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Time reversal transformation of electromagnetic four-potential

  1. Nov 12, 2012 #1
    Consider the time-reversal Lorentz transformation given by the 4x4 matrix:

    [itex]\Lambda_T = \begin{pmatrix}
    -1 & 0 & 0 & 0\\
    0 &1 & 0 & 0\\
    0 & 0 & 1 & 0 \\
    0 & 0 & 0 &1
    \end{pmatrix}.
    [/itex]

    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
    [itex]
    \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*}
    [/itex]
    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 [itex]A\mapsto \Lambda A[/itex], the minus sign should be applied to [itex]A_0[/itex].
     
  2. jcsd
  3. Nov 12, 2012 #2

    dextercioby

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    A coupling (scalar, not pseudoscalar) [itex] j^{\mu}A_{\mu} [/itex] is invariant, so that if j0 is invariant, then the space components of j and the ones of A have the same sign, namely -.
     
  4. Nov 13, 2012 #3
    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?
     
  5. Nov 13, 2012 #4
    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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