In our physics lecture, we did the following example of constructing potentials $\vec A$ and $V$ that supposedly satisfy both the coulomb ($\nabla \vec A=0)$ and the lorenz condition $(\nabla \vec A+ \frac{1}{c^2} \dot V =0)$. Let $\vec E = \vec E_{0} \sin(\vec k \vec r-\omega t), \vec B = \frac{1}{c|\vec k|}\vec k \times \vec E $. Furthermore, we are in vacuum $(\rho = 0 = \vec j )$. The "ansatz" is the following: Let $\vec A = \vec \alpha \cos(\vec k \vec r- \omega t)$. After that I fail to make sense of my notes... it says something like it follows that $\vec \alpha = \frac{\vec E_0}{c|k|} $. Okay so I took the curl of $\vec A$ with that $\vec \alpha$ and using (one of the two) product rules for cross products I get the $\vec B$. But when I take the divergence of $\vec A$ I don't get $0$. We did this example at the very end of lecture so I can't quite make sense of it... It seems to hinge upon the equation $\vec k \vec E_0 =0$, implying that $\vec k, \vec E_0$ are orthogonal... But why? And lastly we conclude $V =0$. I see that it helps us satsify the Lorenz condition, but why do we have the freedom to choose $V =0$? I know we don't have that freedom for $\vec A$.


1 Answer 1


The desired results can all be achieved by consequent applications of the rules of use of curl, divergence and gradient and moreover the knowledge that the scalar potential is zero for an electromagnetic wave.

Normally in a constellation with charges we have the poisson equation for the scalar potential $$\Delta V = 4\pi \rho$$ However, in the given constellation $\rho=0$. $V$ could be still non-zero, if the given boundary conditions at the border of the considered area $\Omega$ are non-zero $$V|_{\partial\Omega}\neq 0$$ This happens typically in electrostatical constellations, but not when an EM-wave is considered. So $$V|_{\partial\Omega} = 0 \,\, \rightarrow V=0$$

Then we can calculate the electrical field from the vector potential by

$$ \vec{E} = -\frac{1}{c}\frac{\partial \vec{A}}{\partial t} = -\vec{\alpha} \omega \sin(\vec{k}\cdot \vec{r}-\omega t) \equiv -\vec{E_0}\sin(\vec{k}\cdot \vec{r}-\omega t) $$

with $\vec{\alpha} = \frac{\vec{E_0}}{c|\vec{k}|}$ (Remember $\omega = c |\vec{k}|$).

In the result of $\vec{E}$ an unexpected minus sign appears, which can be removed by changing the definition of $\vec{\alpha}$.

Now we come to the computation of the B-field: We'll use the following rule for the curl: $$ \nabla\times (\phi \vec{F}) =\nabla \phi \times \vec{F} +\phi \nabla\times \vec{F} $$ In our case $\vec{F} = \vec{\alpha} = const. $. Therefore the 2. term in the rule for the use of the curl-operator vanishes.

$$\vec{B}=\nabla\times \vec{A} = \nabla cos(\vec{k}\cdot \vec{r}-\omega t) \times \alpha = - \vec{k}\times \frac{E_0}{c|\vec{k}|} \sin(\vec{k}\cdot \vec{r}-\omega t) = \frac{1}{c|\vec{k}|} \vec{k}\times \vec{E}$$

Finally evaluating the divergence of $\vec{A}$ we use: $$\nabla\cdot (\phi\vec{F}) =\nabla \phi \cdot F + \phi \nabla\cdot \vec{F} = \nabla \phi \cdot \vec{F} $$ for the same reason as above.

$$0=\nabla\cdot\vec{A} = \nabla cos(\vec{k}\cdot \vec{r}-\omega t) \cdot \vec{\alpha} = - \sin(\vec{k}\cdot \vec{r}-\omega t) \nabla(\vec{k}\cdot \vec{r})\cdot\vec{\alpha} = - \sin(\vec{k}\cdot \vec{r}-\omega t) \vec{k}\cdot \vec{\alpha} = \frac{1}{c|\vec{k}|}\vec{k}\cdot \vec{E}$$

from which we conclude that the electrical field vector $\vec{E}$ is orthogonal to the wave vector $\vec{k}$. This should clear up most of the questions.
EDIT : I corrected a couple of signs.


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