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Divergence what am I doing wrong

  1. Mar 29, 2006 #1
    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,
     
  2. jcsd
  3. Mar 29, 2006 #2

    Tom Mattson

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    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].
     
  4. Mar 29, 2006 #3
    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
     
  5. Mar 29, 2006 #4

    J77

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    ...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])
     
  6. Mar 29, 2006 #5

    Tom Mattson

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    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.
     
  7. Mar 29, 2006 #6

    J77

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    I'm having a very bad brain day :frown:
     
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