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I've been explained that a vector field, when seen as "arrows" in the plane, has 0 divergence when its magnitude doesn't change, i.e. when the "arrows" keep same length. But the following examples puzzle me:

$F(x)=x/|x|$ has always norm 1 but its divergence is not 0

$F(x)=x/|x|^2$ has not constant norm but its divergence is 0

Is there some contradiction or do I have a wrong/incomplete picture?

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migrated from Feb 17 '13 at 2:14

This question came from our site for active researchers, academics and students of physics.

marked as duplicate by JohnD, Amzoti, Henry T. Horton, 5pm, Asaf Karagila Feb 17 '13 at 5:16

This question was marked as an exact duplicate of an existing question.

Just want to mention that looking on divergence in that way is really a very narrow point of view and losses it's actual physical meaning which is the net field flow through closed surface. – TMS Feb 17 '13 at 0:21

I don't think your examples contradict your view on the divergence. Take your first example, the divergence is

$$\frac{\partial}{\partial x}\left(\frac{x}{\sqrt{x^2}}\right) = \frac{\sqrt{x^2} - x\frac{2x}{2\sqrt{x^2}}}{x^2} = \frac{\sqrt{x^2} - \frac{x^2}{\sqrt{x^2}}}{x^2} = 0.$$

And similarly we find for your second example:

$$\frac{\partial}{\partial x}\left(\frac{x}{x^2}\right) = \frac{x^2 - x\cdot2x}{x^4} = -\frac{x^2}{x^4} = -\frac{1}{x^2},$$

which is clearly not zero.

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