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Time reversal symmetry breaking in EM

  1. May 5, 2010 #1
    I have come across a problem I am trying to understand, and hoping someone here has some insight. Basically, when writing down different solutions for an EM field from given sources, there seems to be a problem from the standpoint of time symmetry. From my understanding, if you reverse time, the electric field should remain unchanged, and the magnetic field should have the same magnitude but opposite sign. You can see that must be true by several methods (ie just looking at the force on other particles), but fails here:

    For instance, from Liendard-Wiechert potentials:

    [tex]\vec{E} = \frac{q(\hat{r} - \vec{\beta})}{\gamma^2(1-\vec{\beta}\cdot\hat{r})^3R^2}\bigg|_{ret}[/tex]

    (assuming uniform motion)

    if you reverse time beta changes sign and so E here clearly may not necessarily obey the symmetry.

    Also, if you assume

    [tex] \vec{B} = \hat{r}\times\vec{E}[/tex]

    neither does B.

    Perhaps this is a relativistic effect, if someone knows how to resolve it.
  2. jcsd
  3. May 5, 2010 #2
    I think the problem is to do with the choice of a retarded Green's function and hence retarted time solutions. Einstein apparently believed that choosing retarted solutions meant imposing time-asymmetry by hand and preferred a half-retarded half-advanced solution instead.
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