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Feynmann diagrams, bhabha scattering

  1. Jun 10, 2007 #1
    There is nowone who can help me on the homework forum, so i'll try it here....

     
  2. jcsd
  3. Jun 10, 2007 #2
    Thanks for your reply!

    I still don't get it exactly

    First Question:

    If p3=p4, then p3-p4=0 and the moment of q=p1+p2=p3+p4, I don't get why it should be 0....

    Second Question:

    The '*' is the innerproduct (inproduct), so I don't know if then the rest of your reply holds?
     
  4. Jun 10, 2007 #3

    Hans de Vries

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    First Question: (edited for the annihilation-creation diagram)

    The answer follows from the on-shell condition.
    E1+m = E3+E4, p1 = p3+p4
    This means that E1 and p1 can only be on-shell if p3=p4=0

    Second Question:

    General advice: First try to simplify things with the given relations before
    starting to take squares and things like that. Then use the on-shell
    relations to prove the given restrictions.

    I suppose the '*' is the cross product here. The cross product means that
    p3 and p4 are in the same direction, thus (p4x,p4y,p4z) = a(p3x,p3y,p3z)

    This gives us:

    [tex]E_4+E_3\ =\ \sqrt{ a^2p_3^2 + m^2} + E_3[/tex]

    Since p2= (m,0,0,0) we have q = (E1+m, p1x, p1y, p1z)

    It follows that (p1x,p1y,p1z) = (1-a)(p3x,p3y,p3z)

    This gives us:

    [tex]E_1 = \sqrt{ (1-a)^2p_3^2 + m^2) }[/tex]

    Thus:

    [tex]E_3 = m + \sqrt{ (1-a)^2p_3^2 + m^2} - \sqrt{ a^2p_3^2 + m^2) }[/tex]

    This means that either [itex]a[/itex] must be 0 and thus p4=0 or p3=0.
    There is another solution and that is [itex]a=1[/itex]. In this trivial case p4=p3
    and they can be non-zero......


    Regards, Hans
     
    Last edited: Jun 10, 2007
  5. Jun 10, 2007 #4

    Hans de Vries

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    To which of the two diagrams contributing to Bhabha scattering are you
    referring to. The electron positron scattering or the annihilation/creation one?

    It holds for the x product....:rofl: (for the annihilation/creation diagram)


    Regards, Hans
     
  6. Jun 10, 2007 #5
    To which of the two diagrams contributing to Bhabha scattering are you
    referring to. The electron positron scattering or the annihilation/creation one?

    Annihahilition: p1+p2(incoming momenta)=p3+p4(outgoing momenta)
    Scattering: p1-p3=p4-p2 (toch?)

    It holds for the x product....:rofl: (for the annihilation/creation diagram)

    Doesn't it hold for the innerproduct?
     
  7. Jun 10, 2007 #6

    Hans de Vries

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    OK, The answer then follows from the on-shell condition.

    E1+m = E3+E4, p1 = p3+p4

    This means that E1 and p1 can only be on-shell if p3=p4=0


    Regards, Hans
     
  8. Jun 10, 2007 #7
    Mmm....
    I still don't get it completely:blushing:

    :confused:
     
  9. Jun 10, 2007 #8

    Hans de Vries

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    So p3 and p4 are under 90 degrees. It follows that:

    [tex]p_{3\bot} = -p_{4\bot} [/tex]

    [tex]p_{3\|} + p_{4\|} = q = p_1[/tex]

    The latter is the same as

    [tex]{p_1 \over |p_1|} \cdot p_3\ +\ {p_1 \over |p_1|} \cdot p_4\ =\ p_1[/tex]

    The first left hand term is zero if the second one is at a maximum
    and visa versa. For the energy we have:

    [tex]E_1+m = E_3+E_4[/tex]

    Now, p1 should be on shell so you have to prove that this is only
    true in the two cases given above.


    Regards, Hans
     
    Last edited: Jun 10, 2007
  10. Jun 10, 2007 #9
    Ok, I think I'm getting it now:

    E1, E2, E3 are the rest-energies, p1, p3, p4 the three vectors (p2 is defined 0)
     
  11. Jun 10, 2007 #10
    I'm understanding this one:cool:!
     
  12. Jun 10, 2007 #11

    Hans de Vries

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    The factor 2 in 2p3 is true because p3 and p4 must be parallel to q and p1
    otherwise it would be a variable.


    Regards, Hans
     
    Last edited: Jun 10, 2007
  13. Jun 10, 2007 #12
    About the second: is it true that (threevectors:)
    if p2=0, p1=p3+p4
    so (p1)^2=(p3)^2+(p4)^2+2(p3)(p4)

    now (p3)(p4)=0, so (p1)^2=(p3)^2+(p4)^2, and this can only be true if p4=0 or p3=0?
     
  14. Jun 10, 2007 #13
    small angle approximation:wink:
     
  15. Jun 10, 2007 #14

    Hans de Vries

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    no, the 2 is true anyway :smile:, since p3 and p4 must be parallel to q and p1.
    (I modified my post)

    Regards, Hans.
     
  16. Jun 10, 2007 #15

    Hans de Vries

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    You've got an expression for p1 now, as well as E1+m = E3+E4.
    Next thing is to prove that

    [tex]E_1^2 = p_1^2 + m^2 [/tex]

    is true only in the two given cases.

    Regards, Hans
     
  17. Jun 11, 2007 #16
    It's completely clear to me now!

    Thank you very much for your help!:smile::smile::smile:
     
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