Flux tubes at Finite Temperature

1.6 MB·

by  Nuno Cardoso

∗
Flux tubes at Finite Temperature
Nuno Cardoso, Marco Cardoso and Pedro Bicudo
CFTP, Departamento de Física, Instituto Superior Técnico, Universidade de
7 Lisboa, Av. Rovisco Pais, 1049-001 Lisbon, Portugal
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0
2

n
Inthiswork, weshowthefluxtubesofthequark-antiquarkandquark-
a
quark at finite temperature for SU(3) Lattice QCD. The chromomagnetic
J
and chromoelectric fields are calculated above and below the phase transi-
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tion.

]
t
a PACS numbers: 11.15.Ha; 12.38.Gc
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-
p
e 1. Introduction
h
[ The study of the chromo fields distributions inside the flux tubes formed
QQ and QQ¯ are presented in this study. How the flux tube evolves when
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v the distance between quarks or the temperature increase beyond respective
6 critical values are addressed in this paper. In section 2, we describe the lat-
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tice formulation. We briefly review the Polyakov loop for these systems and
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show how to compute the color fields as well as the Lagrangian distribution.
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0 Insection3, thenumericalresultsareshown. Finally, weconcludeinsection
. 4.
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2. Computation of the chromo-fields in the flux tube
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:
v The central observables that govern the event in the flux tube can be
Xi extracted from the correlation of a plaquette, (cid:3)µν, with the Polyakov loops,
L,
r
a β (cid:20)(cid:104)O(cid:3) (x)(cid:105) (cid:21)
f (r,x) = µν −(cid:104)(cid:3) (x)(cid:105) (1)
µν a4 (cid:104)O(cid:105) µν
where O = L(0)L†(r) for the QQ¯ system or O = L(0)L(r) for the QQ
system, x denotes the distance of the plaquette from the line connecting
∗ Presented by N. Cardoso at the International Meeting "Excited QCD", Costa da
Caparica, Portugal, 6 - 12 March, 2012
(1)
2 proceeding printed on January 9, 2017
quark sources, r is the quark separation, L(r) = 1 TrΠNt U (r,t) where
3 t=1 4
N is the number of time slices of the lattice and using the periodicity in
t
the time direction for the plaquette, (cid:3) (x) = 1 (cid:80)Nt (cid:3) (x,t), allows
µν Nt t=1 µν
averaging over the time direction.
To reduce the fluctuations of the O(cid:3) (x), we measure the following
µν
quantity, [1],
β (cid:20)(cid:104)O(cid:3) (x)(cid:105)−(cid:104)O(cid:3) (x )(cid:105)(cid:21)
µν µν R
f (r,x) = (2)
µν a4 (cid:104)O(cid:105)
where x is the reference point placed far from the quark sources.
R
Therefore, using the plaquette orientation (µ,ν) = (2,3),(1,3),(1,2),
(1,4),(2,4),(3,4), we can relate the six components in Eq. (2) to the com-
ponents of the chromoelectric and chromomagnetic fields,
f → 1 (cid:0)−(cid:10)B2(cid:11),−(cid:10)B2(cid:11),−(cid:10)B2(cid:11),(cid:10)E2(cid:11),(cid:10)E2(cid:11),(cid:10)E2(cid:11)(cid:1) (3)
µν 2 x y z x y z
andalsocalculatethetotalaction(Lagrangian)density,(cid:104)L(cid:105) = 1 (cid:0)(cid:10)E2(cid:11)−(cid:10)B2(cid:11)(cid:1)
2
In order to improve the signal over noise ratio, we use the multihit tech-
nique, [2, 3], replacing each temporal link by it’s thermal average, and the
extended multihit technique, [4], which consists in replacing each temporal
link by it’s thermal average with the first N neighbors fixed. Instead of
taking the thermal average of a temporal link with the first neighbors, we
fix the higher order neighbors, and apply the heat-bath algorithm to all the
links inside, averaging the central link,
U → U¯ = (cid:82) [DU4]ΩU4eβ(cid:80)µ,sTr[Uµ(s)F†(s)] (4)
4 4 (cid:82) [DU4]Ω eβ(cid:80)µ,sTr[Uµ(s)F†(s)]
By using N = 2 we are able to greatly improve the signal, when compared
with the error reduction achieved with the simple multihit. Of course, this
technique is more computer intensive than simple multihit, while being sim-
pler to implement than multilevel. The only restriction is R > 2N for this
technique to be valid.
3. Results
In this section, we present the results for different β values suing a fixed
latticevolumeof483×8,Table1. AllthecomputationsweredoneinNVIDIA
GPUs using CUDA.
The QQ and QQ¯ are located at (0,−R/2,0) and (0,R/2,0) for R =
4,6,8,10 and 12 lattice spacing units. In Figs. 2 and 4, we show the results
proceeding printed on January 9, 2017 3
√
β T/T a σ # config.
c
5.96 0.845 0.235023 5990
6.055 0.988 0.200931 5990/4775*
6.1237 1.100 0.180504 3669
6.2 1.233 0.161013 1868
6.338 1.501 0.132287 3688
6.5 1.868 0.106364 1868
Table 1: Lattice simulations for a 483 ×8 volume. The lattice spacing was
computedusingtheparametrizationfrom[5]inunitsofthestringtensionat
zero temperature. The ∗ means without configurations in the wrong phase
transition.
for the QQ¯ system. As expected the strength of the fields decrease with the
temperature. Also, in the confined phase the width in the middle of the flux
tube increases with the distance between the sources, while above the phase
transition the width decreases with the distance.
Just below the phase transition, we need to make sure that we don’t
have contaminated configurations as already mentioned in [6]. By plotting
the histogram of Polyakov loop history for β = 6.055, Fig. 1, we were able
to identify a second peak which then we were able to remove all the config-
urations that lie on the second peak. Therefore, in Table 1 the value with
asteriskcorrespondstotheconfigurationsafterremovingthesecontaminated
configurations. In Figs. 3a and 3b, we show the results of this effect for the
QQ system below the phase transition.
4000 4000
3500 3500
3000 3000
Frequency 122505000000 Frequency 122505000000
1000 1000
500 500
0 0
-0.015 -0.003 0.009 0.021 0.033 0.045 0.000 0.009 0.018 0.027 0.036 0.045
Re(L) |L|
Figure1: Histogram of the Polyakov loop history for β = 6.055.
4. Conclusions
As the distance increase between the sources, the field strength at the
flux tube decreases as already seen in studies at zero temperature. Below
the phase transition, the fields strength decreases as the temperature in-
creases. However, above the phase transition the fields rapidly decrease to
4 proceeding printed on January 9, 2017
<E2>/2 <B2>/2 /2 <E2>/2 <B2>/2 /2
35 − L 20 − L
30
25 15
2/σ 1250 2/σ 10
O10 O5
5 0
0
5 5
2 1 0 1 2 2 1 0 1 2 2 1 0 1 2 2 1 0 1 2 2 1 0 1 2 2 1 0 1 2
y σ y σ y σ y σ y σ y σ
R = 4 R =p 6 R = 8 p R =( 1a0) βp=5R. =9 612,RT = 4=0.845RT =pc 6. R = 8 p R = 10 p R = 12
<E2>/2 <B2>/2 /2 <E2>/2 <B2>/2 /2
40 − L 20 − L
35
30 15
25
2σ 20 2σ 10
/O15 /O5
10
5 0
0
5 5
2 1 0 1 22 1 0 1 22 1 0 1 2 2 1 0 1 22 1 0 1 22 1 0 1 2
y σ y σ y σ y σ y σ y σ
R = 4 R =p 6(b) β =R =6 8.0p55, TR == 100.988pTcR, =w 1i2tRh = c4ontamRin =pa 6ted conRfi =g 8urpationRs =. 10 p R = 12
<E2>/2 <B2>/2 /2 <E2>/2 <B2>/2 /2
50 − L 25 − L
40 20
30 15
2/σ 20 2/σ 10
O O
10 5
0 0
10 5
2 1 0 1 22 1 0 1 22 1 0 1 2 2 1 0 1 22 1 0 1 22 1 0 1 2
y σ y σ y σ y σ y σ y σ
R = 4 R =( p6c) β =R6 =.0 855p, T =R =0 1.9088Tpc,Rw =it 1h2Ro u= t4contamR =ipn 6ated coR n=fi 8gupratioRn =s 1.0 p R = 12
<E2>/2 <B2>/2 /2 <E2>/2 <B2>/2 /2
70 − L 20 − L
60
50 15
2/σO234000 2/σO105
10 0
0
10 5
1.51.00.50.00.51.01.51.51.00.50.00.51.01.51.51.00.50.00.51.01.5 1.5 1.0 0.50.0 0.5 1.0 1.51.5 1.0 0.50.0 0.5 1.0 1.51.5 1.0 0.50.0 0.5 1.0 1.5
R = 4 R =y 6σ R = 8 y σ R =< (1Ed0)2>βy/=2σR6 =.2 1,2RT == 4<1B.223>3R/T 2=cy .6σ R = 8/2y σ R = 10 y σR = 12
p 6p0 p − p L p p
<E2>/2 <B2>/2 /2 <E2>/2 <B2>/2 /2
300 − 50 L 60 − L
250 40 50
200 40
2/σO110500 2/σO 2300 2/σO2300
50 10 10
0 0
0
50 10
1.0 0.5 0.0 0.5 1.01.0 0.5100.0 0.5 1.01.0 0.5 0.0 0.5 1.0 1.0 0.5 0.0 0.5 1.01.0 0.5 0.0 0.5 1.01.0 0.5 0.0 0.5 1.0
0 5 0 5 0 0 5 0 5 0 0 5 0 5 0
R = 4 R =y 6σ R = 8 y σ1. R =0. 1(0e)0.βy=0.σR6 =.15. 1,12.TR == 04. 10..8680RT. =cy .61.σ1. 0.R = 08. y0.σ R1. = 10 y σR = 12
p p p p p p
R = 4 R =y 6 σ R = 8 y σ R = 10 y σR = 12
p p p
Figure2: The results for the QQ¯ system. The results in the left column
correspond to the fields along the sources (plane XY) and the right column
to the results in the middle of the flux tube (plane XZ). R is the distance
between the sources in lattice units.
zero as the quarks are pulled apart. The width of the flux tube below the
proceeding printed on January 9, 2017 5
<E2>/2 <B2>/2 /2 <E2>/2 <B2>/2 /2
40 − L 20 − L
30 15
10
2σ 20 2σ 5
/O10 /O 0
5
0 10
10 15
2 1 0 1 22 1 0 1 22 1 0 1 2 2 1 0 1 22 1 0 1 22 1 0 1 2
y σ y σ y σ y σ y σ y σ
R = 4 R = p6(a) β =R =6 8.05p5, TR == 100.988pTcR, =w 1i2tRh = c4ontamRin =a p6ted conRfi =g 8uraptionRs =. 10 p R = 12
<E2>/2 <B2>/2 /2 <E2>/2 <B2>/2 /2
600 − L 1200 − L
400 1000
200 800
2/σO 4200000 2/σO2460000000
600 200
800 400
1000 600
2 1 0 1 22 1 0 1 22 1 0 1 2 2 1 0 1 22 1 0 1 22 1 0 1 2
y σ y σ y σ y σ y σ y σ
R = 4 R =( 6bp) β =R6 =.0 855,pT =R =0 1.0988Tpc,Rw =i t1h2Ro =u t4contamR =in 6pated coR n=fi 8gupratioR n=s 1.0 p R = 12
<E2>/2 <B2>/2 /2 <E2>/2 <B2>/2 /2
70 − L 20 − L
60
50 15
2/σO234000 2/σO105
10 0
0
10 5
1.51.00.50.00.51.01.51.51.00.50.00.51.01.51.51.00.50.00.51.01.5 1.5 1.0 0.50.0 0.5 1.0 1.51.5 1.0 0.50.0 0.5 1.0 1.51.5 1.0 0.50.0 0.5 1.0 1.5
R = 4 R =y 6σ R = 8 y σ R =< 1(E0c)2>βy/=2σR6 =.2 1,2TR == 4<1B.223>3R/T 2=cy. 6σ R = 8/2y σ R = 10 y σR = 12
p 6p0 p − p L p p
<E2>/2 <B2>/2 /2 <E2>/2 <B2>/2 /2
300 − 50 L 60 − L
250 40 50
200 40
2/σO110500 2/σO 2300 2/σO2300
50 10 10
0 0
0
50 10
1.0 0.5 0.0 0.5 1.01.0 0.5100.0 0.5 1.01.0 0.5 0.0 0.5 1.0 1.0 0.5 0.0 0.5 1.01.0 0.5 0.0 0.5 1.01.0 0.5 0.0 0.5 1.0
0 5 0 5 0 0 5 0 5 0 0 5 0 5 0
R = 4 R =y 6σ R = 8 y σ1. R =0. (10d0).βy=0.σR6 =.15. 1,12.RT = =04. 10..8680R.T =cy .61.σ1. 0.R = 08. y0.σ R1. = 10 y σR = 12
p p p p p p
R = 4 R =y 6 σ R = 8 y σ R = 10 y σR = 12
p p p
Figure3: The results for the QQ system. The results in the left column
correspond to the fields along the sources (plane XY) and the right column
to the results in the middle of the flux tube (plane XZ). R is the distance
between the sources in lattice units.
phase transition increases with the separation between the quark-antiquark,
however above the phase transition the width seems to decrease.
Acknowledgments
NunoCardosoandMarcoCardosoaresupportedbyFCTunderthecon-
tracts SFRH/BPD/109443/2015 and SFRH/BPD/73140/2010 respectively.
WealsoacknowledgetheuseofCPUandGPUserversofPtQCD,supported
6 proceeding printed on January 9, 2017
<E2>/2 <B2>/2 /2 <E2>/2 <B2>/2 /2
25 − L 7 − L
20 6
5
2/σO11055 2/σO234
1
0 0
5 1
0.0 0.5 1.0 1.5 20..00 0.5 1.0 1.5 20..00 0.5 1.0 1.5 2.0 0.0 0.5 1.0 1.5 20..00 0.5 1.0 1.5 20..00 0.5 1.0 1.5 2.0
r σ r σ r σ r σ r σ r σ
T=0.988Tpc R=0.804 σ p T=1.867Tc Rp = 0.851 σ T=0.988Tpc R=1.206 σ p T=1.501Tc Rp = 1.323 σ
T=1.501Tc R = 0.794p σ p T=1.233Tc R = 1.288p σ T=1.867Tc R = 1.276pσ
p p p
Figure4: Results for the fields of the QQ¯ system in the middle of the flux
tube in the plane XZ.
<E2>/2 <B2>/2 /2 <E2>/2 <B2>/2 /2
3.5 − L 3.5 − L
3.0 3.0
2.5 2.5
2/σO112...050 2/σO112...050
0.5 0.5
0.0 0.0
0.5 0.5
024680246802468024680246802468 024680246802468024680246802468
0.0.0.0.0.1.1.1.1.10..0.0.0.0.1.1.1.1.10..0.0.0.0.1.1.1.1.1. 0.0.0.0.0.1.1.1.1.10..0.0.0.0.1.1.1.1.10..0.0.0.0.1.1.1.1.1.
r σ r σ r σ r σ r σ r σ
T=1.1Tpc R=1.083 σ p T=1.501Tc Rp = 1.058 σ T=1.1Tpc R=1.083 σ p T=1.501Tc Rp = 1.058 σ
(ap) QQ¯. p (bp) QQ. p
Figure5: Results for the fields in the middle of the flux tube in the plane
XZ.
by NVIDIA, CFTP and FCT grant UID/FIS/00777/2013.
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