We define $$x^\alpha=(ct,x,y,z)$$ $$\partial_\alpha=\frac\partial {\partial x^\alpha}=\biggl(\frac 1 {c^2} \frac\partial {\partial t} ,\nabla \biggr)$$ $$F_{\alpha \beta}=\begin{pmatrix} 0 & -\frac{E_x} c & -\frac{E_y} c & -\frac{E_z} c \\ \frac{E_x} c& 0 & B_z & -B_y\\ \frac{E_y} c & -B_z & 0 & B_x \\ \frac{E_z} c & B_y & -B_x & 0 \end{pmatrix}$$ Now, I read on my notes that Maxwell's equation can be expressed as it follows: $$\partial_\alpha F^{\alpha \beta} = -\mu_0J^\beta$$ $$\partial_\alpha F_{\beta\gamma} + \partial_\gamma F_{\alpha\beta}+\partial_\beta F_{\alpha\gamma},$$ where $F^{\alpha \beta}$ is the transpose matrix of $F_{\alpha\beta}$, and $J^\beta=(\rho c,J_x, J_y,J_z)$.

Basically I don't understand the way that that $\partial_\alpha$ acts on $F_{\alpha\beta}$, and so I don't see where is the connection beetwen these equations and Maxwell's ones. Can you explain to me? Thank you in advance

  • $\begingroup$ In $\partial_\alpha F^{\alpha\beta}$ there is an implicit summation over the “contracted” index $\alpha$. You sum over the four possible values for an index, namely 0,1,2,3. $\endgroup$ – G. Smith Oct 4 '19 at 18:58
  • $\begingroup$ In your expression for $\partial_\alpha$, it should be $c$, not $c^2$. $\endgroup$ – G. Smith Oct 4 '19 at 18:59
  • $\begingroup$ In $\partial_\alpha F^{\alpha \beta} = -\mu_0J^\beta$, write out the four equations for $\beta=0,1,2,3$, summing over $\alpha=0,1,2,3$ in each one. $\endgroup$ – G. Smith Oct 4 '19 at 19:01
  • $\begingroup$ Your second Maxwell equation isn’t an equation. You forgot the $=0$. $\endgroup$ – G. Smith Oct 4 '19 at 19:03

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