Green's function for the sphere

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Homework Statement


If a hollow spherical shell of radius a is held at potential [itex]\Phi(a, \theta ', \phi ')[/itex], then the potential at an arbitrary point is given by,

[tex]\Phi(r, \theta, \phi)=\frac{1}{4 \pi} \oint \Phi(a, \theta ', \phi ') \frac{\delta G(r, r')}{\delta n '}dS'[/tex]

where [tex]G(r, r')=\frac{1}{(r^2+a^2-2arcos \gamma)^{3/2}}-\frac{a}{r'(r^2+\frac{a^4}{r'^2}-2r\frac{a^2}{r'}cos \gamma)^{3/2}}[/tex]

and [tex]dS'=a^2 sin(\theta ') d \theta ' d \phi '[/tex]

Suppose the sphere is held at a fixed potential, [itex]V_0[/itex].
(a) Calculate the potential outside the sphere.
(b) Use the formula above to calculate the potential inside the sphere.

The Attempt at a Solution



[tex]\Phi(r, \theta, \phi)=\frac{V_0}{4 \pi} \oint \frac{\delta G(r, r')}{\delta n '}dS'=\frac{V_0 a^2}{4 \pi}\int_0^{2 \pi} d \phi ' \int_0^{\pi}\frac{\delta G(r, r')}{\delta n '} sin(\theta ') d \theta '=\frac{V_0 a^2}{2}\int_0^{\pi}\frac{\delta G(r, r')}{\delta n '} sin(\theta ') d \theta '[/tex]
The problem I am having is that when I calculate the potential outside the sphere then set r=a, I do NOT get the potential as being [itex]V_0[/itex] on the surface as I should.
One spot where I may have messed up is in calculating the derivative of the Green function,

[tex]\frac{\delta G(r, r')}{\delta n'}=-\frac{\delta G(r, r')}{\delta r'} |_{r'=a}[/tex]

I believe this is correct since the normal component of the large sphere points inward along r'..anyway after some algebra I get the answer to be

[tex]\frac{2a^2-r^2-racos \gamma}{a(r^2+a^2-2racos \gamma)^{5/2}}[/tex]

I am thinking either I took this derivative wrong, or I later integrated the expression wrong.

Thanks for your comments.
 
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Answers and Replies

  • #2
Hi kreil,

I am pretty sure this is your problem. You have defined your Green function with powers of 3/2 in your denominators when they should rather be powers of 1/2.

I am assuming you have G defined such that on the surface of the sphere, G=0. Well, to prove to yourself which form is correct, try plugging R into x to see which power (1/2 or 3/2) gives you G(x=R)=0.

Good Luck!
 

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