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How can we show that for a spherically symmetric charge distribution, the dipole, quadrupole and all higher moments about the centre of the distribution are identically zero.

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As we already know that for a spherically symmetric charge distribution potential is first term in equation 1. From that how can we conclude that higher order terms are zero, Why can't they just cancel each other?

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    $\begingroup$ I would look into spherical harmonics, which are at the heart of the multipole expansion. I'll let the more math oriented users answer the question properly, but the idea is that spherical harmonics are orthogonal just like sine waves of different frequencies in the Fourier transform. $\endgroup$
    – KF Gauss
    Feb 13 '20 at 16:31
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What you have done is correct. In my answer I will show why the terms with $l>0$ vanish. As you have shown, $$\Phi = \frac{1}{4\pi\epsilon_0r}\sum^{\infty}_{l=0}\int \rho(r') P_{l}(\cos{\alpha})\bigg(\frac{r'}{r}\bigg)^{l} dV'$$ Since the distribution is spherically symmetric I have considered the origin to be the centre of the sphere. The cosine of the angle made by $r'$ and $r$ will be, $$\cos(\alpha) = \cos(\phi' - \phi) \sin(\theta)\sin(\theta') + \cos(\theta)\cos(\theta')$$ Using the addition theorem of spherical harmonics, $$P_{l}(\cos(\alpha)) = \frac{4\pi}{2l+1} \sum^{l}_{m=-l} Y_{ml}(\theta', \phi')Y_{ml}^{*}(\theta, \phi)$$ Substituting this in the first equation, $$\Phi = \frac{1}{4\pi\epsilon_0r}\sum^{\infty}_{l=0}\int \rho(r') \bigg(\frac{4\pi}{2l+1}\bigg)\times \sum^{l}_{m=-l} Y_{ml}(\theta', \phi')Y_{ml}^{*}(\theta, \phi)\bigg(\frac{r'}{r}\bigg)^{l} dV'$$ If you look only at the angular part of the volume integration, $$\int^{\pi}_{0}\int^{2\pi}_{0} Y_{ml}(\theta', \phi') d\Omega'$$ It comes out to be $\delta_{0l}\delta_{0m}$. Since there is sum over both $m$ and $l$ the only non vanishing term will be $l=0,m=0$. Hence $$\Phi = \frac{1}{4\pi\epsilon_0r}\int^{R}_{0}\rho(r') (4\pi r'^2dr')$$ Which is the same as, $$\Phi = \frac{Q}{4\pi\epsilon_0r}$$


Proof of Orthonormality

To solve the integral above you will require the orthonormality of spherical harmonics. Explicitly writing the spherical harmonics, $$Y_{ml} (\theta, \phi) = A_{ml} P_{ml}(\cos(\theta))e^{im\phi}$$ Where $$A_{ml}= \sqrt{\frac{(2l+1)(l-m)!}{(l+m)!}}$$ Since it's obvious that, $$\int^{2\pi}_{0} e^{i\phi(m-m')} d\phi = \delta_{m m'}$$ We can write, $$\int^{\pi}_{0}\int^{2\pi}_{0} Y_{ml}(\theta, \phi) Y_{m'l'}(\theta, \phi) d\phi d\theta = A_{ml}A_{m'l'}\delta_{mm'} \int^{1}_{-1} P_{ml}(x)P_{m'l'}(x)dx$$ $P_{m'l'}$ are known to be orthogonal for same $m$ (see: this proof). Hence, $$\int^{\pi}_{0}\int^{2\pi}_{0} Y_{ml}(\theta, \phi) Y_{m'l'}(\theta, \phi) d\phi d\theta = \delta_{mm'}\delta_{ll'}$$ For $m'=0, l'=0$, $$\int^{\pi}_{0}\int^{2\pi}_{0} Y_{ml}(\theta', \phi') d\Omega' = \delta_{m0}\delta_{l0}$$

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    $\begingroup$ Perhaps needs a few words to say how one proves that the spherical harmonic functions are orthogonal. $\endgroup$ Sep 20 '20 at 9:45
  • $\begingroup$ @AndrewSteane Thanks for the suggestion. $\endgroup$ Sep 20 '20 at 12:01
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Dipole, quadrupole, octupole etc electric moments arise from non-radial components of an electric field. By symmetry, any spherically symmetric charge distribution can have only a radial electric field. All non-radial components are zero. Hence all multi-pole electric moments are zero.

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    $\begingroup$ Surely the first sentence here merely assumes the thing which the questioner wishes to derive. $\endgroup$ Sep 20 '20 at 9:44

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