The electromagnetic tensor is given as:

\begin{equation} F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu \end{equation}

How do you derive this? And how come there is a partial derivative in front of $A_\mu$? Do you multiply the derivatives or what?


1 Answer 1


I will assume that you know that we can write the magnetic and electric field as:\begin{equation} \mathbf{B}= \mathbf{\nabla} \times \mathbf{A} \end{equation} \begin{equation} \mathbf{E}=-\mathbf{\nabla}\Phi-\frac{\partial \mathbf{A}}{\partial t} \end{equation} where $c=1$ and I am using Heaviside-Lorentz conventions (i.e. "lazy physics" conventions). The important properties the fields have is that they are invariant under a $\mathrm{U(1)}$ gauge transformation of the form: \begin{equation} \mathbf{A}'=\mathbf{A}-\mathbf{\nabla} \alpha = \mathbf{A} + i e^{-i \alpha} \mathbf{\nabla} e^{i \alpha} \end{equation} \begin{equation} \Phi'=\Phi+\frac{\partial \alpha}{\partial t} = \Phi - i e^{-i \alpha} \frac{\partial}{\partial t} e^{i \alpha} \end{equation} We now define the define the four-vector potential, $A^\mu$, and the four-vector electric current, $j^\mu$, as follows (and I am using Minkowski signature $(+,-,-,-)$): \begin{equation} \begin{array}{cc} \{A^\mu\} = \begin{pmatrix} \Phi \\ \mathbf{A} \end{pmatrix} \; ,& \{j^\mu\} =\begin{pmatrix} \rho_e \\ \mathbf{J}_e \end{pmatrix} \end{array} \end{equation} and we define (i.e. we do not derive) the antisymmetric electromagnetic field strength tensor as: \begin{equation} F^{\mu \nu} = \partial^\mu A^\nu - \partial^\nu A^\mu \end{equation} Now, we can evaluate: \begin{equation} \begin{aligned} F^{0 i } & = \partial^0 A^i - \partial^i A^0 \\& = \partial_0 A^i + \partial_i A^0 \\& = - E^i \end{aligned} \end{equation} and: \begin{equation} \begin{aligned} F^{ij} & = \partial^i A^j - \partial^j A^i \\& = (\delta^i{}_l \delta^j{}_m - \delta^j{}_l \delta^i{}_m) \partial^l A^m \\& = \varepsilon_{klm}\varepsilon^{kij}\partial^l A^m \\& = -\varepsilon_{kij} [\mathbf{\nabla} \times \mathbf{A}]^k \\& = -\varepsilon_{ijk} B^k \end{aligned} \end{equation} Therefore, we can write: \begin{equation} \left\{ F^{\mu \nu} \right\} = \begin{pmatrix} 0 & -E^1 & -E^2 & -E^3 \\ E^1 & 0 & -B^3 & B^2 \\ E^2 & B^3 & 0 & -B^1 \\ E^3 & -B^2 & B^1 & 0 \end{pmatrix} \end{equation} Note that the observable electromagnetic field strength tensor is invariant under a $\mathrm{U(1)}$ gauge transformation: \begin{equation} A'^\mu=A^\mu + \partial^\mu \alpha \end{equation} as is required.

Furthermore, we can define the dual of the electromagnetic field tensor, ${}^*F^{\mu \nu}$, as: \begin{equation} {}^*F_{\mu \nu} = \frac{1}{2} \varepsilon_{\mu \nu \sigma \tau} F^{\sigma \tau} \end{equation} such that the Maxwell equations can be written in a more compact and manifestly Lorentz invariant manner: \begin{equation} \begin{array}{cc} \partial_\mu F^{\mu \nu } = j^\nu \; , & \partial_\mu {}^* F^{\mu \nu} =0 \end{array} \end{equation} To see that these last two equations indeed satisfy the Maxwell equations is a bit of a tedious (but worthwhile) exercise that I will leave up to you.


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