The QCD crossover from Lattice QCD July 25, 2018 Patrick - PowerPoint PPT Presentation

The QCD crossover from Lattice QCD July 25, 2018 Patrick Steinbrecher HotQCD collaboration

The QCD phase diagram July 25, 2018 Patrick Steinbrecher Slide 1

Quantum Chromodynamics from first principles #configurations m s /m l =27, N τ =16 Lattice QCD 12 1M 8 HISQ action 6 N σ = 4 N τ sim. at µ = 0 100k physical quarks 10k 2 light quarks 1 strange quark m s / m l = 27 135 145 155 165 175 T [MeV] m π ≃ 138 MeV everything continuum extrapolated July 25, 2018 Patrick Steinbrecher Slide 2

Chiral observables in two-flavor formulation subtracted condensate Σ sub ≡ m s (Σ u + Σ d ) − ( m u + m d )Σ s Σ f = T ∂ with ln Z V ∂ m f subtracted susceptibility � ∂ χ sub ≡ T ∂ � V m s + Σ sub ∂ m u ∂ m d χ disc is defined as χ sub without connected part July 25, 2018 Patrick Steinbrecher Slide 3

Start of the QCD crossover line: T 0 d 2 Σ sub d χ sub ≡ 0 and ≡ 0 dT 2 f 4 f 4 dT K K two crossover temperatures: T 0 (Σ sub ) and T 0 ( χ sub ) July 25, 2018 Patrick Steinbrecher Slide 4

Pseudo-critical temperatures for m l → 0: pseudo-critical temperatures converge to the chiral transition temperature T 0 c at finite quark mass it is given by maximum of O ( 4 ) universal scaling functions (Thursday talk, Sheng-Tai Li, Chiral phase transition) χ m = m 1 /δ − 1 0.4 f χ ( z ) + reg . l 0.35 χ t = m ( β − 1 ) /βδ 0.3 f ′ G ( z ) + reg . l 0.25 0.2 0.15 O(4): f χ (z) for m l → 0 0.1 f’ G (z) 0.05 χ t ∼ ∂ T Σ sub and χ t ∼ ∂ 2 µ B Σ sub 0 χ m ∼ χ sub and χ m ∼ χ disc -3 -2 -1 0 1 2 3 0 )/m l 1/( β δ ) z ∼ (T-T c July 25, 2018 Patrick Steinbrecher Slide 5

The subtracted chiral susceptibility 250 4 χ sub /f k m s /m l =27, N τ =16 12 200 8 6 150 100 50 T [MeV] 0 135 145 155 165 175 July 25, 2018 Patrick Steinbrecher Slide 6

The subtracted chiral susceptibility 250 4 χ sub /f k m s /m l =27, N τ =16 12 200 8 6 150 100 50 T [MeV] 0 135 145 155 165 175 July 25, 2018 Patrick Steinbrecher Slide 6

The 2nd µ B derivative of chiral condensate Σ sub Σ /2 -c 2 m s /m l =27, N τ =12 1.6 8 1.4 6 1.2 µ Q = µ S =0 1 0.8 0.6 HotQCD preliminary 0.4 0.2 T [MeV] 0 135 145 155 165 175 July 25, 2018 Patrick Steinbrecher Slide 7

The 1st T derivative of chiral condensate Σ sub 120 Σ /dT -T dc 0 m s /m l =27, N τ =12 6 100 8 80 60 40 HotQCD preliminary 20 T [MeV] 0 135 145 155 165 175 July 25, 2018 Patrick Steinbrecher Slide 8

The T 0 continuum extrapolation 166 T c (µ B =0) [MeV] χ disc 164 χ sub 162 160 Σ sub 2 Σ sub 158 ∂ µ B 156 2 χ disc ∂ µ B 154 HotQCD preliminary 156.5 ± 1.5 MeV 152 2 1/N τ 150 c N N N N o n = = = = τ τ τ τ t 1 1 8 6 i n 6 2 u u m July 25, 2018 Patrick Steinbrecher Slide 9

Crossover temperature T 0 170 T 0 [MeV] 165 160 155 150 HotQCD preliminary 145 140 Σ sub χ disc χ sub ∂ µ B ∂ µ B Σ sub , Bonati 2015 χ tot , Bazavov 2012 Σ sub , Borsanyi 2010 2 Σ 2 χ s d u i b s c July 25, 2018 Patrick Steinbrecher Slide 10

The QCD crossover at µ � = 0 d 2 Σ sub ( T , µ B ) d χ disc ( T , µ B ) ≡ 0 and ≡ 0 dT 2 f 4 f 4 dT K K need Taylor expansion in T and µ B around ( T 0 , 0 ) July 25, 2018 Patrick Steinbrecher Slide 11

Taylor expansion in chemical potentials (just notation) simplest case µ Q = µ S = 0 subtracted condensate � ∞ n = ∂ Σ sub / f 4 c Σ Σ sub � � n µ n c Σ K = n ! ˆ with � B µ n f 4 ∂ ˆ � B K � n = 0 µ = 0 disconnected susceptibility � c χ ∞ n = ∂χ disc / f 4 χ disc � c χ � n µ n K = n ! ˆ with � B µ n f 4 ∂ ˆ � B K � n = 0 µ = 0 July 25, 2018 Patrick Steinbrecher Slide 12

Coefficients for a strangeness neutral system 0 30 Σ /2 Σ /dT c 2 T dc 2 m s /m l =27, N τ =12 m s /m l =27, N τ =12 -0.2 8 6 20 6 8 -0.4 10 -0.6 n S =0, n Q /n B =0.4 -0.8 0 -1 -10 n S =0, n Q /n B =0.4 -1.2 -20 -1.4 HotQCD preliminary HotQCD preliminary T [MeV] T [MeV] -1.6 -30 135 145 155 165 175 135 145 155 165 175 15 1000 χ /2 c 2 χ /dT m s /m l =27, N τ =12 T dc 2 m s /m l =27, N τ =12 8 6 10 6 8 500 5 HotQCD preliminary HotQCD preliminary 0 0 n S =0, n Q /n B =0.4 -5 n S =0, n Q /n B =0.4 -500 -10 T [MeV] T [MeV] -15 -1000 135 145 155 165 175 135 145 155 165 175 July 25, 2018 Patrick Steinbrecher Slide 13

The curvature of the crossover line � 2 � 4 � µ B � µ B T c ( µ B ) + O ( µ 6 = 1 − κ 2 − κ 4 B ) T 0 T 0 T 0 Taylor expansion in µ and T of: d χ disc ( T , µ B ) = ( ... ) µ 2 B + ( ... ) µ 4 B + ... = 0 f 4 dT K 0.025 has to be zero order by order n S =0, n Q /n B =0.4, m s /m l =27 0.02 0.015 ∂ c χ � ( T 0 , 0 ) − 2 c χ � 0.01 T 0 2 � 2 � χ disc : κ 2 1 ∂ T ( T 0 , 0 ) HotQCD preliminary � 0.005 κ 4 κ 2 = 2 T 2 ∂ 2 c χ � 0 0 0 � ∂ T 2 � ( T 0 , 0 ) -0.005 2 1/N τ -0.01 0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 July 25, 2018 Patrick Steinbrecher Slide 14

The QCD crossover line STAR: arxiv:1701.07065 ALICE: arxiv:1408.6403 170 4 ) T c [MeV] crossover line: O(µ B 165 constant: ε s 160 freeze-out: STAR ALICE 155 150 145 140 n S =0, n Q /n B =0.4 HotQCD preliminary 135 µ B [MeV] 130 0 50 100 150 200 250 300 350 400 July 25, 2018 Patrick Steinbrecher Slide 15

The curvature κ n for strangeness neutral system 0.02 0.015 κ 2 0.01 n S =0, n Q /n B =0.4 0.005 0 κ 4 HotQCD preliminary -0.005 -0.01 Σ sub χ disc Σ sub , Bellwied 2015 July 25, 2018 Patrick Steinbrecher Slide 16

The crossover line � 2 � 4 T c ( µ X ) � µ X � µ X = 1 − κ X − κ X + O ( µ 6 X ) 2 4 T 0 T 0 T 0 0.040 0.030 HotQCD preliminary X 0.020 κ 2 Bonati 2018: X = B , µ S = 0 X κ 4 κ 2 = 0 . 0145 ( 25 ) 0.010 0.000 -0.010 X = B B S I Q n S =0 n Q /n B =0.4 July 25, 2018 Patrick Steinbrecher Slide 17

Fluctuations along the QCD crossover T c ( µ B ) Baryon-number fluctuations � σ 2 ∞ c B 1 ∂ ln Z 1 ∂ ln Z � � n µ n c B B = = n ! ˆ with n = � B Vf 3 Vf 3 µ 2 Vf 3 µ n + 2 ∂ ˆ � ∂ ˆ K K B K � n = 0 B µ = 0 σ 2 B couples to condensate − → diverges at a critical point study increase along the crossover line � 2 � 4 σ 2 B ( T c ( µ B ) , µ B ) − σ 2 B ( T 0 , 0 ) � µ B � µ B = λ 2 + λ 4 + · · · σ 2 T 0 T 0 B ( T 0 , 0 ) July 25, 2018 Patrick Steinbrecher Slide 18

Baryon-number fluctuations � along T c ( µ B ) 1.2 2 (T c ( µ B ), µ B )/ σ B 2 (T 0 ,0) - 1 σ B 1 4 ) O( µ B n S =0, n Q /n B =0.4 0.8 2 ) O( µ B 0.6 HRG 0.4 HotQCD preliminary 0.2 0 µ B [MeV] -0.2 0 50 100 150 200 250 300 July 25, 2018 Patrick Steinbrecher Slide 19

Susceptibility fluctuations � along T c ( µ B ) 0.6 100 4 χ disc (T c ( µ B ), µ B )/ χ disc (T 0 ,0) - 1 χ disc /f k µ B = 0.0 MeV 0.4 125.0 MeV 80 200.0 MeV HotQCD preliminary 0.2 60 0 HotQCD preliminary 40 -0.2 4 ) O( µ B n S =0, n Q /n B =0.4 6 ) n S =0, n Q /n B =0.4 N τ =8, O( µ B 2 ) O( µ B 20 -0.4 µ B [MeV] T [MeV] -0.6 0 0 50 100 150 200 250 300 135 145 155 165 175 185 195 σ 2 B and χ disc show no indication for a narrowing crossover July 25, 2018 Patrick Steinbrecher Slide 20

Critical point from Taylor expansions e.g. expansion of the pressure around µ B = 0 (for µ Q ≡ µ S ≡ 0) ∂ n ln Z � P 1 1 � n ! χ B µ n χ B � T 4 = n ˆ B , n = � µ n VT 3 ∂ ˆ � B µ B = 0 n analysis of convergence radius can determine bound on the location of a critical point: 1 / 2 1 / 2 � � � � ( 2 n + 2 )( 2 n + 1 ) χ B 2 n ( 2 n − 1 ) χ B � � � � r χ r P 2 n 2 n 2 n = , 2 n = � � � � χ B χ B � � � � 2 n + 2 2 n + 2 � � � � only if coefficients are positive for all n ≥ n 0 if not → no critical point on real axis July 25, 2018 Patrick Steinbrecher Slide 21

Critical point from Taylor expansions 9 9 χ 2017: lower bound for r 4 χ 8 8 estimator r 2 χ D’Elia et al., 2016, r 4 crit /T 7 7 Datta et al., 2016 χ -- estimator for µ B Fodor, Katz, 2004 χ ,HRG 6 6 r 6 5 5 χ ,HRG 4 4 r 4 3 3 2 2 χ ,HRG r n r 2 1 1 disfavored region for the location of a critical point 0 0 135 135 140 140 145 145 150 150 155 155 T [MeV] July 25, 2018 Patrick Steinbrecher Slide 22

Summary crossover starts at T 0 = 156 . 5 ± 1 . 5 MeV crossover curvature for strangeness neutral system � 2 � 4 � � T c ( µ B ) + O ( µ 6 = 1 − κ 2 µ B − κ 4 µ B B ) T 0 T 0 T 0 κ 2 = 0 . 0123 ± 0 . 003 κ 4 = 0 . 000131 ± 0 . 0041 for µ B < 250 MeV and n s = 0 , n Q / n B = 0 . 4 crossover along const. entropy density and energy density chemical freeze-out might be close to crossover no indication for critical point July 25, 2018 Patrick Steinbrecher Slide 23

Thank you for your attention! July 25, 2018 Patrick Steinbrecher Slide 24