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)

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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


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): $$\phi(\vec{r}):= \frac{1}{4 \pi \varepsilon_0}\int_{\mathbb{R}^3} \frac{\rho(\vec{r'})}{||\vec{r}-\vec{r'}||} dr' = \frac{1}{4 \pi \varepsilon_0}\int_{\mathbb{R}^3} \frac{\rho(\vec{Rr'})}{||\vec{r}-\vec{Rr'}||} dRr'= \frac{1}{4 \pi \varepsilon_0}\int_{\mathbb{R}^3} \frac{\rho(\vec{r'})}{||\vec{r}-\vec{Rr'}||} dr'= \frac{1}{4 \pi \varepsilon_0}\int_{\mathbb{R}^3} \frac{\rho(\vec{r'})}{||\vec{R^{-1}r}-\vec{r'}||} dr' = \frac{1}{4 \pi \varepsilon_0}\int_{\mathbb{R}^3} \frac{\rho(\vec{r'})}{||r\vec{e}_z-\vec{r'}||} dr' = \frac{1}{4 \pi \varepsilon_0}\int_{\mathbb{R}^3} \frac{\rho(\vec{r'})}{||\vec{r}-\vec{r'}||_2} dr'\:.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.