| New Reply |
Divergence in spherical polar coordinates |
Share Thread | Thread Tools |
| Aug18-11, 06:17 PM | #1 |
|
|
Divergence in spherical polar coordinates
I took the divergence of the function 1/r2[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/r3.
for [itex]\widehat{r}[/itex] I used (xi+yj+zk)/(x2+y2+z2)1/2 what am i missing? |
| PhysOrg.com |
physics news on PhysOrg.com >> Promising doped zirconia >> New X-ray method shows how frog embryos could help thwart disease >> Bringing life into focus |
| Aug18-11, 06:55 PM | #2 |
|
|
What you're missing is the handling of the singularity at r=0.
|
| Aug18-11, 06:58 PM | #3 |
|
Recognitions:
|
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/r3? |
| Aug19-11, 01:53 AM | #4 |
|
|
Divergence in spherical polar coordinates
sorry I made a silly error in my calculation
[itex]\frac{\partial}{\partial x}[/itex] x(x2+y2+z2)-[itex]\frac{3}{2}[/itex] = (x2+y2+z2)-[itex]\frac{3}{2}[/itex] - 3x2(x2+y2+z2)-[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? |
| Aug19-11, 02:02 AM | #5 |
|
Recognitions:
|
And in practice there is no singularity, since the charge would have to be a point charge for that. |
| Aug19-11, 02:11 AM | #6 |
|
|
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? |
| Aug19-11, 02:25 AM | #7 |
|
Recognitions:
|
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 non-zero divergence. |
| Aug19-11, 02:35 AM | #8 |
|
|
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? |
| Aug19-11, 02:38 AM | #9 |
|
Recognitions:
|
Right on both counts. :)
My point about point charge is unrelated to divergence. Only that a real electric field has no singularity. |
| Aug19-11, 03:00 AM | #10 |
|
|
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?
|
| Aug19-11, 03:10 AM | #11 |
|
Recognitions:
|
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.
|
| Aug19-11, 03:16 AM | #12 |
|
|
|
| Aug19-11, 03:19 AM | #13 |
|
|
![]() I know not relevant but... |
| Aug19-11, 03:23 AM | #14 |
|
|
greek to me :) |
| Aug19-11, 03:24 AM | #15 |
|
|
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. |
| Aug19-11, 03:25 AM | #16 |
|
|
|
| Aug19-11, 03:47 AM | #17 |
|
|
A vector in this case is simply a matrix with a scalar. ![]() So to steal more wiki imagery a tensor is: |
| New Reply |
| Thread Tools | |
Similar Threads for: Divergence in spherical polar coordinates
|
||||
| Thread | Forum | Replies | ||
| divergence and curl of spherical polar coordinates | Calculus & Beyond Homework | 0 | ||
| Divergence in spherical coordinates | Differential Equations | 7 | ||
| Deriving the divergence in polar coordinates | Calculus & Beyond Homework | 0 | ||
| Divergence in Polar Coordinates | Calculus & Beyond Homework | 1 | ||
| Proof of Divergence Formula in Spherical Coordinates | Calculus & Beyond Homework | 2 | ||