In classical electrodynamics, if the electric field (or magnetic field, either of the two) is fully known (for simplicity: in a vacuum with $\rho = 0, \vec{j} = 0$), is it possible to unambiguously calculate the other field from Maxwell's equations?
For example, let's assume that $\vec{E}(\vec{r}, t)$ is known with $\vec{E} = 0$. From Maxwell's equations, we know that
$$\nabla \times \vec{E} = – \frac{\partial \vec{B}}{\partial t} \Leftrightarrow \vec{B} = – \int \nabla \times \vec{E} \; \mathrm{d}t$$
However, this (as far as I can tell) results in $\vec{B}(\vec{r}, t) = \left( \begin{smallmatrix}
C_1\\C_2\\C_3
\end{smallmatrix} \right)$ with unknown constants $C_i$. This result satisfies the Maxwell equations, since
$$\nabla \cdot \left( \begin{smallmatrix}
C_1\\C_2\\C_3
\end{smallmatrix} \right) = 0 \quad\text{ and }\quad \nabla \times \left( \begin{smallmatrix}
C_1\\C_2\\C_3
\end{smallmatrix} \right) = \varepsilon_0 \mu_0 \frac{\partial \vec{E}}{\partial t} = 0$$
Does this really mean that if we are given $\vec{E}$ and the source-free Maxwell equations that we cannot determine whether there is no magnetic field at all or whether there is a constant magnetic field filling all of space using the theory?
Note: I am asking this question because in my physics class, when considering planar electromagnetic waves of the form $\vec{E} = \vec{E}_0 e^{i(\vec{k} \cdot \vec{r} – \omega t)}$, we were often asked to calculate $\vec{B}$ given the electric field of the wave using Maxwell's equations. We wondered about the integration constants, but since it was always assumed that the fields are of the form
$$\vec{E}(\vec{r}, t) = \vec{E}_0 e^{i(\vec{k} \cdot \vec{r} – \omega t)}$$
$$\vec{B}(\vec{r}, t) = \vec{B}_0 e^{i(\vec{k} \cdot \vec{r} – \omega t)}$$
the constants were naturally set to zero. However, I'm wondering if this assumption is safe and what the reasoning behind it is.
Best Answer
The most important statement in this answer to your question is: Yes, you can superimpose a constant magnetic field. The combined field remains a solution of Maxwell's equations. $\def\vB{{\vec{B}}}$ $\def\vBp{{\vec{B}}_{\rm p}}$ $\def\vBq{{\vec{B}}_{\rm h}}$ $\def\vE{{\vec{E}}}$ $\def\vr{{\vec{r}}}$ $\def\vk{{\vec{k}}}$ $\def\om{\omega}$ $\def\rot{\operatorname{rot}}$ $\def\grad{\operatorname{grad}}$ $\def\div{\operatorname{div}}$ $\def\l{\left}\def\r{\right}$ $\def\pd{\partial}$ $\def\eps{\varepsilon}$ $\def\ph{\varphi}$
Since you are using plane waves you even cannot enforce the fields to decay sufficiently fast with growing distance to the origin. That would make the solution of Maxwell's equations unique for given space properties (like $\mu,\varepsilon,\kappa$, and maybe space charge $\rho$ and an imprinted current density $\vec{J}$). But, in your case you would not have a generator for the field. Your setup is just the empty space. If you enforce the field to decay sufficiently fast with growing distance you just get zero amplitudes $\vec{E}_0=\vec{0}$, $\vec{B}_0=\vec{0}$ for your waves. Which is certainly a solution of Maxwell's equations but also certainly not what you want to have.
For my point of view you are a bit too fast with the integration constants. You loose some generality by neglecting that these constants can really depend on the space coordinates.
Let us look what really can be deduced for $\vB(\vr,t)$ from Maxwell's equations for a given $\vE(\vr,t)=\vE_0 \cos(\vk\vr-\om t)$ in free space.
At first some recapitulation: We calculate a particular B-field $\vBp$ that satisfies Maxwell's equations: $$ \begin{array}{rl} \nabla\times\l(\vE_0\cos(\vk\vr-\om t)\r)&=-\pd_t \vBp(\vr,t)\\ \l(\nabla\cos(\vk\vr-\om t)\r)\times\vE_0&=-\pd_t\vBp(\vr,t)\\ -\vk\times\vE_0\sin(\vk\vr-\om t) = -\pd_t \vBp(\vr,t) \end{array} $$ This leads us with $\pd_t \cos(\vk\vr-\om t) = \om \sin(\vk\vr-\om t)$ to the ansatz $$ \vBp(\vr,t) = -\vk\times\vE_0 \cos(\vk\vr-\om t)/\om. $$ The divergence equation $\div\vBp(\vr,t)=-\vk\cdot(\vk\times\vE_0)\cos(\vk\vr-\om t)/\om=0$ is satisfied and the space-charge freeness $0=\div\vE(\vr,t) = \vk\cdot\vE_0\sin(\vk\vr-\om t)$ delivers that $\vk$ and $\vE_0$ are orthogonal. The last thing to check is Ampere's law $$ \begin{array}{rl} \rot\vBp&=\mu_0 \eps_0 \pd_t\vE\\ \vk\times(\vk\times\vE_0)\sin(\vk\vr-\om t)/\om &= -\mu_0\eps_0 \vE_0 \sin(\vk\vr-\om t) \om\\ \biggl(\vk \underbrace{(\vk\cdot\vE_0)}_0-\vE_0\vk^2\biggr)\sin(\vk\vr-\om t)/\om&= -\mu_0\eps_0 \vE_0 \sin(\vk\vr-\om t) \om \end{array} $$ which is satisfied for $\frac{\omega}{|\vk|} = \frac1{\sqrt{\mu_0\eps_0}}=c_0$ (the speed of light).
Now, we look which modifications $\vB(\vr,t)=\vBp(\vr,t)+\vBq(\vr,t)$ satisfy Maxwell's laws. $$ \begin{array}{rl} \nabla\times\vE(\vr,t) &= -\pd_t\l(\vBp(\vr,t)+\vBq(\vr,t)\r)\\ \nabla\times\vE(\vr,t) &= -\pd_t\vBp(\vr,t)-\pd_t\vBq(\vr,t)\\ 0 &= -\pd_t\vBq(\vr,t) \end{array} $$ That means, the modification $\vBq$ is independent of time. We just write $\vBq(\vr)$ instead of $\vBq(\vr,t)$. The divergence equation for the modified B-field is $0=\div\l(\vBp(\vr,t)+\vBq(\vr)\r)=\underbrace{\div\l(\vBp(\vr,t)\r)}_{=0} + \div\l(\vBq(\vr)\r)$ telling us that the modification $\vBq(\vr)$ must also be source free: $$ \div\vBq(\vr) = 0 $$ Ampere's law is $$ \begin{array}{rl} \nabla\times(\vBp(\vr,t)+\vBq(\vr)) &= \mu_0\eps_0\pd_t \vE,\\ \rot(\vBq(\vr))&=0. \end{array} $$ Free space is simply path connected. Thus, $\rot(\vBq(\vr))=0$ implies that every admissible $\vBq$ can be represented as gradient of a scalar potential $\vBq(\vr)=-\grad\ph(\vr)$.
From $\div\vBq(\vr) = 0$ there follows that this potential must satisfy Laplace's equation $$ 0=-\div(\vBq(\vr)) = \div\grad\ph = \Delta\ph $$
That is all what Maxwell's equations for the free space tell us with a predefined E-field and without boundary conditions:
The B-field can be modified through the gradient of any harmonic potential.
The thing is that with problems in infinite space one is often approximating some configuration with finite extent which is sufficiently far away from stuff that could influence the measurement significantly.
How are plane electromagnetic waves produced?
One relatively simple generator for electromagnetic waves is a dipole antenna. These do not generate plane waves but spherical curved waves as shown in the following nice picture from the Wikipedia page http://en.wikipedia.org/wiki/Antenna_%28radio%29.
Nevertheless, if you are far away from the sender dipol and there are no reflecting surfaces around you then in your close neighborhood the electromagnetic wave will look like a plane wave and you can treat it as such with sufficiently exact results for your practical purpose.
In this important application the plane wave is an approximation where the superposition with some constant electromagnetic field is not really appropriate.
We just keep in mind if in some special application we need to superimpose a constant field we are allowed to do it.