ATAR Notes: Forum

Uni Stuff => Science => Faculties => Mathematics => Topic started by: /0 on November 27, 2009, 10:17:05 pm

Title: counter-intuitive divergence
Post by: /0 on November 27, 2009, 10:17:05 pm: Given the function

$\mathbf{v}(x,y,z)= \frac{x\mathbf{i}+y\mathbf{j}+z\mathbf{k}}{(x^2+y^2+z^2)^{\frac{3}{2}}}$

We have $\nabla \cdot \mathbf{v}=0$ , and yet, at www.pierce.ctc.edu/dlippman/g1/fullgrapher3d.html, the field is clearly diverging away from the origin AND slowing down.

Why is this?
Title: Re: counter-intuitive divergence
Post by: zzdfa on November 27, 2009, 10:29:11 pm: (http://)copying from wikipedia:
Quote
More rigorously, the divergence is defined as derivative of the net flow of the vector field across the surface of a small region relative to the volume of that region. Formally,

(http://upload.wikimedia.org/math/8/7/e/87e8309ca8f72f663519fa66885ca556.png)

so if you take smaller and smaller spheres around the centre, i guess the flux across the surface of the sphere gets smaller faster than the volume of the sphere shrinks. hence the ratio tends to 0.

just to clarify; divV=0 only at the origin, right?
Title: Re: counter-intuitive divergence
Post by: /0 on November 27, 2009, 10:34:04 pm: Yeah it seems like divF = 0 everywhere...
I dfon't know if divF is defined at the origin...
Title: Re: counter-intuitive divergence
Post by: zzdfa on November 27, 2009, 10:43:00 pm: Intuitive explanation:
this is just gauss' law.
take any closed surface not containing the origin in that space (i am pretty sure divF is undefined at 0, maybe not; i cbf doing the calculations); any arrows going into it (flux) is balanced by stuff going out of it. hence the flux is zero and using the definition above, div=0 as well.
Title: Re: counter-intuitive divergence
Post by: /0 on November 27, 2009, 10:46:16 pm: But does it matter that the arrows get shorter as you go out? It's like if you have a closed gaussian surface, the field lines will be stronger at one side than the other
Title: Re: counter-intuitive divergence
Post by: zzdfa on November 27, 2009, 10:49:08 pm: Quote from: /0 on November 27, 2009, 10:46:16 pm
But does it matter that the arrows get shorter as you go out? It's like if you have a closed gaussian surface, the field lines will be stronger at one side than the other

the fact that they do get shorter is what makes this work; have a look at this:
http://en.wikipedia.org/wiki/Inverse_square_law

Quote
. The density of flux lines is inversely proportional to the square of the distance from the source because the surface area of a sphere increases with the square of the radius. Thus the strength of the field is inversely proportional to the square of the distance from the source.
Title: Re: counter-intuitive divergence
Post by: /0 on November 27, 2009, 10:52:12 pm: Oh ok, but what if you go back and apply your reasoning to a simple divergence:

-> --> ---> ----> -----> ------> -------->
-> --> ---> ----> -----> ------> -------->
-> --> ---> ----> -----> ------> -------->
-> --> ---> ----> -----> ------> -------->

According to 'gaussian' logic, the divergence should be zero here too
Title: Re: counter-intuitive divergence
Post by: zzdfa on November 27, 2009, 10:53:34 pm: no. take

->|--> ---> ----> ----->| ------> -------->
->|--> ---> ----> ----->| ------> -------->
->|--> ---> ----> ----->| ------> -------->
->|--> ---> ----> ----->| ------> -------->

clearly there's more going out than there is going in* (area is the same but field strength is different).
Title: Re: counter-intuitive divergence
Post by: /0 on November 27, 2009, 11:02:30 pm: thanks heaps zzdfa, think i've got it now :)