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How would you prove that?

  1. Jun 15, 2006 #1
    Hi, I'm wondering how to prove the following...can you help me? :redface:

    F^{\mu \rho} G_{\rho \nu} = \eta^\mu_{\phantom{\mu}\nu} \mathbf{E} \cdot \mathbf{B}

    F^{\mu \nu} F_{\mu \nu} = -2\left(\mathbf{E}^2-\mathbf{B}^2\right)

    G^{\mu \nu} F_{\mu \nu} = -4\,\mathbf{E} \cdot \mathbf{B}

    G^{\mu \nu} G_{\mu \nu} = F^{\mu \nu} F_{\mu \nu}

    [itex]F[/itex] is the electromagnetic tensor, [itex]G[/itex] is it's dual, [itex]\eta[/itex] is the metric tensor, [itex]\mathbf{E}[/itex] and [itex]\mathbf{B}[/itex] the electric and magnetic field respectively.

    Thank you for your patience! :wink:
    Last edited: Jun 15, 2006
  2. jcsd
  3. Jun 15, 2006 #2


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    You need to specify the definition of [tex]\vec E[/tex] and [tex]\vec B[/tex] in terms of your field tensors.

    Up to sign, signature, and index-placement conventions, one can write
    [tex]E_b=u^aF_{ab}[/tex] and [tex]B_b=u^aG_{ab}[/tex] for an observer with 4-velocity [tex]u^a[/tex]. By contracting with [tex]u^b[/tex], you can verify that these vectors are orthogonal to, i.e. "spatial according to" the observer with 4-velocity [tex]u^b[/tex]. You will probably need the spatial metric as well... up to conventions, write [tex]g_{ab}=u_au_b+h_{ab}[/tex] with the condition that [tex]u^a h_{ab}=0_b[/tex].
    Last edited: Jun 15, 2006
  4. Jun 15, 2006 #3
    It's quite straightforward. If we assume a Minkowski metric [tex]\eta_{ab}[/tex] of signature [tex](-+++)[/tex] then the components of [tex]F_{ab}[/tex] can be expressed simply as

    0 & -E_{x} & -E_{y} & -E_{z}\\
    E_{x} & 0 & B_{z} & -B_{y}\\
    E_{y} & -B_{z} & 0 & B_{x}\\
    E_{z} & B_{y} & B_{x} & 0\end{array}\right)

    with inverse

    0 & E_{x} & E_{y} & E_{z}\\
    -E_{x} & 0 & B_{z} & -B_{y}\\
    -E_{y} & -B_{z} & 0 & B_{x}\\
    -E_{z} & B_{y} & -B_{x} & 0\end{array}\right)

    The thing about the field tensor is that it is actually a closed exact 2-form field. If we define a one-form field [tex]A=(-\phi,\mathbf{A})=A_adx^a[/tex], then the electromagnetic two-form is defined as [tex]F=dA[/tex] (you can show that the components of [tex]F[/tex] are exactly those given above. Now define the Hodge dual through its action on basis forms as

    [tex]\star\wedge_{i=1}^{p}\omega^{a_{i}}=\frac{\sqrt{|g|}}{(m-p)!}\epsilon_{\phantom{a_{1}\ldots a_{p}}b_{1}\ldots b_{m-p}}^{a_{1}\ldots a_{p}}\wedge_{i=1}^{m-p}\omega^{b_{i}}.[/tex]

    This immediately gives you an expression for the components of [tex]\star F[/tex], or what you call [tex]G[/tex]. You can then find the required results simply by direct calculation.
    Last edited: Jun 15, 2006
  5. Jun 16, 2006 #4
    Thank you everyone! :approve:
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