Gauss's Theorem

  • Thread starter hhhmortal
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  • #1
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Homework Statement



By computing ∇(1/r) show that

∫∇²(1/r).dV = -4Pi if the volume 'V' is a sphere containing the origin.





The Attempt at a Solution



∫∇²(1/r).dV = ∫∇(1/r).dS But I cant differentiate (1/r) since the sphere contains the origin. Im wondering if there's any other way to tackle this.
 
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Answers and Replies

  • #2
5,585
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Homework Statement



By computing ∇(1/r) show that

∫∇²(1/r).dV = -4Pi if the volume 'V' is a sphere containing the origin.





The Attempt at a Solution



∫∇²(1/r).dV = ∫∇(1/r).dS But I cant different (1/r) since the sphere contains the origin. Im wondering if there's any other way to tackle this.

You try to use this?

[tex]\nabla(\frac{1}{R})=\frac{\hat{R}}{R^{2}}[/tex]
 
  • #3
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You try to use this?

[tex]\nabla(\frac{1}{R})=\frac{\hat{R}}{R^{2}}[/tex]

How did you get this? I don't think I can differentiate it since the origin lies in the sphere, so the gradient of 1/0 cant happen.
 
  • #4
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[tex]\vec{R}=\hat{x}(x-x')+\hat{y}(y-y')+\hat{z}(z-z')[/tex]

[tex]\frac{1}{R}=\frac{1}{\sqrt{(x-x')^{2}+(y-y')^{2}+(z-z')^{2}}}\Rightarrow \nabla(\frac{1}{R}) = \hat{x}\frac{\partial \frac{1}{R}}{\partial x}+\hat{y}\frac{\partial \frac{1}{R}}{\partial y}+\hat{z}\frac{\partial \frac{1}{R}}{\partial z}[/tex]

[tex]\nabla(\frac{1}{R}) = -\frac{\hat{x}(x-x')+\hat{y}(y-y')+\hat{z}(z-z')}{((x-x')^{2}+(y-y')^{2}+(z-z')^{2})^{\frac{3}{2}}} = -\frac{\hat{R}}{R^{2}}[/tex]

Yes this won't work at the origin. That is the reason usually problem specify not including origin. Origin is treated differently.....something like Direc Delta function or something like that. I am not familiar with that.
 
  • #5
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[tex]\vec{R}=\hat{x}(x-x')+\hat{y}(y-y')+\hat{z}(z-z')[/tex]

[tex]\frac{1}{R}=\frac{1}{\sqrt{(x-x')^{2}+(y-y')^{2}+(z-z')^{2}}}\Rightarrow \nabla(\frac{1}{R}) = \hat{x}\frac{\partial \frac{1}{R}}{\partial x}+\hat{y}\frac{\partial \frac{1}{R}}{\partial y}+\hat{z}\frac{\partial \frac{1}{R}}{\partial z}[/tex]

[tex]\nabla(\frac{1}{R}) = -\frac{\hat{x}(x-x')+\hat{y}(y-y')+\hat{z}(z-z')}{((x-x')^{2}+(y-y')^{2}+(z-z')^{2})^{\frac{3}{2}}} = -\frac{\hat{R}}{R^{2}}[/tex]

Yes this won't work at the origin. That is the reason usually problem specify not including origin. Origin is treated differently.....something like Direc Delta function or something like that. I am not familiar with that.

Oh ok thanks, that way is much more clear! I still dont know how I could possibly get -4Pi without differentiating (1/r), can anyone help on this please!
 
  • #6
Dick
Science Advisor
Homework Helper
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You aren't supposed to actually integrate the laplacian of 1/r over the interior of the sphere. You can't, it's not even defined at the origin. You are just supposed to show that integrating grad(1/r) over the surface of any sphere centered at the origin is -4*pi, as yungman said. That shows DEFINING laplacian of 1/r to be a dirac delta function at the origin makes sense in terms of Gauss' theorem. Note that laplacian of 1/r is equal to zero EXCEPT at the origin.
 
  • #7
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You aren't supposed to actually integrate the laplacian of 1/r over the interior of the sphere. You can't, it's not even defined at the origin. You are just supposed to show that integrating grad(1/r) over the surface of any sphere centered at the origin is -4*pi, as yungman said. That shows DEFINING laplacian of 1/r to be a dirac delta function at the origin makes sense in terms of Gauss' theorem. Note that laplacian of 1/r is equal to zero EXCEPT at the origin.

Oh right! so,

∫∇(1/r).δ(r-r).dS = ∫∇(1/r).dS

we know the dirac delta function will be 1, so now I can take the gradient of (1/r) and integrate it over the surface of the sphere to get -4Pi.

I'm confused as to what would happen if the dirac delta function is zero?
 
  • #8
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Just for future reference, here's a handy formula:

[tex]\vec{\nabla}= \hat{r}\dfrac{\partial}{\partial r} + \hat{\phi}\dfrac{1}{r}\dfrac{\partial}{\partial\phi} + \hat{\theta}\dfrac{1}{r sin(\phi)}\dfrac{\partial}{\partial\theta}[/tex]

You can just "multiply" this operator by whatever function you want to take the gradient of, and you'll be able to do it in spherical coordinates. You can see that in the case of a spherically symmetric function like [itex]\frac{1}{r}[/itex], only the first term matters, which makes things even easier.
 
  • #9
Dick
Science Advisor
Homework Helper
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Oh right! so,

∫∇(1/r).δ(r-r).dS = ∫∇(1/r).dS

we know the dirac delta function will be 1, so now I can take the gradient of (1/r) and integrate it over the surface of the sphere to get -4Pi.

I'm confused as to what would happen if the dirac delta function is zero?

No, no, no, no. The delta function isn't in the surface integral, it's in the volume integral. You are supposed to show that the surface integral of grad(1/r) is the same as the volume integral of a 'laplacian' if you choose to define the 'laplacian' to be a delta function.
 
  • #10
I think you can assume that the origin is not on the surface of the sphere, so you can differentiate. And then just integrate.
([tex]
dS=r^2\sin{\theta}d\theta d\phi
[/tex])
 

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