Why does the volume element cancel the singularity in the electric field integral?

[kiwakwok]

I am currently reading a book about the electromegnetism. When I went through the introductory chapter, there is a small part that I do not understand clearly. Therefore, I hope that I can seek help here.


Quote:

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$\mathbf{E(r)}=\int\;d\mathbf{r'}\;\frac{\mathbf{r-r'}}{|\mathbf{r-r'}|^3}\rho(\mathbf{r'})$​

where $d\mathbf{r'}$ represents the three-dimensional volume element. Note that in spite of the singularity at $\mathbf{r=r'}$, the integral is finite for a finite charge distribution, even when the point [\itex]\mathbf{r}[\itex] is in the region containing charge. This is because the volume element $d\mathbf{r'}$ in the neighbourhood of a point $\mathbf{r'}$ goes like $|\mathbf{r-r'}|^2$ for small $\mathbf{r-r'}$, thereby cancelling the singularity.

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I have highlighted the part that I do not fully understand. Does that sentence means in the spherical coordinate, one can write the volume element as $d\mathbf{r'}=\tilde{r}^2\sin\theta d\tilde{r}d\theta d\phi$ where $\tilde{r}=|\mathbf{r-r'}|$, and the $\tilde{r}^2$ terms cancel? If so, what if I am not using the spherical coordinate but others such as Cartesian coordinate? There is no $\tilde{r}^2$ term to do the cancellation.

Thanks in advance for giving me a helping hand.


Reference: P.3, Classical Field Theory by Francis E. Low

 

[tiny-tim]

hi kiwakwok!

Does that sentence means in the spherical coordinate, one can write the volume element as $d\mathbf{r'}=\tilde{r}^2\sin\theta d\tilde{r}d\theta d\phi$ where $\tilde{r}=|\mathbf{r-r'}|$, and the $\tilde{r}^2$ terms cancel?

yes

If so, what if I am not using the spherical coordinate but others such as Cartesian coordinate? There is no $\tilde{r}^2$ term to do the cancellation.

then its dx' = dx' dy' dz'

(the dθ and dφ of course are dimensionless, which is why there*was an added r²)

 

[kiwakwok]

Thanks tiny-tim. I have one follow-up question:

In spherical coordinate, we can clearly see that the integral above will not blow up even at the singularity. The situation should be the same no matter which coordinate system we use. In the Cartesian coordinates, the integral can be rewritten as

$\mathbf{E(x)}=\int dx'dy'dz'\;\rho(\mathbf{x'})\frac{(x-x')\hat{x'}+(y-y')\hat{y'}+(z-z')\hat{z'}}{\left[(x-x')^2+(y-y')^2+(z-z')^2\right]^{3/2}}$​

Is there any way that we can know that this integral will not blow up, as what we did previously?

 

[tiny-tim]

why bother?

it's a lot easier with spherical coordinates​

 

[kiwakwok]

I was just interested in such a stupid question xP

 

[Born2bwire]

Thanks tiny-tim. I have one follow-up question:

In spherical coordinate, we can clearly see that the integral above will not blow up even at the singularity. The situation should be the same no matter which coordinate system we use. In the Cartesian coordinates, the integral can be rewritten as

$\mathbf{E(x)}=\int dx'dy'dz'\;\rho(\mathbf{x'})\frac{(x-x')\hat{x'}+(y-y')\hat{y'}+(z-z')\hat{z'}}{\left[(x-x')^2+(y-y')^2+(z-z')^2\right]^{3/2}}$​

Is there any way that we can know that this integral will not blow up, as what we did previously?

Note that if you integrated in spherical coordinates, the limits of integration over a volume centered about the origin would be from 0 to A in terms of r. So if we had a uniform distribution of charge about the origin, we would need the integrand with respect to r to be at least O(1) so that we lost the singularity when evaluating the integration from 0 to A. That is, if the integral was something like 1/r, then we would try to evaluate ln(A)-ln(0) and so on.

But the limits of integration in Cartesian coordinates are different. For a volume about the origin, we would integrate, say z', from -A to A. In this case, an integrand like 1/z^2 is still integrable despite the singularity in the integrand at the origin.