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Need help doing this simple differentiation. Consider 4 d Euclidean(or Minkowskian) spacetime. \begin{equation} \partial_{\mu}\frac{(a-x)_\mu}{(a-x)^4}= ? \end{equation} where $a_\mu$ is a constant vector and the indices are summed over since one really doesn't need to bother about upper and lower indices in flat space. Also $(a-x)^2=(a-x)_\mu(a-x)_\mu$. A simple minded calculation gives me the result $0$. But I think the answer may contain Dirac Delta function like the relation below in 3 dimension \begin{equation} {\bf{\nabla}} . \frac{\hat{r}}{r^2}=4\pi \delta ^3(r) \end{equation} A corollary question. What will be $\partial_{\mu}\frac{1}{(a-x)^2}$ ? Is it $\frac{2(a-x)_\mu}{(a-x)^4}$ or does it also involve the Delta function?

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Related: physics.stackexchange.com/q/9255/2451 –  Qmechanic Jan 29 '13 at 15:31
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2 Answers

up vote 3 down vote accepted
  1. In a $d$-dimensional Euclidean space (with positive definite norm), one has $$ \vec{\nabla} \cdot \frac{\vec{r}}{r^d} ~=~{\rm Vol}(S^{d-1})~\delta^d(\vec{r}), $$ cf. the divergence theorem and arguments involving either test functions and integration by part, or $\epsilon$-regularization, similar to methods applied in this Phys.SE answer. Here ${\rm Vol}(S^{d-1})$ is the surface area of the $(d-1)$-dimensional units sphere $S^{d-1}$.

  2. By similar arguments one may show that the identity $$ \vec{\nabla}(r^{2-d}) ~=~(2-d)\frac{\vec{r}}{r^d} , \qquad d\neq 2, $$ contains no distributional contributions in $d$-dimensional Euclidean space.

  3. For the related questions in Minkowski space, one suggestion is to introduce an $\epsilon$-regularization in the Euclidean formulation, and then perform a Wick rotation, and at the end of the calculation, let $\epsilon\to 0^+$.

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Thanks Qmechanic. Writing down your equation in the index notation gives \begin{equation} \partial_\mu \frac{x_\mu}{x^2}=2\pi^2\delta^4(x) \end{equation} in 4d which is the result I wanted. Now it looks a bit silly to have asked the question in the first place.

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