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I want to show the following equality $$\int_{\left|\vec{r}\right|<R}d^3r\vec{E}\left(\vec{r}\right)=-\frac{\vec{p}}{3\epsilon_0}$$

where $\vec{p}$ is the dipole moment of a charge distribution $\rho\left(\vec{r}^{\prime}\right)$ with respect to the origin. The charge distribution is located inside a sphere with radius $R$.

I started with Gauss's Theorem: $$\int_{\left|\vec{r}\right|<R}d^3r\vec{E}\left(\vec{r}\right)=-\int_{\left|\vec{r}\right|<R}d^3r\nabla\Phi\left(\vec{r}\right)=-\oint_{\left|\vec{r}\right|=R}d\Omega\frac{\vec{r}}{R}R^2 \Phi\left(\vec{r}\right)$$

Now I can use the Poisson integral: $$\Phi\left(\vec{r}\right)=\frac{1}{4\pi\epsilon_0}\int d^3r^{\prime}\frac{\rho\left(\vec{r}^{\prime}\right)}{\left|\vec{r}-\vec{r}^{\prime}\right|}$$

And thus: $$\int_{\left|\vec{r}\right|<R}d^3r\vec{E}\left(\vec{r}\right)=-\frac{R}{4\pi\epsilon_0}\int d^3r^{\prime}\rho\left(\vec{r}^{\prime}\right)\oint_{\left|\vec{r}\right|=R}d\Omega\frac{\vec{r}}{\left|\vec{r}-\vec{r}^{\prime}\right|}$$

I know that the dipole moment is: $$\vec{p}=\int d^3r^{\prime}\vec{r}^{\prime}\rho\left(\vec{r}^{\prime}\right)$$

Therefore, to obtain the top equation, the solid angle integration has to be: $$\oint_{\left|\vec{r}\right|=R}d\Omega\frac{\vec{r}}{\left|\vec{r}-\vec{r}^{\prime}\right|}\overset{!?}{=}\frac{4\pi}{3R}\vec{r}^{\prime}$$

Does anyone know how to solve this integral?

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We start with the integral $$\oint_{|\vec{r}|=R}\mathrm{d}\Omega\frac{\vec{r}}{|\vec{r}-\vec{r}'|}.$$ Since we are integrating over $\vec{r}$, we can without loss of generality, arrange for $\vec{r}'$to lie along the $+Z$ axis, so that $\vec{r}'=r'\hat{z}$. Then the angle between $\vec{r}$ and $\vec{r}'$ is the standard angle (in spherical coordinates) $\theta$ measured from the $+Z$ axis. Then we can write \begin{align} \mathrm{d}\Omega &= \sin\theta\,\mathrm{d}\theta\,\mathrm{d}\varphi\\ \vec{r} &= x\hat{x} + y\hat{y} + z\hat{z}\\ &=r\sin\theta\cos\varphi\hat{x}+r\sin\theta\sin\varphi\hat{y}+r\cos\theta\hat{z}\\ \mbox{and by the law of cosines}\\ |\vec{r}-\vec{r}'| &= \sqrt{r^{2}+r'^{2}-2rr'\cos{\theta}}\\ \end{align} So our integral has become: \begin{align} \oint_{|\vec{r}|=R}\mathrm{d}\Omega\frac{\vec{r}}{|\vec{r}-\vec{r}'|} &= \oint_{|\vec{r}|=R}\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\varphi\frac{r\sin\theta\cos\varphi\hat{x}+r\sin\theta\sin\varphi\hat{y}+r\cos\theta\hat{z}}{\sqrt{r^{2}+r'^{2}-2rr'\cos{\theta}}} \end{align} But when doing the $\varphi$ integrals, the $\cos\varphi\,\hat{x}$ and $\sin\varphi\,\hat{y}$ terms go to zero, so all you are left with is $$\oint_{|\vec{r}|=R}\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\varphi\frac{r\cos\theta\hat{z}}{\sqrt{r^{2}+r'^{2}-2rr'\cos{\theta}}}$$ We choose $u=-2rr'\cos{\theta}$, covert the integral into a standard form, and do it, keeping in mind while evaluating that that $r'<r=R$, so that $\sqrt{r^{2}+(r')^{2}-2rr'}=r-r'$, to finally get: $$\frac{4\pi r'\hat{z}}{3r}|_{|\vec{r}|=R} = \frac{4\pi r'\hat{z}}{3R}$$. But we had chosen $\vec{r}'$ to lie along the $+Z$ axis, so $r'\hat{z}=\vec{r}'$. So our final answer is: $$\frac{4\pi \vec{r}'}{3R}$$

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  • $\begingroup$ what do you mean by the standard form? $\endgroup$ – Andy Nov 14 '14 at 16:35
  • $\begingroup$ $\int \frac{x\mathrm{d}x}{\sqrt{a+x}}$, which I then looked up in an integral table. $\endgroup$ – G. Paily Nov 14 '14 at 23:50
  • $\begingroup$ ok, it's $\frac{2\left(x-2a\right)}{3}\sqrt{x+a}$ $\endgroup$ – Andy Nov 15 '14 at 13:48
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To solve the integral, I came across the multipole expansion $$\frac{1}{\left|\vec{r}-\vec{r}_{0}\right|}=\sum_{l=0}^{\infty}\frac{r_{<}^{l}}{r_{>}^{l+1}}P_{l}\left(\cos\sphericalangle\left(\vec{r},\vec{r}_{0}\right)\right),\;\; r_{<}=\min\left(r,r_{0}\right),\; r_{>}=\max\left(r,r_{0}\right)$$

With $\vec{r}_0=\vec{r}^{\prime}$ one has $r_>=R$ and $r_<=r^{\prime}$. Defining $\sphericalangle\left(\vec{r},\vec{r}^{\prime}\right)\equiv \gamma$, one obtains:

$$\oint_{\left|\vec{r}\right|=R}d\Omega\frac{\vec{r}}{\left|\vec{r}-\vec{r}^{\prime}\right|}=\oint_{\left|\vec{r}\right|=R}d\Omega R\vec{e}_r\sum_{l=0}^{\infty}\frac{r^{\prime l}}{R^{l+1}}P_{l}\left(\cos\gamma\right)=$$

$$=\sum_{l=0}^{\infty}\frac{r^{\prime l}}{R^{l}}\frac{4\pi}{2l+1}\sum_{m=-l}^{l}Y_{lm}\left(\vartheta^{\prime},\varphi^{\prime}\right)\oint_{\left|\vec{r}\right|=R}d\Omega\vec{e}_rY_{lm}^{\star}\left(\vartheta,\varphi\right)=\star$$

In the last step I used the addition theorem of spherical harmonics.

First, one has to evaluate the solid angle integral by orthogonality of spherical harmonics:

$$\oint_{\left|\vec{r}\right|=R}d\Omega\vec{e}_rY_{lm}^{\star}\left(\vartheta,\varphi\right)=\oint_{\left|\vec{r}\right|=R}d\Omega Y_{lm}^{\star}\left(\vartheta,\varphi\right)\left(\begin{array}{c} \sin\vartheta\cos\varphi\\ \sin\vartheta\cos\varphi\\ \cos\vartheta \end{array}\right)=$$ $$=\left(\begin{array}{c} \sqrt{\frac{2\pi}{3}}\left(-\delta_{1l}\delta_{1m}+\delta_{1l}\delta_{-1m}\right)\\ i\sqrt{\frac{2\pi}{3}}\left(\delta_{1l}\delta_{1m}+\delta_{1l}\delta_{-1m}\right)\\ \sqrt{\frac{4\pi}{3}}\delta_{1l}\delta_{0m} \end{array}\right)$$

Then the two sums can be easily performed by the Kronecker $\delta$s.

$$\star=\frac{4\pi r^{\prime}}{3R}\left(\begin{array}{c} \sqrt{\frac{2\pi}{3}}\left(-Y_{11}\left(\vartheta^{\prime},\varphi^{\prime}\right)+Y_{1,-1}\left(\vartheta^{\prime},\varphi^{\prime}\right)\right)\\ i\sqrt{\frac{2\pi}{3}}\left(Y_{11}\left(\vartheta^{\prime},\varphi^{\prime}\right)+Y_{1,-1}\left(\vartheta^{\prime},\varphi^{\prime}\right)\right)\\ \sqrt{\frac{4\pi}{3}}Y_{10}\left(\vartheta^{\prime},\varphi^{\prime}\right) \end{array}\right)=$$ $$=\frac{4\pi r^{\prime}}{3R}\left(\begin{array}{c} \sin\vartheta^{\prime}\cos\varphi^{\prime}\\ \sin\vartheta^{\prime}\cos\varphi^{\prime}\\ \cos\vartheta^{\prime} \end{array}\right)=\frac{4\pi}{3R}\vec{r}^{\prime}$$

If anyone has a faster solution, please let me know.

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