NONLOCAL DENSITY CORRELATIONS AS A SIGNATURE OF HAWKING RADIATION FROM ACOUSTIC BLACK HOLES IN BOSE-EINSTEIN CONDENSATES: THE ANALOGY Part 2

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1 NONLOCAL DENSITY CORRELATIONS AS A SIGNATURE OF HAWKING RADIATION FROM ACOUSTIC BLACK HOLES IN BOSE-EINSTEIN CONDENSATES: THE ANALOGY Part 2 Paul R. Anderson Wake Forest University Collaborators: Roberto Balbinot, University of Bolgona and Alessandro Fabbri, University of Valencia

2 Summary QFT calculation of density-density correlation function for a BEC analogue of a black hole Take into account the effective potential for the quantum field Discuss the effects of an infrared cutoff including the potential

3 Effective metric Lab frame ds 2 = n mc [ c 2 dt 2 + (d x vdt) (d x vdt) ] Let v = vˆx Transformation to static spherically symmetric coordinates dt = dt + v c 2 v 2dx [ ds 2 = n (c2 v 2 ) dt 2 + m c + dy2 + dz 2 ] c c c 2 v 2dx2

4 Solutions to the Gross Pitaevskii Equation Background Solution with ˆΨ = e iˆθ ˆn ˆn = n + ˆn 1 ˆθ = θ + ˆθ 1 n = constant v = h m θ = vˆx v = constant

5 Linearized equation in 3+1 dimensions ˆθ 1 = 0 ˆn 1 = h g ( T ˆθ 1 + h ) θ ˆθ 1 m Dimensional reduction to 1+1 dimensions ˆθ 1 = ˆθ (2) mc 1 hnl 2 ( (2) + V )ˆθ (2) 1 = 0 If c = c(x), then with dx dx(c 2 v 2 )/c (2) = m n V = m n c c 2 v 2( 2 t + x 2 ) [ c 2 ( 1 v2 c 2 ) c 2 4c + 5 v 2 ] 4 c 2 c 2

6 ( t 2 + x 2 ) + c2 v 2 n c m V ˆθ (2) 1 = 0 V = m n [ c 2 ( 1 v2 c 2 ) c 2 4c + 5 v 2 4 c 2 c 2 ] If the term containing V is zero then solutions are ˆθ (2) 1 e iωu and e iωw u = t x w = t + x These solutions can be used to define the Unruh state

7 Unruh State Use positive frequency in static spherically symmetric time, t, for left moving modes at I Define Kruskal coordinates and use positive frequency in Kruskal time for right moving modes on the past horizon Then use a Bogolubov transformation to write in terms of e ±iωu on the past horizon

8 If left-moving modes are ignored, the twopoint function for ˆθ (2) 1 if x l < 0 and x r > 0 is ˆθ (2) 1 (t l, x l )ˆθ (2) 1 (t r, x r ) = 1 8π 0 dω 1 ω sinh ( πω κ ) +e iω(t r t l ) χ ω(x l )χ ω(x r ) ] [ e iω(t r t l ) χ ω (x l )χ ω (x r ) d 2 χ ω dx 2 + ω2 χ ω + V eff χ ω = 0 V eff = c2 v 2 n c m V (x) Potential should be important for modes with ω < V eff max

9 Density-Density Correlation Function G 2 (x l, x r ) = ˆn(x l )ˆn(x r ) ˆn(x l ) ˆn(x r ) = hn ml 2 c2 (x l )c 2 (x r ) ( D c(x l )c(x r ) ˆθ (2) 1 (t l, x l )ˆθ (2) 1 (t r, x r ) ) D = Tl Tr v xl Tr v Tl xr + v 2 xl xr Recall dt = dt + v c 2 v 2dx

10 Numerical Computations Same parameters as the QM calculation by Carusotto, Fagnocchi, Recati, Balbinot, and Fabbri Slight differences: Horizon is at x = 0 Black hole is to left of horizon and condensate moves to the left

11 Values h = ξ = c l = 1 v = 3/4, c r = 1/2 σ x = 1 2 Sound speed is c(x) = c 2 l ( c 2 r c 2 l ) [ π tan 1 ( x σ x )] Surface gravity κ = [ 1 d 2dx ( c(x) )] v2 c(x) x=0 = c (0) = 1 8πσ x (2 3) 0.30

12 Thetopplotisc x andthebottomplotisveff x

13 Effect of the Potential on the Modes Recall the mode equation is d 2 χ ω dx 2 + ω2 χ ω + V eff χ ω = 0 V eff = c2 v 2 c n m V (x) Expect no significant effect for ω 2 V eff max

14 This discussion on initial conditions for the spatial modes is revised from the original talk. A mistake for the modes which initially go into the region outside the event horizon was pointed out by R. Parentani during the talk and in subsequent discussions. It is corrected here. For solutions on the past horizon which initially go into the region outside the event horizon Fix the behavior so they are right moving in the limit x χ ω = e iωu Fix the amplitude so that the right moving modes originating from the past horizon have unit amplitude on that horizon.

15 For the solutions on the past horizon which are always inside the event horizon, fix the behavior so that on the past horizon. χ ω = e iωu For ω = 0 nonzero V eff implies that at large x χ ax

16 Cutoff Recall the 2-point function is ˆθ (2) 1 (t 1, x 1 )ˆθ (2) 1 (t 2, x 2 ) = 1 8π 0 dω 1 ω sinh ( πω κ ) +e iω(t 2 t 1 ) χ ω (x 1)χ ω (x 2) ] [ e iω(t 2 t 1 ) χ ω (x 1 )χ ω (x 2 ) In V eff = 0 case Balbinot, et. al. subtracted from the two-point function κ 8π 2 0 dω ω 2 Won t work for V eff 0 so need to impose a cutoff 1 1 dω 8π ω c ω sinh ( )... πω κ

17 In principle fix by setting λ c = 2πc/ω c 2πc r /ω c the size of the system Effective UV cutoff scale is λ uv 2π 2 c/κ 60 Need λ c λ uv Specific cases tried: λ c 3000, 300

18 Regularization using the subtraction term Regularization using a cutoff of Ω = Regularization using a cutoff of Ω = 0.02

19 Lowest curve at the minimum: Regularization via the subtraction term Next lowest curve: Regularization with a cutoff ω c = 0.002, λ c 3000 Upper curve: Regularization with a cutoff ω c = 0.02, λ c 300

20 Explanations Recall G 2 (x, x hn ) = ml 2 c2 (x)c 2 (x ) ( D c(x)c(x ) ˆθ (2) 1 (t 1, x 1 )ˆθ (2) 1 (t 2, x 2 ) ) D = T T v x T v t x + v 2 x x For V = 0 ˆθ (2) 1 (t 1, x 1 )ˆθ (2) 1 (t 2, x 2 ) = 1 8π ω c dω cos(ω u) ω sinh ( πω κ ) For u = 0 high frequency modes contribute more so cutoff is less important Infrared effects most important for term c (x 1 )c (x 2 ) ˆθ (2) 1 (t 1, x 1 )ˆθ (2) 1 (t 2, x 2 ) For c (x) ω c this term is small

21 Conclusions Including the potential V requires implementation of a cutoff ω c For physically motivated values find relatively weak dependence of peak of correlation function on the cutoff Cutoff dependence is stronger farther from the peak

22 Future Work Consider the case when both points are inside the horizon - QM calculation shows an interesting effect Investigate what happens when the event horizon forms in the sudden approximation