Using divergence theorem to prove Gauss's law

[demonelite123]

i need to prove that div(R/r^3) = 4πδ where R is a vector and r is the magnitude of the vector R. also δ is the dirac delta function.

so div(R/r^3) is 0 everywhere except for the origin. i need to show that the volume integral of div(R/r^3) = 4π as well.

using the divergence theorem we have that the volume integral of div(R/r^3) over the solid sphere is equal to the flux of R/r^3 over the surface of the sphere. however the vector field is not defined at the origin so we cannot apply the divergence theorem in this case. however, if we take the sphere and an inner sphere that surrounds the origin then we can apply the divergence theorem on the volume enclosed by the concentric spheres.

then the volume integral would be 0 since div(R/r^3) is 0 except at the origin and we get that the flux through the surface any sphere surrounding the origin is 4π.

however it seems that my book went on to say that volume integral of div(R/r^3) is also 4π and with that div(R/r^3) = 4πδ. but i am confused about how they can say that the volume integral is also 4π since if the solid sphere includes the origin then the divergence theorem cannot be applied.

 

[BruceW]

The sphere should include the origin. The divergence is not 'zero except at the origin'.
The integral of the divergence is zero if the origin is not included in the volume being integrated. You have misunderstood the meaning of the Dirac delta function here.
Why do you think the divergence theorem can't be used for a volume containing the origin? (I have done the calculation, using a sphere centred at the origin, and the answer comes out as the book says.)

 

[demonelite123]

The sphere should include the origin. The divergence is not 'zero except at the origin'.
The integral of the divergence is zero if the origin is not included in the volume being integrated. You have misunderstood the meaning of the Dirac delta function here.
Why do you think the divergence theorem can't be used for a volume containing the origin? (I have done the calculation, using a sphere centred at the origin, and the answer comes out as the book says.)

to apply the divergence theorem on the vector field R/r^3, doesn't R/r^3 have to be defined on a neighborhood of the solid sphere? however R/r^3 is not defined at the origin which means the divergence theorem cannot be applied in this case? so the integral of the flux over the surface of the sphere is not equal to the volume integral of the div(F). please correct me if I'm wrong.

 

[BruceW]

Its true that R/r^3 goes to infinity at the origin, but wikipedia says that it simply needs to be continuously differentiable for divergence theorem to work.
If you look up the definition of the divergence in spherical polar co-ordinates, you will see that R/r^3 is continuously differentiable at the origin, so divergence theorem does apply.

 

[demonelite123]

Its true that R/r^3 goes to infinity at the origin, but wikipedia says that it simply needs to be continuously differentiable for divergence theorem to work.
If you look up the definition of the divergence in spherical polar co-ordinates, you will see that R/r^3 is continuously differentiable at the origin, so divergence theorem does apply.

ah i see. thanks for your answer. however it seems odd to me that a vector field can be continuously differentiable in one coordinate system and not in another. is this something that happens often?

 

[BruceW]

Yes, this is an important feature of co-ordinate systems.
For example, in quantum mechanics, the solutions of the Schrödinger equation must be solutions in all co-ordinate systems.
This appears not to be a necessary condition for your problem though.

 

[vanhees71]

Not all coordinate systems cover the whole space. E.g., spherical coordinates in \mathbb{R}^3 cover the whole space with the polar axis taken out since the spherical coordinates are singular along this axis. This you can see when taking the Jacobian for the transformation from spherical coordinates to Cartesian coordinates,

\begin{split}<br /> x&amp;=r \cos \varphi \sin \vartheta \\<br /> y&amp;=r \sin \varphi \sin \vartheta \\<br /> z&amp;=r \cos \vartheta.<br /> \end{split}

The Jacobian is

\mathrm{det} \frac{\partial(x,y,z)}{\partial (r, \vartheta,\varphi)}=r^2 \sin \vartheta,

which vanishes for all r&gt;0 for \vartheta \in \{0,\pi \} and at r=0. Thus there's no one-to-one mapping in the neighborhood of the z axis, which I have chosen as the polar axis of the spherical coordinate system.

 

[demonelite123]

so for the same vector field F = R/r^3 as I've defined before, what if i defined F so that its domain is R^3 - {0}. would the divergence theorem apply for cartesian coordinates? does it still have to be continuously differentiable at the origin? i think I'm getting confused about the subtleties of this problem and want to really get a handle on this.

 

[BruceW]

If the vector field is defined to be zero at the origin, then maybe it is continuously differentiable. I'm not so good at the maths specifics, sorry I can't help on that.
But I don't think your question was written to test your understanding of when the divergence theorem can/can't be applied. I think the point of the question is that you see the field has spherical symmetry, so then you know you should use spherical co-ordinates.

 

[samalkhaiat]

i need to prove that div(R/r^3) = 4πδ where R is a vector and r is the magnitude of the vector R. also δ is the dirac delta function.

so div(R/r^3) is 0 everywhere except for the origin. i need to show that the volume integral of div(R/r^3) = 4π as well.

using the divergence theorem we have that the volume integral of div(R/r^3) over the solid sphere is equal to the flux of R/r^3 over the surface of the sphere. however the vector field is not defined at the origin so we cannot apply the divergence theorem in this case. however, if we take the sphere and an inner sphere that surrounds the origin then we can apply the divergence theorem on the volume enclosed by the concentric spheres.

then the volume integral would be 0 since div(R/r^3) is 0 except at the origin and we get that the flux through the surface any sphere surrounding the origin is 4π.

however it seems that my book went on to say that volume integral of div(R/r^3) is also 4π and with that div(R/r^3) = 4πδ. but i am confused about how they can say that the volume integral is also 4π since if the solid sphere includes the origin then the divergence theorem cannot be applied.

See post #10 in

www.physicsforums.com/showthread.php?t=200580

 

[demonelite123]

ah thanks for directing me to that post. i understand that you have to take a smaller sphere (or any smaller closed surface) inside the larger sphere and apply the divergence theorem to the volume in between the 2 spheres. this shows that no matter what surface you take around the origin you get that the flux is 4 pi.

however i don't understand how from that you can say that the integral over the solid sphere centered at the origin is equal to the flux integral over its surface. i thought the reason why we did the trick involving a small sphere that encloses the origin was because we could not apply the divergence theorem to the whole solid sphere in the first place. the vector field in the flux integral is not defined at the origin. the hypothesis of the divergence theorem says the vector field F, in the case of your post F = grad(1/x), should be defined on a neighborhood of the solid sphere but it is not because it isn't defined at the origin.