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While trying to understand the divergence of a vector through the geometrical representation of the vector field, I found that pictures can be misleading. Even a vectors field which looks to be diverging while one observes the direction of arrows may be actually having a zero divergence (example $F(x,y,z)= \frac{(ix+jy+kz)}{(x^2+y^2+z^2)^(3/2)}$) (at points other than origin)or even a negative divergence (example $F(x,y)= \frac{(ix+jy)}{(x^2+y^2)^(3/2)}$). While trying to understand it through visualization of flux entering into and going out form a volume element at the position concerned, I found that the magnitude of the vector is represented by the length of the arrows and the direction is shown by the direction of the arrows.

  • What about the density of points (tail point of the vectors). What do they represent?

Maybe these points do not represent anything and the flux can be calculated by taking dot product of the vector with the surface elements which are parts of the total surface covering the volume element. But in that case the surface offered for incoming flux and for outgoing flux are same. So the flux only depends on the magnitude of the length of the incoming and outgoing vectors. Consider the case when the vectors away from the origin are shorter than the vectors nearer to origin as in the examples stated above. In that case the divergence will be always negative.

  • How the case of zero divergence arise then?
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closed as unclear what you're asking by ACuriousMind, Sebastian Riese, Norbert Schuch, Daniel Griscom, user36790 Jan 15 '16 at 2:41

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  • $\begingroup$ Maybe you have a look at this. Your example is not zero everywhere, i.e. at the origin. $\endgroup$ – mikuszefski Jan 14 '16 at 11:20
  • $\begingroup$ I am talking about the positions other than the origine. I admit that the first examples coincidentally represents the electric field of a point charge. It can be replaced by another example. My question is regarding the visualization of divergence of a vector field in general. How the inward and outward flux are calculated so that the difference between them says whether the divergence is negative positive or zero. $\endgroup$ – user103515 Jan 14 '16 at 11:30
  • $\begingroup$ Well, you basically write it. (Coming back to the point charge)The total flux through any closed surface not containing the origin is zero and such is the divergence inside. $\endgroup$ – mikuszefski Jan 14 '16 at 11:40
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    $\begingroup$ Why do you say "the flux only depends on the magnitude of the length of the incoming and outgoing vectors." That is not correct. The sign of the flux depends on the direction of the vector with respect to the "direction" of the surface. The "direction" of a closed surface is taken to point from inside toward outside of the surface, by convention. $\endgroup$ – garyp Jan 14 '16 at 11:47
  • $\begingroup$ Suppose the elemental volume is a small sphere. Then half of it is subjected to incoming flux and exactly the other half is subjected to the outgoing flux. Angle between incoming vectors and the surface varies from -pi/2 through 2pi to pi/2 and that for the outgoing vector ranges form pi/2 through pi to -pi/2. So the contribution from the angles is only to change the sign of the flux. But the magnitude of the flux becomes dependent on the length of the vectors only. $\endgroup$ – user103515 Jan 14 '16 at 12:02