Time reversal transformation of electromagnetic four-potential

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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.
 
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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 -.
 
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?
 
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?
 
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