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Divergence in spherical polar coordinates 
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#1
Aug1811, 06:17 PM

P: 162

I took the divergence of the function 1/r^{2}[itex]\widehat{r}[/itex] in spherical coordinate system and immediately got the answer as zero, but when I do it in cartesian coordiantes I get the answer as 5/r^{3}.
for [itex]\widehat{r}[/itex] I used (xi+yj+zk)/(x^{2}+y^{2}+z^{2})^{1/2} what am i missing? 


#2
Aug1811, 06:55 PM

P: 159

What you're missing is the handling of the singularity at r=0.



#3
Aug1811, 06:58 PM

HW Helper
P: 6,189

Hi Idoubt!
If I take the cartesian derivative with respect to x I get: [tex]{\partial \over \partial x}{x \over (x^2+y^2+z^2)^{3/2}} = {2 x^2+y^2+z^2 \over (x^2+y^2+z^2)^{5/2}}[/tex] Of course the derivatives wrt to y and z have to be added yet... How did you get 5/r^{3}? 


#4
Aug1911, 01:53 AM

P: 162

Divergence in spherical polar coordinates
sorry I made a silly error in my calculation
[itex]\frac{\partial}{\partial x}[/itex] x(x^{2}+y^{2}+z^{2})^{[itex]\frac{3}{2}[/itex]} = (x^{2}+y^{2}+z^{2})^{[itex]\frac{3}{2}[/itex]}  3x^{2}(x^{2}+y^{2}+z^{2})^{[itex]\frac{5}{2}[/itex]} When i add y & z parts it cancels and becomes zero, so the div of [itex]\frac{1}{r^2}[/itex][itex]\widehat{r}[/itex] is zero. The reason that confused me was Electric fields can have a non zero divergence. Is that because of the singularity at r=0? 


#5
Aug1911, 02:02 AM

HW Helper
P: 6,189

And in practice there is no singularity, since the charge would have to be a point charge for that. 


#6
Aug1911, 02:11 AM

P: 162

Isn't the field due to just one charged assumed to be a field created by a point charge? ( I mean it makes no difference if we make that assumption right? )
But, how can an electric field have a non zero divergence then? 


#7
Aug1911, 02:25 AM

HW Helper
P: 6,189

Yes. It would only make a difference when you get very close to where the point charge is supposed to be.
Take for instance 2 point charges and you'll have nonzero divergence. 


#8
Aug1911, 02:35 AM

P: 162

But when we take two charges we no longer have the [itex]\frac{1}{r^2}[/itex] function ( unless we approximate at large distances)
It is my understanding that even the electric field of a point charge has non zero divergence is it not so? 


#9
Aug1911, 02:38 AM

HW Helper
P: 6,189

Right on both counts. :)
My point about point charge is unrelated to divergence. Only that a real electric field has no singularity. 


#10
Aug1911, 03:00 AM

P: 162

my question is , how can a field which is scalar multiple of the function [itex]\frac{1}{r^2}[/itex][itex]\hat{r}[/itex] have a divergence when i just proved that the divergence of this function is zero?



#11
Aug1911, 03:10 AM

Sci Advisor
Thanks
P: 2,548

The divergence is not 0, it's [itex]4 \pi \delta^{(3)}(\vec{x})[/itex]. It's not too tricky to prove with help of Green's first integral theorem applied to the whole space with a little sphere around the origin taken out and then taking its radius to 0.



#12
Aug1911, 03:16 AM

P: 162




#13
Aug1911, 03:19 AM

P: 45

I know not relevant but... 


#14
Aug1911, 03:23 AM

P: 162

greek to me :) 


#15
Aug1911, 03:24 AM

P: 45

The Greek symbol that looks like an O with a hat means Tensor. E(X) I would assume means the Energy of the system in question in terms of the vector. 


#16
Aug1911, 03:25 AM

P: 162




#17
Aug1911, 03:47 AM

P: 45

A vector in this case is simply a matrix with a scalar. So to steal more wiki imagery a tensor is: 


#18
Aug1911, 04:22 AM

P: 162

hmm ok ok so thats a little advanced for me, but my question was about the divergence of the [itex]\frac{1}{r^2}[/itex][itex]\hat{r}[/itex] function. In simpler terms can you tell me the reason for nonzero divergence in the electric field of a point charge?



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