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A Mandelstam variables

  1. Oct 21, 2016 #1
    For ##2\rightarrow 2## scattering with equal masses in the centre-of-mass frame, are all the four three-momenta equal to each other, or is it that the incoming three-momenta and the outgoing three-momenta sum to 0 separately?
     
  2. jcsd
  3. Oct 21, 2016 #2

    PeterDonis

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    Have you tried to use conservation laws to obtain the answer? Remember that energy must be conserved as well as momentum.
     
  4. Oct 21, 2016 #3
    I have. I get my second answer.
     
  5. Oct 21, 2016 #4

    PeterDonis

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    So what's the problem? Do you think that answer is wrong? Can you show your work?
     
  6. Oct 21, 2016 #5
    This is about calculating one of the incoming momenta and the angle of scattering in terms of the Mandelstam variables. I get

    ##|\vec{k}| = \frac{1}{2}\sqrt{2s+t+u}##

    which is one of the incoming three-momenta ##|\vec{k}|## in terms of the Mandelstam variables and

    ##t-u = 4\ |\vec{k}|\ |\vec{p}|\ \text{cos}\ \theta,##

    where ##|\vec{p}|## is one of the final momenta and ##\theta## is the angle of scattering. How do I proceed with the second equation to remove ##|\vec{p}|## and express ##\cos\ \theta## in terms of the Mandelstam variables?
     
  7. Oct 21, 2016 #6

    PeterDonis

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    Ok, after seeing your work, you weren't doing what I suggested: using conservation laws to obtain the answer. In the center of momentum frame, it should be obvious that the following two facts are true:

    (1) The magnitudes of all four 3-momentum vectors are the same (since the masses are equal and both momentum and energy are conserved).

    (2) The directions of the incoming pair of 3-momentum vectors are opposite, and the directions of the outgoing pair are opposite; but the two pairs can be pointing in different directions (the difference in the incoming and outgoing directions is of course the scattering angle).

    Does that help?
     
  8. Oct 21, 2016 #7
    Why should all the four three-momenta have the same magnitude just because their masses are the same?

    Conservation law gives us ##E_{1}+E_{2}=E_{\text{cm}}=E_{3}+E_{4}## and ##\vec{k}_{1}+\vec{k}_{2}=0=\vec{k}_{3}+\vec{k}_{4}##, where ##1## and ##2## are incoming and ##3## and ##4## are outgoing, so that ##\vec{k}_{1}=-\vec{k}_{2}## and ##\vec{k}_{3}=-\vec{k}_{4}##, but not ##|\vec{k}_{1}|=|\vec{k}_{3}|##.
     
  9. Oct 22, 2016 #8

    PeterDonis

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    The conservation laws you wrote down should tell you. Try writing the energies in terms of the (squared) magnitudes of the momentum vectors.
     
  10. Oct 22, 2016 #9
    Right, I get it.
     
  11. Oct 22, 2016 #10
    Do you also have some suggestion for managing this calculation:

    ##\displaystyle{\frac{d\sigma}{d\theta} = \frac{|\mathcal{M}|^{2}}{64\pi^{2}E_{\text{cm}}^{2}}}##

    ##\displaystyle{= \frac{g^{4}}{64\pi^{2}E_{\text{cm}}^{2}}\left|\frac{i}{s^{2}-m^{2}} + \frac{i}{t^{2}-m^{2}} + \frac{i}{u^{2}-m^{2}}\right|^{2}}##

    ##\displaystyle{= \frac{g^{4}}{64\pi^{2}E_{\text{cm}}^{2}}\bigg[\frac{1}{(s^{2}-m^{2})^{2}}+\frac{1}{(t^{2}-m^{2})^{2}}+\frac{1}{(u^{2}-m^{2})^{2}}+\frac{2}{(s^{2}-m^{2})(t^{2}-m^{2})}+\frac{2}{(s^{2}-m^{2})(u^{2}-m^{2})}+\frac{2}{(t^{2}-m^{2})(u^{2}-m^{2})}\bigg]}##

    ##\displaystyle{=\frac{g^{4}}{64\pi^{2}E_{\text{cm}}^{2}}\bigg[\frac{[(t^{2}-m^{2})(u^{2}-m^{2})]^{2}+[(s^{2}-m^{2})(u^{2}-m^{2})]^{2}+[(s^{2}-m^{2})(t^{2}-m^{2})]^{2}}{[(s^{2}-m^{2})(t^{2}-m^{2})(u^{2}-m^{2})]^{2}}+\frac{2[(u^{2}-m^{2})+(t^{2}-m^{2})+(s^{2}-m^{2})]}{(s^{2}-m^{2})(t^{2}-m^{2})(u^{2}-m^{2})}\bigg]}##
     
  12. Oct 22, 2016 #11

    PeterDonis

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    This is for the same problem?
     
  13. Oct 22, 2016 #12
    Yes, for ##2\rightarrow 2## scattering, but now specialised to the case of ##\phi^{3}## theory.
     
  14. Oct 22, 2016 #13
    I computed

    ##|\vec{k}_{1}| = \frac{1}{2}\sqrt{2s+t+u}##

    and

    ##\text{cos}\ \theta = \frac{t-u}{2s+t+u}##.

    They don't help, do they?
     
  15. Oct 22, 2016 #14

    PeterDonis

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    Have you obtained expressions for ##s##, ##t##, and ##u## taking into account the conservation laws? For example, ##s = E_{cm}^2##, correct?
     
  16. Oct 22, 2016 #15
    Yes, they are

    ##s=E_{\text{cm}}^{2}##

    ##t=-2|\vec{k}_{1}|^{2}(1-\cos\theta)##

    ##u=-2|\vec{k}_{1}|^{2}(1+\cos\theta)##.
     
    Last edited: Oct 22, 2016
  17. Oct 22, 2016 #16

    PeterDonis

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    So each one has units of energy squared--if so, it seems to me that the formula in post #10 should have factors like ##s - m^2##, ##t - m^2##, ##u - m^2##, instead of having the squares of ##s##, ##t##, and ##u##.
     
  18. Oct 22, 2016 #17
    Yes, they do. But how do you proceed?
     
  19. Oct 22, 2016 #18

    PeterDonis

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    You can also use the relativistic energy-momentum relation to write ##E_{cm}^2 - m^2 = s - m^2 = |\vec{k}|^2 = |\vec{p}|^2##. You can use that to get a relation between ##s## and ##t## (and/or ##s## and ##u##).
     
  20. Oct 22, 2016 #19
    I see. So, you do have to expand the brackets in the formula and simplify using the relations for ##s+t##, ##s+u##, ##t+u##, etc?
     
  21. Oct 22, 2016 #20

    PeterDonis

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    That seems like a good thing to try, yes.
     
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