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For evaluating the electric field of some charge distribution one can use $$\phi(r):= \frac{1}{4 \pi \varepsilon_0}\int_{\mathbb{R}^3} \frac{\rho(r')}{||r-r'||_2} dr'.$$

My question is: What symmetry do we need to have that we can write in spherical coordinates $$||r-r'||_2 = \sqrt{||r||_2^2+||r'||_2^2-2||r||_2||r'||_2\cos(\theta')}~?$$

This is of course not the most general way to express this distance, as the $\phi$ dependence is missing. So, under what conditions can the distance be expressed like this?

Notice that $\theta'$ is the respective angle in spherical coordinates, so it's NOT the angle between $r$ and $r'$.

So in particular, your answer should clarify, why we can evaluate for example the electric potential of a sphere by integrating: $$\frac{1}{4 \pi \varepsilon_0} \int_0^{2\pi}\int_0^\pi \int_0^{\infty} \frac{\rho(r')||r'^2|| sin(\theta')}{\sqrt{||r||_2^2+||r'||_2^2-2||r||_2||r'||_2\cos(\theta')}}d||r'||_2 d\theta'd\phi',$$ but need to refer to a more general equation in this example, where the $\phi$ angle is used too: excercise 14b)

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    $\begingroup$ I'm not sure I understand the question so I'll just leave a comment. First, the $\phi$ is the electrostatic potential for a stationary distribution of charges $\rho$. The formula for the distance is simply the cosine theorem, which is proved in Euclid's Elements :-) How do you define $||r-r'||_2$? $\endgroup$
    – pppqqq
    Jun 13, 2014 at 18:41
  • $\begingroup$ $||r-r'||_2 = \sqrt{(x-x')^2+(y-y')^2+(z-z')^2}$. $\endgroup$
    – Xin Wang
    Jun 13, 2014 at 18:42
  • $\begingroup$ It is assumed that either $r$ is directed along $z$ or that $\theta'$ is the angle between $r$ and $r'$ and not the polar angle. $\endgroup$ Jun 13, 2014 at 18:44
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    $\begingroup$ Try to give a look at en.wikipedia.org/wiki/Law_of_cosines - As V.Moretti points out maybe your confusion arises from the fact that $\theta '$ in the formula is the angle beetween $r$ and $r'$... in that case, the law of cosines is the theorem. $\endgroup$
    – pppqqq
    Jun 13, 2014 at 18:46
  • $\begingroup$ @V.Moretti ah, so it works for the sphere, because in that case, the potential in $z-$ direction is the same as in any other direction? $\endgroup$
    – Xin Wang
    Jun 13, 2014 at 18:47

1 Answer 1

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The answer is that $\rho$ must be spherically symmetric (necessary and sufficient condition).

To show it, let me change notation. Now $r$ ad $r'$ are the absolute values of the vectors $\vec{r}$,$\vec{r'}$ and we know that $$\phi(\vec{r}):= \frac{1}{4 \pi \varepsilon_0}\int_{\mathbb{R}^3} \frac{\rho(\vec{r'})}{||\vec{r}-\vec{r'}||_2} dr'.\tag{0}$$ where $$||\vec{r}-\vec{r'}||_2 = \sqrt{r^2+{r'}^2-2rr'\cos(\theta')}\tag{1}\:,$$ $\theta'$ being the polar angle of $\vec{r'}$.

From the general theory we know that it also hold $$\phi(\vec{r}):= \frac{1}{4 \pi \varepsilon_0}\int_{\mathbb{R}^3} \frac{\rho(\vec{r'})}{||\vec{r}-\vec{r'}||} dr'\tag{2}$$ where $$||\vec{r}-\vec{r'}|| = \sqrt{r^2+{r'}^2-2rr'\cos(\alpha)}\tag{3}\:,$$ $\alpha$ being the angle between $\vec{r'}$ and $\vec{r}$. Comparing (1) and (3), we conclude that $$\phi(\vec{r})= \phi(r \vec{e}_z)= f(r)$$ Consequently we have that $$\phi(\vec{r}) = f(r)\:.$$ Since, for some constant depending on the unit system, $\kappa \Delta \phi(\vec{r}) = \rho(\vec{r})$, we have that $$\rho(\vec{r}) = \kappa \Delta f(r)\:.$$ In other words $\rho$ must necessarily be a spherically symmetric function.

The found condition is also sufficient. Indeed, if $\rho$ is spherically symmetric, using the rotational invariance of the measure and the standard distance, it easily arises that the right-hand side of (2) can be re-written as the right-hand side of (0): $$\phi(\vec{r}):= \frac{1}{4 \pi \varepsilon_0}\int_{\mathbb{R}^3} \frac{\rho(\vec{r'})}{||\vec{r}-\vec{r'}||} dr' = \frac{1}{4 \pi \varepsilon_0}\int_{\mathbb{R}^3} \frac{\rho(\vec{Rr'})}{||\vec{r}-\vec{Rr'}||} dRr'= \frac{1}{4 \pi \varepsilon_0}\int_{\mathbb{R}^3} \frac{\rho(\vec{r'})}{||\vec{r}-\vec{Rr'}||} dr'= \frac{1}{4 \pi \varepsilon_0}\int_{\mathbb{R}^3} \frac{\rho(\vec{r'})}{||\vec{R^{-1}r}-\vec{r'}||} dr' = \frac{1}{4 \pi \varepsilon_0}\int_{\mathbb{R}^3} \frac{\rho(\vec{r'})}{||r\vec{e}_z-\vec{r'}||} dr' = \frac{1}{4 \pi \varepsilon_0}\int_{\mathbb{R}^3} \frac{\rho(\vec{r'})}{||\vec{r}-\vec{r'}||_2} dr'\:.$$

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