Let's say I have a spatially constant but time-varying electric field in vacuum of the form $$\textbf{E} = E_0 e^{i \, \omega t} \, \hat{x}$$ that describes some E-M wave. What is the magnetic vector potential $\textbf{A}$ associated with this field?

In my attempt, I used the relation $\nabla \times \textbf{A} = \textbf{B}$ along with the Maxwell equation $$\nabla \times \textbf{B} = \mu_0 \epsilon_0 \frac{\partial \textbf{E}}{\partial t}$$ and gauge freedom $\nabla \cdot \textbf{A} = 0$ to get $$\nabla^2 \textbf{A} = -\mu_0 \epsilon_0 E_0 i \omega e^{i \, \omega t} \, \hat{x}. $$ Since the right-hand-side is spatially constant and solely in the x-direction, I simply integrated the x-component of $\textbf{A}$ twice and set the y and the z components to 0 in order to get

$$\textbf{A} = -\frac{1}{2}\mu_0 \epsilon_0 E_0 i \omega e^{i \, \omega t} \, x^2 \, \hat{x}.$$

Is this correct? If not, where am I going wrong? If the context helps, I am trying to obtain the vector potential of an E.M. wave so I can determine the time-dependent perturbation to a time-independent Hamiltonian.

  • $\begingroup$ Is this correct? If not, where am I going wrong? Check-my-work questions are off-topic here. $\endgroup$
    – G. Smith
    Nov 1 '20 at 6:54
  • $\begingroup$ Is the divergence of this vector potential zero? ;) $\endgroup$
    – Philip
    Nov 1 '20 at 7:07
  • 1
    $\begingroup$ Are you sure the $\vec{E}$ has no spatial variation? Faraday's law would then imply $\vec{B}$ has no time variation since $\vec{E}$ is curl-free. $\endgroup$
    – Puk
    Nov 1 '20 at 8:05

If you write the Electric and Magnetic fields in terms of the scalar and vector potentials $\phi$ and $\mathbf{A}$, you should be able to show that

\begin{align} \mathbf{B} &= \nabla \times \mathbf{A}, \\ \mathbf{E} &= -\nabla\phi - \frac{\partial \mathbf{A}}{\partial t}.\\ \end{align}

Since your field is constant all over space at any given instant of time, it can be described using a constant $\phi$, and therefore the second equation just gives you $$\mathbf{E} = -\frac{\partial\mathbf{A}}{\partial t}.$$

The rest should be straightforward.


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