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Parity conservation and the Field-Strength Tensor‏

  1. Mar 9, 2014 #1
    In reexamining chapter 11 of Jackson's Classical Electrodynamics, especially equations 11.148, it seems obvious that in placing the E and B transformation values into the electro-magnetic field-strength tensor one is ignoring the standard rules which do not allow combining polar vectors and axial vectors, or in this case, scalars and pseudoscalars. The result is that one loses parity information.

    That usage apparently originated with Minkowski. How did he motivate that apparent lapse of rigor? How did the receivers of that usage justify its acceptance?

    On page 558 Jackson states "Transformation (11.149) shows that E and B have no independent existence. A purely electric or magnetic field in one coordinate system will appear as a mixture of electric and magnetic fields in another coordinate frame."

    Isn't he ignoring the fact that the electro-magnetic field-strength tensor obliterates the parity designations implicit in the separation of E and B fields?

    Pauli, on p. 78 and 79 of his monograph on relativity delves a little deeper and seems to say that one may choose one of two different forms for the electro-magnetic field-strength tensor, one of which he calls the dual tensor. But the issue remains. The parity conservation rules from classical EM evaporate leaving an indeterminacy, don't they?
     
  2. jcsd
  3. Mar 9, 2014 #2
    Look at the transformation carefully and you will see that parity is NOT violated by the transformation.
     
  4. Mar 9, 2014 #3
    I didn't say it was violated, but rather becomes indeterminate. But yes, further details in its handling seem necessary.
     
  5. Mar 9, 2014 #4
    No, it doesn't become indeterminate. Look harder.
     
  6. Mar 9, 2014 #5

    PeterDonis

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    No, one doesn't, because B is not really a vector; it's an antisymmetric 2nd-rank tensor, which can simply be incorporated directly into the EM field tensor. Parity only comes into play when you insist on working with axial vectors instead of antisymmetric 2nd-rank tensors, because converting the latter into the former forces you to make a parity choice, corresponding to which direction the standard "normal" vector to a plane points. In other words, when you work with axial vectors, the vector by itself doesn't contain all the information that was in the corresponding antisymmetric 2nd-rank tensor; some of that information now resides in the definition of the standard "normal" to a plane. If you work with the tensors directly, all the information is in the tensor.
     
  7. Mar 9, 2014 #6
    For reference, here are Jackson's equations (11.149) [tex]{\bf E'} = \gamma ({\bf E} + \boldsymbol{\beta \times} {\bf B} ) - \frac{\gamma ^2}{\gamma+1}\boldsymbol{\beta}(\boldsymbol{\beta} \cdot {\bf E}),[/tex] and [tex]{\bf B'} = \gamma ({\bf B} - \boldsymbol{\beta \times} {\bf E}) - \frac{\gamma ^2}{\gamma+1}\boldsymbol{\beta}(\boldsymbol{\beta} \cdot {\bf B}).[/tex]
     
  8. Mar 9, 2014 #7

    PeterDonis

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    PhilDSP, note a key property of these equations:

    In the first equation, ##\bf{B}## only appears in a cross product; while in the second equation, ##\bf{E}## only appears in a cross product. The cross product converts polar vectors to axial vectors and vice versa; so all the terms in each equation are the same kind of vector, hence parity information is not lost.

    In other words, taking a cross product involves making a parity choice, just as converting a 2nd-rank antisymmetric tensor to an axial vector does (since a cross product actually *is* a 2nd-rank antisymmetric tensor "under the hood", so writing it as a vector requires exactly such a conversion).
     
  9. Mar 9, 2014 #8
    Okay thanks! That's pretty obvious from 11.149. And in 11.148 a lot of terms from the cross product drop out because the motion is specified to be along the x axis.

    I need to ponder the subtleties of this a bit more.
     
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