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Electric / magnetic field transformations

  1. Sep 23, 2007 #1


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    Hi. I thought I had tensors and Lorentz transformations under control, but now I'm in doubt again.

    For example, consider the electromagnetic field tensor
    [tex] F_{\mu\nu} = \begin{pmatrix}
    0 & -E_1 & -E_2 & -E_3 \\
    E_1 & 0 & B_3 & -B_2 \\
    E_2 & -B_3 & 0 & B_1 \\
    E_3 & B_2 & -B_1 & 0 \\
    \qquad\text{ so }
    F^{\mu\nu} = \begin{pmatrix}
    0 & E_1 & E_2 & E_3 \\
    -E_1 & 0 & B_3 & -B_2 \\
    -E_2 & -B_3 & 0 & B_1 \\
    -E_3 & B_2 & -B_1 & 0 \\
    in the (-1, 1, 1, 1) metric.

    Now we apply a Lorentz transformation, and to keep it simple we take a (counter clockwise) rotation around an angle [itex]\theta[/itex] about the [itex]z[/itex]-axis. Now I thought I'd write this as
    [tex] R^\mu_\nu = \begin{pmatrix}
    1 & 0 & 0 & 0 \\
    0 & \cos\theta & -\sin\theta & 0 \\
    0 & \sin\theta & \cos\theta & 0 \\
    0 & 0 & 0 & 1 \\
    as it works on a vector and produces a vector ([tex](v')^\mu = R^\mu_\nu v^\nu[/tex]).
    Did I get this right? In this case wrong placement of the indices doesn't introduce errors yet, as far as I can see, but this will generally not be the case for boosts (which do not have just zeros in the first column and row).

    Now the components of the electric field [itex]E_i = F_{i0}[/itex] transform as
    [tex] E_i' = F'_{i0} = R_i^\mu R_0^\nu F_{\mu\nu}. [/tex]
    Working out the transformation yields
    [tex] E_1' = E_1 \cos\theta - E_2 \sin \theta; \quad
    E_2' = E_1 \sin\theta + E_2 \cos \theta; \quad
    E_3' = E_3,
    which can be written in vector notation as
    [tex] \vec E' = \mat R \vec E
    \qquad\text{ where }
    \mat R = \begin{pmatrix}
    \cos\theta & -\sin\theta & 0 \\
    \sin\theta & \cos\theta & 0 \\
    0 & 0 & 1 \\
    which is, I think, the transformation rule for a vector hence what one would expect.

    Similarly, the components of the magnetic field are [itex]B_i = \frac12 \epsilon_{ijk} F^{jk}[/itex].
    As raising both the indices on [itex]F_{\mu\nu}[/itex] does not affect the components in the
    lower right [itex]3 \times 3[/itex] block -- that is, [itex]F_{ij} = F^{ij}[/itex] for [itex]i, j = 1, 2, 3[/itex] --
    we can calculate
    [tex] B_i' = \frac12 \epsilon_{ijk} F'^{jk} = \frac12 \epsilon_{ijk} R^j_\mu R^k_\nu F^{\mu\nu}. [/tex]
    Explicit calculation yields
    [tex] B_1' = B_1 \cos\theta - B_2 \sin\theta; \quad
    B_2' = B_1 \sin\theta + B_2 \cos\theta; \quad
    B_3' = B_3,
    which is exactly the same as the electric field. Yet the magnetic field is not a vector, but a pseudo-vector; therefore I doubt my answer.

    I'd like to get this right, especially with the indices etc., before I proceed to boosts, e.g.
    [tex] R^\mu_\nu \to \Lambda^\mu_\nu = \begin{pmatrix}
    \cosh\theta & \sinh\theta & 0 & 0 \\
    \sinh\theta & \cosh\theta & 0 & 0 \\
    0 & 0 & 1 & 0 \\
    0 & 0 & 0 & 1 \\

    Thanks a lot.
  2. jcsd
  3. Sep 23, 2007 #2
    Even though it is a pseudo-vector, it is supposed to transform like a vector in rotations. You should check space reflections to see if you get the desired difference in the transformation of E and B.
  4. Sep 23, 2007 #3


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    Using GRtensor, I take

    F_{ab} =\left[ \begin {array}{cccc} 0&-{\it Ex}&-{\it Ey}&-{\it Ez}\\\noalign{\medskip}{\it Ex}&0&{\it Bz}&-{\it By}\\\noalign{\medskip}{
    \it Ey}&-{\it Bz}&0&{\it Bx}\\\noalign{\medskip}{\it Ez}&{\it By}&-{
    \it Bx}&0\end {array} \right]

    L^a{}_b = \left[ \begin {array}{cccc} 1&0&0&0\\\noalign{\medskip}0&\cos \left( \theta \right) &-\sin \left( \theta \right) &0\\\noalign{\medskip}0&
    \sin \left( \theta \right) &\cos \left( \theta \right) &0
    \\\noalign{\medskip}0&0&0&1\end {array} \right]

    to compute

    F^{\prime}_{ab} = F_{cd} L^c{}_a L^d{}_b[/tex]

    which is

    \left[ \begin {array}{cccc} 0&-{\it Ex}\,\cos \left( \theta \right) -{\it Ey}\,\sin \left( \theta \right) &{\it Ex}\,\sin \left( \theta
    \right) -{\it Ey}\,\cos \left( \theta \right) &-{\it Ez}
    \\\noalign{\medskip}{\it Ex}\,\cos \left( \theta \right) +{\it Ey}\,
    \sin \left( \theta \right) &0&{\it Bz}\, \left( \cos \left( \theta
    \right) \right) ^{2}+{\it Bz}\, \left( \sin \left( \theta \right)
    \right) ^{2}&-{\it By}\,\cos \left( \theta \right) +{\it Bx}\,\sin
    \left( \theta \right) \\\noalign{\medskip}-{\it Ex}\,\sin \left(
    \theta \right) +{\it Ey}\,\cos \left( \theta \right) &-{\it Bz}\,
    \left( \sin \left( \theta \right) \right) ^{2}-{\it Bz}\, \left(
    \cos \left( \theta \right) \right) ^{2}&0&{\it By}\,\sin \left(
    \theta \right) +{\it Bx}\,\cos \left( \theta \right)
    \\\noalign{\medskip}{\it Ez}&{\it By}\,\cos \left( \theta \right) -{
    \it Bx}\,\sin \left( \theta \right) &-{\it By}\,\sin \left( \theta
    \right) -{\it Bx}\,\cos \left( \theta \right) &0\end {array} \right]

    I didn't think this quite matched some of the signs in your result, but I thought it might be helpful.
    Last edited: Sep 23, 2007
  5. Sep 23, 2007 #4


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    jostpuur, of course you are right about the (pseudo)-vector remark. Thanks.

    pervect: I think you calculated [itex]L^T F L[/itex] whereas I did [itex]L F L^T[/itex]. One of us should be wrong then (and it's probably me), which would mean I messed up the indices... just what I was afraid of.

    [edit]I found it, there was an error in my Mathematica code (it read
    Code (Text):
    e[i_] := Sum[R[[i, \[Mu]]] R[[1, \[Nu]]] F[[\[Mu], \[Nu]]], {\[Mu], 1, 4}, {\[Nu], 1, 4}]
    instead of
    Code (Text):
    e[i_] := Sum[R[[\[Mu], i]] R[[\[Nu], 1]] F[[\[Mu], \[Nu]]], {\[Mu], 1, 4}, {\[Nu], 1, 4}]
    -- note the indices of the rotation matrix.
    So apparantly we agree now.[/edit].
    Last edited: Sep 23, 2007
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