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In Classical Electrodynamics, Jackson derives the electric potential for a surface with a dipole charge.

Here is his derivation. I will omit constants for brevity.

Letting $D(\textbf{x}) := \lim_{d(\textbf{x}) \to 0} \sigma(\textbf{x}) d(\textbf{x})$ where $d(\textbf{x})$ is the local separation of $S$ and $S'$ with $S$ having charge density $\sigma(x)$ and $S'$ having equal and opposite charge density.

The potential due to the two surfaces is:

$$ \phi(\textbf{x}) = \int_S \frac{\sigma(\textbf{x}')}{|\textbf{x} - \textbf{x}'|} da' - \int_{S'} \frac{\sigma(\textbf{x'})}{|\textbf{x} - \textbf{x}' + \textbf{n}d|} da'' \tag{1}$$

where $\textbf{n}$ is the unit normal to the surface $S$ pointing away from $S'$.

He uses a Taylor expansion

$$ \frac{1}{|\textbf{x} + \textbf{a}|} = \frac{1}{x} + \textbf{a} \cdot \nabla \Big( \frac{1}{x} \Big) \tag{2}$$

He says this is valid when $|\textbf{a}| \ll |\textbf{x}|$ (and I assume $x := |\textbf{x}|$). Then as $d \to 0$ (and I believe he redefines $\textbf{x} := \textbf{x} - \textbf{x}'$ and $\textbf{a} := \textbf{n}d$) he arrives at

$$ \phi(\textbf{x}) = \int_S D(\textbf{x}') \textbf{n} \cdot \nabla'\Big( \frac{1}{|\textbf{x} - \textbf{x}'|} \Big) da' \tag{3}$$

and since $\textbf{p} = \textbf{n}\ D\ da'$ then the potential at $\textbf{x}$ caused by a dipole at $\textbf{x}'$ is

$$ \phi(\textbf{x}) = \frac{\textbf{p} \cdot (\textbf{x} - \textbf{x}')}{|\textbf{x} - \textbf{x}'|^3} \tag{4}$$

There are many steps I don't understand:

  1. From (1) Jackson used $\sigma(\textbf{x}')$ at $S$ and $- \sigma(\textbf{x}')$ at $S'$. But, if $\textbf{x}'$ traces out $S$ wouldn't this be starting with the assumption that $S$ and $S'$ are the same surface?

  2. Why is $|\textbf{a}| \ll |\textbf{x}|$ a necessary assumption to use the Taylor expansion? The 1D case would be analogous to expanding the function $1/(x+a)$ and I do not see a reason that $a \ll x$ is necessary to do this.

  3. After substituting the Taylor expansion into (1) (and using $\textbf{x} := \textbf{x} - \textbf{x}'$ and $\textbf{a} := \textbf{n}d$) we get

$$ \phi(\textbf{x}) = \int_S \frac{\sigma(\textbf{x}')}{|\textbf{x} - \textbf{x}'|} da' - \int_{S'} \sigma(\textbf{x'}) \Big( \frac{1}{|\textbf{x} - \textbf{x}'|} + \textbf{n}d \cdot \nabla \Big( \frac{1}{|\textbf{x} - \textbf{x}'|} \Big) \Big) da''$$

$$ \phi(\textbf{x}) = \int_S \frac{\sigma(\textbf{x}')}{|\textbf{x} - \textbf{x}'|} da' - \int_{S'} \sigma(\textbf{x'}) \frac{1}{|\textbf{x} - \textbf{x}'|} da'' - \int_{S'} \sigma(\textbf{x'}) \textbf{n}d \cdot \nabla \Big( \frac{1}{|\textbf{x} - \textbf{x}'|} \Big) da''$$

which somehow reduces to (3). It seems like Jackson cancelled the first two terms but how is this valid when we are integrating over $S$ in one and $S'$ in the other? Also, it seems like Jackson is missing a negative sign from the third term above. Also, the third term above differs from (3) in that he switches from integrating over $S'$ to $S$. Is his change justified because after we do the limiting process the two surfaces coincide, allowing using swap?

EDIT: I just realized these issues (besides the missing negative sign) can be resolved by doing the limiting process first and then expanding the integral into two terms. But this just begs the question; how do we know which order to do these steps when modeling with differentials and limiting processes?

  1. Should $\textbf{n}d$ be $\textbf{n}d(\textbf{x}')$?

  2. I just don't see the jump from (3) to (4) equation.

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  • $\begingroup$ For (2) the condition $|a| << |x|$ is only necessary insofar as it justifies neglecting higher order terms as a good approximation. In the 1D case the expansion is $\frac{1}{x+a} = \frac{1}{x} \frac{1}{1+a/x} = \frac{1}{x} \left[1 - \frac{a}{x} + \left(\frac{a}{x}\right)^2 + \ldots \right]$. For the infinite series to converge we must have $|a/x| < 1$. For a good approximation to be obtained with only two terms we want $a$ much smaller than $x$. $\endgroup$
    – RRL
    Aug 19, 2022 at 20:29
  • $\begingroup$ @RRL When I take the Taylor series I am getting $a$'s in the denominator. How are you getting them in the numerator? Also, how do you know from the problem what point to take the Taylor series around? $\endgroup$
    – ngc1300
    Aug 20, 2022 at 20:48
  • $\begingroup$ For the taylor series expansion I am getting $\frac{1}{x_0+a} - \frac{x - x_0}{(x_0+a)^2} + \frac{(x - x_0)^2}{(x_0+a)^3} - ...$ $\endgroup$
    – ngc1300
    Aug 20, 2022 at 21:16
  • $\begingroup$ Wait, so you're expanding it using the geometric series, not taylor series. Did Jackson mean to write geometric instead of taylor? $\endgroup$
    – ngc1300
    Aug 20, 2022 at 21:31
  • $\begingroup$ The 1D series $1 - y + y^2 - y^3 + \ldots$ is called a geometric series and also is the Taylor series around $y=0$ for $f(y) = \frac{1}{1+y}$. Jackson is referring to the multidimensional Taylor series. $\endgroup$
    – RRL
    Aug 21, 2022 at 0:07

1 Answer 1

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He uses a Taylor expansion

$$ \frac{1}{|\textbf{x} + \textbf{a}|} = \frac{1}{x} + \textbf{a} \cdot \nabla \Big( \frac{1}{x} \Big) $$

This should be something like

$$ \frac{1}{|\textbf{x} + \textbf{a}|} = \frac{1}{x} + \textbf{a} \cdot \nabla \Big( \frac{1}{x} \Big) +...$$

but he is silently discarding the second-order terms.

He says this is valid when $|\textbf{a}| \ll |\textbf{x}|$

Discarding the second order terms is valid in that case. The Taylor series is always valid, and the above is the standard Taylor expansion of $f(x+a)$. For example see Lang "Calculus of Several Variables" chapter 6.

(and I assume $x := |\textbf{x}|$).

Your assumption is correct.

Then as $d \to 0$ (and I believe he redefines $\textbf{x} := \textbf{x} - \textbf{x}'$ and $\textbf{a} := \textbf{n}d$)

It's not really redefining so much as recycling the same symbol $\textbf{x}$ to mean "a general vector". That can be a bit annoying I suppose but it's quite a common practice so you'll just have to be wary about it.

  1. From the first equation Jackson used $\sigma(\textbf{x}')$ at $S$ and $- \sigma(\textbf{x}')$ at $S'$. But, if $\textbf{x}'$ traces out $S$ wouldn't this be starting with the assumption that $S$ and $S'$ are the same surface?

This means that the charge distribution $\sigma$ at the point $x'+nd$ on $S'$ is the same as the charge distribution at the point $x'$ on $S$. That is the reason he's using the normal vector $n$, because $\sigma$ on $S'$ at the point normal to $x'$ on $S$ is the same as the value of $\sigma(x')$ on $S$.

  1. Why is $|\textbf{a}| \ll |\textbf{x}|$ a necessary assumption to use the Taylor expansion? The 1D case would be analogous to expanding the function $1/(x+a)$ and I do not see a reason that $a \ll x$ is necessary to do this.

As I mentioned above this is not justifying using the Taylor expansion but just the discarding of the higher-order terms of the expansion. The "dipole moment" of an electric field basically is the second term in the Taylor expansion anyway.

  1. After substituting the Taylor expansion into the first equation (and using $\textbf{x} := \textbf{x} - \textbf{x}'$ and $\textbf{a} := \textbf{n}d$) we get

$$ \phi(\textbf{x}) = \int_S \frac{\sigma(\textbf{x}')}{|\textbf{x} - \textbf{x}'|} da' - \int_{S'} \sigma(\textbf{x'}) \Big( \frac{1}{|\textbf{x} - \textbf{x}'|} + \textbf{n}d \cdot \nabla \Big( \frac{1}{|\textbf{x} - \textbf{x}'|} \Big) \Big) da''$$

That should be $\nabla'$ which is the gradient with respect to $x'$. See below for why that is important.

$$ \phi(\textbf{x}) = \int_S \frac{\sigma(\textbf{x}')}{|\textbf{x} - \textbf{x}'|} da' - \int_{S'} \sigma(\textbf{x'}) \frac{1}{|\textbf{x} - \textbf{x}'|} da'' - \int_{S'} \sigma(\textbf{x'}) \textbf{n}d \cdot \nabla \Big( \frac{1}{|\textbf{x} - \textbf{x}'|} \Big) da''$$

which somehow reduces to Jackson's third equation.

By Jackson's third equation, you mean (1.24), right?

It seems like Jackson cancelled the first two terms but how is this valid when we are integrating over $S$ in one and $S'$ in the other?

The value $x'$ is the same for the two surfaces, so these cancel out. The difference between the two surfaces is in the final term of the integral.

Also, it seems like Jackson is missing a negative sign from the third term above.

The $\nabla'$ is differentiation with respect to $x'$ so there is a negative sign coming from that, because the denominator is $x-x'$, which is absorbed into Jackson's answer.

Also, the third term above differs from Jackson's third equation in that he switches from integrating over $S'$ to $S$. Is his change justified because after we do the limiting process the two surfaces coincide, allowing using swap?

$x'$ is the same for either surface, the difference between the surfaces is all in the $nd$ part.

  1. Should $\textbf{n}d$ be $\textbf{n}d(\textbf{x}')$?

It looks like for this particular calculation $d$ is meant to be constant with respect to $x'$.

  1. I just don't see the jump from the third to fourth equation.

Jackson does not derive your equation (4) from (3). That equation is in the book for the purpose of comparision.

It is the dipole moment equation for point charges. Jackson is a graduate-level text, in other words it is for people who've already done undergraduate physics. The electric field of an electric dipole, equation (4) in your question, is a standard part of an undergraduate electricity and magnetism course, so Jackson doesn't derive it.

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