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Electric Field Strength Comservation in Direction of Motion

  1. Aug 29, 2013 #1
    (Sorry for the typo in the title. I don't know how to get rid of it)

    Recently I was looking at the Lorentz co-variance of F=dA (the Faraday 2-form) and *F=*dA (the Maxwell 2-form). I realized that Lorentz co-variance implies that the electric field strength and magnetic field strength are invariant in the direction of motion. In Einstein's notation in Zur Elektrodynamik bewegter Körper (Engl trans.:ON THE ELECTRODYNAMICS OF MOVING BODIES), X=X' and L=L'. It struck me as interesting that he had to assume the invariance in order to prove the co-variance of Maxwell's equations:

    Maxwell's eqns:
    img78.gif

    transformed:

    img79.gif

    and now in the new variables:
    img86.gif

    Notice how he assumes X=X' and L=L'. This was no small assumption since it was at the time something Poincare was wanting to prove. The proof is, it seems, because it is necessary to establish co-variance, an established fact by then - otherwise you could measure the absolute velocity of the Earth through the luminiferous aether.

    Further along he states explicitly:

    img93.gif
     
  2. jcsd
  3. Aug 29, 2013 #2

    vanhees71

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    It's simply the transformation law for a 2nd-rank tensor,
    [tex]F'^{\mu \nu}={\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma} F^{\rho \sigma}.[/tex]
    Plugging in the components in 1+3-dimensional notation, i.e., for [itex]\mu \in \{1,2,3 \}[/itex]
    [tex]F^{0 \mu}=\partial^{0} A^{\mu}-\partial^{\mu} A^0=\partial_0 A^{\mu} + \partial_{\mu} A^0=\dot{A}^{\mu}+\partial_{\mu} \Phi=-E^{\mu},[/tex]
    and for [itex]\mu,\nu \in \{1,2,3 \}[/itex]
    [tex]F^{\mu \nu}=\partial^{\mu} A^{\nu} - \partial^{\nu} A^{\mu}=-\partial_{\mu} A^{\nu} + \partial_{\nu} A^{\mu}=\epsilon^{\nu \mu \rho} B^{\rho},[/tex]
    you find the transformation law in terms of [itex](\vec{E},\vec{B})[/itex] for a boost,
    [tex]
    \begin{split}
    \vec{E}'&=\gamma(\vec{E}+\vec{\beta} \times \vec{B})-\frac{\gamma^2}{1+\gamma} \vec{\beta}(\vec{\beta} \cdot \vec{E}),\\
    \vec{B}'&=\gamma(\vec{B}-\vec{\beta} \times \vec{E})-\frac{\gamma^2}{1+\gamma} \vec{\beta}(\vec{\beta} \cdot \vec{B}),
    \end{split}
    [/tex]
    where [itex]\vec{\beta}=\vec{v}/c[/itex] and [itex]\gamma=1/\sqrt{1-\vec{\beta}^2}[/itex]. As you see, the components in boost direction are indeed unchanged.

    Projecting to the parts in direction of [itex]\vec{\beta}[/itex] and perpendicular to it gives
    [tex]
    \begin{split}
    \vec{E}_{\parallel}'&=\vec{E}_{\parallel},\\
    \vec{B}_{\parallel}'&=\vec{B}_{\parallel},\\
    \vec{E}_{\perp}'&=\gamma(\vec{E}_{\perp}+\vec{\beta} \times \vec{B}),\\
    \vec{B}_{\perp}'&=\gamma(\vec{B}_{\perp}-\vec{\beta} \times \vec{E}).
    \end{split}
    [/tex]
     
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