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Born approximation

  1. Jan 2, 2008 #1
    1. The problem statement, all variables and given/known data
    A particle is charge +e is incident on an electric dipole of charge +e and a charge of -e separated by a vector d (which runs from -e to +e). Use the Born approximation to calculate the differential scattering cross section as a function of the initial wave vector, the scatter wave vector, and the standard Rutherford scattering cross section.


    2. Relevant equations



    3. The attempt at a solution
    So they gave me the relative positions of the two charges. But I do not see how this is possible without knowing the absolute position of one of charges. Don't I need that in order to write down the potential?
     
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  3. Jan 2, 2008 #2

    malawi_glenn

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    you only need the relative position of the electrons in order to get a potential. the cross section here is just respect to the dipol, not the lab or anything like that.

    When you calculate the cross section for electron scattering on a nucleus, you dont have to know where in the room the nuclues are, only the form factor etc (the distrubution of protons).
     
  4. Jan 2, 2008 #3
    But the cross-section needs to be with respect to a point i.e. a solid angle is defined as a projection from a point in space. A dipole is not a point in space. Do I just arbitrarily choose the plus or minus charge for my point?
     
  5. Jan 2, 2008 #4

    malawi_glenn

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    Have you tried do forumulate a potential for the dipole? The potential for this diploe is the superposition of two columb potentials, separeted by the vector d of yours. i.e take the origin in one of the particles in the dipole.

    Field from an electric dipole:
    http://en.wikipedia.org/wiki/Dipole
     
    Last edited: Jan 2, 2008
  6. Jan 2, 2008 #5
    Of course.

    [tex]\frac{d\sigma}{d\Omega}=\left(\frac{m}{2\pi\hbar^2} \right)^2\left|\int V(\vec{r}) e^{-i \vec{K}\cdot\vec{r}}d^3\vec{r} \right|^2[/tex]

    where K is the momentum transfer and m is the mass.

    The problem I am having is how to write down the potential. The standard Coulomb potential for two positive charges would be is [itex]V(r) = \frac{e^2}{4\pi\epsilon_0 r}[/itex] where r is the distance between the two charges.
    But now we have three charges, two in the dipole and one incident on the dipole and we want to calculate the potential for the latter. There are two contributions:

    [tex]V(r_1) = \frac{e^2}{4\pi\epsilon_0 r_1}[/tex]

    [tex]V(r_2) = \frac{-e^2}{4\pi\epsilon_0 r_2}[/tex]

    where r_1 is the distance from the positive charge and r_2 is the distance from the negative charge. So there are basically two coordinate systems: the r_1 coordinate system and the r_2 coordinate system. Which one should I use? Or should I construct a third one?
     
    Last edited: Jan 2, 2008
  7. Jan 2, 2008 #6

    malawi_glenn

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    see my post above, i did some editing.
     
  8. Jan 2, 2008 #7
    I read the Wikipedia article and I still have the same questions from above. They seem to assume that the absolute positions of the dipole are known.
     
  9. Jan 2, 2008 #8

    malawi_glenn

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    when you are given the relative vector, then it is assumed that the "origin" is in one of the particles...

    if A and B are separated by the vector (3,2,-4) in the 3dim space, then it you can put the origin in A or B. You can put it anywere, the important thing is that you have the relative position vector. The cross section does not depend where in space (relative to the beam of course) the scatterer are.
     
  10. Jan 2, 2008 #9

    malawi_glenn

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    [tex] \vec{r} [/tex] is the vector from the incoming e+ to the e- in the dipole. Then [tex] \vec{r} + \vec{d} [/tex] is the vector from the incoming e+ and the e+ in the dipole. How would you construct a potential for this one?
     
  11. Jan 2, 2008 #10
    [tex] V(\vec{r}) = \frac{-e^2}{4\pi\epsilon_0 r} + \frac{e^2}{4\pi\epsilon_0 \left|\vec{r}-\vec{d} \right|}=\frac{-e^2}{4\pi\epsilon_0 r} + \frac{e^2}{4\pi\epsilon_0 \sqrt{r^2+d^2 -2\vec{r}\cdot\vec{d}}}[/tex]


    [tex]\frac{d\sigma}{d\Omega}=\left(\frac{m}{2\pi\hbar^2 } \right)^2\left|\int \frac{-e^2}{4\pi\epsilon_0 r}e^{-i \vec{K}\cdot\vec{r}}d^3\vec{r} + \int \frac{e^2}{4\pi\epsilon_0 \sqrt{r^2+d^2 -2\vec{r}\cdot\vec{d} }} e^{-i \vec{K}\cdot\vec{r}}d^3\vec{r} \right|^2[/tex]

    Is that right?
    I am not really sure how to evaluate the second integral.
     
    Last edited: Jan 2, 2008
  12. Jan 2, 2008 #11

    malawi_glenn

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    go to spherical coordinates, remember that the angle between r and d is different from K and r, and that you have the same angel in the spherical coordinat integration element as the angle bewteen K and r.


    Your potential seems ok now, but it should be [tex] \vec{r} + \vec{d} [/tex]. And the magnitude of distance is the square root of the dot-product.

    EDIT: I have not a clue what the answer should be, I remember we did this one in QM scattering a while ago, but have not the answer at home :)
     
    Last edited: Jan 2, 2008
  13. Jan 2, 2008 #12
    The distance between two vectors is found by taking the norm of their difference.
     
  14. Jan 2, 2008 #13

    malawi_glenn

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    I dont know what you are talking about now.

    [tex] \vec{r} [/tex] is the vector from the incoming e+ to the e- in the dipole. Then [tex] \vec{r} + \vec{d} [/tex] is the vector from the incoming e+ and the e+ in the dipole. So the distance from the incoming e+ and the e+ in the potential is [tex] \left|\vec{r} + \vec{d} \right| [/tex], right?

    You have the answer for this one?
     
    Last edited: Jan 2, 2008
  15. Jan 2, 2008 #14
    Yes. I have the answer. I can give it to you if that will help.

    When my book did the monopole Coulomb potential case, they added a convergence factor of e^{-r/d} to the integral and then sent d to infinity after the integration was done. Do I need to do that for the dipole case?
     
  16. Jan 2, 2008 #15

    malawi_glenn

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    I have never added a convergence factor when having monopole scattering.

    But in some way, there should not be an angle dependance of the angle between r and d in the final expression.
     
  17. Jan 2, 2008 #16
    Both of the books I am using put in convergence factors.

    EDIT: this trick is called regularization
    EDIT: http://en.wikipedia.org/wiki/Regularization_(physics)

    By the way, the answer is,

    [tex] 4\sin^2\left((\vec{k_f}-\vec{k_i})\cdot \vec{d}/2\right)\left( \frac{d\sigma}{d\Omega}\right)_R [/tex]

    where the last term is the standard Coulomb scattering differential cross-section. I have no idea how that is going to come out of my integral. We don't even have K dot d.
     
    Last edited: Jan 2, 2008
  18. Jan 3, 2008 #17

    malawi_glenn

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    K = k_i - k_f in your expression right?

    regularization I have used in other problems.

    Have you tried work out the integrals?
    k dot d = |vec(k)| |vec(d)| cos(angle)
     
    Last edited: Jan 3, 2008
  19. Jan 3, 2008 #18
    I am saying that

    [tex]\frac{d\sigma}{d\Omega}=\left(\frac{m}{2\pi\hbar^2 } \right)^2\left|\int \frac{-e^2}{4\pi\epsilon_0 r}e^{-i \vec{K}\cdot\vec{r}}d^3\vec{r} + \int \frac{e^2}{4\pi\epsilon_0 \sqrt{r^2+d^2 -2\vec{r}\cdot\vec{d} }} e^{-i \vec{K}\cdot\vec{r}}d^3\vec{r} \right|^2[/tex]

    does not contain a K dot d. So I don't see how it will contain one after we integrate.
     
  20. Jan 3, 2008 #19

    malawi_glenn

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    So you dont think that it can fall out? I dont think I can help you more on this one=/
     
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