I'm reading a Physics paper (https://pubs.aip.org/aapt/ajp/article-abstract/35/9/832/1045177/Causality-in-the-Coulomb-Gauge?redirectedFrom=fulltext), in which it is claimed an identity, which I find as far as I can understand not totally rigorous. The identity is the following: $$T_{ij}(\vec{r},\vec{r'})=\frac{\partial^2}{\partial x_i \partial x_j^{'}}\left(\frac{1}{|\vec{r}-\vec{r'}|}\right)=\frac{\delta_{ij}}{|\vec{r}-\vec{r'}|^3}-\frac{3(x_i-x_i^{'})(x_j-x_j^{'})}{|\vec{r}-\vec{r'}|^5}+\frac{4\pi}{3}\delta_{ij}\delta({\vec{r}-\vec{r'}}).$$ Here $\vec{r}=(x,y,z)$ and $\vec{r^{'}}=(x^{'},y^{'},z^{'})$ and i,j=1,2,3. I'm wrong If I claim that if $\vec{r} \neq\vec{r'}$ and at the same time $i\neq j$, then the identity is ill-defined? In fact the second term in the sum loses meaning since the relative limit for $\vec{r} \rightarrow \vec{r'}$ doesn't exist.enter image description here
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1Why did you added the arrows above $r$ and $r'$ if they are not vectors? – jjagmath Oct 26 '24 at 19:38
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Do you mean $|\vec r|=\sqrt {x^2+y^2+z^2}$ and so on? – lulu Oct 26 '24 at 19:40
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Yes, I'm so sorry! – Antonino Roccaforte Oct 26 '24 at 19:40
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$\frac{4\pi}{3}\delta{ij}\delta({\vec{r}-\vec{r'}})$ is that $\delta i j$ supposed to be $\delta_{ij}$ ? – GEdgar Oct 26 '24 at 19:42
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Yes, I'm editing the question to correct these mistakes! – Antonino Roccaforte Oct 26 '24 at 19:43
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δij here is supposed to be the Kronecker delta- – Antonino Roccaforte Oct 26 '24 at 19:47
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In the case of $i\neq j$ you can forget about the second term since $\delta_{ij}=0$ – K.defaoite Oct 26 '24 at 19:59
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With "second term", I mean the term with the three at the beginning. – Antonino Roccaforte Oct 26 '24 at 20:03
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See this reference – Jean Marie Oct 26 '24 at 20:14
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https://math.stackexchange.com/q/3525923/168433 – md2perpe Oct 27 '24 at 06:47
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Let's write $r = (x,y,z), r' = (u,v,w)$. Consider the case $i=1,j=2$, Then the claim is $$ \frac{\partial^2}{\partial x \partial v}\left(\frac{1}{\sqrt{(x-u)^2+(y-v)^2+(z-w)^2}}\right) \\= 0 + \frac{-3(x-u)(y-v)}{\big((x-u)^2+(y-v)^2+(z-w)^2\big)^{5/2}} + 0 . $$ I think this is true.
GEdgar
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I think it is required a Kronecker delta (r, r') to exclude this case. – Antonino Roccaforte Oct 26 '24 at 20:06
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If $r \to r'$, then even $$\left(\frac{1}{\sqrt{(x-u)^2+(y-v)^2+(z-w)^2}}\right)$$ is undefined, so we had better not try to use this formula in that case. – GEdgar Oct 26 '24 at 20:06
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The Dirac Delta in the third term "includes" the case r=r', I think it derives from $\nabla^2(1/r)=-4 \pi \delta (r)$. So I think that in the formula in the end only a factor(another Kronecker delta) $\delta(r,r')$ is missing in the second term. – Antonino Roccaforte Oct 26 '24 at 20:22