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Simulation of pp-collision and Z boson production

  1. Dec 11, 2013 #1
    Hi all,

    I have a question about simulating (Monte Carlo) proton-proton collisions resulting in, for example, a Z boson. Assume two quarks (quark and antiquark) from each proton collide head-on along the z-axis. The quark momenta are distributed according to the Parton Density Functions.

    If proton 1 has momentum [itex]\vec{p}_1 = p \vec{e}_z[/itex] and proton 2 [itex]\vec{p}_2 = -p \vec{e}_z[/itex], the momenta of the quarks inside the protons can be written as [itex]\vec{p}_1 = x_1 p \vec{e}_z[/itex] resp. [itex]\vec{p}_1 =- x_2 p \vec{e}_z[/itex]. Here, [itex]x_1[/itex] and [itex]x_2[/itex] represents the momentum fraction according to the Parton Density Functions. Both values must be generated in the simulation, but the kinematics of the collision set constraints on both values. Indeed, enough energy must be available to create a Z boson (at rest).

    In essence, we have a head-on quark-quark interaction. The four-vectors in the lab-frame of the quarks can be written as:

    [itex]p_1 = (E_1,0,0,x_1 p) \approx (x_1 p,0,0,x_1 p) [/itex]
    [itex]p_2 = (E_2,0,0,-x_2 p) \approx (x_2 p,0,0,-x_2 p) [/itex],

    where I have neglected the quark mass. The four-vector of the Z boson can be written as:

    [itex]p_Z = (E_Z,0,0,p_z) \approx (\sqrt{p_z^2 + m_Z^2c^2}/c,0,0,p_z) [/itex]

    Conservation of momentum yields the following equations:

    [itex] (x_1+x_2)p = \sqrt{p_z^2 + m_Z^2c^2} [/itex] and [itex] (x_1-x_2)p = p_z [/itex]

    The minimum requirement to create at least a Z boson, is when the boson is at rest. This gives the following constraints on [itex]x_1[/itex] and [itex]x_2[/itex]:

    [itex] x_1+x_2 \geq \frac{m_Zc}{p} [/itex]

    However, I always thought that the center-of-mass energy [itex] \sqrt{s} [/itex] is the total energy available to create new particles. We can write then: [itex] \sqrt{s} > m_Zc^2 [/itex]. The center-of-mass energy in this system can be calculated as:

    s = (p_1c+p_2c)^2 = ((x_1+x_2)pc, 0,0,(x_1-x_2)p)^2c^2 = (x_1+x_2)p^2c^2 - (x_1-x_2)^2p^2c^2 = 4x_1 x_2 p^2c^2

    The condition [itex] \sqrt{s} > m_Zc^2 [/itex] becomes: [itex] \sqrt{x_1 x_2} > \frac{m_Zc}{2p} [/itex].

    Which condition or reasoning is the correct one?


  2. jcsd
  3. Dec 11, 2013 #2

    One could produce a z boson at rest in the lab frame. So in this case each quark would have to have exactly mz/2 momentum. That is if the z is produced on shell.

    One could produce a z off shell which then decayed and then the quark momentum fraction could be less.

    The probability of the incoming quarks having equal and opposite momentum in the lab frame is very small. Since you have the whole range of quark momentum from the Parton density function.

    Sorry, I realise I didn't answer this the first time. Your answers are equivalent here due to how you set up the problem. Think about the situation if the z is not produced at rest, e.g. X1 =3/4 mz and x2 = mz/2. Or change frames to x1 and x2 are equal and opposite.
    Last edited: Dec 11, 2013
  4. Dec 11, 2013 #3
    Sorry, I have forgot to mention that off shell Z boson production can be neglected.
  5. Dec 11, 2013 #4


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    For the first condition, you require that the Z boson is at rest and the incoming protons have the same momenta (just in opposite directions of course), so x1=x2. This makes the first condition a special case of the second condition (plug it in and test it!), where you do not require this.
  6. Dec 12, 2013 #5
    Thanks for your reply. But why do I require in the first condition that the Z boson must be at rest? The only constraint I apply is that the energy of both partons must be at least the mass of the Z boson. It doesn't matter is x1 or x2 are equal or not.
  7. Dec 12, 2013 #6


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    If the Z boson is not at rest, this inequality is not sufficient, and you need more energy.
  8. Dec 12, 2013 #7
    Can you explain why this equation is not sufficient? I do not see it immediately..
  9. Dec 12, 2013 #8


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    Your equation assumes you just need enough total energy to produce a Z boson at rest.
    That is true - but only for Z bosons at rest.
  10. Dec 12, 2013 #9
    Yes indeed, I understand now! In principle, the first constraint must be written as:

    x_1 + x_2 \geq \frac{E_Z}{p} = \frac{\sqrt{p_Z^2c^2+m_Z^2c^2}}{p} = \frac{\sqrt{(x_1-x_2)^2p^2c^2+m_Z^2c^2}}{p}

    After rewriting this equation, the second constraint is obtained: [itex] \sqrt{x_1x_2} \geq \frac{M_zc}{2p}[/itex]


  11. Dec 13, 2013 #10
    I'm still confused with the center-of-mass (com) energy I think..

    The com-energy is the total energy available to create new particles. If the Z boson is created, the com-energy equals the total boson energy:

    [itex]\sqrt{s}=\sqrt{4x_1x_2p^2} = \sqrt{p_z^2+m_Z^2}[/itex]

    (assume [itex]c=1[/itex]). This is correct, right? From this equation, the Z boson momentum can be calculated:

    [itex]p_z = \sqrt{4x_1x_2p^2 - m_z^2}[/itex]

    But, from the conservation of four-momentum in the LAB frame (see first post), the Z boson momentum equals:

    [itex]p_z = |x_1-x_2|p[/itex]

    Both expressions for [itex]p_z[/itex] are clearly not equal to each other, so I must have made a mistake when interpreting the com energy I guess..

    Thanks again
  12. Dec 13, 2013 #11
    Let me try and understand what is going wrong here. (bare with me)

    So we collide two partons, x1 and x2 and produce only a Z which in this case must travel only along the z-direction (here we are assuming we don't see all the other hadronic crap and jets which could be radiated to put if off the z-direction) giving us a 2-d problem (Energy and z-momentum).

    E_Z = \sqrt{ |p_Z|^2 + m_Z^2 }
    p_z = x1 + (-x2 )

    What you are calculating is the 'invariant mass' of the incoming partons.
    (P1 + P2 )^2 = 4 x1 x2

    And this must be greater than m_Z^2 (like you say). So I think the confusion is that in producing a Z, the invariant mass of just a Z alone must be just the Z. While in reality, this almost never happens at a hadron collider, where what you produce is Z+jet. Which means the Z boson has some x and y component of momentum, in this case what you have to calculate is the invariant mass of the jet+Z, which of course does not have to be equal just to the Z mass.

    Does this help?
  13. Dec 13, 2013 #12
    I understand the limitations of the model, that there are no jets are produced and the problem is 1D. I do not understand your argumentation I think. If you say that the invariant mass is just the Z boson, do you mean the rest mass, or the total energy of the boson?

    As I understand it, because the com-energy is Lorentz invariant, the magnitude of the four-momentum in each frame must be equal to the com energy. Especially in the LAB frame, but then I get this contradiction.
  14. Dec 13, 2013 #13
    What I think is happening is the following.

    Your assumption is that the Z is produced on shell, i.e. a real Z of mass 91 GeV say, and it therefore must have only z-momentum. It's 4-momentum is therefore,
    P = ( x1 +|x2|, 0, 0, x1 - x2 }
    and P^2 - the invariant - must be 4x1 x2 like you say. But for a single particle which is on shell, the invariant mass must be its on-shell mass.

    So by the assumptions you are making about it being on shell and just 1 particle in the final state. You are choosing 1 special solution that satisfies this condition.

    the centre of mass energy invariant is {P1 + P2}^2 , which if you want to produce the 1 Z particle in the final state P3, must have this same invariant, i.e. P3 ^2 = mz^2 = {P1 + P2}^2

    So, for a larger centre of mass energy than the Z mass, one has to produce something else with the Z, or make it of shell for nature to let the process work. Unless we violate lorentz invariance!
  15. Dec 13, 2013 #14
    Thanks for your reply. It makes the problem a little more clear to me.

    Now, I understand why on-shell production implies the fact that the com-energy must be equal to the Z boson mass. But this implies, as get it correct, that the x-values must satisfy the condition:

    [itex]s^2 = (p_1 + p_2)^2 = m_z^2 \Rightarrow 4x_1x_2 p^2 = m_z^2[/itex].

    So If I generate x_1 according to the PDFs, I can calculate x_2 from this equation, without generating it?
    And the probability that, in pp-collisions, both x values satisfy this relationship is very small, hence always s > M_Z and jets are produced?
  16. Dec 13, 2013 #15

    I agree with this. Though, if you wanted to obtain the cross-section for the process, you would have to also evaluate the fx2{x2,Q^2} of the quark, or anti-quark content at Q^2 = Mz^2. The probability of the quark being at that x value for the given Q^2 momentum exchange of the interaction.

    Yes, I *think* the cross-section to produce an on shell Z+ jet is larger. Especially if the jet is not very energetic.

    I looked at this results,http://cms-physics.web.cern.ch/cms-physics/public/EWK-10-002-pas.pdf

    If you scroll down to figure 15. It shows the pT of the reconstructed mu mu pair (i.e. the Z mass}. You see the tallest bin, highest cross section, is not the lowest one. Though, this region is actually very difficult to simulate theoretically cleanly.
  17. Dec 13, 2013 #16
    Thank you for the explanation! I've implemented in my code, but the results are, as expected, very strange. Suppose, I generate an x1 value from the up-quark PDF. This PDF increases as x decreases, and the generated x1 value is always very very small (order of 10^-4 - 10^-6). The corresponding x2 value is then very high (order of 10^4).
    These results are clearly not correct..
  18. Dec 13, 2013 #17
    How do you generate the up quark? Is it a random number between zero and 1? I guess you can generate the random number logarithmically to get the low x values you need for 4TeV protons or whatever.

    Is your \sqrt{4 x1 * x2 } value giving you 91.8 GeV? Are you using Q^2 = [91.8] GeV^2 ?

    If so then, the normalisation just needs to be fixed. Are you using LHAPDF?
  19. Dec 13, 2013 #18
    I have a set of PDFs from the PDF4LHC program (see http://arxiv.org/abs/0802.0007).

    My program is written in ROOT, where I load the PDFs (for u,d anti-u and anti-d quarks), and then select a random value between 0 and 1 according to this distribution (thus non uniform).

    I'll give you an overview of the algorithm:

    1) select random x1 from e.g. u-quark PDF
    2) calculate x2 = m_Z^2/(4*p^2) [this value corresponds to the anti-up quark]
    3) calculate p_z = (x1-x2)*p
    4) construct Z boson Lorentzvector
  20. Dec 13, 2013 #19
    This looks good to me. What energy collisions, LHC collisions?
    Both x1 and x2 must always be in the range 0-1 to avoid violating the momentum sum rules. (the quark cant have more energy than its mother hadron).

    If this is now working, hopefully it can help you to calculate whatever it is you want to calculate.
  21. Dec 13, 2013 #20
    Yes, p = 4000 GeV, or LHC collisions. Here is an image of the u-PDF, randomly chosen over 1E6 times: http://postimg.org/image/zc6xga97n [Broken]

    It is most likely to produce an x1 value smaller then 10E-1, 1E-2. The correspondig x2 values are then always > 1. Maybe I have to force x2 to be smaller than 1, which implies a constraint on x1:

    [itex]x_2 = \frac{m_Z^2}{4p^2x_1} \leq 1 \Rightarrow x_1 \geq \frac{m_Z^2}{4p^2} [/itex]

    If x1 is too small, it has not enough energy (even with a 4000 GeV quark) to produce a Z boson. Can this be correct?
    Last edited by a moderator: May 6, 2017
  22. Dec 13, 2013 #21
    Okay nice, the image looks like everything is working.

    So I tried a little example.

    x1 = 1e-2 for the up quark. {i.e. 40 GeV for 4TeV collisions}
    x2 p = Mz^2 / 4 x1 p
    x2 p = {91.8}^2/ {4 * 1e-2 4000 }
    x2 p = 52 GeV
    x2 = 52/4000 = 0.01316 for the anti-up.

    Does this agree with the code?
  23. Dec 13, 2013 #22
    Yes indeed, that is exactly the result here from my program. One last question about the randomness. Now, I sample always from the u-PDF distribution. But, off course, it is also possible to sample from the anti-u PDFs. I guess it is important to incorporate both PDFs?
  24. Dec 13, 2013 #23
    Great! Because in this case we have a fixed invariant mass ( the Mz ) by samplying the u-PDF you are simultaneously sampling the ubar since there is only one solution.

    I *guess* in practice, one should scan over the entire x1-range. For each point in x1, in this case there is only 1 viable x2, so multiplying these two together gives you the flux.

    If you were generating Zjets, or off shell Z's, I think what you have to do is integrate over the entire x2-range of values for each x1 piece. Which is alot more computer intensive..

    I think the Monte-Carlo programs which do event generation do not do this, but have clever ways to improve efficiency. Maybe someone else knows about this.

    good luck!
  25. Dec 13, 2013 #24
    Yes, indeed. Thank you for the answers!

  26. Dec 13, 2013 #25


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    Right (for Z bosons flying along the Z-direction).
    At least 2 jets are always produced from the remaining parts of the protons. The partons of the high-energetic collision can produce jets as well, or photons, or whatever, but it would probably need a calculation to see this influence on the cross-sections.

    Symmetry should give that for free.
    Don't forget the other quarks, however.
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