Simulation of ppcollision and Z boson productionby Jan Eysermans Tags: boson, center of mass, collider beams, partons, ppcollision, production, quarks, simulation 

#1
Dec1113, 03:00 PM

P: 17

Hi all,
I have a question about simulating (Monte Carlo) protonproton collisions resulting in, for example, a Z boson. Assume two quarks (quark and antiquark) from each proton collide headon along the zaxis. 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 headon quarkquark interaction. The fourvectors in the labframe 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 fourvector 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_1x_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 centerofmass 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 centerofmass energy in this system can be calculated as: [itex] s = (p_1c+p_2c)^2 = ((x_1+x_2)pc, 0,0,(x_1x_2)p)^2c^2 = (x_1+x_2)p^2c^2  (x_1x_2)^2p^2c^2 = 4x_1 x_2 p^2c^2 [/itex] 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? Thanks! Jan 



#2
Dec1113, 03:51 PM

P: 81

Hello,
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. 



#3
Dec1113, 04:10 PM

P: 17





#4
Dec1113, 07:22 PM

Mentor
P: 10,809

Simulation of ppcollision and Z boson production
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 x_{1}=x_{2}. This makes the first condition a special case of the second condition (plug it in and test it!), where you do not require this.




#5
Dec1213, 11:37 AM

P: 17





#6
Dec1213, 11:57 AM

Mentor
P: 10,809





#7
Dec1213, 12:10 PM

P: 17





#8
Dec1213, 12:18 PM

Mentor
P: 10,809

That is true  but only for Z bosons at rest. 



#9
Dec1213, 12:44 PM

P: 17

[itex] x_1 + x_2 \geq \frac{E_Z}{p} = \frac{\sqrt{p_Z^2c^2+m_Z^2c^2}}{p} = \frac{\sqrt{(x_1x_2)^2p^2c^2+m_Z^2c^2}}{p} [/itex] After rewriting this equation, the second constraint is obtained: [itex] \sqrt{x_1x_2} \geq \frac{M_zc}{2p}[/itex] Thanks! Jan 



#10
Dec1313, 03:52 AM

P: 17

I'm still confused with the centerofmass (com) energy I think..
The comenergy is the total energy available to create new particles. If the Z boson is created, the comenergy 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 fourmomentum in the LAB frame (see first post), the Z boson momentum equals: [itex]p_z = x_1x_2p[/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 Jan 



#11
Dec1313, 04:15 AM

P: 81

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 zdirection (here we are assuming we don't see all the other hadronic crap and jets which could be radiated to put if off the zdirection) giving us a 2d problem (Energy and zmomentum). 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? 



#12
Dec1313, 04:38 AM

P: 17

As I understand it, because the comenergy is Lorentz invariant, the magnitude of the fourmomentum in each frame must be equal to the com energy. Especially in the LAB frame, but then I get this contradiction. 



#13
Dec1313, 04:52 AM

P: 81

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 zmomentum. It's 4momentum 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 onshell 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! 



#14
Dec1313, 05:30 AM

P: 17

Now, I understand why onshell production implies the fact that the comenergy must be equal to the Z boson mass. But this implies, as get it correct, that the xvalues 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 ppcollisions, both x values satisfy this relationship is very small, hence always s > M_Z and jets are produced? 



#15
Dec1313, 05:47 AM

P: 81

Hello,
I agree with this. Though, if you wanted to obtain the crosssection for the process, you would have to also evaluate the fx2{x2,Q^2} of the quark, or antiquark 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 crosssection to produce an on shell Z+ jet is larger. Especially if the jet is not very energetic. I looked at this results,http://cmsphysics.web.cern.ch/cmsp...10002pas.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. 



#16
Dec1313, 06:19 AM

P: 17

These results are clearly not correct.. 



#17
Dec1313, 06:58 AM

P: 81

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? 



#18
Dec1313, 07:27 AM

P: 17

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 antiu and antid 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. uquark PDF 2) calculate x2 = m_Z^2/(4*p^2) [this value corresponds to the antiup quark] 3) calculate p_z = (x1x2)*p 4) construct Z boson Lorentzvector 


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