Point-Source derivative form Gauss's law valid?

  • #1
I reviewed Maxwell's equations today and I was upset to find that the volume integral of the derivative form of Gauss's equation seems to be 0 for point sources when it should come out to be Q/(electro constant). This comes from the fact that the gradient is equal to 0 and anything else you do to 0 appears to make it stay zero.

By the way I checked both versions of Gauss's law on valid field equations that did not have singularities and everything seemed to work out fine. (For instance, I used the integral form of gauss's law to create a field equation for gravity inside a sphere of uniform density to check the derivative form of Gauss's law)

What I did (and sorry I don't know how to make the math look pretty like you other guys do):

It looks like the equation for the field of a point source is -kq(x/r^3, y/r^3, z/r^3) where r^2 = x^2 + y^2 + z^2 and k = 1/4/pi/(electro constant). You can use Pythagoras's theorem to check that the magnitude of this vector is kq/r^2. When I took the divergence for x I got -1/(x^2 + y^2 + z^2)^1.5 + 3x^2/(x^2 + y^2 + z^2)^2.5 .... (pattern continues for y and z). Which equals -3/r^3 + 3(x^2 + y^2 + z^2)/r^5 = 3(1/r^3 - 1/r^3) = 0. Maybe you should check to make sure I did this right. This is what you would expect from Gauss's theorem I think because the charge density is zero everywhere except at the point source. However, taking the volume integral of this should give you kq4pi = q/(electro constant).

1. Have I made a mistake?

2. If not, is there a way to resolve this issue with the singularity to make Gauss's law valid in this case?
 
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Answers and Replies

  • #2
rude man
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The problem is there is a singularity at the origin. The diivergence is not definable there.

Using spherical coordinates to make life easier, with vectors in bold,

D = Dr r where r = unit vector in r direction,
so
div D = (1/r2)∂/∂r(r2Dr)
which at r = 0 gives ∞ for the 1/r2 part and zero for the partial derivative, and of course Er= Dr0.

So what is zero times infinity? Trouble, that's what ...

Possibly one can fool around with the δ function to get around this. Any takers?

Somehow you did not take div D correctly. I suspect you "used" the ∞ times zero thing and got beached. As you probably know, you can "prove" 1 = 2 by falling in that trap.

P.S. the divergence theorem is OK because it multiplies div D by volume element dv = (4/3)πdr3 surrounding the point charge. dv is infinitesmally small but not zero, whereas a point charge has identically zero volume.
By the same token, the surface element dA surrounding dv is non-zero.
 
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