Simulation of pp-collision and Z boson production

In summary, the conversation discusses the simulation of proton-proton collisions resulting in a Z boson and the constraints on the momentum fractions of the quarks involved. Two conditions are proposed, one requiring the Z boson to be at rest and the other not requiring this. It is determined that the first condition is a special case of the second and that the minimum requirement for creating a Z boson is when it is at rest. It is also discussed that the center-of-mass energy is the total energy available to create new particles and can be calculated in different ways but leads to different expressions for the Z boson momentum.
  • #1
Jan Eysermans
17
0
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:

[itex]
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
[/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
 
Physics news on Phys.org
  • #2
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 realize 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:
  • #3
RGauld said:
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 have forgot to mention that off shell Z boson production can be neglected.
 
  • #4
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
mfb said:
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.

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.
 
  • #6
Jan Eysermans said:
The minimum requirement to create at least a Z boson, is when the boson is at rest. This gives the following constraints on x1 and x2:
If the Z boson is not at rest, this inequality is not sufficient, and you need more energy.
 
  • #7
mfb said:
If the Z boson is not at rest, this inequality is not sufficient, and you need more energy.

Can you explain why this equation is not sufficient? I do not see it immediately..
 
  • #8
Jan Eysermans said:
Can you explain why this equation is not sufficient? I do not see it immediately..
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
mfb said:
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.

Yes indeed, I understand now! In principle, the first constraint must be written as:

[itex]
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}
[/itex]

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

Thanks!

Jan
 
  • #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
Jan
 
  • #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?
 
  • #12
RGauld said:
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.
Does this help?

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
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
RGauld said:
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!

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?
 
  • #15
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
RGauld said:
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.

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
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
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
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 can't 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
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:
  • #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?
 
  • #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?
 
  • #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 a lot 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!
 
  • #24
Yes, indeed. Thank you for the answers!

Jan
 
  • #25
Jan Eysermans said:
So If I generate x_1 according to the PDFs, I can calculate x_2 from this equation, without generating it?
Right (for Z bosons flying along the Z-direction).
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?
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.

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?
Symmetry should give that for free.
Don't forget the other quarks, however.
 
  • #26
mfb said:
Symmetry should give that for free.
Don't forget the other quarks, however.

What do you mean by symmetry? When I compare the u-PDF generating x1 values with the antiu-PDFs, I notice a slightly difference (not a statistical one because the difference is "reproducible").

The other quarks are incorporated by weighting over all the PDFs.
 
  • #27
You should get two different answers. Because of the valence content of the proton. If you compared s, c, b to sbar cbar bbar you expect a symmetry.

Presumably just means interchange of x1 x2 labels to the quark or anti quark
 
  • #28
RGauld said:
You should get two different answers. Because of the valence content of the proton. If you compared s, c, b to sbar cbar bbar you expect a symmetry.

Presumably just means interchange of x1 x2 labels to the quark or anti quark

I indeed interchanged the x1 and x2 labels, but still, if I sample only from the u-PDF, there is a difference when I sample only from the anti-u PDF. From this, it is not correct to sample only from the u-PDFs I guess.

Is there any statistical procedure to incorporate all the distributions for selecting the x1?
 
  • #29
For your procedure of producing the Z on shell. It is correct to only sample the u-PDF in my opinion. You automatically sample all possibilities at the same time here.

My guess is that when you find a suitable solution of x1 and x2. i.e. one where you get the z mass you can store these x1' values, then calculate the sum of,
u ubar @ x1'
d dbar @ x1'
s sbar @ x1'
c cbar @ x1'
b bbar @ x1'

If you want to make things more detailed etc. that is always possible. In event generators, I think the hard process is sampled first, then evolution back to the parton distribution function is done. This way you don't have to veto lots and lots of processes which don't give you the right energy exchange to produce the Z {though you are sort of doing this by forcing Z on shell}.

But at this point I think you may need to consult publications and textbooks - my normal approach is to look here first, http://books.google.co.uk/books/about/QCD_and_Collider_Physics.html?id=TqrPVoS6s0UC&redir_esc=y or search online to try and find something useful.
 
  • #30
Your explanation was for improving the error on handling the PDFs at q = M_z, right? What I mean is the difference between selecting x1 out of the quark-distributions or the antiquark-distributions.

To make things clear, let say we want to produce an on-shell Z boson from u and d quarks (and the anti-quarks). Now, my algorithm is the following:

1. select x1 out of (u+d)-PDF @ q = M_z
2. calculate x2
3. calculate p_Z = (x1-x2)*p, and calculate, for example, the boost parameter beta
4. plot beta in an histogram

When I repeat this algorithm, but changing step 1 into selecting x1 out of (anti-u + anti-d)-PDF at q= M_Z, I get a slightly different result. In my opinion, I have to incorporate both u+d and antiu+antid PDFs for generating the x1 value.

Thanks again!
Jan
 
  • #31
This isn't going to work. That procedure forces x2 to have a particular distribution, and that distribution may or may not (in fact, doesn't) match the correct x2 distribution.

What you need to do instead, if you want to go down this path, is once you have the x2, you calculate the probability of getting this x2, and then toss a random number. If the random number matches this probability, you keep the event, otherwise, you start over. This is called reweighting.
 
  • #32
Jan Eysermans said:
What do you mean by symmetry? When I compare the u-PDF generating x1 values with the antiu-PDFs, I notice a slightly difference (not a statistical one because the difference is "reproducible").
"quark from proton 1 + antiquark from proton 2" should give the same result as "antiquark from proton 1 + quark from proton 2". If it does not, there is something wrong.

From this, it is not correct to sample only from the u-PDFs I guess.
Sure, you have to account for all quarks. Just add their contributions afterwards?

I don't see how sampling the u-PDF covers processes like anti-s + s -> Z.


I agree with Vanadium in terms of reweighting.
 
  • #33
Indeed, including the distributions for x2 are needed.. I have implemented the method of Vanadium 50 and the results are more or less ok! Thanks everyone.
 

1. What is a pp-collision?

A pp-collision is a type of collision that occurs between two protons in a particle accelerator. It is used to study the behavior and properties of subatomic particles.

2. How is the simulation of pp-collision and Z boson production done?

The simulation of pp-collision and Z boson production is done using computer programs that use mathematical models and algorithms to simulate the behavior of subatomic particles in a particle accelerator.

3. Why is it important to simulate pp-collision and Z boson production?

Simulating pp-collision and Z boson production allows scientists to study the fundamental forces and particles that make up our universe. It also helps to validate and improve our understanding of the Standard Model of particle physics.

4. What is the role of Z bosons in pp-collision?

Z bosons are a type of subatomic particle that plays a crucial role in pp-collision. They are responsible for mediating the weak nuclear force and are produced in high-energy collisions between protons.

5. How do scientists analyze the data from simulations of pp-collision and Z boson production?

Scientists analyze the data from simulations of pp-collision and Z boson production by comparing it to experimental data from particle accelerators. They also use statistical analysis and mathematical models to interpret the data and make predictions about the behavior of subatomic particles.

Similar threads

  • High Energy, Nuclear, Particle Physics
Replies
11
Views
1K
  • High Energy, Nuclear, Particle Physics
Replies
3
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
661
  • Advanced Physics Homework Help
Replies
10
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • High Energy, Nuclear, Particle Physics
Replies
1
Views
1K
  • High Energy, Nuclear, Particle Physics
Replies
9
Views
3K
  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
1K
  • General Math
4
Replies
125
Views
16K
  • Math Proof Training and Practice
Replies
28
Views
5K
Back
Top