# Representing electromagnetic tensor as the curl of four vector potential and Lorentz gauge condition

[Question is from electrodynamics section of 2nd chapter in Weinberg's Cosmology and gravitation text book]

This section first introduces electromagnetic tensor and reduces Maxwell's four equations into just two. $$\partial_{\alpha} F^{\alpha\beta} = -J^\beta \tag{2.7.6}$$

$$\epsilon^{\alpha\beta\gamma\delta} \partial_{\beta} F_{\gamma\delta} = 0 \tag{2.7.7}$$ Where F is the electromagnetic tensor, J is the current density four vector and $$\epsilon$$ is the levi civita symbol.

Later in this section it says that equation (2.7.7) allows us to write $$F_{\gamma\delta}$$ as the curl of a four vector A.

$$F_{\gamma\delta} = \partial_{\gamma}A_{\delta} - \partial_{\delta}A_{\gamma} \tag{2.7.11}$$

How exactly is equation (2.7.7) allowing us to write F as curl of A ? (2.7.11) satisfies equation (2.7.7) but does that alone make it a valid assumption?

Then the textbook reads, "we can change $$A_\gamma$$ by a term $$\partial_{\gamma} \phi$$ without affecting $$F_{\gamma\delta}$$. So $$A_\gamma$$ may be defined so that $$\partial^\alpha A_\alpha =0 \tag{2.7.12}$$ Here changing $$A_\gamma$$ by a term $$\partial_{\gamma} \phi$$ means adding this term, right?

And how are we defining $$A_\gamma$$ as $$\partial^\alpha A_\alpha =0$$ ? [ I understand this is called gauge condition]

1. For your first question, you're right that $$(2.7.11)$$ satisfies $$(2.7.7)$$. The converse holds locally (on contractible domains such as open balls, or ellipsoids or the whole of $$\Bbb{R}^4$$ for example), and this is known as Poincare's lemma. In terms of differential forms, $$F$$ is a $$2$$-form and $$(2.7.11)$$ says "$$F$$ is exact" while $$(2.7.7)$$ says "$$F$$ is closed", and Poincare's lemma says "all closed forms are locally exact". Note that this is the higher dimensional analogue of facts you're probably well aware of from vector calculus: that "vanishing divergence implies locally there exists a vector potential" ($$\nabla \cdot \mathbf{B}=0$$ implies $$B=\nabla \times \mathbf{A}$$ locally), and that "vanishing curl implies locally is a gradient" (i.e $$\nabla \times \mathbf{E}=0$$ implies $$\mathbf{E}=\nabla \phi$$ locally). For the special case that we work in a star-shaped open neighborhood of the origin in $$\Bbb{R}^n$$ (yes this works for $$\Bbb{R}^n$$ even for $$n\neq 4$$), by explicitly carrying out the proof of Poincare's lemma we have \begin{align} A_{\beta}(x)&=\int_0^1tF_{\alpha\beta}(tx)\,dt\cdot x^{\alpha}. \end{align} I leave it to you to carry out the explicit differentiation under the integral sign etc to show that $$(2.7.11)$$ follows from the assumption $$(2.7.7)$$.
2. "Here changing $$A_{\gamma}$$ by a term $$\partial_{\gamma}\phi$$ means adding this term right?" Yes. In terms of differential forms, this is also succinctly stated as "if $$F=dA$$ and $$\phi$$ is a smooth function then $$d(A+d\phi)=dA+d(d\phi)=dA+0=F$$".
3. One basically has to choose $$\phi$$ to solve an appropriate non-homogeneous wave equation. Now, I'm not too sure about PDE theory but I would guess that under nice conditions, simple constant-coefficient PDEs (such as the wave equation) are guaranteed to have local solutions. So, for simplicity, let me suppose we're working in $$\Bbb{R}^4$$ with the flat Lorentzian metric, say signature $$(-,+,+,+)$$ (I set speed of light $$c=1$$ for convenience). Suppose we have found $$A$$ such that $$F=dA$$ (i.e $$2.7.11$$ holds). Then consider $$A'=A+d\phi$$. Then the divergence-free condition $$(2.7.12)$$ holds for $$A'$$ if and only if \begin{align} \partial^{\alpha}(A_{\alpha}+\partial_{\alpha}\phi)&=\partial^{\alpha}A_{\alpha}+(\partial^{\alpha}\partial_{\alpha}\phi) =0, \end{align} i.e if and only if \begin{align} \partial^{\alpha}\partial_{\alpha}\phi&=-(\partial^{\alpha}A_{\alpha}) \end{align} Note that we are given $$A$$ so we know the RHS; let us call it $$-f$$. This is a wave-equation for the unknown function $$\phi$$: \begin{align} -\frac{\partial^2\phi}{\partial t^2}+\frac{\partial^2\phi}{\partial x^2}+\frac{\partial^2\phi}{\partial y^2}+\frac{\partial^2\phi}{\partial z^2}&=-f \end{align} (We can also write it as $$\Delta_g\phi=-f$$, i.e the "Laplacian with respect to the metric tensor $$g$$" equal to $$-f$$). If we manage to solve this equation, then we can find the desired $$A'$$ which satisfies the divergence-free condition $$(2.7.12)$$.