In have a question about the following exercise. I am given the following expressions for the scalar and vector potential, $$\vec{A}(\vec{r},t) = \frac{\sigma}{\epsilon_0}\left[\frac{x}{c}\hat{x} + tU(z)\hat{z}\right],$$ $$V(\vec{r},t) = 0$$

Here $U(z)$ is the unit step function. The question is to derive a gauge transformation $\lambda$ such that the Coulomb Gauge is satisfied ($\nabla\cdot\vec{A} = 0$). I denote the new potential as $\vec{A'}$ and write, $$\vec{A'} = \vec{A} + \nabla \lambda$$, $$\nabla \cdot\vec{A'} = \nabla \cdot \vec{A} + \nabla^2 \lambda.$$ So then, $$\nabla^2\lambda = -\frac{\sigma}{\epsilon_0c} - \frac{\sigma t}{\epsilon_0} \delta(z).$$ So to find a suitable gauge I need to integrate this equation. The second term however gives me some trouble. I would assume that integrating a delta function twice gives you a ramp function, so I would find for $\lambda$, $$\lambda = -\frac{\sigma}{2\epsilon_0c} z^2 - \frac{\sigma t}{\epsilon_0} z U(z). $$ $zU(z)$ is then the analytical representation of the ramp function. The answer sheet however states the following, $$\lambda = -\frac{\sigma}{2\epsilon_0c} z^2 - \frac{2\sigma t}{\epsilon_0} |z|. $$ This looks very different from my answer, especially for $z<0$ where my second term vanishes while theirs grows linearly. If anyone can point out to me where I am going wrong it would be very much appreciated!


1 Answer 1


First of all, I assume that in the second term in the "right" answer 2 should be in denominator; if this is the case, it actually satisfies the equation necessary (as $\frac{\partial^2}{\partial z^2} |z| = 2 \delta(z)$ in terms of distributions)

Both your answer and the other one are actually OK; different gauge transformations can lead to Couloumb gauge. This is because the solution to $∇^2λ=−σϵ_0c−σtϵ_0δ(z).$ is not unique; any function satisfying the homogeneous equation ($\nabla^2 \lambda = 0$) can be added to a given solution, and the result would be a solution as well. Difference between your answer and the "right" one is a function linear in $z$ (provided that 2 is in the denominator, as I mentioned earlier), thus, it satisfies the homogeneous equation.


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