More on the Z lineshape at LHC December 19, 2008

Posted by dorigo in personal, physics, science.
Tags: , , ,

Yesterday I posted a nice-looking graph without abounding in explanations on how I determined it. Let me fill that gap here today.

A short introduction

Z bosons will be produced copiously at the LHC in proton-proton collisions. What happens is that a quark from one proton hits an antiquark of the same flavour in the other proton, and the pair annihilates, producing the Z. This is a weak interaction: a relatively rare process, because weak interactions are much less frequent than strong interactions. Quarks carry colour charge as well as weak hypercharge, and most of the times when they hit each other what “reacts” is their colour, not their hypercharge. Similarly, when you meet  John at the coffee machine you discuss football more often than chinese checkers: in particle physics terms, that is because your football coupling with John is stronger than your chinese-checkers coupling.

Quarks do “discuss” weak hypercharge occasionally, and if they are a particle-antiparticle pair they produce Z bosons. The Z has a distinct, well-known mass (91.1876 GeV), but being an unstable particle, its mass is not bound to be exactly that number: a Z boson may have a mass of 85, or 100, or even 120 GeV, but the farther is its mass from 91.2 GeV, the rarer it is. The probability is described by a Breit-Wigner distribution, shown below. The precise form of the distribution is

$F(x) = \frac{\Gamma^2 M^2}{(x^2-M^2)^2 + \Gamma^2 M^2}$

where $M=91.1876 GeV$, $\Gamma=2.4952 GeV$ are the mass and natural width of the Z boson.

PDF effects

As you see, the most probable value (the peak) is indeed sitting where it should, but the function has non-zero values for all positive values of the x axis, because of the finite value of $\Gamma$. Because of that crucial fact, when two quarks annihilate through weak interaction they may create a Z boson even if their combined energy is different from 91.1876 GeV. And it usually is!

The energy of quarks colliding inside LHC protons is a variable quantity. It is governed by functions called Parton Density Functions (PDF), which have been studied by dozens of experiments in the past. The meaningful quantity on which PDF depend is called “Bjorken x“: it is the fraction of the proton energy carried by the quark which participates in the collision.  There is another variable on which PDFs depend, the squared energy at which they are computed, usually labeled as $Q^2$; I will neglect that detail here, although it does enter my calculations.

Now, if one wants to determine the shape of the mass distribution of Z bosons produced at the LHC and measured by the CMS experiment, as is my goal, one needs to put together several factors. There is a production phase and a detection phase, and I will treat them in succession.

Z production details

At production, we need to compute the convolution of the Breit-Wigner formula with the PDF of the quarks which create the Z boson: a thing called a factorization integral, which can be written in simplified form as

$\int f_q (x_1) f_{\bar q}(x_2) \hat \sigma (x_1 x_2 \sqrt{s}) dx_1 dx_2$

where $f_q()$ is the PDF of quark q, $x_1, x_2$ are the Bjorken-x fractions of energy carried by the quarks participating in the collision, and  $\hat \sigma()$ is the cross-section, which is given by the Breit-Wigner distribution; finally, s is the square of the nominal LHC energy, 14 TeV (or 10, at start-up in 2009).

Now, things are murkier than that, for two reasons. One, you need to consider all possible quark-antiquark pairs in the reaction, so the integral becomes a sum of five integrals, accounting for $u \bar u$, $d \bar d$, $s \bar s$, $c \bar c$, and $b \bar b$ collisions (we omit the top quark, which has a really tiny probability of being found inside a proton). And two, different quark pairs “couple” weakly with a different strength. It all boils down to weigh down-type quark pairs slightly more than up-type pairs: 28.9% more so. The full expression then becomes:

$F(M)= \Sigma_{i=1}^5 \int_0^1 f_{q_i} (x_1) f_{\bar q_i}(x_2) K_i \hat \sigma (M) dx_1 dx_2$

where the $K_i$ account for the different couplings, where the sum runs on the five light flavors, and where I have now substituted the produced Z mass M for its expression in terms of the Bjorken’s x values.

If you compute that integral, you end up with a function of the Z mass which is still similar to a Breit-Wigner, but it shows a higher tail to low mass and a lower tail to high mass. If you want to be more accurate (it is important for the use I will make of F(M) ),  you can complicate things slightly further, and derive the F(M) form for different values of the rapidity of the produced Z boson.

What is rapidity ? It is a measure of the boost of the produced boson in the laboratory frame, where our detector sits. Since the quark and the antiquark which produced the Z do not in general carry the same fraction of proton energy, the Z boson will not be at rest, but will be moving along the proton axis. The momentum of the Z boson is simply given by $(x_1-x_2)\sqrt{s}/2$, but a better quantity is rapidity, labeled by the letter Y, and expressed by the formula

$Y = 0.5 \log ( \frac{E+p_z}{E-p_z} )$.

Now, since rapidity depends on the difference between the two x values of the colliding quarks, one might guess that the form of F(M) above might in fact depend on the value of Y of the produced Z boson. That is entirely correct: indeed, the larger the motion of the Z after production, the larger is the deformation of the Breit-Wigner distribution due to the PDF, because a large Z rapidity means a large imbalance in the Bjorken x of the two quarks, and that in turn means that one of them was very small (otherwise their product $x_1 x_2 \sqrt{s}/2$ does not make the energy needed to create a Z). At very small x values, the PDF f(x) becomes very steep, and its variations drive a variation in the Z lineshape that can be computed with the factorization integral.

So, to account for Z rapidity, we need to compute several different factorization integrals. One simple way is to multiply the integrand by a step function -a function equal to 1.0 in a small range, and 0.0 elsewhere. If we want to compute F(M) for Z rapidity in the range $0.5<|Y|<0.6$, for instance, we only consider x1, x2 values such that the resulting rapidity is in the quoted range.

Measurement of decay products

So far we have only discussed Z production. The calculation given above, while technically correct, is only an approximation, because we have totally neglected “next-to-leading-order” effects (NLO), which are caused by the radiation of one gluon by one of the two colliding quarks. Such phenomena, called “initial state radiation”, affect the motion of the produced Z boson in the plane transverse to the beam direction: not the $P_z$, thus, but the x and y components of the motion.

Once the Z boson is produced, it decays. In about 3.3% of cases, it gives rise to two energetic muons, which are then measured by detecting the ionization tracks they leave in the silicon tracker and the muon drift chambers of the CMS detector. If we neglect small effects due to “final state radiation” -the emission of photons by the muons, in this case-, the precise measurement of the muon momenta allow a reconstruction of the mass of the produced Z boson.

The CMS detector is excellently designed to measure muon tracks, but no measurement is free of uncertainty. In fact, the muon-track trajectories are determined with a precision limited by the strength of the magnetic field, the dimensions of the tracker, and the position precision provided by the silicon microstrip detectors: all this affects the estimate of track momentum and direction.

When one measures the two tracks, three quantities are sufficient to determine the kinematics of each, if one agrees that they originate from the same spatial position on the collider beam axis. Usually one picks transverse momentum $P_t$ -the particle momentum projected onto the plane transverse to the beam direction-, which is the quantity directly related to the curvature of the charged particle in the strong axial magnetic field; plus the angle made by the particle with the beam axis $\theta$ and the angle in the plane transverse to that axis $\phi$.

What do we do with those six variables ? We try and reconstruct their combined invariant mass: mass and energy are related by Einstein’s formula $E=m c^2$, and so we know that if the two leptons are all what’s left of a Z boson disintegration, the mass of the Z boson must survive in the form of energy of its decay products. One needs some (easy, but lengthy to explain) rudiments in relativistic kinematics to get the invariant mass, but in these days of object-oriented programming one may well forget formulas and concentrate on the superfluous art of code design: a simple line of code does it:

double M = (lepton1+lepton2).m();

…Duh! I am not so sure this means less time for coding and more for thought…

Anyway, with the six kinematic variables we can do another thing: since we know the uncertainty of each of them -from the characteristics of our detector and our measurement technique- we can propagate the six uncertainties to the expected uncertainty on the Z boson mass, $\sigma_M$. The formulas needed for that purpose are quite a bit more complicated than the others I have discussed so far, so I will spare you from those murky details. All what is important is to notice that, given a pair of muons from Z decay, we have a means to compute their invariant mass, and the expected resolution on that quantity.

Now, the crucial question to ask is “What is the probability that a Z boson produced with a rapidity Y is measured at a mass M, if the expected uncertainty on M is $\sigma_M$ ?“. A precise answer to this question is the input I need for the calibration algorithm which uses Z bosons to determine the muon momentum scale in the CMS detector.

To answer the question above, we need to take a convolution of the factorization integral with a Gaussian function, the latter representing the smearing of the true Z mass due to the measurement procedure. The Gaussian, of course, has a width equal to $\sigma_M$.

The end result is shown below -well, two posts below, in fact. It represents the probability to observe a Z boson of measured mass M, as a function of the resolution $\sigma_M$. Using this information it is possible to put together a likelihood function which depends on the measured kinematical characteristics of the muons. The maximization of the likelihood with respect to parameters describing a suitable form of the measurement bias allow us to correct the track momentum scale.

A description of the algorithm, I am afraid, belong to a different post. You can check a rough description I gave in this blog here; and a more general discussion is here.

Instead, here I will conclude this longish and uninformative post with a last fancy graph, which shows the difference between two probability functions, both extracted as discussed above, one using the CTEQ PDF set and the other using the MRST set. As you see, the difference is never larger than a part in ten thousand: as far as shapes are concerned, PDF systematics do not have a large impact in the observed peak of the Z boson. Other characteristics, in fact, are better suited to be exploited if one wants to be sensitive to differences in the PDF of the proton.

(As in the graph posted yesterday, the x axis has the reconstructed mass of the Z boson from 71 to 111 GeV, the y axis has the expected mass resolution from 0 to 50 GeV (a unnaturally large value given that the typical resolution is of the order of 1-2 GeV, but the algorithm requires the function to be defined in a wide area), and the z axis is the difference between the probability values extracted with CTEQ and MRST sets.)