In Jackson's book about classical electrodynamics, this formula comes up:

$$q_{lm} = \int \mathrm d^3 x' \, Y^*_{lm}\left(\theta', \phi'\right) r'^l \rho\left(\vec x'\right)$$

What does that $^*$ mean?

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# [Physics] What does the * mean in spherical harmonics

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classical-electrodynamics

In Jackson's book about classical electrodynamics, this formula comes up:

$$q_{lm} = \int \mathrm d^3 x' \, Y^*_{lm}\left(\theta', \phi'\right) r'^l \rho\left(\vec x'\right)$$

What does that $^*$ mean?

First, $\vec{r}^\prime$ is a vector that goes from the origin to the source of charge. If the source is a volumetric distribution, one must sum all contributions of charge, that's why one integrates over all the volume, say $\mathcal{V}$; the (correct) expression for the potential should be
$$V( \vec{r}) = \frac{1}{4 \pi \epsilon _{0}} \int_\mathcal{V} \frac{\rho (\vec{r}^\prime)}{ℛ} d\mathcal{V}^\prime$$
so that all dependence of $V$ remains on $\vec{r}$. Then, $r^\prime$ is just the magnitude $|\vec{r}^\prime|$, being the *distance* from the origin to *the* source of charge.

Second, usually, the series expansion of a function $f(x)$ about some point $x_0$ is useful because if you want to know the value of $f$ near $x_0$, you may just take some few terms of the expansion; it is as seeing the plot of $f$ with a magnifying glass. You should remember this from your first calculus courses, it is done a lot in physics. Here the expansion about $\epsilon=0$ will be useful since $\epsilon\to0$ implies $r\to\infty$ (just *really* big, if you will). The (correct) expression
$$V(\vec{r}) = \frac{1}{4 \pi \epsilon _{0}} \sum ^{\infty}_{n=0}\frac{1}{r^{n+1}} \int(r')^n\,P_{n}(\cos \theta^\prime)\,\rho( \vec{r}')\,d\mathcal{V}'$$
is just another way of writing the series expansion in terms of $r$, $r^\prime$ and $\theta^\prime$, where $P_n$ are the Legendre polynomials (Griffiths defines them there, ain't he?). This expression is useful, as it means, explicitly, that
$$V(\vec{r})=\frac{1}{4\pi\epsilon_0}\left[\frac{1}{r}\int\rho(\vec{r}')\,d\mathcal{V}'+\frac{1}{r^2}\int{r'}\cos\theta'\,\rho(\vec{r}')\,d\mathcal{V}'+\frac{1}{r^3}\left(\cdots\right)+\ldots\right]$$
so that if you want to evaluate the potential for points *far* from the source (big $r$), then you may just neglect higher order terms in $r$ and just take the $1/r$ (monopole) term; and so on if you're considering a better approximation, you may take the $1/r^2$ (dipole) term, etc... That's the real *usefulness* of the series expansion; in a lot of situations evaluating $V( \vec{r}) = \frac{1}{4 \pi \epsilon _{0}} \int \frac{\rho (\vec{r}')}{ℛ} d\mathcal{V}'$ will get really ugly, and then, mostly, is when the multipole approximation will be useful.

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.

## Best Answer

The superscript $*$ is a common notation for complex conjugate. Going back to check, (3.53) in the blue English edition states $$Y_{l,m} = \sqrt{\frac{2l+1}{4\pi}\frac{(l-m)!}{(l+m)!}}P^m_l(\cos\theta)e^{im\phi}$$ which is followed by (3.54) $$Y_{l,-m}(\theta,\phi) = (-1)^m Y^*_{l,m}(\theta,\phi),$$ making is clear that it has to be complex conjugation.