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?
$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}$
$F$ 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?
$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}$
$F$ 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

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

with inverse

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

The thing about the field tensor is that it is actually a closed exact 2-form field. If we define a one-form field $$A=(-\phi,\mathbf{A})=A_adx^a$$, 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!