# Simulation of pp-collision and Z boson production

1. ### Jan Eysermans

17
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 $\vec{p}_1 = p \vec{e}_z$ and proton 2 $\vec{p}_2 = -p \vec{e}_z$, the momenta of the quarks inside the protons can be written as $\vec{p}_1 = x_1 p \vec{e}_z$ resp. $\vec{p}_1 =- x_2 p \vec{e}_z$. Here, $x_1$ and $x_2$ 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:

$p_1 = (E_1,0,0,x_1 p) \approx (x_1 p,0,0,x_1 p)$
$p_2 = (E_2,0,0,-x_2 p) \approx (x_2 p,0,0,-x_2 p)$,

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

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

Conservation of momentum yields the following equations:

$(x_1+x_2)p = \sqrt{p_z^2 + m_Z^2c^2}$ and $(x_1-x_2)p = p_z$

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

$x_1+x_2 \geq \frac{m_Zc}{p}$

However, I always thought that the center-of-mass energy $\sqrt{s}$ is the total energy available to create new particles. We can write then: $\sqrt{s} > m_Zc^2$. 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 $\sqrt{s} > m_Zc^2$ becomes: $\sqrt{x_1 x_2} > \frac{m_Zc}{2p}$.

Which condition or reasoning is the correct one?

Thanks!

Jan

2. ### RGevo

90
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.

Last edited: Dec 11, 2013
3. ### Jan Eysermans

17
Sorry, I have forgot to mention that off shell Z boson production can be neglected.

### Staff: Mentor

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.

5. ### Jan Eysermans

17
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.

### Staff: Mentor

If the Z boson is not at rest, this inequality is not sufficient, and you need more energy.

7. ### Jan Eysermans

17
Can you explain why this equation is not sufficient? I do not see it immediately..

### Staff: Mentor

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.

9. ### Jan Eysermans

17
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: $\sqrt{x_1x_2} \geq \frac{M_zc}{2p}$

Thanks!

Jan

10. ### Jan Eysermans

17
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:

$\sqrt{s}=\sqrt{4x_1x_2p^2} = \sqrt{p_z^2+m_Z^2}$

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

$p_z = \sqrt{4x_1x_2p^2 - m_z^2}$

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

$p_z = |x_1-x_2|p$

Both expressions for $p_z$ 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. ### RGevo

90
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?

12. ### Jan Eysermans

17
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.

13. ### RGevo

90
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!

14. ### Jan Eysermans

17
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:

$s^2 = (p_1 + p_2)^2 = m_z^2 \Rightarrow 4x_1x_2 p^2 = m_z^2$.

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?

15. ### RGevo

90
Hello,

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.

16. ### Jan Eysermans

17
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..

17. ### RGevo

90
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. ### Jan Eysermans

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 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

19. ### RGevo

90
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.

20. ### Jan Eysermans

17
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

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:

$x_2 = \frac{m_Z^2}{4p^2x_1} \leq 1 \Rightarrow x_1 \geq \frac{m_Z^2}{4p^2}$

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?