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Lorentz transformation of electric and magnetic fields

  1. Apr 30, 2016 #1
    1. The problem statement, all variables and given/known data

    Using the tensor transformation law applied to ##F_{\mu\nu}##, show how the electric and magnetic field ##3##-vectors ##\textbf{E}## and ##\textbf{B}## transform under

    (a) a rotation about the ##y##-axis,
    (b) a boost along the ##z##-axis.

    2. Relevant equations

    3. The attempt at a solution

    The tensor transformation law applied to ##F_{\mu\nu}## is the following:

    ##F_{\mu'\nu'}=\frac{\partial x^{\mu}}{\partial x^{\mu'}}\frac{\partial x^{\nu}}{\partial x^{\nu'}}F_{\mu\nu}={{(\Lambda^{-1})}_{\mu'}}^{\mu}{{(\Lambda^{-1})}_{\nu'}}^{\nu}F_{\mu\nu}={\Lambda^{\mu}}_{\mu'}{\Lambda^{\nu}}_{\nu'}F_{\mu\nu}##

    (a) A rotation of angle ##\theta## about the ##y##-axis in the counterclockwise sense from the unprimed frame to the primed frame is specified by

    ##{\Lambda^{\mu'}}_{\mu}=
    \left( \begin{array}{cccc}
    1 & 0 & 0 & 0 \\
    0 & \text{cos}\ \theta & 0 & -\text{sin}\ \theta \\
    0 & 0 & 1 & 0 \\
    0 & \text{sin}\ \theta & 0 & \text{cos}\ \theta \end{array} \right)
    ##

    Therefore, the same transformation of a rotation of angle ##\theta## about the ##y##-axis in the clockwise sense from the primed frame to the unprimed frame is given by

    ##{\Lambda^{\mu}}_{\mu'}=
    \left( \begin{array}{cccc}
    1 & 0 & 0 & 0 \\
    0 & \text{cos}\ (-\theta) & 0 & -\text{sin}\ (-\theta) \\
    0 & 0 & 1 & 0 \\
    0 & \text{sin}\ (-\theta) & 0 & \text{cos}\ (-\theta) \end{array} \right)=
    \left( \begin{array}{cccc}
    1 & 0 & 0 & 0 \\
    0 & \text{cos}\ \theta & 0 & \text{sin}\ \theta \\
    0 & 0 & 1 & 0 \\
    0 & -\text{sin}\ \theta & 0 & \text{cos}\ \theta \end{array} \right)
    ##

    Have I made a mistake?
     
  2. jcsd
  3. May 4, 2016 #2

    TSny

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    Homework Helper
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    Looks good.
     
  4. May 5, 2016 #3
    Thanks!

    So, we have ##F_{\mu'\nu'}={\Lambda^{\mu}}_{\mu'}{\Lambda^{\nu}}_{\nu'}F_{\mu\nu}##
    ##=
    \left( \begin{array}{cccc}
    1 & 0 & 0 & 0 \\
    0 & \text{cos}\ \theta & 0 & \text{sin}\ \theta \\
    0 & 0 & 1 & 0 \\
    0 & -\text{sin}\ \theta & 0 & \text{cos}\ \theta \end{array} \right)
    \left( \begin{array}{cccc}
    1 & 0 & 0 & 0 \\
    0 & \text{cos}\ \theta & 0 & \text{sin}\ \theta \\
    0 & 0 & 1 & 0 \\
    0 & -\text{sin}\ \theta & 0 & \text{cos}\ \theta \end{array} \right)
    \left( \begin{array}{cccc}
    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 \end{array} \right)=
    \left( \begin{array}{cccc}
    1 & 0 & 0 & 0 \\
    0 & \text{cos}^{2}\ \theta - \text{sin}^{2}\ \theta & 0 & 2\ \text{sin}\ \theta\ \text{cos}\ \theta \\
    0 & 0 & 1 & 0 \\
    0 & -2\ \text{sin}\ \theta\ \text{cos}\ \theta & 0 & \text{cos}^{2}\ \theta - \text{sin}^{2}\ \theta \end{array} \right)
    \left( \begin{array}{cccc}
    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 \end{array} \right)=
    \left( \begin{array}{cccc}
    1 & -E_{1} & -E_{2} & -E_{3} \\
    E_{1}(\text{cos}^{2}\ \theta - \text{sin}^{2}\ \theta)+E_{3}(2\ \text{sin}\ \theta\ \text{cos}\ \theta) & -2\ \text{sin}\ \theta\ \text{cos}\ \theta\ B_{2} & -B_{3}(\text{cos}^{2}\ \theta - \text{sin}^{2}\ \theta)+(2\ \text{sin}\ \theta\ \text{cos}\ \theta)B_{1} & (\text{cos}^{2}\ \theta - \text{sin}^{2}\ \theta)B_{2} \\
    E_{2} & B_{3} & 0 & -B_{1} \\
    E_{3}(\text{cos}^{2}\ \theta - \text{sin}^{2}\ \theta)-E_{1}(2\ \text{sin}\ \theta\ \text{cos}\ \theta) & -(\text{cos}^{2}\ \theta - \text{sin}^{2}\ \theta)B_{2} & B_{1}(\text{cos}^{2}\ \theta - \text{sin}^{2}\ \theta)+B_{3}(2\ \text{sin}\ \theta\ \text{cos}\ \theta) & -2\ \text{sin}\ \theta\ \text{cos}\ \theta\ B_{2} \end{array} \right).##

    I'm not really sure how to figure out the transformation of the ##\textbf{E}## and ##\textbf{B}## fields since the transformed electromagnetic strength tensor looks really weird. For example, there are non-zero-valued entries in ##F_{\mu'\nu'}## for zero-valued corresponding entries in ##F_{\mu\nu}##.
     
  5. May 5, 2016 #4

    TSny

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    Homework Helper
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    Due to the placement of the indices, this does not express the matrix multiplication ##\Lambda \Lambda F##.
     
  6. May 5, 2016 #5
    Alright, then!

    ##F_{\mu'\nu'}={\Lambda^{\mu}}_{\mu'}{\Lambda^{\nu}}_{\nu'}F_{\mu\nu}={\Lambda^{\mu}}_{\mu'}F_{\mu\nu}{\Lambda^{\nu}}_{\nu'}={(\Lambda^{-1})_{\mu'}}^{\mu}F_{\mu\nu}{\Lambda^{\nu}}_{\nu'}##.

    But, ##{(\Lambda^{-1})_{\mu'}}^{\mu}={\Lambda^{\mu}}_{\mu'}## is the same matrix as ##{\Lambda^{\nu}}_{\nu'}##, which is given by ##\Lambda^{-1}=
    \left( \begin{array}{cccc}
    1 & 0 & 0 & 0 \\
    0 & \text{cos}\ \theta & 0 & \text{sin}\ \theta \\
    0 & 0 & 1 & 0 \\
    0 & -\text{sin}\ \theta & 0 & \text{cos}\ \theta \end{array} \right)##.

    So, ##F_{\mu'\nu'}={\Lambda^{\mu}}_{\mu'}{\Lambda^{\nu}}_{\nu'}F_{\mu\nu}## becomes ##\Lambda^{-1}F\Lambda^{-1}##?
     
  7. May 5, 2016 #6

    TSny

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    yes [edit: no, I believe one of the ##\Lambda##'s should not be an inverse.]
     
    Last edited: May 5, 2016
  8. May 5, 2016 #7
    Thanks!

    On a different note, I've confused myself now over the correctness of the tensor transformation law

    ##F_{\mu'\nu'}=\frac{\partial x^{\mu}}{\partial x^{\mu'}}\frac{\partial x^{\nu}}{\partial x^{\nu'}}F_{\mu\nu}={{(\Lambda^{-1})}_{\mu'}}^{\mu}{{(\Lambda^{-1})}_{\nu'}}^{\nu}F_{\mu\nu}={\Lambda^{\mu}}_{\mu'}{\Lambda^{\nu}}_{\nu'}F_{\mu\nu}##.


    Consider the following:

    ##x^{\mu'}={\Lambda^{\mu'}}_{\nu}x^{\nu} \implies {(\Lambda^{-1})^{\rho}}_{\mu'}x^{\mu'}={(\Lambda^{-1})^{\rho}}_{\mu'}{\Lambda^{\mu'}}_{\nu}x^{\nu} \implies x^{\rho}={(\Lambda^{-1})^{\rho}}_{\mu'}x^{\mu'}##

    so that

    ##\frac{\partial x^{\mu}}{\partial x^{\mu'}}={(\Lambda^{-1})^{\mu}}_{\mu'}##.

    This contradicts with the order of the indices in my tensor transformation law. Where have I gone wrong exactly?
     
  9. May 5, 2016 #8
    [tex]{(\Lambda^{-1})^{\rho}}_{\mu'}x^{\mu'} \neq x^{\rho}[/tex]
    In fact, I would dispense with the use of [itex]\Lambda^{-1}[/itex] altogether - the inverse can be succinctly obtained just by swapping the upper and lower (primed and unprimed) indices.
     
  10. May 6, 2016 #9
    Since ##{(\Lambda)^{\rho}}_{\mu'}## is the Lorentz transformation from the unprimed to the primed coordinates, does ##{(\Lambda^{-1})^{\rho}}_{\mu'}## represent the Lorentz transformation from the primed to the unprimed coordinates?


    Also, how may I derive the relation ##\frac{\partial x^{\mu}}{\partial x^{\mu'}}={(\Lambda^{-1})_{\mu'}}^{\mu}## and then show that ##{(\Lambda^{-1})_{\mu'}}^{\mu}={(\Lambda)^{\mu}}_{\mu'}##?
     
  11. May 6, 2016 #10
    This is very strange notation - why should they be equal? Usually when I see people explicitly using ##\Lambda^{-1}##, it is because they do not have primes on the indices, and hence they need the explicit inverse written there to indicate the 'direction' of the transform. When your indices are already primed, there is no longer any ambiguity, so the usage of ##\Lambda^{-1}## doesn't make much sense.
     
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