## Camels and dromedaries – rapidity at a hadron collider May 12, 2008

Posted by dorigo in physics, science.

Today we had our meeting of the CMS analysis group in Padova, a monthly recurrence where we get adjourned of the various efforts going on. It was my turn to chair the meeting (I am co-convener of the meeting with Ezio Torassa and we alternate), and I had put together a tightly packed agenda, which included updates on the global cosmic runs (weeks of data taking when muons from cosmic rays are collected and used to understand the detector response), the tracker checkout (issues with the final commissioning of the silicon tracker), the trigger studies for SLHC (or how to measure muon momenta accurately enough to prevent being overwhelmed by the huge rate of fake muons of low transverse momentum, when we will take data with CMS at a luminosity of $10^{35} cm^{-2} s^{-1}$), plus analyses of the $H \to WW$ decay, ttH production, and dimuon mass spectra.

Ignazio Lazzizzera, from the associated group of Trento, presented some kinematical distributions of muon tracks extracted from minimum bias Monte Carlo that will be used for SLHC studies. Minimum bias is a jargon that particle physicists use to describe events that withstood no selection whatsoever: events which suffered the minimum possible bias by the fact of having been collected by the detector. Such a collection of events is useful to understand what our “priors” are: at the full LHC luminosity (just a factor 10 below SLHC ones), every 25 nanoseconds we will have 20 proton-proton collisions to deal with, and only very rarely these interactions originate a high-momentum muon, which tags a potentially very interesting event. We have to rely on these minimum bias simulations to understand how easy it is for a light hadron -a pion or a kaon- to fool our detection system and be identified as a muon by our trigger, if we want to understand our chances of tuning trigger cuts and select good muons with high efficiency without being drowned in impossibly high rates from fake muons.

As Ignazio showed the plot below, which is the distribution of rapidity of simulated muon tracks in minimum bias data, I jumped on my chair. What was going on ? The two-humped distribution resembled a camel’s back!

To let you understand why such a distribution is unphysical, I need to take a step back. When you collide protons with other protons at high energy, what you are actually doing is creating hard interactions about proton constituents: quarks and gluons. Each of these constituents of a high-energy proton carries a fraction of the proton momentum: the two streams of “partons” (i.e. quarks or gluons) travel together in the positive and negative direction along the z axis – the beam direction- inside each proton; but some carry a larger, and many a smaller fraction of the total protons momentum.

Because of the variable amount of momentum carried by each parton, the collision center-of-momentum reference frame is not at rest in the detector reference frame: if a 90mph truck hits a 50mph compact car head on the debris will fly away following the truck direction!

What governs the probability that quarks and gluons carry a certain momentum fraction of the proton containing them are some functions called “Parton Distribution Functions“. They are shown below for the different constituents of protons.

As you see, it is increasingly probable (in a measured described by the PDF xf(x)) to find a parton carrying a smaller and smaller momentum fraction x (forget the u-distribution, which has a local maximum due to valence quarks: we are discussing the low-x tail of these shapes, since we are discussing not-so-high-energy interactions which constitute the bulk of collisions). Is this enough to figure out what will be the distribution of the debris, and in particular, the motion of the most energetic particles produced in the collision in the detector frame ?

Well, basically yes. If we label $x_1, x_2$ the momentum fractions of the colliding partons (which can be assumed massless for all practical purposes at LHC), the center-of-mass energy will be their geometric average $E=\sqrt x_1 x_2$ times the 14 TeV globally possessed by the colliding protons. The motion of the center-of-momentum frame in the detector frame will instead be described by rapidity – the quantity $y = 0.5 \log (E+P_z)/(E-P_z)$, which reduces to $0.5 \log (x_1/x_2)$.

Rapidity is, for the muons, the quantity plotted in the two-humped histogram above. Can there be a hole at zero in this distribution ? Not really! It does not take complicated math to realize that if you pick at random two values $x_1, x_2$ from a monotonous function, their values are most likely to be close to each other, and so their ratio will be close to one more often than not. The logarithm of one is zero, and at zero there cannot be a minimum! The distribution has to have a single maximum at zero rapidity instead!

You might find the above reasoning rather complicated. It is. However, had you worked at a hadron collider for 16 years, you would not need the math at all: the rapidity distribution of any physics process is (with very few exceptions) a broad distribution with a maximum at zero, unless the data have been biased by selection cuts.

I could thus explain what was going on in the distribution Ignazio was showing: the data he was plotting had been stripped of events which could fire the CMS trigger -that is, events with high-Pt, central muons in our case. Take a dromedar, substract stuff in the middle (the muons which are central), and you are left with a camel!

It remains to be seen why the minimum bias Monte Carlo had been selected this way. I suppose one such sample is rather useless for trigger studies!

1. Seth Zenz - May 12, 2008

I looked around a bit to try to understand this, because it’s not so obvious to me that x1 ~ x2… if they’re not both small, the next most likely outcome is for one to be big and the other small. My question would then be if there’s some reason to expect they can’t both be small, but that’s as far as my reasoning gets.

I assume that’s a generator-level plot, but if it were reconstruction then page 10 of the following paper (“Effect of Magnetif Field”) would shed some light on the situation…

http://home.fnal.gov/~sceno/jpg/results/Minbi.pdf

2. Stefan - May 12, 2008

Nice to see a rapidity distribution again 🙂

So the plot shows the rapidity distribution of all those muons with transverse momentum below a certain cutoff? What was this cutoff momentum, just to get an impression of the momenta involved at the LHC?

And since I am too lazy right now to check it out for myself, what is the rapidity of the incoming proton beams of the LHC? Seems it is something around +/- 6?

Cheers, Stefan

3. dorigo - May 13, 2008

Hi Stefan,

the plot shows the rapidity of muons that did not fire the trigger (or so I believe – one never knows with a plot crafted by somebody else until it is verified etc.). The trigger collects muons with Pt>20 GeV, but the lower-Pt muons have a fat chance to be mismeasured at trigger level, and so they may trigger even if they have 5, or 10 GeV. The momenta of muons from interesting physics processes is larger than these values, fortunately – so we can keep thresholds higher and still collect interesting stuff. In any case, for W and Z bosons decaying to muons, the typical Pt remains of the order of M/2, i.e. 30-40 GeV, because most bosons are produced at low transverse momentum themselves, providing no additional boost.

For the proton beams, the crossing angle of the two beams is very, very small – I think it is 96 microradians. That means a rapidity of 14 if I’m not screwing something up.

Cheers,
T.

4. Stefan - May 13, 2008

Hi Tommaso,

thanks for the explanation about the cutoff!

Concerning proton rapidity in the beam – your argument using the angle shows that you are an experimentalist, equating pseudo-rapidity and rapidity! In the heavy-ion simulation codes I have worked with, one has of course easy access to energy and longitudinal momentum and can use the formula you have given in the post, or y = arcosh γ.

Now plugging in γ = 7000/0.94 for the protons in the beam yields y = 9.6, or 19.2 for the rapidity interval between the colliding protons… Which makes me wonder, why is the muon distribution not much wider? Why are there no muons closer to beam rapidity? Can they not be measured with the CMS detector, which may be just OK for the purpose of the experiment? Or is the plot for a lower beam energy?

Cheers, Stefan

5. dorigo - May 15, 2008

Hi again,

yes, ignoring the mass is a good idea – unless one deals with W or Z bosons (it did happen to me once to produce a screwed Z rapidity distribution by using the massless approximation, lol).

As for muons, they are the result of decays from light hadrons. To produce one, you have to provide some Q^2 to the interaction. This is in effect a cutoff on the rapidity of the center of momentum.

For an example, take $x_1=0.1, x_2=0.0000001$. This makes a hard interaction with $E=\sqrt x_1 x_2 s =1.4 GeV$, which is close to the minimum value which will produce mesons. For such a system the rapidity is $y=0.5 \log x_1/x_2 = 6.9$
And this totally neglects the fact that the plot above might have been constructed by asking for a minimal requirement on the transverse momentum to the muons!

Ciao,
T.

6. dorigo - May 15, 2008

Hi Seth,

the plot is indeed at generation level. As for a real demonstration from first principles on the fact that at zero rapidity there should be a maximum, given monotonous PDF functions, I am unable to provide one, although my intuition tells me it is a correct statement. Maybe there is somebody willing to fiddle with it around ?

The problem can be stated as follows:
Given f(x) a continuous function in [0,1] such that $df(x)/dx<0$, and the function $F(y)=\int dx_1 dx_2 f(x_1) f(x_2) \delta(y-0.5 \log (x_1/x_2))$, show that in y=0 one has $dF(y)/dy=0$, $d^2F(y)/dy^2<0$.

This is my conjecture 😉

Cheers,
T.

7. Why is rapidity maximum at 0 ? « A Quantum Diaries Survivor - May 15, 2008

[…] maximum at 0 ? May 15, 2008 Posted by dorigo in mathematics, physics, science. trackback In a recent post where I shortly discussed a rapidity distribution of muons produced in low-energy proton-proton […]

8. Newcomer - October 9, 2008

i didnt understand why rapidity distribution is so important. but want to know more. does anyone have the patience to explain everything

9. dorigo - October 9, 2008

Hi newcomer,

sure. Just make a question and I will try to answer it. As for your initial one: the rapidity distribution of particles created in a proton-proton collision is important for several reasons.

1) we construct our detector around the collision point in a way that allows us to “see” as many as possible of the particles created in the interaction. The rapidity distribution is basically a distribution of the angle that the emitted bodies make with the proton beams. The fact that the distribution peaks at zero rapidity, and falls off rapidly at larger absolute values, implies that we can construct our detector as a cylinder, instrumenting best the zones which particles emitted at large angle from the beam will hit. A much flatter rapidity distribution would force us to build our detectors as a hour-glass, with two lobes covering best the forward- and backward- going regions, close to the beam axes.

2) a particle rapidity tells us a lot about the production mechanism. By studying it in detail we can figure out details about the interaction which produced it. For instance, by studying the rapidity distribution of muons emitted from a W boson decay ($pp \to W \to \mu \nu$), we can determine the asymmetry with which the W boson was created, and infer information on the distribution of quarks in the proton.

3) Rapidity is one of the angular quantities we measure for each particle, from which we determine the particle energy and direction with good precision. These quantities are crucial to reconstruct the possible origin of the particles we detect, such as, for instance, if a heavy particle has been produced and decayed to those we detect. Take for instance a Higgs boson: $pp \to H \to \mu \mu \mu \mu$. It can decay to four muon tracks, and if we measure their energies and momenta well, we can then determine the Higgs boson mass.

Cheers,
T.

10. Michael Stingo - November 21, 2008

Hello Tommaso,

I am an undergraduate student at the University of Kansas doing research with the physics department. My research is looking into Feynman’s limiting fragmentation. I see that the histogram of the rapidity distribution is done in ROOT. I was wondering if there is anyway I could get a copy of the ROOT file of this distribution as I am trying to find a function, some curve, to fit these plots with. I think it would give me a good start at and a basis for checking my goodness of fit for my curves. Thanks for your time.

Best,

Michael Stingo

11. Chang Wei - August 7, 2009

Yes, this blog is kind of old, but referring to your blog on Z boson rapidity at https://dorigo.wordpress.com/2008/11/25/the-z-mass-at-a-hadron-collider/

the Z rapidity also has two bumps …
I don’t understand why Z is ok but not this one. I’m confused.

12. dorigo - August 8, 2009

Dear Chang,

I am confused too right now. I think that plot is either wrong (for instance it assumed zero Z mass to plot the pseudorapidity instead than the rapidity) or it is showing only some of the sea-sea interactions.

The total Z production _rapidity_ distribution should not have a dip at zero, any more than that of its decay products; please note that in this post above we are discussing the rapidity of muons from the Z, while in the other post we were discussing the Z rapidity itself.

All I can find in my archives about the studies I produced for the Z rapidity is the plot you can find here , which I think is the correct one (it has the total distribution, as well as the one for each quark type -I think ddbar is in red, uubar is in blue, ssbar in green, ccbar in yellow and bbbar in magenta).

Cheers,
T.

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