- #1
Undergrad Question about divergence theorem and delta dirac function
- Thread starter Clara Chung
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The discussion centers on proving the validity of the divergence theorem for any closed surface containing the origin, specifically that the line integral equals 4π. It is noted that by removing a small spherical cavity at the origin, the total surface integral remains zero, while the integral over the cavity yields -4π. The conversation also touches on the substitution of vectors in deriving equations, explaining that the transformation from r to r - r' is permissible through the chain rule. Participants suggest that using spherical coordinates simplifies the proof without loss of generality. The overall focus is on applying mathematical principles to validate the theorem.
- #2
jedishrfu
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I would say by inspection, if you note that for a non-spherical surface the R would in fact be a function: ##R(r,\theta,\phi)## and that ##R(r,\theta,\phi)## factors cancel out using the rules of algebra, leaving an integral of angles only.
Since the ##R(r,\theta,\phi)## factors cancel, then you can choose R to be a spherical surface with no loss in generality.
Perhaps, @Mark44 or @fresh_42 have a more rigorous set of steps.
Since the ##R(r,\theta,\phi)## factors cancel, then you can choose R to be a spherical surface with no loss in generality.
Perhaps, @Mark44 or @fresh_42 have a more rigorous set of steps.
- #3
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Clara Chung said:
You can use the divergence theorem. Take any volume that includes the origin. To avoid the problem at the origin you could remove a small spherical cavity at the origin. The total surface integral is zero and the surface integral of the small cavity is ##-4\pi##.
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- #4
- #5
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Clara Chung said:
That's just a substitution ##\vec{r}## to ##\vec{r} - \vec{r'}##.
- #6
Clara Chung
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How do you do the substitution? Why is it ok to let ##\vec{r}## = ##\vec{r} - \vec{r'}## ? They are not equal..PeroK said:
- #7
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Clara Chung said:
In general, if ##f'(x) = g(x)##, then ##f'(x-a) = g(x-a)##. It's a simple application of the chain rule.
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