How do you prove the relationship between electromagnetic and metric tensors?

[Emanuel84]

Hi, I'm wondering how to prove the following...can you help me?
<br /> F^{\mu \rho} G_{\rho \nu} = \eta^\mu_{\phantom{\mu}\nu} \mathbf{E} \cdot \mathbf{B}<br /><br /> F^{\mu \nu} F_{\mu \nu} = -2\left(\mathbf{E}^2-\mathbf{B}^2\right)<br /><br /> G^{\mu \nu} F_{\mu \nu} = -4\,\mathbf{E} \cdot \mathbf{B}<br /><br /> G^{\mu \nu} G_{\mu \nu} = F^{\mu \nu} F_{\mu \nu}<br />
F is the electromagnetic tensor, G is it's dual, \eta is the metric tensor, \mathbf{E} and \mathbf{B} the electric and magnetic field respectively.Thank you for your patience!

 

[robphy]

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

Up to sign, signature, and index-placement conventions, one can write
E_b=u^aF_{ab} and B_b=u^aG_{ab} for an observer with 4-velocity u^a. By contracting with u^b, you can verify that these vectors are orthogonal to, i.e. "spatial according to" the observer with 4-velocity u^b. You will probably need the spatial metric as well... up to conventions, write g_{ab}=u_au_b+h_{ab} with the condition that u^a h_{ab}=0_b.

 

[coalquay404]

Hi, I'm wondering how to prove the following...can you help me?
<br /> F^{\mu \rho} G_{\rho \nu} = \eta^\mu_{\phantom{\mu}\nu} \mathbf{E} \cdot \mathbf{B}<br /><br /> F^{\mu \nu} F_{\mu \nu} = -2\left(\mathbf{E}^2-\mathbf{B}^2\right)<br /><br /> G^{\mu \nu} F_{\mu \nu} = -4\,\mathbf{E} \cdot \mathbf{B}<br /><br /> G^{\mu \nu} G_{\mu \nu} = F^{\mu \nu} F_{\mu \nu}<br />
F is the electromagnetic tensor, G is it's dual, \eta is the metric tensor, \mathbf{E} and \mathbf{B} the electric and magnetic field respectively.Thank you for your patience!

It's quite straightforward. If we assume a Minkowski metric \eta_{ab} of signature (-+++) then the components of F_{ab} can be expressed simply as

<br /> F_{ab}=\left(\begin{array}{cccc}<br /> 0 &amp; -E_{x} &amp; -E_{y} &amp; -E_{z}\\<br /> E_{x} &amp; 0 &amp; B_{z} &amp; -B_{y}\\<br /> E_{y} &amp; -B_{z} &amp; 0 &amp; B_{x}\\<br /> E_{z} &amp; B_{y} &amp; B_{x} &amp; 0\end{array}\right)<br />

with inverse

<br /> F^{ab}=\left(\begin{array}{cccc}<br /> 0 &amp; E_{x} &amp; E_{y} &amp; E_{z}\\<br /> -E_{x} &amp; 0 &amp; B_{z} &amp; -B_{y}\\<br /> -E_{y} &amp; -B_{z} &amp; 0 &amp; B_{x}\\<br /> -E_{z} &amp; B_{y} &amp; -B_{x} &amp; 0\end{array}\right)<br />

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

\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}}.

This immediately gives you an expression for the components of \star F, or what you call G. You can then find the required results simply by direct calculation.

 

[Emanuel84]

Thank you everyone!