Transport coe ffi cients of QGP in strong magnetic fields Daisuke - - PowerPoint PPT Presentation

Transport coefficients of QGP in strong magnetic fields

Daisuke Satow (Frankfurt U. !) Collaborators: Koichi Hattori, Xu-Guang Huang (Fudan U, Shanghai ") Shiyong Li, Ho-Ung Yee (Uni. Illinois at Chicago #) Dirk Rischke (Frankfurt U. !)

• K. Hattori and D. S.,
• Phys. Rev. D 94, 114032 (2016).
• K. Hattori, S. Li, D. S., H. U. Yee,
• Phys. Rev. D 95, 076008 (2017).

Outline

2

• (Long) Introduction

—quarks and gluons in strong B

• Electrical Conductivity
• Bulk Viscosity
• Summary

Introduction

3

Strong magnetic field (B) may be generated in heavy ion collision due to Ampere’s law. B

Introduction

4

At thermalization time (~0.5 fm), there still may be strong B.

Heavy ion collision may give a chance to investigate QCD matter at finite temperature in strong magnetic field.

lifetime: ~0.3fm ~ 100 [MeV] strength:

√ eB

(Comparable with T)

• 2

4 6 8 10 tfm 104 0.001 0.010 0.100 1 eBmπ

2

• K. Tuchin, Phys. Rev. C 93, 014905 (2016).

/e. σ = 5.8 MeV, z = 0.2 fm, t0 = 0.2 fm. S = 2000 (LHC). = panel: γ = 2000 (LHC).

Introduction

5

How the particles behaves when √eB is much larger than the other energy scales of the system?

(√eB >> T, m, ΛQCD…)

Quark in Strong B

6

One-particle state of quark in magnetic field Classical: Cyclotron motion due to Lorentz force B

+

Quark in Strong B

7

Quantum: Landau Quantization B

En = s (pz)2 + m2 + 2eB ✓ n + 1 2 ◆

The gap (~√eB) is generated by zero-point oscillation.

E

√eB

Longitudinal: Plane wave Transverse: Gaussian

+

pz

1 √ eB

Quark in Strong B

8

For spin-1/2 particle, we have Zeeman effect:

En = s (pz)2 + m2 + 2eB ✓ n + 1 2⌥1 2 ◆

When n=0 (LLL), gap is small (m~1MeV). When n>0, gap is large (~√eB~100MeV) B

+

s E

n=0 n=1

Lowest Landau Level (LLL) Approximation

9

In heavy-ion collision, this condition can be marginally realized (T~√eB~100MeV). But in Weyl semi-metal, it is already realized (T~1meV, √eB~10eV).

When the typical energy of particle (T) is much smaller than gap (√eB), the higher LL does not contribute (~exp(-√eB/T)), so we can focus on the LLL.

En = p (pz)2 + m2

B

+

s pz One-dimension dispersion, no spin degrees of freedom.

• Q. Li, D. Kharzeev, et. al., nature physics 12, 550 (2016)

Gluon in Strong B

10

Gluon does not have charge, so it does not feel B in the zeroth approximation. Massless boson in 3D B

Gluon in Strong B

11

Coupling with (1+1)D quarks generates gluon mass. (Schwinger mass generation)

• K. Fukushima, Phys. Rev. D 83, 111501 (2011).

∼ g2eB

Landau degeneracy Color factor Schwinger mass

M 2 ≡ 1 2 × g2 π X

f

|eB| 2π

B

1 √ eB

(surface density) ~(average distance)-2~eB

Electrical Conductivity

J=σE J E

Motivation to Discuss Electrical Conductivity

13

Electrical conductivity is phenomenologically important because

• Input parameter of magnetohydrodynamics

(transport coefficient)

• May increase lifetime of B (Lenz’s law)

B

∂tB<0

E, j

r ⇥ E = ∂tB

When σ is large

∂tE = r ⇥ B j

Hierarchy of Energy Scale at LLL

14

For ordering of m and M, we consider the both cases. (m<<M and m>>M)

<< E T

√ eB

<< << m M << E T

√ eB

<< << M m

(M~g√eB) LLL approximation Thermally excited

15

Electrical Conductivity Conductivity at weak B (√eB<<T)

ji=σijEj

B=0 E

σij =   σ0 σ0 σ0  

j

j = σ0E

j = σ1E × B

weak B E j B ・

○

σij =   σ0 σ1 σ1 σ0 σ0  

16

Electrical Conductivity Quarks are confined in the direction of B, so there is no current in other directions.

ω) = Π (ω) e ji = σijEj, w

σ33 is finite, other components are zero. (Very different from weak B case)

σij =   σ33  

B

+

E j Strong B (LLL)

Calculation of conductivity

17

B E

Thermal equilibrium in strong B

j

Linear response against E

We linearize the distri as nf(k3, T, Z) = nF (✏L

k ) +

arized version of Eq. (D1

We linearize the distribution function as nf(k3, T, Z) = nF (✏L

k ) + nf(k3, T, Z).

arized version of Eq. (D1) reads

Slightly non-equilibrium, finite j

Evaluation of δnF is necessary.

✏L

k ≡

p (k3)2 + m2

= σ33E3

j3(T, Z) = 2e X

f

qfNc |eqfB| 2π Z dk3 2π v3δnf(k3, T, Z)

Landau degeneracy

hose momentum is k and ), v3 ⌘ @✏L

k /(@k3) = k3/✏L k

Calculation of conductivity

18

Evaluate nF with (1+1)D Boltzmann equation

[@T + v3@Z + eqfE3(T, Z)@k3]nf(k3, T, Z) = C[n],

E

Constant and homogeneous E

19

Possible Scattering Process for Conductivity Decay of a gluon into quark and anti- quark becomes kinematically possible. (1 to 2) strong B

Gluon is effectively massive in (1+1)D

E = q p2

z + p2 ⊥ + M 2

Cut

B=0

1 to 2 scattering is kinematically forbidden;

• ne massless particle can not decay to two

massless particles. 2 to 2 is leading process.

Cut Cut

The scattering process is very different from that in B=0…

Chirality in (1+1)D

20

Spin is always up. When m=0, the direction of pz determines chirality. B

+

s pz χ=+1 B

+

s pz χ=-1

Chirality in (1+1)D

21

B

+

χ=+1+1=2 Chirality is conserved at m=0: 1 to 2 scattering is forbidden at m=0.

• χ=0

22

Possible Scattering Process for Conductivity Vanishes at m=0!

(chirality conservation)

|M|2 = 4g2Cfm2

Collision term for 1 to 2:

2

Cut

l k+l k

cf: 2 to 2

Cut Cut

2

~g4 C[n] = 1 2✏L

k

Z |M|2[nk+l

B (1 − nk F )(1 − nl F ) − (1 + nk+l B )nk F nl F ]

l

23

Calculation of conductivity linearize

We linearize the distribution function as nf(k3, T, Z) = nF (✏L

k ) + nf(k3, T, Z).

arized version of Eq. (D1) reads

linearize

damping rate of quark (=-2ξkδnkF)

C[n] = − 1 2✏L

k

Z

l

|M|2 ⇥ nk

F

• nk+l

B

+ nl

F

• − nl

F

• nk+l

B

+ nk

F

⇤

)v3nF (✏L

k )[1 nF (✏L k )] = C[nf(k3, T, Z)],

(D2) eqfE3

+ eqfE3(T, Z)@k3]nf(k3, T, Z) = C[n],

(β=1/T)

C[n] = 1 2✏L

k

Z |M|2[nk+l

B (1 − nk F )(1 − nl F ) − (1 + nk+l B )nk F nl F ]

l

24

Calculation of conductivity

j3(T, Z) = 2e X

f

qfNc |Bf| 2⇡ Z dk3 2⇡ v3nf(k3, T, Z)

Solution for δnF with damping rate ξk

j3 = e2 X

f

q2

fNc

|eqfB| 2⇡ 4 Z dk3 2⇡ (v3)2 1 2⇠k nF (✏L

k )[1 − nF (✏L k )]E3

σ33

nk

F = − 1

2⇠k eqfE3v3nF (✏L

k )[1 − nF (✏L k )]

Quark Damping Rate

25

leading-log approximation (ln[T/m]>>1)

matrix element soft fermion and hard boson nF(1+nB)+(1-nF)nB=nF+nB log divergence in phase space integral UV cutoff: T IR cutoff: m

l0<<T dominates

✏L

k ⇠k = g2CF m2

4⇡ Z ∞

m

dl0 nF (l0) + nB(l0 + ✏L

k )

p (l0)2 − m2

✏L

k ⇠k ' g2CF m2

4⇡ 1 2 + nB(✏L

k )

Z ∞

m

dl0 1 p (l0)2 m2 ' g2CF m2 4⇡ 1 2 + nB(✏L

k )

• ln

✓ T m ◆

2

Cut

k l k+l

Results

26

(average distance among quarks)~1/T →(quark density in 1D)~T Landau degeneracy Quark damping rate Quark density in 1D

Due to chirality conservation, collision is forbidden when m=0. Thus, σ~1/m2.

When M>>m, ln(T/m)→ln(T/M).

σ33 = e2 X

f

q2

fNc

|eqfB| 2π 4T g2CF m2ln(T/m) B

1 √ eB

1/T

Other Term Does Not Contribute

27

Other Term

C[n] = −2g2CF m2 ✏L

k

Z

l

[nk

F (nk+l B

+ nl

F ) − nl F (nk+l B

+ nk

F )]

nl

F = −eqf

2⇠l E3@l3nF (✏L

l ) :odd in l3

(Other term)~

Z

l

(nk+l

B

+ nk

F )δnl F

even in l3

Our result (only retaining quark damping rate term) is correct.

function of (εLk+εLl)

Equivalent Diagrams

28

Our calculation is based on (unestablished) (1+1)D kinetic theory, but actually we can reproduce the same result by field theory calculation.

• J. -S. Gagnon and S. Jeon, Phys. Rev. D 75, 025014 (2007); 76, 105019 (2007).

e jµ ≡ e

f qfψfγµψf

gitudinal and transverse

f: flavor index, qf: electric charge

ΠRµν(x) ⌘ iθ(x0)h[jµ(x), jν(0)]i,

Kubo formula:

σij ⌘ lim

ω→0

ΠRij(ω) iω

p→0 limit

Because of m-2 dependence, s contribution is very small.

Possible Phenomenological Implications

29

• 1. Order Estimate

Lattice: B. B. Brandt, A. Francis, B. Jaeger and H. B. Meyer, Phys. Rev. D 93, 054510 (2016). BAMPS: M. Greif, I. Bouras, C. Greiner and Z. Xu, Phys. Rev. D 90, 094014 (2014).

√eB<T T<√αseB M=160MeV>>m

αs = g2 4π = 0.3, mf = 3MeV(u, d), 100MeV(s), eB = 10m2

π = (440MeV)2.

σ33 = e2 X

f

q2

fNc

|eqfB| 2π 4T g2CF m2 ln(T/M)

Possible Phenomenological Implications

30

√eB<T

σ33 = e2 X

f

q2

fNc

|eqfB| 2π 4T g2CF m2 ln(T/M)

Beyond LLL approximation, there are also spin down particles, so the scattering is not suppressed by m2.

σ33 is expected to be smaller at large T, so that it smoothly connects with B=0 result. E

n=0 n=1

Possible Phenomenological Implications

31

• 2. Soft Dilepton Production

σ33 is large Soft dilepton production is enhanced by B?

p1 p2

d d4p = ↵ 12⇡4!2 T33 p

• G. D. Moore and J. -M. Robert, hep-ph/0607172.

p=0

→

(quark mean free path)-1 (photon interaction energy w leptons)

e p eB ⌧ ω ⌧ g2m2 T ln ✓ T M ◆

p ∵(virtual photon emission rate)~nB(ω)ImΠµµ~Tσ33

Possible Phenomenological Implications

32

• 3. Back Reaction to EM Fields

B

∂tB<0

E, j Bad news: In LLL approximation, we have no current in transverse plane, so Lenz’s law does NOT work! The lifetime of B does not increase…

Bulk Viscosity

ΔP=-ζ ∇・u u

Linearized Boltzmann equation

34

• @t + v3@z
• f(k, x, t) = C[f],

' nF (k)[1 nF (k)]X(x) ⇥ Θβ✏L

k v3k3⇤

] ' ⌧ 1

k f(k, x, t)

expansion decreases T

Expansion in direction of B: Boltzmann eq. without E linearize

F

feq(k, x, t) + f(k, x, t),

3

re feq(k, x, t) ⌘ [exp{(t)u(✏L

k k3u3(x))} + 1]1 2 1

nonequilibrium deviation (responsible for viscosity)

⇥ ≡

✓@P @" ◆

n,B

✓ ◆

ǁ

: (speed of sound)2

d X(x) ⌘ @zu3(x) : expansion (source term)

y u3(x).

1

B

Linearized Boltzmann equation

35

f(k, x, t) ' ⌧knF (k)[1 nF (k)]X(x) ⇥ Θβ✏L

k v3k3⇤

solution: In conformal case, the system is at equilibrium even after the expansion. No nonequilibrium deviation, no bulk viscosity.

1 (not 1/3, since the quarks live in one-dimension)

k3 1

Conformal case (m=0): δf =0.

Linearized Boltzmann equation

36

f(k, x, t) ' ⌧knF (k)[1 nF (k)]X(x) ⇥ Θβ✏L

k v3k3⇤

solution:

X-G. Huang, A. Sedrakian, D. Rischke, Annals Phys. 326 3075 (2011).

Two conformal breaking factor [(k3)2-Θβ(εLk)2]2~(m2)2

B expansion decreases T. So even in no-dissipative case, the pressure changes, and thus this contribution needs to be subtracted.

Δ(Pǁ-Θβε)=-3ζǁX(x)

[Pk − Θβ✏] = Nc |eqfB| 2⇡ 1 ⇡ Z 1

1

dk3 (k3)2 − Θβ(✏L

k )2

✏L

k

f(k, x, t)

[(k3)2-Θβ(εLk)2]

• [(k3)2-Θβ(εLk)2]/εLk

Results

37

Same as the conductivity, except for the extra m4 dependence, due to the conformal breaking factor.

(Same as B=0 case)

P . Arnold, C. Dogan, G. Moore,

• Phys. Rev. D 74 085021 (2006).

ζ ∼ m4 g4T ln(g−1)

s quark contribution would dominates over u/d contribution, in contrast to the electrical conductivity.

ζk = Nc |eqfB| 2π 9.6m2 g2π2CfT ln (T/m)

Conformal breaking

(m2)2 m2 = m2

Chirality non-conservation

Possible Phenomenological Implications

38

• 1. Order Estimate

M=160MeV>>m

αs = g2 4π = 0.3, mf = 3MeV(u, d), 100MeV(s), eB = 10m2

π = (440MeV)2.

Contribution from s quark

' 0.13m4

s

T .

(B=0)

P . Arnold, C. Dogan, G. Moore, Phys. Rev. D 74 085021 (2006).

ζk = ζ? '

ζk ζs

B=0

' 4.7 1 ln (T/M).

B enhances ζǁ.

ζk = Nc |eqfB| 2π 9.6m2

s

g2π2CfT ln (T/m)

39

B small ∂P, uǁ large ∂P, u⊥ ζǁ

Possible Phenomenological Implications

• 2. possible effect on flow

Enhances v2? ζǁ suppresses uǁ

The conductivity is enhanced by large B, and small m. The sensitivity to m was explained in terms

• f chirality conservation.

Summary (electrical conductivity)

40

Landau degeneracy Quark damping rate Quark density in 1D When M>>m, ln(T/m)→ln(T/M).

σ33 = e2 X

f

q2

fNc

|eqfB| 2π 4T g2CF m2ln(T/m)

Summary (bulk viscosity)

41

The bulk viscosity is proportional to m2, due to the conformal-breaking effect.

When M>>m, ln(T/m)→ln(T/M).

ζk = Nc |eqfB| 2π 9.6m2 g2π2CfT ln (T/m)

Conformal breaking

(m2)2 m2 = m2

Chirality non-conservation

Future Perspective

42

• Go beyond LLL approximation... (more realistic B)
• Ask hydro guys to simulate MHD with our transport

coefficients