# Maxwell Equations in Tensor notation

#### Karliski

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

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).

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The equation
$$0 = \epsilon^{\alpha \beta \gamma \delta} \frac{\partial F_{\alpha \beta}}{\partial x^\gamma}$$
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 $$\epsilon_{\alpha\beta\gamma0}F_{\alpha\beta,\gamma}$$.

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