energy and number density created at rhic
DESCRIPTION
Energy and Number Density Created at RHIC. What’s in the PHENIX White Paper, and a little bit more. Paul Stankus, ORNL PHENIX Focus, Apr 11 06. Energy Density, Take 1. Just divide energy by volume, in some frame. 2R/ g ~ .13 fm. 2R ~ 14 fm. Boosted Frame - PowerPoint PPT PresentationTRANSCRIPT
Energy and Number Density Created at RHIC
What’s in the PHENIX White Paper, and a little bit more
Paul Stankus, ORNL PHENIX Focus, Apr 11 06
2
Energy Density, Take 1
2R ~ 14 fm 2R/ ~ .13 fm
Rest Frame
= E/V = M/V0
~ 0.14 GeV/fm3 =
Boosted Frame
= E/V = M/(V0/) = 02
RHIC = 106
~ 1570 GeV/fm3 (!!)
Just divide energy by volume, in some frame.
3
Energy Density, Take 2Examine a box with total momentum zero.
= 0 ~ 3150 GeV/fm3
= ?
Very high, but very short-lived!
4
Energy Density, Take 3Count up energy in produced particles/matter.
Define produced as everything at velocities/rapidities intermediate between those of the original incoming nuclei.
Two extremes:
All particles Bjorken
All fluid Landau
5
The Bjorken Picture: Pure Particles
Key ideas:
• Thin radiator
• Classical trajectories
• Finite formation time
6
Particles in a thin box with random velocities
Release them suddenly, and let them follow
classical trajectories without interactions
Strong position-momentum
correlations!
7
J.D. Bjorken, Phys. Rev. D 27 (1983) 140“Highly relativistic nucleus-nucleus
collisions: The central rapidity region”
Key idea: Use the space-momentum correlation to translate between spatial density dN/dz and momentum density dN/dpZ
Thin radiator
The diagram is appropriate for any frame near mid-rapidity, not just the A+A CMS frame specifically.
8
x
z
pT = pX= p pZ
= 0 y = 0
E=√m2+pT2mT
x’
z’
pT = pX mT =√m2+pT2
pZ = mT sinh(y) E = √m2+p2 = mT
cosh(y) Z= pZ/E = sinh(y)/cosh(y) y =tanh-1(Z) y
Z for Z<<1
Useful relations for particles in different Lorentz frames
9
dZ
Exercise: Count particles in the green box at some time t, add up their energies, and divide by the volume.
Particles in the box iff 0<Z<dZ/t (limit of infinitely thin source)
€
Number of particles dN =dN
dβ Z
dZ
t=
dN
dy
dZ
t
Average energy per particle E = mT
Volume of the box V = A ⋅dZ
A ≡ xy cross section area (limit t << R)
R
€
(t) =Energy
Volume=
E dN
V=
1
t A
dN(t)
dymT (t) =
1
t A
dET (t)
dy
Valid for material at any rapidity and for any shape in dET(t)/dy! A plateau in dET(t)/dy is not required.
10
€
(t) =1
t A
dET (t)
dyHow low can t go? Two basic limits:
€
t >> R 1 γ A +1 γ B( ) Crossing time
t ≥ τ Form Formation time
€
(τ Form ) =1
τ Form A
dET (τ Form )
dy≈
1
τ Form A
dETFinal State
dy
Bjorken
For many years this Bjorken formula was used with a nominal Form=1.0 fm/c with no real justification, even when it manifestly violated the crossing time limit for validity!
2R/ = 5.3 fm/c for AGS Au+Au, 1.6 fm/c for SPS Pb+Pb.
11
Better formation time estimates
Generic quantum mechanics: a particle can’t be considered formed in a frame faster than hbar/E
Translation: Form 1/mT ~ 1/<mT>
€
mT =dET (τ Form ) /dy
dN(τ Form ) /dy≈
dET /dη
dN /dη (Final State)
PHENIX Data: (dET/d)/(dNch/d) ~ 0.85 GeV
Assuming 2/3 of particles are charged, this implies Form ~ 0.35 fm/c
12
13
Some assumptions we’ve used
• Transverse energy density dET/dy only goes down with time.
• The number density of particles does not go down with time (entropy conservation).
• We can estimate, or at least bound, thermalization time from other evidence.
An unanswered question: What are the initially produced particles? (Bj: “quanta”)
14
Identifying the intial “quanta”
Multiplicities in Au+Au at RHIC were lower than initial pQCD predictions. Indicates need for “regularization”. Good candidate is CGC.
CGC identifies intial quanta as high-ish pT gluons (~1 GeV), which is consistent with our particle picture.
15
The Landau Picture: Pure Fluid
Key ideas:
• Complete, instant thermalization
• Fluid evolves according to ideal relativistic fluid dynamics (1+1)
• Very simple √s dependences for multiplicity and dN/dy (Gaussian)
16
Courtesy of P. Steinberg; see nucl-ex/0405022
Multiplicities
Widths
17
€
dE = TdS − PdVBasic
Thermodynamics
Sudden expansion, fluid fills empty space without loss of energy.
dE = 0 PdV > 0 therefore dS > 0
Gradual expansion (equilibrium maintained), fluid loses energy through PdV work.
dE = -PdV therefore dS = 0Isentropic Adiabatic
Hot
Hot
Hot
Hot
Cool