Etele Molnar 2012adas

Introduction Perfect Fluids Dissipative Fluids Fluid dynamics from kinetic theory Outlook

From the Boltzmann equation to fluid dynamics and itsapplications

Etele Molnar

MTA-DE Particle Physics Research Group

09.10.2012

in collaboration with:I. Bouras, G. Denicol, P. Huovinen, H. Niemi, D. H. Rischke

Etele Molnar From the Boltzmann equation to fluid dynamics and its applications


Stages of evolution in heavy-ion collisions I.

Initial stage (Belensky and Landau 1955)

When two nucle(i)ons collide, a compound systemis formed, and energy is released in a small volumeV subject to a Lorentz contraction in thelongitudinal direction.At the instant of the collision,a large number of particles are formed: themean free path (m.f.p.) in the resulting system issmall compared with its dimension, andstatistical equilibrium is set up.

Initial particle production 1 fm/c:Two nuclei fly through each other (Bjorken1976, 1983), producing highly excited matter.

Non-equilibrium evolution of the matter(thermalization) . 1 fm/c.



Stages of evolution in heavy-ion collisions II.

Fluid dynamical evolution of theQGP 5 fm/cTransition back to hadronic matter(QCD phase transition)

Evolution of the hadron gas

Hydrodynamical stage (Belensky and Landau 1955)

The second stage of the collision consists in theexpansion of the system. Here thehydrodynamical approach must be used, and theexpansion may beregarded as the motion of an ideal fluid (zeroviscosity and zero thermal conductivity). During theprocess of expansion the m.f.p., remains small incomparisons with the dimensions of the system, andthis justifies the use of hydrodynamics.Since the velocities in the system arecomparable with that of light, we must use notordinary but relativistic hydrodynamics. Particlesare formed and absorbed in the system throughoutthe first and second stages of collision. The highdensity of energy in the system is of importancehere. In this case, the number of particles is not anintegral of the system, on account of the stronginteraction between the individual particles.



Stages of evolution in heavy-ion collisions III.

Transition to free particles(Cooper-Frye 1976)

Freeze-out stage (Belensky and Landau 1955)

As the system expands,the interaction becomes weaker and the mean freepath becomes larger. The number of particlesappears as the physical characteristic when theinteraction is sufficiently weak. When the mean freepath becomes comparable with the lineardimensions of the system, the laterbreaks up into individual particles. This may becalled the break-up stage. It occurs with atemperature of the system of the order T mpic2,where mpi is the mass of the pion.



Stages of evolution in heavy-ion collisions III+.

The space-time picture of evolution.dV /V = d/ in 1+1D case.



Perfect Liquid



Notation

For simplicity we choose the flat space-time, the metric isg g = diag(1,1,1,1).The normalized 4-flow of matter is denoted by u(t, x), whereuu = 1, (c2 = 1).

The local rest frame (LRF) is defined as, uLRF = (1, 0, 0, 0).

The projection tensor is defined perpendicular to the 4-flow of matter, = g uu , where u = 0, and = 3.The comoving time-derivative or proper-time derivative in LRF of an arbitrary4-vector A is denoted by, A = uA while, the comoving spatial-derivativeor gradient is denoted by, = .The orthogonal projection of A A = A.The symmetric, traceless and orthogonal part of any tensor is denoted by,

A A =[

12

(

+

) 1

3

]A .



Perfect Fluids I.

Conservation laws for a simple (single component) perfect fluid (no dissipation)

N0 = 0 charge conservation 1 eq.

T0 = 0 energy-momentum conservation 4 eqs.

Perfect fluid decomposition with respect to u

N0 = n0u

T0 = e0uu p0

n0 = N0 u (net)charge density

e0 = T0 uu energy density

p0 = 13

T0 isotropic pressure

We have 5 equations for 6 unknowns not closed: n0(1), e0(1), p0(1) and u(3).These equations are postulated!



Perfect Fluids II.

The assumption of local thermal equilibrium! Provides closure:

Equation of State (EoS)

p0 = p0(e0, n0) EoS 1 eq.

and/or p(T , ) or s = s(e, n).

Auxiliary, S0 = s0u, where s0 = S

0 u, and for continuous solutions

S0 = 0

entropy is maximum in local thermal equilibrium

Thermodynamics

Ts = e + p nT s = e np = sT + n

The fundamental thermodynamic relations are derived from,T(su) = (eu) + p(u) (nu)



Dissipative Fluids I.

Conservation laws for a simple (single component) dissipative fluid

N = 0 charge conservation 1 eq.

T = 0 energy-momentum conservation 4 eqs.

General decomposition with u

N = nu + V

T = euu (p + ) + Wu + W u + pin = Nu charge density

e = Tuu energy density in LRF

p + = 13

T isotropic (equilibrium + bulk viscous) pressure

V = N charge flow

W = Tu energy-momentum flow

q = W e + pn

V heat flow 1 eq.

pi = T stress tensor

We only have 5 equations for 18 unknowns, n(1), e(1), p(1), u(3) and(1),V(3),W(3), pi(5).



Dissipative Fluids II.

Generally, N = N0 + N and T = T0 + T

Simplifications (I): Matching to equilibrium and the EOS

n = n0 + n

e = e0 + e

p(e, n) = p0(e0, n0) + p

Convenient choice, n = 0, e = 0, p = hence, n = n0, e = e0,p0 = p0(e0, n0), such that p(e, n) = p0 + , while T = T0 and = 0, buts = s0 + s!

Still with 14 + 3 unknowns! n(1), e(1),(1), V(3),W(3), pi(5) and u(3).

Simplifications (II): Fixing the Local Rest Frame

uE = N/n V = 0 q = W Eckart

uL = TuL/e W = 0 q =

e + p

nV Landau & Lifsitz

We eliminated (3) unknowns, such that uL = uE + u

.

We are left with 14 unknowns! n(1), e(1), u(3) and p(1), q(3), pi(5).



Dissipative Fluids III. - The relativistic Navier-Stokes theory

The entropy current is also modified S S0 + S = (s0 + s)u + , wheres Su = (s0 + s) and = S .

2nd law of thermodynamics (Eckarts frame)

S =

[ q

T

] q

T

(1

TT u

)

Tu

+pi

Tu 0

Assuming that s = s0, i.e., s = 0, while = q/T

For small gradients, linear relations between thermodynamical forces and fluxes!

The relativistic Navier-Stokes relations

NS = upiNS = 2 u

qNS = TT n

e + p

( T

) 0, 0 bulk and shear viscosity coefficients, 0 coefficient of thermalconductivity.

Now, the equations of fluid dynamics are closed, but the relativistic Navier-Stokestheory leads to accausal signal propagation and stability issues.



Dissipative Fluids III+.

(Gas)Kinetic theory

pixzNS = vz (x)

x ' 1.2T/ = 1.2Tmfpn

/s Tmfp

(Dilute) Gases may have or 0 hence ideal gas!Air 1.8 105Pa.s (T 20Co)Fluids (liquids), 0 or hence 0 ideal fluid (liquid)!Water 50 Air = 9 104Pa.sgas(T ) increases with increasing Tliquid (T ) decreases with increasing T

Ref. Csernai et.al. PRL (2006)(s

)pi

=15

16pi

f 4piT 4(

s

)QCD

=5.12

g4 ln(2.42g)> 1(

s

)ADS/CFD

=1

4pi

fpi 130MeV pion decay constantg QCD coupling constant



Dissipative Fluids IV.

2nd order theories of Muller (1967), Israel (1976) and Stewart (1971, 1977)The previously mentioned issues are cured if:

Entropy current generalization (Eckarts frame)

S s0u + q

T (02 1qq + 2pipi) u

2T

0q

T+1pi

q

T+ O3

where the newly introduced coefficients 0, 1, 2, 0, 1 are related to therelaxation time/length of bulk viscosity, heat conductivity and shear viscosity

= 0

q = T1

pi = 22

lq = 0

lq = T0

lqpi = T1

lpiq = 21

The 0, 1, 2, 0, 1 coefficients are frame dependent and remain undeterminedin phenomenological theories!



Dissipative Fluids V.

2nd order equations from entropy production

Relaxation equations in Eckarts frame (Israel 1976)

+ = NS + lqqq

q

+ q = qNS + lq lqpipi

pipi + pi = piNS + lpiqq

The equations determine the time evolution of , q, and pi

The Navier-Stokes theory appears if the relaxation times and length scalesi 0, li 0 with , and q fixedLater O2 corrections ( i , i , . . ., etc.) were added by Israel andStewart (1977-1979), Hiscock and Lindblom (1983), Relativistic ExtendedThermodynamics of Liu, Muller and Ruggeri (1983) etc.

We need kinetic theory to motivate and determine the above introducedphenomenological equations self-consistently!



The relativistic Boltzmann equation and conservation equations

In a dilute gas, the space-time evolution of the single-particle distribution functionfk = f (t, x, k

0, k) due to particle motion and binary collisions is given by the

The relativistic Boltzmann equation

kfk C [fk] =1

2

dK dPdPWkkpp

(fpfp fk fk fkfk fp fp

)where k = (k0, k) is the four-momenta of particles with mass m and energyk0 =

k2 + m2. fk = 1 afk, with a = 0 for Boltzmann, a = 1 for Fermi, a = 1 for

Bose statistics. The inv. phase-space element is, dK = gd3 k/[(2pi)3k0

]. The

transition rate W is invariant, to final state momenta Wkkpp = Wkkpp, andsatisfies detailed balance, Wkkpp = Wppkk.

Conservation laws from the Boltzmann equation

N

dK kfk =

dK C [fk] = 0 charge cons.

T

dK kkfk =

dK kC [fk] = 0 energy-momentum cons.

Conservation laws are obtained, 5-collisional invariants, but we still need the solutionof the Boltzmann equation fk!



The hierarchy of moments and balance equations

Moments of the single-particle distribution function and collision integral

N(t, x)

dK k fk = k charge current

T(t, x)

dK kk fk = kk energy-momentum tensor

F1...n (t, x)

dK k1 . . . kn fk = k1 . . . kn n-rank moment

P1...n (t, x)

dK k1 . . . kn C [fk] = C1...n n-rank production term

where p1p2 . . . pn1pn dK p1 . . . pn fk.The mass-shell condition kk = m2 gives

F1...n = m2F1...n

P1...n = m2P1...n

F1...n = P1...n balance of fluxes

These balance equations are never closed but the solution of the Boltzmann equationis also a solution of the hierarchy!



Local thermal equilibrium

In local thermal equilibrium S0 = 0, hence fk f0k, where f0k = f (0, 0,Ek)

f0k [exp (0 + 0Ek) + a]1 Juttner distribution

where 0 = 0/T0 is the chemical potential, 0 = 1/T0 is the inverse temperature,Ek = k

u, and a = 0 for Boltzmann, a = 1 for Fermi, a = 1 for Bose statistics.

Moments of the equilibrium distribution function

N0 (t, x)

dK k f0k = k0 charge current

T0 (t, x)

dK kk f0k = kk0 energy-momentum tensor

S0 (t, x)

dK k f0k (ln f0k 1) = k0 entropy current

Conservation eqs. for and ideal gas

N0 = 0, T

0 = 0, S

0 0

(a = 0) hence p0 = n0T0 ideal gas law!



General kinetic decompositions and definitions

Decompose the momenta, k = Eku + k, where Ek = ku and k =

k

N = Eku + kT = E2k uu +

1

3kk+ Ekku + Ekku + kk

where kk = kk and k1k2 . . . kn dK k1 . . . kn fk for any fk

n Ek e E2k p 1

3kk

V k W Ekk pi kk

In equilibrium k1k2 . . . kn 0 dK k1 . . . kn f0k, hence

n0 = Ek0 e0 = E2k 0 p0 = 1

3kk0

V k0 = 0 W Ekk0 = 0 pi kk0 = 0



Out of equilibrium

For a system out of equilibrium

fk = f0k + fk

Let us define the following irreducible moment

1...`(r)

(Ek)r k1 . . . k `

where . . . . . . . . .0 =dK . . . fk

dK . . . f0k =

dK . . . fk

n = (1) e = (2) = m2

3(0)

V = (0)

W = (1)

pi = (0)

Matching: Non-equilibrium to equilibrium

n Ek = 0 n = n0e E2k = 0 e = e0

= 13kk p(e, n) = p0(e0, n0) +



General equations of motion I.

Equations of motion for the irreducible moments

1...`(r)

= 1...`1...`

d

d

dK (Ek)

r k1 . . . k `fk

where A uA dA/d and 1...`(r) 1...`1...`

1...`(r)

and

k1 k m = 1m1mk1 km

Now, using the Boltzmann equation kfk = C [f ] in the following form

fk = f0k E1k k f0k E1k kfk + E1k C [f ]

where = , we obtain the exact equations for the comoving derivatives of theirreducible moments,

1...`(r)



General equations of motion II.

Scalar

r = Cr1 + (0)r +

(rr1 +

G2r

D20W

)u

r1 +G3r

D20V

G2rD20

W

+1

3

[(r 1)m2r2 (r + 2) r 3

G2r

D20

]

+

[(r 1) r2 +

G2r

D20pi

] .

Inq =(1)q

(2q + 1)!!

dKE n2qk

(kk

)qf0k,

Jnq =(1)q

(2q + 1)!!

dKE n2qk

(kk

)qf0k f0k,

Gnm = Jn0Jm0 Jn1,0Jm+1,0,Dnq = Jn+1,qJn1,q (Jnq)2 .

Vector

r = C

r1 +

(1)r 0 hr W + rr1u

+1

3

[rm2r1 (r + 3) r+1 3hr

]u

13

(m2r1 r+1

)+ hr

(r1 + hr pi

)+

1

3

[(r 1)m2r2 (r + 3) r 4hrW

]

+1

5

[(2r 2)m2r2 (2r + 3) r 5hrW

]

+(r +

hrW

) + (r 1) r2 .

C1`r 1`1` C

1...`r

= 1`1`

dKE rkk

1 k`C [f ] .



General equations of motion III.

Tensor

r = C

r1 + 2

(2)r

+2

15

[(r 1)m4r2

(2r + 3)m2r + (r + 4)r+2]

+2

5

[rm2

r1 (r + 5) r+1

]u

+ rr1 u 2

5

(m2

r1 r+1

)+

1

3

[(r 1)m2r2 (r + 4) r

]

+2

7

[(2r 2)m2r2 (2r + 5) r

]

+ 2r r1

+ (r 1)r2 .

0 =1

D20

[J30

(n0 + V

)

+ J20(0 + P0 +

)

+J20

(W

W u pi)],

0 =1

D20

[J20

(n0 + V

)

+ J10(0 + P0 +

)

+J10

(W

W u pi)],

u = 10

(h100 0

)

h10

n0

(u

)

h10

n0

[ 43W + W

(

)+ W + pi

],

(0)r = (1 r) Ir1 Ir0

n0

D20

(h0G2r G3r

),

(1)r = Jr+1,1 h

10

Jr+2,1,

(2)r = Ir+2,1 + (r 1) Ir+2,2,

hr =

0

0 + P0

Jr+2,1,



General equations of motion IV.

Polynomial expansion

fk f0k + fk = f0k(

1 + f0kk),

k =`=0

1`k k1 k`,

1`k =

N`n=0

c1`n P

(`)kn ,

P(`)kn =

nr=0

a(`)nr Erk , orthogonal poly

c1`n

N (`)`!

P

(`)kn k1 k `

=N (`)`!

nr=0

1`r a

(`)nr .

fk = f0k

1 + f0k `=0

N`n=0

H(`)kn 1`n k1 k`

,H(`)kn =

N (`)`!

N`m=n

a(`)mnP(`)km .



The 14-moment method

Truncation

k N0n=0

cnP(0)kn ' c0P

(0)k0 + c1P

(0)k1 + c2P

(0)k2 ,

k

N1n=0

cn P(1)kn ' c

0 P

(1)k0 + c

1 P

(1)k1 ,

k

N2n=0

cn P(2)kn ' c

0 P

(2)k0 ,

Matching

r N0n=0

nm=0

cna(0)nm Jr+m,0 =

r ,

r

N1n=0

nm=0

cn a(1)nm Jr+m+2,1 =

Vr V

+ Wr W,

r

N2n=0

nm=0

cn a(2)nm Jr+m+4,2 =

pir pi

.



The 14-moment method - Collision integral in a Boltzmann gas

Collision integral for Boltzmann particles

C1...`r =

1

2

dKdK dPdPWkkppf0kf0k (Ek)r k1 . . . k`

[(p + p k k)+ (pp kk)]Linearized collision integral

Cr1 u1 u`C1`r`1 = CXr3,1

Cr1 1u2 u`C

1`r` = BVVXr2,3 + BWWXr2,3,

Cr1 12u3 u`C

1`r`+1 = ApipiXr1,4.

Collision integral

Xr 1

dKdK dPdPWkkppf0kf0k f0p f0pE

rkkk

(pp + pp kk kk

)=(Xr,1u

u + Xr,2)(uu 1

3

)+ 4Xr,3u

( )( u ) + Xr,4

.



The 14-moment method - (r ?)

Scalar

= r

r

r + rWWu

+ rVVu

`rW W `rV V r+ rWW

0 + rVV0 + rpipi .

Tensor

pi = pi

rpi+

2r

rpi + 2rpi

+ 2 rpiWW u + 2 rpiVV

u

+ 2`rpiWW + 2`rpiV V

2rpiWW 0 2rpiVV 0 2rpipipi + 2pi 2rpipipi.

Vector

V + Wr W = V

rV Wr

W

rW+

rq

rV20h

20

0

+ Wr W + V

Wr rWWW + rVVV

Wr rWWW + rVVV rqu rqpipi u+ `rq `rqpipi

+ rq0 + rqpipi0,



The exact relaxation equations with a linearized collision integral

The original Israel-Stewart derivation neglected several terms; corrections added later.More terms from the non-linearized collision integral!

+ = NS + q q u `q q 0 + q q + pi pi

q q + q = qNS q u qpi pi u

+ `q `qpi pi + q q 1 q

qq q + q + qpi pi

pi pi + pi = piNS + 2 piq q

+ 2 `piq + 2 pi pi 2 2 pi 2 pi pi 2piq q + 2pi

W. Israel, J.M. Stewart, Ann. Phys. 118 (1979) 341W. Israel, J.M. Stewart, Ann. Phys. 118 (1979) 341A. Muronga, PRC 76 (2007) 014909; A. Muronga, ibid.B. Betz, D. Henkel, D. H. Rischke, Prog. Part. Nucl. Phys. 62:556 (2009); B. Betzet.al. EPJ Web of Conf (2011).



So Many Jokes, So Little Time ...

Seriously...

The new method leads to equations and coefficients that are in better agreementwith the non-equilibrium solution of the Boltzmann equation, (than the othersout there)!

Hard to see the fine details, though !

Different choices (of methods, moments, coefficients etc.) lead to the sameasymptotic (Navier-Stokes) solutions, (but not the other way around)!

New terms from the second order collision integral,(pp kk

)(to be

published soon)

References

B. Betz, G. S. Denicol, T. Koide, E. Molnar, H. Niemi and D. H. Rischke,Second order dissipative fluid dynamics from kinetic theory,EPJ Web Conf. 13, 07005 (2011).

G. S. Denicol, H. Niemi, E. Molnar and D. H. Rischke, Derivation of transientrelativistic fluid dynamics from the Boltzmann equation,Phys. Rev. D 85, 114047 (2012).

G. S. Denicol, E. Molnar, H. Niemi and D. H. Rischke, Derivation of fluiddynamics from kinetic theory with the 14moment approximation,Eur. Phys. J. A (2012).



Backup slides

Backup



Approximate solutions: Grads method of moments

The previous definitions from kinetic theory are only useful if we can specify fk or fk

fk = f0k + fk ' f0k (1 + k)

where k = fk/f0k 1 close to equilibrium, f0k [exp (0 + 0Ek) + a]1, where0 = 0/T0 is the chemical potential, 0 = 1/T0 is the inverse temperature,Ek = k

u, and a = 0 for Boltzmann, a = 1 for Fermi, a = 1 for Bose statistics.

Relativistic Grads approach: Stewart (1972), Israel and Stewart (1977-1979)

k l=0

1...l(l)

k1 ...kl = (0) + (1)k +

(2)

kk + . . .

Assuming that in k only l = 0, 1, 2-rank tensors appear (1), (4) and (9)

where = 0, we provide a closure for the hierarchy (truncation) and be able todetermine the 14 moments of dissipative fluid dynamics!



Applying the 14-method

Non-equilibrium 14-moment ansatz

fk f0k[1 + (0) + Ek

(1)u +

(1)

k + E2k (2)

uu +1

3

(kk

)

(2)

+ Ek(2)

ku + Ek(2)

ku + (2)

kk]

which is basically an expansion in Enk and k1 . . . kn

Decomposition and matching

(0) 0 = A0

(1)u 0 = A1

(2)uu = A2

(1)

= B0q + B1V

(2)

u = C0q + C1V

(2)

= A3(3uu ) + 2C0u(q) + 2C1u(V ) + D0pi

(2)

= D0pi

The coefficients (0), (1),

(2)expressed in terms of dissipative fields, , q(V), pi



The relaxation equations

Using the 14-moment ansatz, the 3rd moment, kkk = F (e, n, p, u,, q, pi),therefore, using the Boltzmann transport eq., kkk = C, we get:

Israel-Stewart (1979)

uukkk = uuC Bulk eq.ukkk = uC Heat-flow eq.

kkk = C Shear viscosity eq.

The linearised collision integral

uuC = CuC = Cqq

C = Cpipi

where C,Cq ,Cpi



Further corrections

Take into account 2nd order corrections in the collision integral, i.e.,C [f0 + f ] f + (f )2

Full collision integral: Rischke et.al. (2010)

uuC C + A02 + A1Vq + A2qq + A3pipiuC Cqq + B0q + B1V + B2qpi + B3Vpi

C Cpipi + C0pi + C1qV + C2qq + C3pipi

some terms also by Moore (2009)

Problem with the traditional moment method

u1 . . . unk1 . . . knk = u1 . . . un C1...n 1u2 . . . unk1 . . . knk = 1u2 . . . un C1...n

12u3 . . . unk1 . . . knk = 12u3 . . . un C1...n

Parts of higher order moments may contribute for , q, pi , so why stop ?


IntroductionPerfect FluidsDissipative FluidsFluid dynamics from kinetic theoryGeneral equations of motion for the momentsGeneral equations of motionGeneral equations of motionGeneral equations of motionThe 14-moment methodThe 14-moment methodThe 14-moment method

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Etele Molnar 2012adas