Physics Asked by photonQ on September 3, 2020
I am having a problem with an expansion that should be simple. Let’s say I solve Maxwell equations in vacuum, but in spherical coordinates. The solutions of the TM family can be found easily to be
$$
E^{TM}_{lm}(k;mathbf{r}) propto sqrt{l}j_{l+1}(kr)mathbf{V}^m_l(theta,phi)-sqrt{l+1}j_{l-1}(kr)mathbf{W}^m_l(theta,phi),
$$
where $l=1,2,3,…$, $m=-l,…,l$, $k=omega/c$ is the wavevector, $j_l(z)$ is the spherical Bessel function of order $l$, and the angular dependence is given by combinations of vector spherical harmonics (not important for this post). A similar solution can be found for the TE modes.
Now, assume I have a plane wave $mathbf{E} = mathbf{E}_0e^{imathbf{k}cdotmathbf{r}}$ that I want to expand in terms of the above eigenmodes in spherical coordinates. More specifically, if the electric energy of such plane wave is $U = int dV epsilon_0vert mathbf{E}vert^2/2 = epsilon_0 vert mathbf{E}_0vert^2V/2$, I wonder how much of such energy lies in the spherical eigenmode $TM_{klm}$. Assuming the plane wave is decomposed as
$$
mathbf{E} = sum_{lm}sum_sigma c_{lm} E^{sigma}_{lm}(k;mathbf{r}),
$$
with $sigma = TM,TE$, we can find easily that the fraction of the energy going to the $TM_{klm}$ mode is
$$
F = frac{1}{vert mathbf{E}_0vert^2V}left(int dV mathbf{E_0}e^{-imathbf{k}mathbf{r}}cdot frac{E^{TM}_{lm}(k;mathbf{r})}{sqrt{int dV vert E^{TM}_{lm}(k;mathbf{r})vert^2}}right)^2.
$$
Now, the angular integrals are $r-$independent, and the radial integral dependence is easy to find as
$$
lim_{Rtoinfty}int_0^R dr r^2 j_lambda^2(kr)to frac{pi R}{2 k^2} propto V^{1/3},
$$
$$
lim_{Rtoinfty}int_0^R dr r^2 int dtheta dphi e^{-imathbf{k}mathbf{r}}cdot E^{TM}_{lm}(k;mathbf{r})propto lim_{Rtoinfty}int_0^R dr r^2 j_lambda^2(kr)to frac{pi R}{2 k^2} propto V^{1/3},
$$
for any $lambda$. Then, the fraction we seek is
$$
F propto V^{-frac{2}{3}} to 0
$$
in the limit of infinite volume.
This happens for all the modes, and is related to the fact that the mode normalization in Cartesian coordinates is proportional to $V$ whereas in spherical coordinates it is proportional to $V^{1/3}$. This makes me think that it is not possible to expand a Cartesian plane wave in spherical waves, but this has to be wrong based on physical arguments.
Does anyone know where am I making a mistake? or have I bumped into some real limitation of the spherical vs cartesian coordinates? If so, what is the physical intuition behind it being not possible to expand an electromagnetic plane wave in spherical waves?
Thanks!
$$defbk{{bf k}} defbr{{bf r}} letth=vartheta letphi=varphi letd=delta$$ You have already seen that the problem is present for a scalar wave too, so let's examine this case, which leads to much simpler formulas. The spherical waves expansion of a plane wave is $$e^{i,bkcdotbr} = 4pi sum_{l=0}^infty,sum_{m=-l}^l i^l j_l(kr),{Y_l^m}^*(alpha,beta),Y_l^m(th,phi).$$ If you try to normalize the plane wave $e^{i,bkcdotbr}$ and the spherical wave $j_l(kr),Y_l^m(th,phi)$ you get into trouble as you take limits.
The physics behind is the following. A plane wave has a uniform energy density all over the space, and it makes sense to speak of the energy contained in a volume $V$, e.g. a cube of side $L$. But this is not the case for the spherical wave, which has a center and gets weaker as it gets away from it. Asymptotically, $j_l(kr)sim1/r$. You can constrain the wave in a spherical cavity of radius $R$, then let $R$ go to infinity. But the two boxes (cubical and spherical) do not fit together.
The way out is to use $delta$ normalization. I don't know whether you are familiar with this mathematical device, and cannot dwell into the matter. The usual normalizations are:
If $psi_{bk_1}(br)$, $psi_{bk_2}(br)$ are plane waves with wave vectors $bk_1$, $bk_2$ $$int!psi_{bk_1}(br)^*,psi_{bk_2}(br),d^{(3)}br = d^{(3)}(bk_1-bk_2)$$ and $$psi_{bk}(br) = (2pi)^{-3/2} e^{i bkcdotbr}.$$
If $chi_{l_1}^{m_1}(k_1;r,th,phi)$, $chi_{l_2}^{m_2}(k_2;r,th,phi)$ are spherical waves with wave vectors $k_1$, $k_2$ and multipole modes $l_1,m_1$, $l_2,m_2$, then $$int!chi_{l_1}^{m_1}(k_1;r,th,phi)^*, chi_{l_2}^{m_2}(k_2;r,th,phi)> r^2 dr,sinth,dth,dphi = d(k_1-k_2),d_{l_1l_2},d_{m_1m_2}.$$ and $$chi_l^m(k;r,th,phi) = {i^l over k},j_l(kr),Y_l^m(th,phi).$$
Answered by Elio Fabri on September 3, 2020
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