# Maxwell Equations in Tensor notation

1. Feb 19, 2009

### Karliski

http://en.wikipedia.org/wiki/Formul...s_in_special_relativity#Maxwell.27s_equations

Why does
$$0 = \epsilon^{\alpha \beta \gamma \delta} \frac{\partial F_{\alpha \beta}}{\partial x^\gamma}$$
reduce to
$$0 = {\partial F_{\alpha\beta}\over\partial x^\gamma} + {\partial F_{\beta\gamma}\over\partial x^\alpha} + {\partial F_{\gamma\alpha}\over\partial x^\beta}$$
?

2. Feb 19, 2009

$$0 = \epsilon^{\alpha \beta \gamma 0} \frac{\partial F_{\alpha \beta}}{\partial x^\gamma} = {\partial F_{\alpha\beta}\over\partial x^\gamma} + {\partial F_{\beta\gamma}\over\partial x^\alpha} + {\partial F_{\gamma\alpha}\over\partial x^\beta} - {\partial F_{\beta\alpha}\over\partial x^\gamma} - {\partial F_{\gamma\beta}\over\partial x^\alpha} - {\partial F_{\alpha\gamma}\over\partial x^\beta} = 2 \left( {\partial F_{\alpha\beta}\over\partial x^\gamma} + {\partial F_{\beta\gamma}\over\partial x^\alpha} + {\partial F_{\gamma\alpha}\over\partial x^\beta} \right)$$
because $$F_{\alpha\beta}[/itex] is antisymmetric. 3. Feb 21, 2009 ### Karliski Why was delta set to 0? 4. Feb 21, 2009 ### HallsofIvy What "delta" are you talking about? There was no "delta" in your original question nor in Adriank's response. 5. Feb 21, 2009 ### confinement Actually, delta is the fourth index on the anti-symmetric epsilon tensor in both Karliski's post and in the wikipedia equation he linked to. It is basically an error, we should have delta = 0 as in adriank's post, but if delta is not zero then we just get four copies of the same equation. 6. Feb 21, 2009 ### Fredrik Staff Emeritus The "equation" in #1 is actually four equations, one for each value of $\delta$. The one with $\delta=0$ is the scalar equation [tex]\nabla\cdot\vec B=0$$

The one with $\delta=i\neq 0$ is the ith component of the vector equation

$$\nabla\times\vec E+\frac{\partial\vec B}{\partial t}=0$$

(Edit: ...except for a factor of two).

Last edited: Feb 21, 2009
7. Feb 21, 2009

$$0 = \epsilon^{\alpha \beta \gamma \delta} \frac{\partial F_{\alpha \beta}}{\partial x^\gamma}$$
is written with a $$\delta$$ only on one side, which means we can "plug in" any specific value for it. So I put $$\delta = 0$$ and explicitly wrote out the sum $$\epsilon_{\alpha\beta\gamma0}F_{\alpha\beta,\gamma}$$.