First let me ensure that the problem I am having is from a Physics book. But the problem is totally related to mathematics. Hence I think that this question here may not be illegitimate.

Using the concept of inverse square law of magnetic monopoles, this book derives an equation for PE between two magnetic monopoles:

$$M=-\int X dx=-PP'\int \dfrac{1}{r^2} \dfrac{x'-x}{r}dx=-PP'\int \dfrac{x'-x}{r^3} dx=\dfrac{PP'}{r}+C$$

If we define reference point of magnetic monopole at infinity, $C=0$ and PE at a point becomes: $M=\dfrac{PP'}{r}$

Now by adding all the dipoles of two magnets, the book derives an equation for PE between two magnets which is the same as PE between two closed circuits.

$$M=-\int X dx= \left( -\oint^{s'}_0 \oint^{s}_0 \dfrac{\vec{ds}.\vec{ds'}}{r}+C \right) ii'$$

Since reference position of all magnetic monopoles are defined at infinity, reference position of magnet is also at infinity. Therefore $C=0$ and PE at a position becomes:

$$M=-\int X dx=-\oint^{s'}_0 \oint^{s}_0 \dfrac{\vec{ds}.\vec{ds'}}{r}ii'$$

In another page, this book also derives another equation for PE between two closed circuits.

$$X=-\dfrac{\partial (-\oint^{s'}_0 \oint^{s}_0 \rho \text{ } {\vec{ds}.\vec{ds'})} }{\partial x}ii' =-\dfrac{\partial (-\oint^{s'}_0 \oint^{s}_0 \rho \text{ } {\vec{ds}.\vec{ds'}) +a+C} }{\partial x}ii'$$

where '$a$' is a constant

'$C$' is an arbitrary constant depending on reference point.


$$M=-\int X dx= \left[ \left( -\oint^{s'}_0 \oint^{s}_0 \rho \text{ } {\vec{ds}.\vec{ds'}} \right) +a+C \right] ii'$$

Now while equating the two potential energies:

We note that if we consider the same reference point, the two arbitrary constants (in the two equations of potential energies) are equal. Therefore:

$$-\oint^{s'}_0 \oint^{s}_0 \dfrac{\vec{ds}.\vec{ds'}}{r}ii' =\left[ \left( -\oint^{s'}_0 \oint^{s}_0 \rho \text{ } {\vec{ds}.\vec{ds'}} \right) +a \right] ii'$$

The book gives $\rho=\frac{1}{r}$ after equating the two potential energies. This is only possible if constant $a=0$. How can we ascertain that '$a$' must be zero?

  • 1
    $\begingroup$ What book is this? $\endgroup$ – Yuriy S Feb 25 '18 at 17:34

You can actually begin with \begin{align} M &=-\int Xdx=\dfrac{qq{^\prime}}{r}+C \hspace{4mm}\mbox{ and } \\M& =-\int Xdx=p \text{ } qq{^\prime}+D, \end{align} where $C$ and $D$ are constants. By setting the two potentials equal to each other, we see that $$ \dfrac{qq{^\prime}}{r}+C = p \text{ } qq{^\prime}+D. $$ Since the constant terms must equate to each other, we see that $C$ must equal $D$. It also follows that $p=\frac{1}{r}$.

  • $\begingroup$ If you can tell me which book (name and page number) these equations are coming from, then I can deduce this more clearly with more detail. $\endgroup$ – Mee Seong Im Feb 25 '18 at 17:45
  • $\begingroup$ Equations (31) and (36) (from Maxwell's treatise): Link $\endgroup$ – Joe Feb 25 '18 at 17:48
  • $\begingroup$ @Joe Thanks. I will take a look. $\endgroup$ – Mee Seong Im Feb 25 '18 at 17:48
  • $\begingroup$ @Joe Your book states that $r=\sqrt{(x'-x)^2 + (y'-y)^2+(z'-z)^2}$ is a function of $\vec{x}=\langle x,y,z \rangle$ and $p$ is a function of $r$ (see Equation (24) for the electromagnetic setting). So $p$ is also a function of $\vec{x}$. After computing the two $\textit{integrals}$, it is clear that additional constants cannot drop out of the terms $\frac{qq'}{r}$ and $p\:qq'$ since you are doing improper integrals (if you were to do proper integrals, then constants produced are already grouped in $C$ or $D$, and we would know the values of $C$ and $D$). $\endgroup$ – Mee Seong Im Feb 25 '18 at 18:46
  • $\begingroup$ MeeSeong: I have edited the whole question. Please take a look at the edit. $\endgroup$ – Joe Mar 2 '18 at 5:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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