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Divergence Theorem Q

  1. Jan 26, 2009 #1
    I need to show that, using Gauss' Theorem (Divergence Theorem), i.e. integration by parts, that:

    [itex]\int_V dV e^{-r} \nabla \cdot (\frac{\vec{\hat{r}}}{r^2}) = \int_V dV \frac{e^{-r}}{r^2}[/itex]

    any ideas on where to start?
     
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  3. Jan 26, 2009 #2

    gabbagabbahey

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    Hint:The following vector identity should be useful:

    [tex]\vec{\nabla}\cdot{f\vec{A}}=f(\vec{\nabla}\cdot\vec{A})+\vec{A}\cdot(\vec{\nabla}f)[/tex]
     
  4. Jan 26, 2009 #3
    thanks.
    1, do i set [itex]f=\frac{1}{r^2}[/itex]?
    2, how did you know to do that so quickly?
     
  5. Jan 26, 2009 #4

    gabbagabbahey

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    1. No, set [tex]f=e^{-r}[/tex]

    2. It's a common theme in many EM derivations.
     
  6. Jan 26, 2009 #5
    ok so it then equals
    [itex] \int_V dV \nabla \cdot (\frac{e^{-r} \vec{\hat{r}}}{r^2}) - \int_V dV (\frac{\vec{\hat{r}}}{r^2}) \cdot (\nabla e^{-r}) = \int_S (\frac{e^{-r} \vec{\hat{r}}}{r^2}) \cdot \vec{n} dS - \int_V dV (\frac{\vec{\hat{r}}}{r^2}) \cdot (\nabla e^{-r}) [/itex]
    My instinct is to do the integral over the surface S using S = sphere of radius a as then [itex]\vec{n}=\vec{\hat{r}} [/itex] and i can get rid of that dot product. but then [itex]dS=4 \pi a^2[/itex] which messes stuff up and also what do I do on the right?
     
  7. Jan 26, 2009 #6

    gabbagabbahey

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    I'm assuming that the original integral was over all of space, correct?

    If so, then your surface integral is over the boundary of all space which is at [tex]r\to\infty[/tex]....what is the value of the integrand there?:wink:

    For the second term; Take the gradient of [tex]e^{-r}[/tex] using spherical coords....
     
  8. Jan 26, 2009 #7
    ok. might be getting somewhere now. on the right i took the grad of the exponential term and got [itex] \nabla e^{-r} = -e^{-r} \vec{\hat{r}}[/itex] is that right?
    seems to be so because that causes the integral on the right to simplify substantially to [itex]-\int_V dV (\frac{\vec{\hat{r}}}{r^2}) \cdot (\nabla e^{-r}) = + \int_V dV \frac{e^{-r}}{r^2}[/itex] which is what we're looking for - therefore it appears to just be a matter of showing the integral on the left vanishes?
     
  9. Jan 26, 2009 #8
    indeed it is over all space. i would then imagine it would zero because [itex] e^{-r} \rightarrow 0[/itex] as [itex] r \rightarrow \infty[/itex]. If this is the case, how owuld you recommend I write that in order to be an acceptable answer? Also, I took the grad of the exponential using Cartesians - does that matter?
     
  10. Jan 26, 2009 #9

    gabbagabbahey

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    Formally, you could start by assuming that your volume is a sphere of radius [itex]a[/itex] and then take the limit as [itex]a[/itex] approaches infinity.

    And it doesn't matter that you used Cartesian coords to take the gradient; although it would probably have been quicker to use spherical coords.
     
  11. Jan 26, 2009 #10
    ok
    but then
    [itex]\int_S (\frac{e^{-r} \vec{\hat{r}}}{r^2}) \cdot \vec{n} dS = \int_S (\frac{e^{-r} \vec{\hat{r}} \cdot \vec{\hat{r}}}{r^2}) dS= \frac{e^{-r}}{r^2} 4 \pi a^2 [/itex]
    if i take the limit as [itex]a \rightarrow \infty [/itex] i get [itex]\infty[/itex]
     
  12. Jan 26, 2009 #11
    unless i take [itex]dS=4 \pi r^2[/itex] and then the intergal becomes [itex]4 \pi e^{-r} [/itex] and then the limit as [itex]r \rightarrow \infty [/itex] would be 0. Any advice? Cheers.
     
  13. Jan 26, 2009 #12

    gabbagabbahey

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    Along the surface, [itex]r=a[/itex].
     
  14. Jan 26, 2009 #13
    i dont understand what you mean there sorry
     
  15. Jan 26, 2009 #14
    actually surely if i integrate over a sphere of radius a, [itex]dS=a^2 \sin{\theta} d\theta d\phi [/itex]?

    in which case this integral becomes [itex]\int_S \frac{e^{-r} a^2}{r^2} \sin{\theta} d\theta d\phi = 4 \pi \frac{a^2}{r^2} e^{-r} [/itex]

    would i then let a or r go to infinity, or is this all complete bollocks? HELP im really confused now! arghhhh!
     
  16. Jan 26, 2009 #15

    gabbagabbahey

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    [itex]r[/itex] is the distance from the origin to the infinitesimal area element [itex]dS[/itex]. Since the entire surface is a distance [itex]a[/itex] from the origin, that means that [itex]r=a[/itex] everywhere on the surface.

    [tex]\implies \int_S \frac{e^{-r}}{r^2}dS=\int_S \frac{e^{-a}}{a^2} dS =\int_S \frac{e^{-a} a^2}{a^2} \sin{\theta} d\theta d\phi = 4 \pi e^{-a} [/tex]
     
  17. Jan 27, 2009 #16
    cheers buddy!
     
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