# Proof of the gradient of the inverse of the magnitude of a vector?

In Jackson's Classical Electrodynamics (a physics text), he mentions in passing a sort of identity which I have not been able to prove. For a given vector field $\vec{v}(x,y,z)$ and a constant vector $\vec{v}'$, the negative gradient of the inverse of the magnitude of the difference of these vectors is the difference vector divided by the magnitude cubed. $$-\nabla \left[\frac{1}{||\vec{v} - \vec{v}'||}\right] = \frac{\vec{v} - \vec{v}'}{||\vec{v} - \vec{v}'||^3}$$ Here, the vertical bars imply magnitude, or norm of the vector, and the vectors $\vec{v},\vec{v}' \in \mathbb{R}^3$. I've begun the problem by letting the vector field $\vec{x}$ be the difference vector of any vector field and a constant vector, so the problem becomes: $$-\nabla \left[\frac{1}{||\vec{x}||}\right] = \frac{\vec{x}}{||\vec{x}||^3}$$ Where $$\vec{x} = \vec{v} - \vec{v}'$$ Which is also equivalent to setting $\vec{v}' = \vec{0}$, which is a special case of the original identity.

So naturally I am interested in seeing how this is true, because it is not a simple statement. To prove it, I began by expanding, and calculating the components, like so: $$\nabla \left[\frac{1}{||\vec{x}||}\right] = \frac{\partial}{\partial x} \frac{1}{||\vec{x}||} \mathbf{i} + \frac{\partial}{\partial y} \frac{1}{||\vec{x}||} \mathbf{j} + \frac{\partial}{\partial z} \frac{1}{||\vec{x}||} \mathbf{k}$$ And for the first component: $$\frac{\partial}{\partial x} \frac{1}{||\vec{x}||} = \frac{\partial}{\partial x} (x_1^2 + x_2^2 + x_3^2)^\frac{-1}{2}$$ $$= \frac{-1}{2||\vec{x}||^3}\frac{\partial}{\partial x} (x_1^2 + x_2^2 + x_3^2)$$ $$= \frac{-1}{||\vec{x}||^3} (x_1 \frac{\partial}{\partial x} [x_1] + x_2 \frac{\partial}{\partial x} [x_2] + x_3 \frac{\partial}{\partial x} [x_3])$$ Which, continuing for all components, gives the expected factor of $\frac{1}{||\vec{x}||^3}$, implying that, for each component, $$x_i = (x_1 \frac{\partial}{\partial n} [x_1] + x_2 \frac{\partial}{\partial n} [x_2] + x_3 \frac{\partial}{\partial n} [x_3])$$ Where $n$ is $x$, $y$, or $z$ for $i = 1,2,3$ respectively. Written out, we have: $$x_1 = (x_1 \frac{\partial}{\partial x} [x_1] + x_2 \frac{\partial}{\partial x} [x_2] + x_3 \frac{\partial}{\partial x} [x_3])$$ $$x_2 = (x_1 \frac{\partial}{\partial y} [x_1] + x_2 \frac{\partial}{\partial y} [x_2] + x_3 \frac{\partial}{\partial y} [x_3])$$ $$x_3 = (x_1 \frac{\partial}{\partial z} [x_1] + x_2 \frac{\partial}{\partial z} [x_2] + x_3 \frac{\partial}{\partial z} [x_3])$$ Which can be written more succinctly as: $$x_1 = \left(\frac{\partial}{\partial x}\vec{x}\right) \circ \vec{x}$$ $$x_2 = \left(\frac{\partial}{\partial y}\vec{x}\right) \circ \vec{x}$$ $$x_3 = \left(\frac{\partial}{\partial y}\vec{x}\right) \circ \vec{x}$$ Where $\circ$ represents the dot product.

These relations do not hold for many vector fields. For example, $\vec{r} = x^2 \mathbf{i} + x^2 \mathbf{j} + x^2 \mathbf{k}$ has $\left(\frac{\partial}{\partial x}\vec{x}\right) \circ \vec{x} = 6x^3$. It does hold for some vector fields, but most will not fit this criteria. Thus I've made a mistake somewhere, or the result is not general, but I would bet on the former.

Can someone help point me in the right direction, perhaps point out my error?

P.S. This type of math question is closely related to physics, as is clear from the notation and assumptions made such as Euclidean vectors in a normed vector space. If it would be more appropriate in Physics SE I would be happy to move it.

A proof of this result was posted in Physics.SE here including the reason why it works in Classical Electrodynamics.

I think there's a problem with the first identity to begin with, probably only true when $\vec{v} = \vec{r} = x\hat{e_1} + y\hat{e_2} + z\hat{e_3}$, not any field vector as you state,
Please see the factor $\color{red}{\nabla\vec{v}}$. If $\vec{v} = \color{red}{\vec{r}}$, then $\color{red}{\nabla\vec{v} = I}$ adn you recover the identity you show