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Simulation of pp-collision and Z boson production

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RGevo
#19
Dec13-13, 07:40 AM
P: 82
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.
Jan Eysermans
#20
Dec13-13, 08:13 AM
P: 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:

[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?
RGevo
#21
Dec13-13, 08:22 AM
P: 82
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?
Jan Eysermans
#22
Dec13-13, 08:59 AM
P: 17
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?
RGevo
#23
Dec13-13, 09:08 AM
P: 82
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!
Jan Eysermans
#24
Dec13-13, 10:00 AM
P: 17
Yes, indeed. Thank you for the answers!

Jan
mfb
#25
Dec13-13, 06:58 PM
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P: 11,589
Quote Quote by Jan Eysermans View Post
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.
Jan Eysermans
#26
Dec14-13, 02:42 AM
P: 17
Quote Quote by mfb View Post

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.
RGevo
#27
Dec14-13, 04:01 AM
P: 82
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
Jan Eysermans
#28
Dec14-13, 04:50 AM
P: 17
Quote Quote by RGauld View Post
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?
RGevo
#29
Dec14-13, 05:18 AM
P: 82
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 text books - my normal approach is to look here first, http://books.google.co.uk/books/abou...UC&redir_esc=y or search online to try and find something useful.
Jan Eysermans
#30
Dec14-13, 06:36 AM
P: 17
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
Vanadium 50
#31
Dec14-13, 09:17 AM
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P: 16,162
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.
mfb
#32
Dec14-13, 10:23 AM
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P: 11,589
Quote Quote by Jan Eysermans View Post
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.
Jan Eysermans
#33
Dec14-13, 01:07 PM
P: 17
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.


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