# How is $\frac{1}{|\vec{r} - \vec{r}'|} = \frac{1}{r} - \vec{r} \cdot \triangledown \frac{1}{r} + \ldots$ a Taylor expansion?

From pg. 112 of No-Nonsense Electrodynamics, the author uses multivariable Taylor expansion to assert:

In case it matters, the author is also assuming that $$|\vec{r}| \gg | \vec{r}'|$$ (stated elsewhere in the text). Also, $$\vec{r}, \vec{r}'$$ are both 3-dimensional vectors.

Question: How does this identity follow from Taylor? According to Wikipedia

A second-order Taylor series expansion of a scalar-valued function of more than one variable can be written compactly as

$$T(\mathbf{x}) = f(\mathbf{a}) + (\mathbf{x} - \mathbf{a})^\mathsf{T} D f(\mathbf{a}) + \cdots$$

where $$D$$ in this context denotes the gradient $$\triangledown$$ operator. If we plug this into to our context we get

$$f(\mathbf{a}) + (\mathbf{x} - \mathbf{a})^T Df(\mathbf{a}) = \frac{1}{r} + ( \vec{r} - \vec{r}') \cdot \triangledown \frac{1}{| \vec{r} - \vec{r}' |}$$

which doesn't obviously resemble the formula the author derived, unless I'm missing something?

## 1 Answer

HINT:

We can write the term of interest as

\begin{align} \frac1{|\vec r-\vec r'|}&=\frac{1}{\sqrt{|\vec r|^2+|\vec r'|^2-2\vec r'\cdot\vec r}}\\\\ &=\frac1{|\vec r|}\left(1-2\frac{\vec r'\cdot\vec r}{|\vec r|^2}+\left(\frac{|\vec r'|}{|\vec r|}\right)^2\right)^{-1/2} \end{align}

Now applying the binomial theorem (which is a special case of Taylor's Theorem) reveals

\begin{align} \frac1{|\vec r-\vec r'|}&=\frac1{|\vec r|}\left(1+\frac{\vec r'\cdot\vec r}{|\vec r|^2}+O\left(\frac{|\vec r'|^2}{|\vec r|^2}\right)\right)\\\\ &=\frac1{|\vec r|}+\vec r'\cdot \frac{\vec r}{|\vec r|^3}+O\left(\frac{|\vec r'|^2}{|\vec r|^3}\right) \end{align}

and we see that the Equation $$(4.30)$$ in the OP is incorrect by a factor of $$-1$$ on the second term of the expansion. In fact, $$(4.30)$$ is inconsistent with the equation that precedes it since $$\nabla \frac1r=-\frac{\vec r}{|\vec r|^3}$$ and not $$\nabla \frac1r=+\frac{\vec r}{|\vec r|^3}$$

NOTE:

Application of Taylor's Theorem to the function $$\frac1{|\vec r-\vec r'|}$$ that lead to the expression written in the OP is $$o(|\vec r-\vec r'|)$$ and does not lead, therefore, to a useful expansion when $$|\vec r|>>|\vec r'|$$.

Instead, we have used Taylor's Theorem (which is also application of the binomial theorem in this case) to expand the term $$\displaystyle \left(1-2\frac{\vec r'\cdot\vec r}{|\vec r|^2}+\left(\frac{|\vec r'|}{|\vec r|}\right)^2\right)^{-1/2}$$.