Yukawa form of electrostatic potential with $\mu_{\gamma}\ne 0$ (photon mass) is:

\begin{equation} \varphi(r)=a\,\frac{e^{\mu_{\gamma}r}}{r}+b\frac{e^{-\mu_{\gamma}r}}{r} \end{equation}

Since this is a solution of a differential equation, if we impose boundary conditions:

$$ \lim_{r\to\infty}\varphi (r)<+\infty, \tag{1} $$

this condition leads to the conclusion $a=0$ $\color{red}{\text{(Why?)}}$.

So you only have the component of a non-zero space type, and it assumes the shape of Yukawa spherically symmetrical:

\begin{equation} \varphi(r)=b\frac{e^{-\mu_{\gamma}r}}{r} \end{equation}

My questions, in addition to the one marked in red, are:

1) The source being a punctiform charge $q$ resting in origin and why?

2) Why must be valid the condition $ \lim_{r\to\infty}\varphi (r)<+\infty\, ? $

  • $\begingroup$ Why: $\exp$ rises faster than $1/r$, therefore if the potential is bounded in the limit $r\rightarrow \infty$, the constant $a$ has to be $0$. The first question: by construction, if you want to write down the potential of point charge no in the origin just rewrite $r$. The second questions has complete and complicated explanation and also has simple explanation ("general physics"): from the experiment you know that the potential should goes to zero for large distances. $\endgroup$ Mar 20, 2019 at 22:56
  • $\begingroup$ Yes! It is true. I'm not very lucid :-( and I am very tired. For the 2nd question is not finded anything (very simple and clear) from a course of "general physics". $\endgroup$
    – Sebastiano
    Mar 20, 2019 at 23:01

1 Answer 1


(1) It's because it's the form of your equation. Note Culomb's law for point charges

$$ U(r) = \frac{1}{4\pi\epsilon_0} \frac{q}{r} \tag{1} $$

is also only valid for a point source located at the origin. The coordinate free version is

$$ U(\mathbf{r}) = \frac{1}{4\pi\epsilon_0} \frac{q}{|\mathbf{r - r'}|} $$

where $\mathbf{r}$ is the field point and $\mathbf{r'}$ is the vector from the origin to the charge $q$. Indeed, this is the most general form of the potential of a point charge that reduces to (1) in the special case that $\mathbf{r'}=0$. That is,

$$ U(\mathbf{r}) = \frac{1}{4\pi\epsilon_0} \frac{q}{|\mathbf{r}|} = \frac{1}{4\pi\epsilon_0} \frac{q}{r} $$.

Why does a point charge have to be located at the origin?

It doesn't. Note that moving the point charge off the origin amounts to making the variable substitution $\mathbf{r} \to \mathbf{r-r'}$ in the differential equation. This doesn't change the nature of the solution, since $\mathbf{r'}$ is constant, and so we may write with impunity that

$$ \varphi(\mathbf{r})=a\,\frac{e^{\mu_{\gamma}|\mathbf{r-r'}|}}{|\mathbf{r-r'}|}+b\frac{e^{-\mu_{\gamma}|\mathbf{r-r'}|}}{|\mathbf{r-r'}|} $$

is the correct "coordinate free" version. Note that that your boundary condition now changes to $\lim_{|\mathbf{r-r'}|\to \infty} \varphi < \infty$ since the whole idea is that we are getting arbitrarily far away from the point charge.

(2) If the field generated by a point source didn't go to zero at infinity, it will either (a) go to a constant or (b) diverge.

(a) If it goes to a constant, then we can just take that to be our "zero point" for potential, since only changes in potential matter.

$\bullet$ That is, truthfully, the most general form is

\begin{equation} \Delta\varphi(r)=a\,\frac{e^{\mu_{\gamma}r}}{r}+b\frac{e^{-\mu_{\gamma}r}}{r} \end{equation}

where $\Delta\varphi = \varphi(r) - \varphi(r_0)$ where $r_0$ is some reference point. We just usually take $r_0$ to be at $\infty$. Note that if we impose the condition then that $\varphi(\infty) = 0$ then we have that $\varphi(r_0) = 0$ if we take $r_0$ to be at $\infty$, and we can therefore just write $\varphi(r)$ instead of $\Delta\varphi$, since with this choice $\Delta\varphi = \varphi(r)$.

(b) If the actual physical field diverges at infinity, then that just simply would not describe a point particle. So we impose this condition on physical grounds.

this condition leads to the conclusion $a=0$ (Why?).

The equation

\begin{equation} \varphi(r)=a\,\frac{e^{\mu_{\gamma}r}}{r}+b\frac{e^{-\mu_{\gamma}r}}{r} \end{equation}

is the general solution to a second order differential equation, and so $a,b$ are arbitrary constants that must be determined from the boundary conditions. One of your conditions is

$$ \lim_{r\to\infty}\varphi (r) \to \rm{finite} $$

So applying this yields

\begin{align*} \lim_{r\to\infty}\varphi(r)&=a\lim_{r\to\infty}\frac{e^{\mu_{\gamma}r}}{r}+b\underbrace{\lim_{r\to\infty}\frac{e^{-\mu_{\gamma}r}}{r}}_{\searrow 0}\\ &=a\lim_{r\to\infty}\frac{e^{\mu_{\gamma}r}}{r} \\ & \to \infty \end{align*}

The term in the last line diverges, so the only chance we have to satisfy the boundary condition is to eliminate the term all together by taking $a=0$.

  • $\begingroup$ Thank you for your answer and for the upvote :-). I sincerely have understood what you've written. I ask you for a further favor if possible. Could you, depending on what I asked you to do, edit your answer and add mathematical details to your answers? If possible I would like to have a complete picture. Thank you. Greetings. $\endgroup$
    – Sebastiano
    Mar 21, 2019 at 21:26
  • 1
    $\begingroup$ Okay I added some detail.. Is that what you were looking for? $\endgroup$ Mar 21, 2019 at 21:37
  • $\begingroup$ Yes too. But I have understood the (1), but why is necessary that a punctiform charge $q$ must be rest in origin (is there a mathematic explanation)? Can you explain with other simple explanation because: a) If it goes to a constant, then we can just take that to be our "zero point" for potential, since only changes in potential matter. (b) If it diverges then that just simply would not describe the physical configuration. So we impose this condition on physical grounds. $\endgroup$
    – Sebastiano
    Mar 21, 2019 at 21:43
  • 1
    $\begingroup$ I wanted to thank you once again for your wonderful response and for your patience. We read your humility and at least I understand that you are a good and sensitive person. Thank you. $\endgroup$
    – Sebastiano
    Mar 25, 2019 at 22:48

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.