}^i=-j^i$.

But, why is $j^\mu A_\mu$ invariant? And what does $j^\mu A_\mu$ mean physically?

4. Nov 13, 2012

### Backpacker

Hold on, I guess I see where some of my confusion is coming from.

In class, we showed that under a Lorentz transformation $\Lambda$, the current (just like any good 4-vector) transforms as $j'^\mu={\Lambda^\mu}_\nu j^\nu$. But I guess this is only for proper orthchronous transformations. Why is this different for improper transformations?