Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How would you prove that?

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



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


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


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


    [itex]
    G^{\mu \nu} G_{\mu \nu} = F^{\mu \nu} F_{\mu \nu}
    [/itex]



    [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

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    [tex]
    F_{ab}=\left(\begin{array}{cccc}
    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)
    [/tex]

    with inverse

    [tex]
    F^{ab}=\left(\begin{array}{cccc}
    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)
    [/tex]

    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:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?