No-drag frame for anomalous chiral fluid

Transcription

1 No-drag frame for anomalous chiral fluid M. Stephanov University of Illinois at Chicago M. Stephanov (UIC) No-drag frame UCLA / 15

2 Based on arxiv: with Ho-Ung Yee. M. Stephanov (UIC) No-drag frame UCLA / 15

3 Hydrodynamic transport Two kinds of transport in a normal fluid: convective (ideal, reversible) and dissipative (irreversible). Non-convective fluxes of charge and energy-momentum J µ nu µ + J µ diff T µν wu µ u ν + pg µν + T µν heat + T µν visc generate entropy/heat (fluid relaxes to equilibrium) T ( S) 0 + (J µ diff )2 + (T µν heat )2 + (T µν visc )2 M. Stephanov (UIC) No-drag frame UCLA / 15

4 Hydrodynamic transport Two kinds of transport in a normal fluid: convective (ideal, reversible) and dissipative (irreversible). Non-convective fluxes of charge and energy-momentum J µ nu µ + J µ diff T µν wu µ u ν + pg µν + T µν heat + T µν visc generate entropy/heat (fluid relaxes to equilibrium) T ( S) 0 + (J µ diff )2 + (T µν heat )2 + (T µν visc )2 in addition to contributing to entropy/heat transport T S µ = T su µ + µj µ diff + u νt µν heat M. Stephanov (UIC) No-drag frame UCLA / 15

5 CME/CVE equilibrium, non-dissipative transport The CME and CVE non-convective transport of charge energy and entropy (heat) J µ nu µ + J µ anom; T µν wu µ u ν + pg µν + T µν anom; S µ = su µ + S µ anom; which does not generate entropy/heat ( S) = M. Stephanov (UIC) No-drag frame UCLA / 15

6 CME/CVE equilibrium, non-dissipative transport The CME and CVE non-convective transport of charge energy and entropy (heat) J µ nu µ + J µ anom; T µν wu µ u ν + pg µν + T µν anom; S µ = su µ + S µ anom; which does not generate entropy/heat ( S) = How do we separate convective and anomalous flows? M. Stephanov (UIC) No-drag frame UCLA / 15

7 Fluxes are frame dependent J µ anom = ξ Jω ω µ + ξ JB B µ, T µν anom = (ξ T ω ω µ + ξ T B B µ )u ν + (µ ν), S µ anom = ξ Sω ω µ + ξ SB B µ. The coefficients ξ depend on the definition of u µ (comoving frame). leads, for example, to u µ u µ + α ω ω µ + α B B µ, ξ Jω ξ Jω nα ω ; ξ JB ξ JB nα B ; M. Stephanov (UIC) No-drag frame UCLA / 15

8 Fluxes are frame dependent J µ anom = ξ Jω ω µ + ξ JB B µ, T µν anom = (ξ T ω ω µ + ξ T B B µ )u ν + (µ ν), S µ anom = ξ Sω ω µ + ξ SB B µ. The coefficients ξ depend on the definition of u µ (comoving frame). leads, for example, to u µ u µ + α ω ω µ + α B B µ, ξ Jω ξ Jω nα ω ; ξ JB ξ JB nα B ; We can choose comov. frame where, e.g., J µ anom vanishes, or T µν anom or S µ anom. How do we separate convective fluxes from CVE/CME? M. Stephanov (UIC) No-drag frame UCLA / 15

9 For example, one natural choice π i T 0i = 0 (Landau, liquid at rest) ( ξ JB = C µ n µ2 + O(T 2 ) ) ; ξ T B 0. Son-Surowka 2w M. Stephanov (UIC) No-drag frame UCLA / 15

10 For example, one natural choice π i T 0i = 0 (Landau, liquid at rest) ( ξ JB = C µ n µ2 + O(T 2 ) ) ; ξ T B 0. Son-Surowka 2w but weak coupling (HTL or CKT) calculation says ξ JB = Cµ; ξ T B = 1 2 Cµ X BT 2 M. Stephanov (UIC) No-drag frame UCLA / 15

11 For example, one natural choice π i T 0i = 0 (Landau, liquid at rest) ( ξ JB = C µ n µ2 + O(T 2 ) ) ; ξ T B 0. Son-Surowka 2w but weak coupling (HTL or CKT) calculation says ξ JB = Cµ; ξ T B = 1 2 Cµ X BT 2 The fluid as a whole is not at rest in this calculational frame. What is special about that frame? Which frame is more natural. M. Stephanov (UIC) No-drag frame UCLA / 15

12 Hydrodynamics with drag Consider hydrodynamic description of the force on a test impurity, or obstacle (e.g., heavy quark): µ T µν = F νλ J λ F ν ; µ J µ = CE B. M. Stephanov (UIC) No-drag frame UCLA / 15

13 Hydrodynamics with drag Consider hydrodynamic description of the force on a test impurity, or obstacle (e.g., heavy quark): µ T µν = F νλ J λ F ν ; µ J µ = CE B. F 0 = 0 in the rest frame U of the obstacle (no work): U F = 0 (For a heavy quark, P 2 = M 2 0 = d(p P ) = MU Fdτ.) M. Stephanov (UIC) No-drag frame UCLA / 15

14 Hydrodynamics with drag Consider hydrodynamic description of the force on a test impurity, or obstacle (e.g., heavy quark): µ T µν = F νλ J λ F ν ; µ J µ = CE B. F 0 = 0 in the rest frame U of the obstacle (no work): U F = 0 (For a heavy quark, P 2 = M 2 0 = d(p P ) = MU Fdτ.) Does the second law S 0 impose constraints on F? M. Stephanov (UIC) No-drag frame UCLA / 15

15 Second law constraints Anomaly introduces 4 new terms into S: E B, E ω, B and ω. They must vanish independently. This gives 4 conditions: ξ JB µc = K B n/w, ξ Jω 2(T ξ SB + µξ JB ξ T B ) = K ω n/w; T dξ SB + µdξ JB dξ T B = K B dp/w; T dξ Sω + µdξ Jω dξ T ω = K ω dp/w; with K ω 2T ξ Sω + 2µξ Jω 3ξ T ω ; K B T ξ SB + µξ JB 2ξ T B. Two gauge degrees of freedom due to u µ u µ + α ω ω µ + α B B µ. If we fix them also (choose frame), all 6 coefficients ξ are fixed. M. Stephanov (UIC) No-drag frame UCLA / 15

16 Entropy production from drag The only remaining term: ( T ( S) = u K ) ωω + K B B F w M. Stephanov (UIC) No-drag frame UCLA / 15

17 Entropy production from drag The only remaining term: ( T ( S) = u K ) ωω + K B B F ū F w For any choice of u there is a special velocity (frame) ( ū u K ) ωω + K B B w M. Stephanov (UIC) No-drag frame UCLA / 15

18 Entropy production from drag The only remaining term: ( T ( S) = u K ) ωω + K B B F ū F w For any choice of u there is a special velocity (frame) ( ū u K ) ωω + K B B w If the particle moves with the same velocity, i.e., ū = U, then T ( S) = U F = 0. I.e. ū is the no-drag frame. It is invariant w.r.t. u µ u µ + α ω ω µ + α B B µ : K i K i + wα i M. Stephanov (UIC) No-drag frame UCLA / 15

19 No-drag frame We can thus choose u = ū (whence K i = 0) to simplify discussion. The condition U F = 0 is satisfied by F µ = λ F (u µ + U µ (u U)) M. Stephanov (UIC) No-drag frame UCLA / 15

20 No-drag frame We can thus choose u = ū (whence K i = 0) to simplify discussion. The condition U F = 0 is satisfied by F µ = λ F (u µ + U µ (u U)) Thus entropy/heat production: T ( S) = u F = λ F ((u U) 2 1) 0 requires λ F 0. In non-relativistic limit, U (1, V ) and u (1, v): F λ F (v V ) and u F λ F (v V ) 2. M. Stephanov (UIC) No-drag frame UCLA / 15

21 Anomalous coefficients in the no-drag frame The coefficients are now fixed up to 2 integration consts X B and X ω. ξ SB = X B T ξ Sω = 2X B µt + X ω T 2 ξ JB = Cµ ξ Jω = Cµ 2 + X B T 2 ξ T B = 1 2 Cµ X BT 2 ξ T ω = 2 3 Cµ3 + 2X B µt X ωt 3 Remarkable: All coefficients are polynomial in the no-drag frame. Entropy flow is not zero! No-drag frame is not entropy frame. M. Stephanov (UIC) No-drag frame UCLA / 15

22 Is anomalous flow like superfluid flow? Two fluid model of Landau (constrained by second law) J µ nu µ + J µ super; J µ super = v 2 ψ µ, ψ µ = µ φ + A µ, T µν wu µ u ν + pg µν + T µν super; T µν super = v 2 ψ µ ψ ν, S µ = T su µ + 0. (entropy frame) M. Stephanov (UIC) No-drag frame UCLA / 15

23 Is anomalous flow like superfluid flow? Two fluid model of Landau (constrained by second law) J µ nu µ + J µ super; J µ super = v 2 ψ µ, ψ µ = µ φ + A µ, T µν wu µ u ν + pg µν + T µν super; T µν super = v 2 ψ µ ψ ν, S µ = T su µ + 0. (entropy frame) The superfluid flow does not contribute to dissipation: T ( S) = M. Stephanov (UIC) No-drag frame UCLA / 15

24 Is anomalous flow like superfluid flow? Two fluid model of Landau (constrained by second law) J µ nu µ + J µ super; J µ super = v 2 ψ µ, ψ µ = µ φ + A µ, T µν wu µ u ν + pg µν + T µν super; T µν super = v 2 ψ µ ψ ν, S µ = T su µ + 0. (entropy frame) The superfluid flow does not contribute to dissipation: T ( S) = With drag: in the entropy frame T ( S) = u F. I.e., the entropy frame is the no-drag frame. There is no superfluid entropy current! M. Stephanov (UIC) No-drag frame UCLA / 15

25 The coefficient X B is proportional to mixed gauge-gravity anomaly. ξ SB a = X a BT ξ Sω = 2X a Bµ a T M. Stephanov (UIC) No-drag frame UCLA / 15

26 The coefficient X B is proportional to mixed gauge-gravity anomaly. ξ SB a = X a BT ξ Sω = 2X a Bµ a T Vanishes for any physical gauge field in the Standard Model. ξ SB = 0 due to anomaly cancellation. I.e., CME, like the superfluid flow, does not carry entropy. M. Stephanov (UIC) No-drag frame UCLA / 15

27 The coefficient X B is proportional to mixed gauge-gravity anomaly. ξ SB a = X a BT ξ Sω = 2X a Bµ a T Vanishes for any physical gauge field in the Standard Model. ξ SB = 0 due to anomaly cancellation. I.e., CME, like the superfluid flow, does not carry entropy. Example: Rajagopal-Sadofyev AdS/CFT calculation of CME drag: no drag no entropy flow. M. Stephanov (UIC) No-drag frame UCLA / 15

28 The coefficient X B is proportional to mixed gauge-gravity anomaly. ξ SB a = X a BT ξ Sω = 2X a Bµ a T Vanishes for any physical gauge field in the Standard Model. ξ SB = 0 due to anomaly cancellation. I.e., CME, like the superfluid flow, does not carry entropy. Example: Rajagopal-Sadofyev AdS/CFT calculation of CME drag: no drag no entropy flow. But if µ is not a dynamical gauge field (e.g., U(1) A in QCD), ξ Sω 0 is OK. Example: Chen et al CVE from CKT with collisions ξ Sω = 1 6 µt. M. Stephanov (UIC) No-drag frame UCLA / 15

29 CME/CVE drag on a heavy quark Rajagopal-Sadofyev M. Stephanov (UIC) No-drag frame UCLA / 15

30 CME/CVE drag on a heavy quark Rajagopal-Sadofyev Energy flow due to CME/CVE π anom = ξ T B B + ξ T ω ω 0 M. Stephanov (UIC) No-drag frame UCLA / 15

31 CME/CVE drag on a heavy quark Rajagopal-Sadofyev Energy flow due to CME/CVE π anom = ξ T B B + ξ T ω ω 0 But momentum density (= energy flow) cannot be changed by B π = wv + π anom = 0 Thus v = ξ T BB + ξ T ω ω w M. Stephanov (UIC) No-drag frame UCLA / 15

32 CME/CVE drag on a heavy quark Rajagopal-Sadofyev Energy flow due to CME/CVE π anom = ξ T B B + ξ T ω ω 0 But momentum density (= energy flow) cannot be changed by B π = wv + π anom = 0 Thus v = ξ T BB + ξ T ω ω w The drag force: F = λ F (v V ). M. Stephanov (UIC) No-drag frame UCLA / 15

33 CME/CVE drag on a heavy quark Rajagopal-Sadofyev Energy flow due to CME/CVE π anom = ξ T B B + ξ T ω ω 0 But momentum density (= energy flow) cannot be changed by B π = wv + π anom = 0 Entropy flow s = sv + ξ Sω ω Thus v = ξ T BB + ξ T ω ω w The drag force: F = λ F (v V ). M. Stephanov (UIC) No-drag frame UCLA / 15

34 Summary/Conclusions Hydrodynamics with drag is nontrivially constrained by the second law of thermodynamics. No-drag frame allows conceptually clean separation between convective and anomalous flows. simplifies the expression for the force: F = λ F (v V ); the (frame-dependent) coefficients ξ are polynomials of T and µ. CME is similar to superfluid no entropy flux, but for a non-trivial reason gauge-gravity anomaly cancellation. CVE can carry entropy without drag, unlike superfluid. M. Stephanov (UIC) No-drag frame UCLA / 15