I am working on a homework problem that says an electron in a continuous laser field can be modeled as experiencing a homogeneous oscillating electric field $\vec{E}(\vec{r},t)=\cos \omega t \ \hat {z}$.

It then proceeds to say that a suitable gauge choice is magnetic potential $\vec{A}=0$ and electric potential $\phi(\vec r,t) = -\cos \omega t \ \hat {z} \cdot \vec r$.

This means that the magnetic field $\vec B=0$.

But Maxwell's equation $$\mathbf{\nabla \times B} = \mu_0 \mathbf{j} + \frac{1}{c^2}\frac{\partial \mathbf{E}}{\partial t}$$ says that a changing electric field will produce a nonzero curl of the magnetic field so $\vec B$ cannot be zero.

What am I getting wrong here?

  • $\begingroup$ Why would B = 0? $\endgroup$ – ggcg May 8 '19 at 15:34
  • 1
    $\begingroup$ You are correct. You have just proven that the pre-supposed approach $\phi(\vec{r},t) = -\cos(\omega t)\hat{\vec{z}}\cdot\vec{r}$ and $\vec{A}(\vec{r},t)=\vec{0}$ is wrong. $\endgroup$ – Thomas Fritsch May 8 '19 at 16:10
  • 1
    $\begingroup$ @ggcg I guess it is because $\vec{A}$ is zero and $B = \nabla \times A$. $\endgroup$ – Abhay Hegde May 8 '19 at 16:24

Basically, what you're saying is that a homogeneous electric field $\vec{E}(\vec{r},t)=\cos \omega t \ \hat {z}$ in vacuum requires you to have a magnetic field that satisfies $$ \nabla \times \mathbf{B} =\frac{1}{c^2}\frac{\partial \mathbf{E}}{\partial t}, $$ giving you a non-homogeneous magnetic field, which would then require the electric field to satisfy $$ \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}, $$ that therefore needs an inhomogeneous electric field, in conflict with your initial assumption? Or, in other words, you're saying that the electric field assumed by the question, $\vec{E}(\vec{r},t)=\cos \omega t \ \hat {z}$, is inconsistent with the Maxwell equations?

If that is what you're saying, then: good! you're right. That electric field is not, strictly speaking, compatible with the Maxwell equations.

It is, however, an excellent and interesting approximation to useful solutions under suitable approximations (specifically, the dipole approximation). And, under that approximation, you neglect both the magnetic field and the conflict that this creates with the Ampère-Maxwell law. This obviously means that the fields are not precisely correct, and that there is the possibility of non-dipole effects (which come in a hierarchy, starting with magnetic dipole effects and then through electric quadrupole field dependences, and then upwards), but if the system is localized enough then those effects can be completely negligible.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.