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

Transport coe ffi cients 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 • (Long) Introduction —quarks and gluons in strong B • Electrical Conductivity • Bulk Viscosity • Summary 2

Introduction Strong magnetic field ( B ) may be generated in heavy ion collision due to Ampere’s law. B 3

Introduction K. Tuchin, Phys. Rev. C 93 , 014905 (2016). 2 eB � m π strength: /e . σ = 5 . 8 MeV, z = 0 . 2 fm, t 0 = 0 . 2 fm. S 1 = = 2000 (LHC). panel: γ = 2000 (LHC). ~ 100 [MeV] √ 0.100 eB (Comparable with T ) 0.010 0.001 10 � 4 � t � fm � 0 2 4 6 8 10 lifetime: ~ 0.3fm 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 . 4

Introduction How the particles behaves when √ eB is much larger than the other energy scales of the system? ( √ eB >> T , m , Λ QCD …) 5

Quark in Strong B One-particle state of quark in magnetic field Classical: Cyclotron motion due to Lorentz force B + 6

Quark in Strong B Quantum: Landau Quantization B p z + Longitudinal: Plane wave Transverse: Gaussian 1 √ eB E s ✓ n + 1 ◆ ( p z ) 2 + m 2 + 2 eB E n = 2 √ eB The gap (~ √ eB ) is generated by zero-point oscillation. 7

Quark in Strong B s For spin - 1/2 particle, we have Zeeman effect: B + E n =1 n =0 s ✓ ◆ n + 1 2 ⌥ 1 ( p z ) 2 + m 2 + 2 eB E n = 2 When n =0 (LLL), gap is small ( m ~1MeV). When n >0 , gap is large (~ √ eB ~100MeV) 8

Lowest Landau Level (LLL) Approximation 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 . One-dimension dispersion, no spin degrees of freedom. s B + ( p z ) 2 + m 2 p E n = p z 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 ). Q. Li, D. Kharzeev, et. al., nature physics 12 , 550 (2016) 9

Gluon in Strong B Gluon does not have charge, so it does not feel B in the zeroth approximation. B Massless boson in 3D 10

Gluon in Strong B K. Fukushima, Phys. Rev. D 83 , 111501 (2011). Coupling with (1+1)D quarks generates gluon mass. (Schwinger mass generation) B (surface density) ~(average distance) -2 ~ eB 1 Color factor √ eB 2 × g 2 | eB | M 2 ≡ 1 ∼ g 2 eB X 2 π π f Schwinger Landau degeneracy mass 11

Electrical Conductivity J= σ E J E

Motivation to Discuss Electrical Conductivity Electrical conductivity is phenomenologically important because • Input parameter of magnetohydrodynamics (transport coefficient) ∂ t B< 0 B • May increase lifetime of B (Lenz’s law) r ⇥ E = � ∂ t B E, j ∂ t E = r ⇥ B � j When σ is large 13

Hierarchy of Energy Scale at LLL For ordering of m and M , we consider the both cases. ( m << M and m >> M ) ( M~g √ eB ) E E √ √ eB eB << << LLL approximation T T << << Thermally excited M m << << m M 14

・ Electrical Conductivity Conductivity at weak B ( √ eB << T ) weak B B =0 j i = σ ij E j j = σ 0 E j = σ 1 E × B B E E j ○ j   0 0   σ 0 0 σ 0 σ 1 σ ij = σ ij = - 0 0 σ 0 0   σ 1 σ 0   0 0 σ 0 0 0 σ 0 15

Electrical Conductivity Strong B (LLL) Quarks are confined in the direction of B , so there is no current in other directions. j E B + σ 33 is finite, other components are zero. (Very di ff erent from weak B case) ω ) = Π ( ω )   e j i = σ ij E j , w 0 0 0 σ ij = 0 0 0   0 0 σ 33 16

Calculation of conductivity j B E Thermal equilibrium Linear response Slightly non-equilibrium, in strong B against E finite j We linearize the distri We linearize the distribution function as n f ( k 3 , T, Z ) = n F ( ✏ L as n f ( k 3 , T, Z ) = n F ( ✏ L k ) + � n f ( k 3 , T, Z ). k ) + arized version of Eq. (D1) reads arized version of Eq. (D1 ( k 3 ) 2 + m 2 p ✏ L Z dk 3 k ≡ | eq f B | = σ 33 E 3 X j 3 ( T, Z ) = 2 e 2 π v 3 δ n f ( k 3 , T, Z ) q f N c 2 π f hose momentum is k and Landau degeneracy ), v 3 ⌘ @✏ L k / ( @ k 3 ) = k 3 / ✏ L k Evaluation of δ n F is necessary. 17

Calculation of conductivity Evaluate n F with (1+1)D Boltzmann equation E [ @ T + v 3 @ Z + eq f E 3 ( T, Z ) @ k 3 ] n f ( k 3 , T, Z ) = C [ n ] , Constant and homogeneous E 18

Cut Cut Cut 19 (1 to 2) quark becomes kinematically possible. Decay of a gluon into quark and anti- q ⊥ + M 2 Gluon is effectively massive in (1+1)D z + p 2 p 2 E = strong B 2 to 2 is leading process. massless particles. one massless particle can not decay to two 1 to 2 scattering is kinematically forbidden; B =0 The scattering process is very different from that in B =0 … Possible Scattering Process for Conductivity

Chirality in (1+1)D Spin is always up. s s B B + + p z p z χ =+1 χ =-1 When m= 0 , the direction of p z determines chirality. 20

Chirality in (1+1)D Chirality is conserved at m =0 : - B + χ =+1+1=2 χ =0 1 to 2 scattering is forbidden at m =0 . 21

Cut Cut Cut Possible Scattering Process for Conductivity Collision term for 1 to 2: 1 Z | M | 2 [ n k + l F ) − (1 + n k + l B (1 − n k F )(1 − n l B ) n k F n l C [ n ] = F ] 2 ✏ L k l l 2 k+l k | M | 2 = 4 g 2 C f m 2 cf: 2 to 2 Vanishes at m =0 ! 2 (chirality conservation) ~ g 4 22

Calculation of conductivity + eq f E 3 ( T, Z ) @ k 3 ] n f ( k 3 , T, Z ) = C [ n ] , We linearize the distribution function linearize as n f ( k 3 , T, Z ) = n F ( ✏ L k ) + � n f ( k 3 , T, Z ). arized version of Eq. (D1) reads ) � v 3 n F ( ✏ L k )[1 � n F ( ✏ L k )] = C [ � n f ( k 3 , T, Z )] , eq f E 3 (D2) ( β = 1/ T ) 1 Z | M | 2 [ n k + l F ) − (1 + n k + l B (1 − n k F )(1 − n l B ) n k F n l C [ n ] = F ] 2 ✏ L k l linearize C [ � n ] = − 1 Z | M | 2 ⇥ � n k n k + l + n l − � n l n k + l + n k � � � �⇤ F F F F B B 2 ✏ L l k damping rate of quark ( =-2 ξ k δ n kF ) 23

Calculation of conductivity Solution for δ n F with damping rate ξ k F = − 1 � n k eq f E 3 � v 3 n F ( ✏ L k )[1 − n F ( ✏ L k )] 2 ⇠ k Z dk 3 | B f | X j 3 ( T, Z ) = 2 e 2 ⇡ v 3 � n f ( k 3 , T, Z ) q f N c 2 ⇡ f Z dk 3 | eq f B | 2 ⇡ ( v 3 ) 2 1 j 3 = e 2 X n F ( ✏ L k )[1 − n F ( ✏ L q 2 k )] E 3 4 � f N c 2 ⇡ 2 ⇠ k f σ 33 24

Cut Quark Damping Rate dl 0 n F ( l 0 ) + n B ( l 0 + ✏ L Z ∞ k ⇠ k = g 2 C F m 2 2 l k ) ✏ L k+l 4 ⇡ p ( l 0 ) 2 − m 2 m k leading-log approximation ( ln[ T / m ]>>1 ) l 0 << T dominates � Z ∞ k ⇠ k ' g 2 C F m 2  1 1 ✏ L 2 + n B ( ✏ L dl 0 k ) p 4 ⇡ ( l 0 ) 2 � m 2 m ✓ T ' g 2 C F m 2  1 � ◆ 2 + n B ( ✏ L k ) ln 4 ⇡ m log divergence matrix element in phase space integral soft fermion and hard boson UV cuto ff : T n F (1+ n B )+(1- n F ) n B = n F + n B IR cuto ff : m 25

Results (average distance among quarks)~ 1/ T B → (quark density in 1D)~ T 1/ T Quark density in 1D 1 √ eB | eq f B | 4 T σ 33 = e 2 X q 2 f N c 2 π g 2 C F m 2 ln( T/m ) f Landau Quark damping degeneracy rate Due to chirality conservation, collision is forbidden when m =0 . Thus, σ ~1/ m 2 . When M>>m , ln( T/m ) → ln( T/M ). 26

Other Term Does Not Contribute C [ � n ] = − 2 g 2 C F m 2 Z F ( n k + l F ( n k + l [ � n k + n l F ) − � n l + n k F )] B B ✏ L l k Other Term F = − eq f l ) :odd in l 3 � n l E 3 @ l 3 n F ( ✏ L 2 ⇠ l function of ( ε L k + ε L l ) Z (Other term)~ 0 ( n k + l + n k F ) δ n l F B l even in l 3 Our result (only retaining quark damping rate term) is correct. 27

Equivalent Diagrams 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). Kubo formula: e j µ ≡ e � Π Rij ( ω ) f q f ψ f γ µ ψ f σ ij ⌘ lim gitudinal and transverse i ω f : flavor index, q f : electric charge ω → 0 Π Rµ ν ( x ) ⌘ i θ ( x 0 ) h [ j µ ( x ) , j ν (0)] i , p → 0 limit 28

Possible Phenomenological Implications | eq f B | 4 T σ 33 = e 2 X q 2 f N c 1. Order Estimate g 2 C F m 2 ln( T/M ) 2 π f Because of m -2 dependence, s contribution is very small. T < √ α s eB √ eB < T α s = g 2 4 π = 0 . 3 , m f = 3MeV( u, d ) , 100MeV( s ) , eB = 10 m 2 π = (440MeV) 2 . M =160MeV>> m BAMPS: M. Greif, I. Bouras, C. Greiner and Z. Xu, Phys. Rev. D 90 , 094014 (2014). Lattice: B. B. Brandt, A. Francis, B. Jaeger and H. B. Meyer, Phys. Rev. D 93 , 054510 (2016). 29