# Newtonian potential expansion identity

Preliminaries

Consider the Newtonian potential $$\frac{1}{|\vec x - \vec y|}$$ with $$\vec{x}, \vec{y} \in \mathbb{R}^3$$ and $$|\vec{x}| = x > y = |\vec{y}|$$. Its Taylor expansion is given by $$\frac{1}{|\vec x - \vec y|} = \sum_{n=0}^\infty \frac{(-1)^n}{n!} (\vec{y} \cdot\vec{\nabla}_\vec{x})^n \frac{1}{x},$$ which can also be written in terms of Legendre polynomials as $$\frac{1}{|\vec x - \vec y|} = \sum_{n = 0}^\infty P_n (\hat{x} \cdot \hat{y}) \frac{y^n}{x^{n+1}}.$$ The key identity here is $$\frac{(-1)^n}{n!} (\vec{y} \cdot\vec{\nabla}_\vec{x})^n \frac{1}{x} = P_n (\hat{x} \cdot \hat{y}) \frac{y^n}{x^{n+1}}$$ Using the key identity twice, we have $$\frac{(-1)^n}{n!} (\vec{y} \cdot\vec{\nabla}_\vec{x})^n \frac{1}{x} = P_n (\hat{x} \cdot \hat{y}) \frac{y^n}{x^{n+1}} = \left(\frac{y}{x}\right)^{2n+1} P_n (\hat{x} \cdot \hat{y}) \frac{x^n}{y^{n+1}} = \left(\frac{y}{x}\right)^{2n+1} \frac{(-1)^n}{n!} (\vec{x} \cdot\vec{\nabla}_\vec{y})^n \frac{1}{y}.$$

The Question

Is there a way to prove the identity $$\boxed{(\vec{y} \cdot\vec{\nabla}_\vec{x})^n \frac{1}{x} = \left(\frac{y}{x}\right)^{2n+1} (\vec{x} \cdot\vec{\nabla}_\vec{y})^n \frac{1}{y}}$$ directly, i.e., without resorting to Legendre polynomials?

Based on Fabian's correction, here is the proof:

start with noticing that $$|y\vec x+{\Lambda\over y} \vec y|^2=y^2 x^2 +\Lambda ^2 +2\Lambda \vec x \cdot \vec y= |x\vec y+{\Lambda\over x} \vec x|^2$$. Thus $$\frac{1}{y|\vec{x}+\frac{\Lambda}{y^{2}}\vec{y}|}=\frac{1}{x|\vec{y}+\frac{\Lambda}{x^2}\vec{x}|}$$ We need the translation operator $$e^{\vec{a}\nabla_{b}}$$ which has the property $$e^{\vec{a}\nabla_{b}}F(\vec{b})=F(\vec{b}+\vec{a})$$ as long as $$\nabla_b$$ and $$a$$ commute ( this can be seen by noting that applying the expansion of this operator is actually the Taylor series). By applying this operator on $${1 \over y}$$ we can get $$x^{-1}e^{\Lambda(x^{-2}\vec{x}\cdot\vec{\nabla}_{\vec{y}})}\frac{1}{y}=\frac{1}{x|\vec{y}+\frac{\Lambda}{x^2}\vec{x}|}$$ and similarly $$y^{-1}e^{(\Lambda y^{-2}\vec{y}\cdot\vec{\nabla}_{\vec{x}})}\frac{1}{x}=\frac{1}{y|\vec{x}+\frac{\Lambda}{y^{2}}\vec{y}|}$$ Thus $$y^{-1}e^{(\Lambda y^{-2}\vec{y}\cdot\vec{\nabla}_{\vec{x}})}\frac{1}{x}=x^{-1}e^{\Lambda(x^{-2}\vec{x}\cdot\vec{\nabla}_{\vec{y}})}\frac{1}{y}$$ Identifying $$O(\Lambda^n)$$ terms yields the identity.

• I believe, you are mistaken. In fact, I believe the last equation is true in general. Consider applying a mirror operation that maps the ray $\lamabda \vec{x}$ onto the ray $\lambda\vec{y}$ (keeping all the lengths invariant). – Fabian Dec 23 '18 at 16:01
• you are right! looks like we can prove the identity this way.I will edit my answer accordingly. – user619894 Dec 24 '18 at 4:15

$$\color{brown}{\textbf{Edition of 18.12.2018}}$$

Let us present the issue identity $$(\vec{y} \cdot\vec{\nabla}_\vec{x})^n \frac{1}{x} = \left(\frac{y}{x}\right)^{2n+1} (\vec{x} \cdot\vec{\nabla}_\vec{y})^n \frac{1}{y}\tag1$$ in the form of $$f_n(x,y)=f_n(y,x),\tag2$$ where \begin{align} f_n(x,y)=\left(x_1^2+x_2^2+x_3^2\right)^{(2n+1)/2}\left(y_1\dfrac{\partial}{\partial x_1}+y_2\dfrac{\partial}{\partial x_2}+y_3\dfrac{\partial}{\partial x_3}\right)^{(n)} \left(x_1^2+x_2^2+x_3^2\right)^{(-1/2)} \end{align}.\tag3 Then \begin{align} &f_1(x,y) = \left(x_1^2+x_2^2+x_3^2\right)^{3/2}\left(y_1\dfrac{\partial}{\partial x_1}+y_2\dfrac{\partial}{\partial x_2}+y_3\dfrac{\partial}{\partial x_3}\right)\left(x_1^2+x_2^2+x_3^2\right)^{-1/2}\\[4pt] &= - \left(x_1^2+x_2^2+x_3^2\right)^{3/2}\left(y_1 x_1 + y_2 x_2+y_3 x_3\right) \left(x_1^2+x_2^2+x_3^2\right)^{-3/2}, \end{align} $$f_1(x,y) = -\left(y_1x_1 + y_2x_2 + y_3 x_3\right),\tag4$$

\begin{align} &f_{m+1}(x,y)=\left(x_1^2+x_2^2+x_3^2\right)^{(2m+3)/2}\left(y_1\dfrac{\partial}{\partial x_1}+y_2\dfrac{\partial}{\partial x_2}+y_3\dfrac{\partial}{\partial x_3}\right)^{m+1} \left(x_1^2+x_2^2+x_3^2\right)^{(-1/2)}\\[4pt] &=\left(x_1^2+x_2^2+x_3^2\right)^{(2m+3)/2}\left(y_1\dfrac{\partial}{\partial x_1}+y_2\dfrac{\partial}{\partial x_2}+y_3\dfrac{\partial}{\partial x_3}\right)\left(f_m(x,y)\left(x_1^2+x_2^2+x_3^2\right)^{-(2m+1)/2}\right), \end{align} \begin{align} &f_{m+1}(x,y)=\left(x_1^2+x_2^2+x_3^2\right)\left(y_1\dfrac{\partial}{\partial x_1}+y_2\dfrac{\partial}{\partial x_2}+y_3\dfrac{\partial}{\partial x_3}\right)f_m(x,y)\\[4pt] &\quad -(2m+1)\left(y_1x_1 + y_2x_2 + y_3 x_3\right)f_m(x,y). \end{align}\tag5

Using $$(4)-(5),$$ one can get \begin{align} &f_2(x,y)=-\left(x_1^2+x_2^2+x_3^2\right)\left(y_1\dfrac{\partial}{\partial x_1}+y_2\dfrac{\partial}{\partial x_2}+y_3\dfrac{\partial}{\partial x_3}\right)\left(y_1x_1 + y_2x_2 + y_3 x_3\right)\\[4pt] &\quad +3\left(y_1x_1 + y_2x_2 + y_3 x_3\right)^2, \end{align} $$f_2(x,y) = - \left(x_1^2+x_2^2+x_3^2\right)\left(y_1^2 + y_2^2 + y_3^2\right) + 3\left(y_1x_1 + y_2x_2 + y_3 x_3\right)^2,\tag6$$

\begin{align} &f_3(x,y)=-\left(x_1^2+x_2^2+x_3^2\right)\left(y_1^2+y_2^2+y_3^2\right)\\[4pt] &\qquad\times\left(y_1\dfrac{\partial}{\partial x_1}+y_2\dfrac{\partial}{\partial x_2}+y_3\dfrac{\partial}{\partial x_3}\right)\left(x_1^2 + x_2^2 + x_3^2\right)\\[4pt] &\quad +3\left(x_1^2+x_2^2+x_3^2\right)\left(y_1\dfrac{\partial}{\partial x_1}+y_2\dfrac{\partial}{\partial x_2}+y_3\dfrac{\partial}{\partial x_3}\right)\left(y_1x_1 + y_2x_2 + y_3 x_3\right)^2\\[4pt] &\quad +5\left(x_1^2+x_2^2+x_3^2\right)\left(y_1^2+y_2^2+y_3^2\right)\left(y_1x_1 + y_2x_2 + y_3 x_3\right)-15\left(y_1x_1 + y_2x_2 + y_3 x_3\right)^3\\[4pt] &=-2\left(x_1^2+x_2^2+x_3^2\right)\left(y_1^2+y_2^2+y_3^2\right)\left(y_1x_1 + y_2x_2 + y_3 x_3\right)\\[4pt] &\quad +6\left(x_1^2+x_2^2+x_3^2\right)\left(y_1^2+y_2^2+y_3^2\right)\left(y_1x_1 + y_2x_2 + y_3 x_3\right)\\[4pt] &\quad +5\left(x_1^2+x_2^2+x_3^2\right)\left(y_1^2+y_2^2+y_3^2\right)\left(y_1x_1 + y_2x_2 + y_3 x_3\right)-15\left(y_1x_1 + y_2x_2 + y_3 x_3\right)^3, \end{align} $$f_3(x,y) = 9\left(x_1^2+x_2^2+x_3^2\right)\left(y_1^2 + y_2^2 + y_3^2\right) + 3\left(y_1x_1 + y_2x_2 + y_3 x_3\right)^2.\tag7$$

From the formulas $$(4),(6)-(7)$$ should $$f_1(x,y)=f_1(y,x)\in \mathbb F,\quad f_2(x,y)=f_2(y,x)\in \mathbb F,\quad f_3(x,y)=f_3(y,x)\in \mathbb F,\tag8$$ where $$\mathbb F = \left\{\sum_{k=0}^Q C_k\left(x_1^2+x_2^2+x_3^2\right)^{A_k} \left(y_1^2 + y_2^2 + y_3^2\right)^{A_k} \left(x_1y_1+x_2y_2+x_3y_3\right)^{B_k}\right\},$$

Let us proved $$f_m(x,y)\in \mathbb F,$$ then $$f_m(x,y)=f_m(y,x).\tag9$$

Taking in account $$(3)-(4)$$, one can get \begin{align} &f_{m+1}(x,y)=\left(x_1^2+x_2^2+x_3^2\right) \sum_{k=0}^{Q(m)}C_k\left(y_1^2 + y_2^2 +y_3^2\right)^{A_k}\\[4pt] &\qquad\times\left(y_1\dfrac{\partial}{\partial x_1}+y_2\dfrac{\partial}{\partial x_2}+y_3\dfrac{\partial}{\partial x_3}\right)\Big(\left(x_1^2+x_2^2+x_3^2\right)^{A_k} \left(x_1y_1+x_2y_2+x_3y_3\right)^{B_k}\Big)\\[4pt] &\quad -(2m+1)\left(y_1x_1 + y_2x_2 + y_3 x_3\right)\\[4pt] &\quad\times \sum_{k=0}^{Q(m)}C_k\left(x_1^2+x_2^2+x_3^2\right)^{A_k}\left(y_1^2 + y_2^2 + y_3^2\right)^{A_k}\left(x_1y_1+x_2y_2+x_3y_3\right)^{B_k}\\[4pt] &=\sum_{k=0}^{Q(m)}C_k \left(x_1^2+x_2^2+x_3^2\right)^{A_k}\left(y_1^2 + y_2^2 +y_3^2\right)^{A_k}\Big(2A_k\left(x_1y_1+x_2y_2+x_3y_3\right)^{B_k+1}\\[4pt] &\quad + B_k\left(x_1^2+x_2^2+x_3^2\right)\left(y_1^2 + y_2^2 +y_3^2\right) \left(x_1y_1+x_2y_2+x_3y_3\right)^{B_k-1}\\[4pt] &\quad -(2m+1)\left(y_1x_1 + y_2x_2 + y_3 x_3\right)^{B_k+1}\Big)\in\mathbb F, \end{align} $$f_{m+1}(x,y) = f_{m+1}(y,x).\tag{10}$$ Therefore, identity $$(2)$$ is valid for $$n=m+1$$ and, taking in account $$(8)-(10),$$ by induction - for all $$n>1.$$

$$\color{brown}{\textbf{Proved.}}$$