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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

    Staff: Mentor

    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

    Staff: Mentor

    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

    Staff: Mentor

    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

    Staff: Mentor

    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

    Staff: Mentor

    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

    Staff: Mentor

    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

    Staff: Mentor

    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

    Staff: Mentor

    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

    Staff: Mentor

    That seems like a good thing to try, yes.
     
  22. Oct 22, 2016 #21
    Well, I have this so far:

    ##\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-m^{2}} + \frac{i}{t-m^{2}} + \frac{i}{u-m^{2}}\right|^{2}}##

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

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

    ##\displaystyle{= \frac{g^{4}}{64\pi^{2}E_{\text{cm}}^{2}}\bigg[\frac{[tu-(t+u)m^{2}+m^{4}]^{2}+[su-(s+u)m^{2}+m^{4}]^{2}+[st-(s+t)m^{2}+m^{4}]^{2}}{[(s-m^{2})(t-m^{2})(u-m^{2})]^{2}}+\frac{2[s+t+u-3m^{2}]}{(s-m^{2})(t-m^{2})(u-m^{2})}\bigg]}##

    Expanding out the next set of brackets now seems like a foolish thing to try.
     
    Last edited: Oct 22, 2016
  23. Oct 22, 2016 #22

    PeterDonis

    Staff: Mentor

    I'm not sure what your goal is. Sometimes these expressions are complicated. The only things I can see that you haven't done are to write ##s## instead of ##E_{cm}^2## in the denominator of the factor in front, and to try to express ##t## and ##u## in terms of ##s## by using the energy-momentum relations, per one of my previous posts.
     
  24. Oct 22, 2016 #23
    How is ##\displaystyle{\frac{d\sigma}{d\theta}}## usually written?

    Is it usually written in terms of ##s##?

    Actually, I get

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

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

    ##s-m^{2}=E_{1}^{2}+E_{2}^{2}-m^{2}##

    ##s-{m^{2}}=\vec{k}^{1}_{1}+m^{2}+\vec{k}^{2}_{1}+m^{2}-m^{2}##

    ##s-m^{2}=2\vec{k}^{2}_{1}-m^{2}##,

    where ##\vec{k}_{1}## and ##\vec{k}_{2}## are the incoming momenta.
     
    Last edited: Oct 22, 2016
  25. Oct 22, 2016 #24

    PeterDonis

    Staff: Mentor

    You're right, I had misstated the result. But I think your last line should be

    $$
    s - m^2 = 2 \vec{k}_1^2 + m^2
    $$

    In the line before you have ##+ m^2## twice and ##- m^2## once, which adds up to ##+ m^2##. This would imply

    $$
    s = 2 \left( \vec{k}_1^2 + m^2 \right)
    $$

    which is the correct relativistic energy-momentum relation for the incoming pair (or the outgoing pair since all of the squared momenta are the same).
     
  26. Oct 22, 2016 #25
    Yes, thanks.

    Now,

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

    and ##u=-2|\vec{k}_{1}|^{2}(1+\cos\theta)## so that ##u=-2s(1+\cos\theta)##.

    Therefore,

    ##\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-m^{2}} + \frac{i}{t-m^{2}} + \frac{i}{u-m^{2}}\right|^{2}}##

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

    ##\displaystyle{= \frac{g^{4}}{64\pi^{2}s}\left(\frac{1}{s-m^{2}} + \frac{1}{2s(1-\cos\theta)-m^{2}} + \frac{1}{2s(1+\cos\theta)-m^{2}}\right)^{2}}##

    Is this a good form to in which to leave the final answer?
     
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