Time reversal transformation of electromagnetic four-potential

[Backpacker]

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

\Lambda_T = \begin{pmatrix}<br /> -1 &amp; 0 &amp; 0 &amp; 0\\<br /> 0 &amp;1 &amp; 0 &amp; 0\\<br /> 0 &amp; 0 &amp; 1 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp;1<br /> \end{pmatrix}.<br />

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
<br /> \begin{align*}<br /> A_0 (x^0,x^i)\longmapsto &amp; A&#039;_0 (x&#039;^0,x&#039;^i)=A_0 (-x^0,x^i)\\<br /> A_j (x^0,x^i)\longmapsto &amp; A&#039;_j (x&#039;^0,x&#039;^i)=-A_j (-x^0,x^i)<br /> \end{align*}<br />
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.

 

[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 -.

 

[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?

 

[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?