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.


1 Answer 1


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,

\begin{eqnarray} -\nabla\left[\frac{1}{|\vec{v} - \vec{v}'|} \right] &\stackrel{\vec{u} = \vec{v}-\vec{v}'}{=}& -\nabla \frac{1}{|\vec{u}|} = -\sum_i \hat{e}_i\frac{\partial}{\partial x_i} |u|^{-1} = -\sum_i \hat{e}_i\frac{\partial}{\partial x_i} \left(\sum_j u_j^2\right)^{-1/2} \\ &=& \frac{1}{2}\sum_i \hat{e}_i \left(\sum_j u_j^2\right)^{-3/2} \frac{\partial}{\partial x_i} \left(\sum_j u_j^2\right) = \frac{1}{2}\sum_i \hat{e}_i|u|^{-3} \sum_j 2u_j\hat{e}_i\frac{\partial u_j}{\partial x_i} \\ &=& \frac{1}{|u|^3}\sum_{i,j} u_j \color{blue}{\hat{e}_i\frac{\partial }{\partial x_i}} u_j = \frac{1}{|u|^3} \sum_{j}u_j\color{blue}{\nabla} u_j = \frac{1}{|u|^{3}}\sum_{j,k}(u_j\hat{e}_j)\cdot\nabla(u_k\hat{e_k})\\ &=& \frac{\vec{u}\cdot \nabla \vec{u}}{|u|^3} \\ &=& \frac{\vec{v} - \vec{v'}}{|\vec{v} - \vec{v'}|^3}\cdot \color{red}{\nabla\vec{v}} \end{eqnarray}

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

  • $\begingroup$ This is interesting, because I was thinking that the gradient of the vector term could be used, but I'm only just learning tensor analysis and didn't feel up to the task. I'll ask Physics SE and see if there's something physics-related as to why Jackson gives this as true. (For those curious, the book is here: link in the start of section 1.5 on page 7) $\endgroup$ Mar 19, 2018 at 18:33
  • $\begingroup$ @SamGallagher sure, just make sure to include a link to this question $\endgroup$
    – caverac
    Mar 19, 2018 at 18:37
  • $\begingroup$ I realized that the vector v is not a vector field, but a position vector, i.e. v = r, and so the statement works. I've put the question on Physics.SE for sake of sharing the information. I should have made sure I was using the right definitions before assuming it applied to all vector fields! I'll post the link to the P.SE question-answer when I finish the post. $\endgroup$ Mar 19, 2018 at 19:04

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