# Course Outline for Physics 880.05

## II. D. Nucleus-Nucleus Collisions (Chs. 12,13) Part 4

We'll start by finishing our discussion started last time of the experimental evidence for nuclear stopping (see part 3 notes). Then we'll discuss the space-time picture of a heavy-ion collision and Bjorken's estimate of the initial energy density for the formation of a quark-gluon plasma.

### Revisit: Total Inelastic Cross Section

Last time we defined a total inelastic cross section, \sigma^{A}_in, for the collisions of a nucleon with an A-nucleon target nucleus. We found that it was related to \sigma_in, the total inelastic nucleon-nucleon cross section by:
\sigma{A}_in = A * \sigma_in / <n'>
where <n'> is the average number of nucleon-nucleon collisions, given that there is at least one. (So <n'> is at least equal to one.) We can understand this result intuitively:

• Think geometrically: Consider each nucleon in the target to be a disk of area \sigma_in.
• If the nucleons are one deep, the total cross section is clearly the total area, A*\sigma_in.
• But there can also be only one collision at most (since one deep), so <n'> = 1 and the formula is satisfied.
• Now suppose they are two deep. Just consider them stacked in two's. Then <n'> = 2 (if it hits one, it hits the other) and there are A/2 pairs, so \sigma^{AB}_in = (A/2)*\sigma_in, and the formula works again.
• Next consider 5 deep (and so on).

### Set-up for High-Energy Heavy-Ion Collision

• Consider the case relevant to RHIC with the largest amount of energy deposited: a head-on (\vec b = 0) collision of two equal nuclei in the COM frame (e.g., gold on gold)
• Consider extremely high energy, so that effectively the nuclei are Lorentz contracted to lines; that is, the z-coordinates of all nucleons in the nuclei are the same
• Glauber estimates predicts uranium on uranium will have on average about 800 collisions [try this with b=0, comparing <n(b)> = A*B*T(b)*\sigmin_in and <n'(b)> = <n(b)> / [1-(1-T(b)\sigma_in)^{A*B}]
• Each collision has a large loss of energy. The incident baryons slow down but still have large rapidity after collisions => move away from collision region
• The collisions are approximately additive, implying a very high energy density is attained but with small net baryon number

### Space-Time Scenario

• Consider the figure with the space-time evolution of the collision.
• The projectile and target nuclei move along the light cones from z=-\infinity (B) and z=+\infinity (A). They meet at z=0, where we'll define the time (in the lab frame) to be t=0. All collisions start then, because of the extreme Lorentz contraction.
• Soon after the collisions we might have quarks, gluons, or hadrons near z \sim 0.
• The sequence of "eras" in the evolution is formation, then equilibration, then evolution according to equilibrium thermodynamics (QGP hydrodynamics -> later!). As the system expands, it cools (temperature drops) until hadronization. The hadrons stream out when T is below "freeze-out".

### Estimate of Initial Energy Density at \tau=\tau_0

• We'll need the total energy and the relevant volume.
• We'll base our estimate on the produced hadrons
• If we trace them backwards to the (t,z) coordinates at which they were created, we can find the energy content and the volume where particles were created. We don't want to literally trace them, but associate their rapidity (which we know about) to when and where they were created.
• After (0,0), the volume depends on time so the energy density drops. We'll look at z=0 and t=\tau_0, which should be comparable to the time at which particles start being produced near z=0. So we know the time, how many are produced near z=0?
• Question to class: Based on our discussions so far, what do we know (or can find out from experiment) about the produced hadrons?
• The produced particles are mostly pions with <p_T> about 0.35 GeV/c, which implies m_T about 0.38 GeV/c.
• We have information on dN_ch/dy, the rapidity distribution of the produced particles (how many per unit rapidity).
• How do we relate that quantity to space-time position? Remember that rapidity is a sort of relativistic velocity. If we know velocity, we can relate z and t.
• Let's transfrom dN/dy to a spatial distribution. This sounds like a crazy thing to do, at first! But if we can relate y and z, we can do it.
• Recall how p_z and p_0 are related to rapidity:
p_z = m_T (sinh y) and p_0 = m_T (cosh y)
So the velocity in the z-direction is (relativistically valid!) v_z = p_z/p_0 = tanh y.
• Starting from (0,0), z = t (tanh y) or \tau = \sqrt{t^2-z^2} = t \sqrt{tanh^2-1} = t/cosh y. Substituting back:
z = \tau (sinh y) and t = \tau (cosh y).
Note the similarity to p_z and p_0!
• We can invert the equations to solve for y by adding and subtracting: t+z=\tau e^y and t-z=\tau e^-y, which implies:
y = (1/2) ln(t+z/t-z).
• The central rapidity region corresponds to small y. For a given \tau, small y implies small z, so central rapidity also corresponds to z \sim 0.
• Now consider a fluid element at rest in the com at z=0.
• Consider a thickness \Delta z, as in the figure.
• Define the transverse overlap area as A_curl, which implies that the volume of the element is A_curl*\Delta z.
• Focus on the time \tau_0, where the QGP may have formed and equilibrated.
• How many particles are there in volume A_curl*\Delta z? At z=0 and \tau=\tau_0:
\Delta N/(A_curl \Delta z) = (1/A_curl) dN/dy dy/dz |y=0 = (1/A_curl) dN/dy 1/(\tau_0 cosh y)|y=0
• We're able to do this because we can express z in terms of y. So a small region in z around z=0 translates directly into a small region in y around y=0. Let's see how that works.
• Since the energy of a particle with rapidity y is m_T cosh y, the energy density is:
\epsilon_0 = m_T cosh y (\Delta N/A_curl \Delta z) = (m_T/\tau_0 A_curl) dN/dy|y=0.

### Plugging in Numbers

First we need a reasonable estimate for \tau_0. We don't have to be super precise, but we'd like an estimate for \epsilon_0 better than a factor of two, if possible.

• Bjorken estimate \tau_0 = 1 fm/c
• It should be consistent with particle production time, which ranges in various estimates from 0.4 to 1.2 fm/c (quite a large range!) The idea is that the energy from the colliding baryons is initially stored in the field between the separating quarks. Only when it is stretched enough can it show up as hadrons or free quarks and gluons.
• From our string example, we found a minimum distance 2*m_T/\kappa for the string to be stretched before particles were produced. With m_T about 0.4 GeV for a quark and \kappa = 1 GeV/fm, L_min \approx 0.8 fm. For two sources separating at the speed of light, this implies \tau_0 about 0.4 fm/c.
• Other estimates from other models (Chapters 5-8) are consistent.

From experiment we can get some estimates of dN/dy.

• In central collisions of S32 or O16 on Au at 200*A GeV, the peak dN/d\eta (\approx dN/dy) = 160 for S32 and 110 for O16.
• See Figure 13.5. What do we observe?
• dN_ch/d\eta_peak scales linearly with A^(1/3) for a S32 beam. Can you give a reasonable explanation?
• We can identify central collisions by how much energy is deposited in the forward calorimeter. Define central collision by having less energy than some threshold, like 20% of the beam energy.
• Class: how is dN_ch/dy related to dN/dy?
dN/dy = (3/2) dN_ch/dy

Estimate of dN/dy from Glauber model:

• Consider nucleus-nucleus collision at impact parameter b.
• If assume that each collision contributes produced particles the same way, then each produces like a NN collision (for which we have experimental numbers.
• So if there are n'(b) inelastic NN collisions, we'll expect:
dN/dy \approx n'(b) dN_pp/dy.
• This overestimates the result, however, since the earlier collisions are more effective in producing particles, since they have lost less of their energy (less stopped)
• We can account for this with a semi-empirical reduction factor R = 1/(1 + a(A^{1/3} + B^{1/3})), where a is an empirical constant.
• So the reduction factor is more important as the nucleus gets thicker.
• So dN/dy \approx R * n'(b) * dN_pp/dy
• Recall our Glauber result for n'(b):
n'(b) = A*B * T(b)\sigma_in / {1 - (1-T(b)\sigma_in)^{A*B}}
(remember that T(b)\sigma_in is the probability of a collision, so this is the mean of a binomial distribution with that probability, adjusted by the denominator to exclude the n=0 result)
• Using the gaussian approximations, and noting that
dN/dy(b) \approx (3\sigma_in/2\pi(r0')^2) * (A*B/(A^{2/3}+B^{2/3}) * R * e^{-b^2/2\beta^2} * dN_pp/dy

Put it together:

• Take \tau_0=1 fm/c, A_curl = \pi (1.2)^2 A^2/3 fm^2 (from the transverse radius of a nucleus with A nucleons)
• The result is 1 GeV/fm^3 for S32 on Al and 2.1 GeV/fm^3 for S32 on Au.
• Some single events with large maximum dN_ch/d\eta can have values as high as 3.3 GeV/fm^3.
• So now the question is: Will this be enough (scaled for heavier nuclei) for a quark-gluon plasma? Time for some thermodynamics!