Darwin Term (Fine Structure) and the Taylor expansion of the electric potential energy

Question

I am trying to derive the Taylor expansion for the potential $U(\vec r + \delta \vec r)$. The general expression for the Taylor expansion is: $$f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n.$$

I am trying to derive the same expression as in the Wikipedia article on Fine structure, but my derivation is not proper and it's messy, with multiple punctuality mistakes etc. From my understanding the variable is $\delta\vec r$. So we need to take the derivative with respect to it.

\begin{align} U(\vec r + \delta \vec r)=& U(\vec r + \delta \vec r)|_{\delta \vec r=0}(\delta \vec r'-\delta \vec r)^0 \\ &+ \sum_i \frac{\partial}{\partial x_i}U(\vec r + \delta \vec r)|_{\delta \vec r=0}(\delta \vec x'_i-\delta x_i)^1|_{\delta x_i=0}\\ &+\frac 1 2\sum_{i,j}\frac{\partial}{\partial x_i}\frac{\partial}{\partial x_j}U(\vec r + \delta \vec r)|_{\delta \vec r=0}(\delta \vec x'_i-\delta x_i)|_{\delta x_i=0}(\delta \vec x'_j-\delta x_j)|_{\delta x_j=0} +\cdots \end{align}

I have the following questions for what I wrote above:

1. If we want to be precise, shouldn't I write $\frac{\partial}{\partial \delta x_i}$ instead of $\frac{\partial}{\partial x_i}$ ? Because if am writing $\frac{\partial}{\partial x_i}$ than it looks as if I am taking the derivative with respect to $x_i$.

2. In the Wikipedia article, the 2nd term is written as the product of the Nabla operator and the potential, so a gradient. How do I cleanly produce that expression?

Answer 1

1. Loose the prime $\delta r$ terms and the weird $|_{\delta x_j=0}$, they make no sense. Your formula should read something like $$ U(\vec r + \delta \vec r)= U(\vec r + \delta \vec r)|_{\delta \vec r=0}(\delta \vec r)^0 \\ + \sum_i \frac{\partial}{\partial x_i}U(\vec r + \delta \vec r)|_{\delta \vec r=0}(\delta \vec r_i)^1\\ +\frac 1 2\sum_{i,j}\frac{\partial}{\partial x_i}\frac{\partial}{\partial x_j}U(\vec r + \delta \vec r)|_{\delta \vec r=0}(\delta \vec r_i)(\delta \vec r_j) +\cdots $$ In words: To evaluate the function $U$ at a point that is offset from $\vec{r}$ by $\delta \vec{r}$ take $U$ evaluated at $\vec{r}$ (the first term on the RHS) plus the product of first derivatives of $U$ evaluated at $\vec{r}$ and the components $\delta \vec{r}_i$ in the respective directions of the offset/perturbation we want to use (the second term), plus ....

2. To answer your question no. 1: You have stumbled across something which one might call a touchy subject. In physics conventions for partial derivatives are not always clear and it helps to have a little experience to see what's going on. If you're doing analysis in $\mathbb{R}^2$ and you read something like $$ \frac{\partial}{\partial x}f(x,y) $$ what is actually meant by this? The problem is that $x$ appears in a double role: denoting the derivative with respect to the first coordinate in $\frac{\partial}{\partial x}$ and the evaluation at some particular value $x$ of the coordinate in the argument. Now, how would you read $$ \frac{\partial}{\partial x}f(-y,x)\,, $$ is the derivative taken with respect to the first or the second argument? I think a mathematician would be likely to read this as $\frac{\partial}{\partial x}f$ (i.e. the derivative wrt. the first argument) evaluated at $(-y,x)$ (which I think is the correct reading), while some people in physics would read this as "take the expression $f(-y,x)$ and derive wrt. $x$ (in which they are using the partial derivative like a total derivative). So, to come back to your question (I hope I haven't confused you above :) ), the $\frac{\partial}{\partial x_i}$ is fine, it just says "partial derivative with respect to the $i$-th component of the functions argument that is usually denoted by $x$". There are other notations like $\partial_i$ for that where you don't face the problem of putting an unnecessary name tag on the function argument when writing down the derivative.

3. To answer your no. 2: This one is quite easy, for a vector $\vec{a}$ and a scalar function $f$ we have $$ \vec{a} \cdot \nabla f = \sum_i \vec{a}_i \partial_i f $$ and you see that the first-order term in Wikipedia using the gradient conforms with the one written down here.