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Author Topic: counter-intuitive divergence  (Read 1867 times)  Share 

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/0

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counter-intuitive divergence
« on: November 27, 2009, 10:17:05 pm »
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Given the function



We have , 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?

zzdfa

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Re: counter-intuitive divergence
« Reply #1 on: November 27, 2009, 10:29:11 pm »
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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,

(Image removed from quote.)

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?
« Last Edit: November 27, 2009, 10:34:53 pm by zzdfa »

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Re: counter-intuitive divergence
« Reply #2 on: November 27, 2009, 10:34:04 pm »
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Yeah it seems like divF = 0 everywhere...
I dfon't know if divF is defined at the origin...
« Last Edit: November 27, 2009, 10:38:12 pm by /0 »

zzdfa

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Re: counter-intuitive divergence
« Reply #3 on: November 27, 2009, 10:43:00 pm »
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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.



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Re: counter-intuitive divergence
« Reply #4 on: November 27, 2009, 10:46:16 pm »
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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

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Re: counter-intuitive divergence
« Reply #5 on: November 27, 2009, 10:49:08 pm »
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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.

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Re: counter-intuitive divergence
« Reply #6 on: November 27, 2009, 10:52:12 pm »
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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

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Re: counter-intuitive divergence
« Reply #7 on: November 27, 2009, 10:53:34 pm »
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no. take

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


clearly there's more going out than there is going in* (area is the same but field strength is different).
« Last Edit: November 27, 2009, 11:04:26 pm by zzdfa »

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Re: counter-intuitive divergence
« Reply #8 on: November 27, 2009, 11:02:30 pm »
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thanks heaps zzdfa, think i've got it now :)