Re: Divergence Thm, 1/r^2 flux, and the origin

On Jun 21, 6:48 pm, "C...@xxxxxxx" <C...@xxxxxxx> wrote:
On Jun 21, 1:20 pm, J Jensen <jjense...@xxxxxxxxxxx> wrote:



On Jun 21, 7:25 am, David C. Ullrich <ullr...@xxxxxxxxxxxxxxxx> wrote:

On Wed, 20 Jun 2007 22:07:45 -0700, J Jensen <jjense...@xxxxxxxxxxx>
wrote:

I'm not quite seeing why the proof of the divergence theorem should
fail in this simple case:

Let's take a solid ball in R^3 centered at the origin, say B=B(0,R)
and the vector field
F= r^{-3} r or 1/r^2 e_r ( anyway, its the inverse square radial
vector field from the origin).

Obviously, this vector field has a singularity at 0, but it seems like
that should just be ignorable and the proof of the divergence thm
should still work.

Huh? Why would you imagine that?

The div theorem is: Integral over boundary B of F.n dA = Integral
over B of Div(F) and Div(F) = 0 for all points except 0.

Thus the surface integral *should* be 0, but by hollowing out B by
removing a small ball B(0,\epsilon), we apply the divergence thm to
the new manifold and get the answer of 4\pi for the surface integral.

--Jeff

************************

David C. Ullrich

The proof of the Divergence Thm, as given in Marsden & Tromba 'Vector
Calculus' is:

Let M be a region in R^3 of "Type 4" (which seems to mean that it's
boundary can be described
by curves in such a way that M can be integrated over as an iterated
integral in any order
of dx,dy, dz).

Let boundary of M be oriented.

Let the vector field F: M ---> R^3 be smooth. F= [ f^1, f^2, f^3 ]

THEN Integral{boundry M} (F.n dA) = Integral{M}(del.F dV)

For the pf, they expand both sides and show term by term equality:

Int{dM}( f^i n^i dA ) = Int(M)(D_i f^i dV) where n^i is the i-th
component of surface normal.

Let us take i=3, so it is the z component for concreteness.

Since M is a "type 4" region, the RHS integral can be written:

Int{D}( Int{ l(x,y) to h(x,y) } ( d/dz f^3 dz ) dx dy

where D is a region in the x-y plane, and the "top" and "bottom" of M
can be expressed as graphs
over D.

From here we just evaluate this inner integral by the fundamental thm

of calculus and then we
write out the formula for the normal surface vector to a graph and
show that the surface integral
of the 3rd component of the flux equals what we just calculated.

So, the only place I can see where we touch the bad point at 0 is when
we evaluate the
iterated volume integral, and certainly Fubini's thm can handle a
measure 0 singularity...

Yes, for an "ordinary" function. But Div(F) is not such; if anything,
it is a generalized function (i.e., a "distribution"). For more
details, see the books by L Schwartz on "Distributions", or one of the
numerous books that deal with distributions or generalized functions.
See, eg.,http://eom.springer.de/g/g043810.htmhttp://www.claymath.org/programs/outreach/academy/LectureNotes/Vasy.pdfhttp://adsabs.harvard.edu/abs/1994igfa.rept.....F



So what obvious thing am I missing?

Apparently, you missed my entire initial response. It is true that
Div(F) = 0 (except at the origin) implies that the surface integral
over a sphere of radius R is the same as the surface integral over any
"nice" region having the origin in its interior. Basically, this is
the basis of Gauss' Law that is used all the time in electrostatics.

R.G. Vickson

Thanks for the detailed response, R.G. However, I am just working on
a "simple" vector calculus problem, so distribution theory should play
no role. I just want to know precisely what step in the proof I posted
here ( from Marsden & Tromba 'Vector Calculus' ) fails in the case
that I am considering.

--Jeff

.

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