Divergence what am I doing wrong

[FrogPad]

I don't understand what I am doing wrong here.

I'm supposed to show that this function is divergence free.

[tex] \vec v = \left( \frac{x}{x^2+y^2}, \frac{y}{x^2+y^2} \right) [/tex]

I ran the divergence through with my TI-89 at it equals 0. But, I want to calculate it by hand, so it would be easier to do this in polar coordinates (and this is my problem).

[tex] \vec v(r,\theta) = \left ( \frac{\cos \theta}{r}, \frac{\sin \theta}{r} \left) [/tex]

Standard divergence in cylindrical coordinates (dropping the z component)
[tex] div\,\, \vec u = \frac{1}{r} \left( \frac{\partial}{\partial r} (rF_r) +\frac{\partial}{\partial \theta} (F_\theta) \right) [/tex]

[tex] div\,\, \vec v = \frac{1}{r}\left( \frac{\partial}{\partial r} (\cos \theta ) + \frac{\partial}{\partial \theta}\left (\frac{\sin \theta}{r} \right) \right) [/tex]

Now this is obviously not 0, since the [itex] \sin \theta [/itex] is not going anywhere with the partials. So what am I doing wrong?

Thanks,

 

[quantumdude]

[tex] \vec v(r,\theta) = \left ( \frac{\cos \theta}{r}, \frac{\sin \theta}{r} \left) [/tex]

You made a subtle mistake here. Your vector above is still just [itex]
\vec{v}=(v_x,v_y)[/itex]. You just changed the expressions for the components. You do not have [itex]\vec{v}=(v_r,v_{\theta})[/itex] yet. To get that you have to transform the basis vectors from [itex]\hat{e}_x[/itex] and [itex]\hat{e}_y[/itex] to [itex]\hat{e}_r[/itex] and [itex]\hat{e}_{\theta}[/itex].

 

[FrogPad]

I'm pretty sure I follow what you are saying. Well, I at least understand it. I am unfortunatly too tired to tackle it, so I will do it in the morning... But I should be good with what you said :)

thanks man

 

[J77]

...but I think he's pretty much there.

Since the [tex]\cos[/tex] and [tex]\sin[/tex] are bounded, and [tex]1/r\rightarrow 0[/tex] as [tex]r\rightarrow\infty[/tex]

(with the exception at [tex]r=0[/tex])

 

[quantumdude]

No, he is not even close. This has nothing to do with taking a limit. The divergence should vanish identically in polar coordinates, just as it does in rectangular coordinates. His mistake is exactly what I said it was: He is plugging the rectangular components of [itex]\vec{v}[/itex] into the polar form of [itex]\vec{\nabla}[/itex].

[itex]v_x=\frac{x}{x^2+y^2}=\frac{\cos(\theta)}{r}[/itex]
Note that this is still just [itex]v_x[/itex]

[itex]v_y=\frac{y}{x^2+y^2}=\frac{\sin(\theta)}{r}[/itex]
Note that this is still just [itex]v_y[/itex]

He needs to find the components [itex]v_r[/itex] and [itex]v_{\theta}[/itex], which both contain contributions from [itex]v_x[/itex] and [itex]v_y[/itex]. This is done by transforming the basis vectors, just like I said.

 

[J77]

No, he is not even close. This has nothing to do with taking a limit.

I'm having a very bad brain day