Investigating subphotospheric dissipation in gamma-ray bursts by fitting a physical model

∼ 1050−54 −1

∼ 2 5× 1010

0.0 2.5 5.0 7.5 Redshift 0.0 0.1 0.2 0.3 Nor m al is ed fr eq u en cy

1σ

Λ H0 = 67.3 Ωλ= 0.685

∼ 2

T90

0.01 0.1 1 10 100 1000 T90(s) 0 50 100 150 200 250 300 350 Num b er of GRBs T90

∼ 1 1015

∼ ∼ 107

c K K′ t′= Γ ! t₋β cx " , x′= Γ(x− vt) K′ K′

v x K t = 0 β = v/c Γ Γ = # 1 1− β2. L′= x′1−x′0 K′ L = x1−x0= L′/Γ K 1/Γ K′ K 1/Γ T = ΓT′ _{T T}′ K K′ β 1/(1 + z) Eobs_{= E}lab_{/(1 + z) z} Γ >> 1 K′ K θ β t0 t1 ∆tobs_{= ∆t}′_{Γ(1− β cos θ) = ∆t}′_/D D = _Γ(1 1 − β cos θ). ν =Dν′ _{T =DT}′

θ θ =−π D∼ 1/2Γ θ = 0 β(Γ) D∼ 2Γ θ′ =±π/2 sin θ =±1/Γ Γ ≫ 1 θ ∼ ±1/Γ 1052 e± LE = 4πGM mpc/σT G σT LE(1M⊙) ∼ 1038 −1

Γ = η η ≡ L/ ˙M c2 aT04= L/(4πr02c), a T0= ! _L 4πr2 0ca "1/4 . ∆r _{∼ r}0 V′_{= 4π∆r}′_r2_{= 4πr} 0Γ(r)r2 Eobs ∝ Γ(r)V′(r)T′(r)4_{= Γ(r)}2_r2_r 0T′(r)4, dS = (dU + pdV )/T S′ = V′(u′ + p′)/T′ u p p′= u′/3∝ T′4 S′= V′u′1 + 1 3 T′_(r) ∝ r0Γ(r)r 2_T′_(r)3_{= const,} T (r) = Γ(r)T′(r) = const rs rs ≡ ηr0 η_{≡ L/ ˙}M c2 γ_{∼ η} rs rs/r0 rs

T′(r)∝ r−2/3 Γ(r) = $ r r0 r < rs η r > rs, T (r) = $ T0 r < rs T0(r/rs)−2/3 r > rs, V′(r)∝ $ r3 _{r < r} s r2 _{r > r} s, T (r) r < rs r0 r0/c r0 τ = 1 τ τ = (LσT)/(8πrβΓ3mpc3) σT mp r τ = 1 rph= (LσT)/(8πβΓ3mpc3)

∼ r0 δt = r0/c ∆Γ ∼ η η ∆v _{∼ c/2η}2 ∆r = r0 rd = 2r0Γ2 t′ merge= ∆r′/c = Γr0/c

tmerge= r0/c = ∆tobs ∆tobs

m1, m2 Γ1, Γ2 Γ1, Γ2≫ 1 Γf = [(m1Γ1+ m2Γ2)/(m1/Γ1+ m2/Γ2)]1/2. e = 1−_m(m1+ m2)Γf 1Γ1+ m2Γ2 , m1∼ m2 Γ1≫ Γ2

Γf 1_{− 10%} B γe ve Psync= 4 3σTcβ 2_γ2 e B2 8π,

σT B2 8π νcrit= 3 2γ 3 eνBsin α, α νB = νB(γe, B) νcrit P (ν)_{∝ ν}1/3 _ν crit νcrit P (ν)∝ ν−p−12 , p νc= ν(γc) P (ν)∝ ν1/3 ν < νc P (ν)_{∝ ν}−1/2 _{< ν} c < ν < νcrit ν > νcrit γm< γc γm> γc γm γm > γc γc γm < γc γe > γc

γe Eγe ≪ mec2 E EIC= Eγe2, EIC Eγe∼ mec2 γenec2 E−1

10

1

10

2

10

3

Energy (keV)

10

1

10

2

10

3

EF

E

(k

eV

s

− 1

cm

− 2

)

Total

N (E) = % KEα_exp&₋E E0 ' , E≤ (α − β)E0 K [(α− β) E0](α−β)Eβexp (β− α), E > (α− β)E0 , N (E) α β E0 K

Epeak= (2+α)E_α−β 0 Eiso Epeaklab ∝ E η iso, η L_∝(Epeaklab )γ erg s−1, L Elab peak γ N (E) = A ! E 100 keV "α e− (α+2)E Ep _, α Ep νFν A −1 −2 −1 Epeak α β

10

1

10

2

10

3

10

4

Energy (keV)

10

1

10

2

10

3

10

4

EF

E

(k

eV

s

− 1

cm

− 2

)

Band CPL BB α =−2/3 β = −3 E0 = 100 α = /3/2 Ep= 200 50 kBT = 50 kB EFE Epeak α β > 100

0 500 1000 1500 Epeak(keV) 0.000 0.001 0.002 0.003 0.004 Ar b . n or m −1 0 1 2 α 0.00 0.25 0.50 0.75 1.00 1.25 Ar b . n or m 2 4 β 0.0 0.5 1.0 1.5 Ar b . n or m Epeak α β α∼ 1 α = −2/3 α = −3/2

α

α > _−3/2

α < 1

r0

10−14 0.1 ∼ 8 · 10−7 ₁ 1052 −1 L0,52 τ εd εe εb Γ εpl εpl = 0 ∆r ∼ r0 r0 rd rd= L0,52σT 4πτ c3_Γ3_m p = Γ 2_r 0. ∆t = rd/Γ2c t′ dyn = Γ∆t rd→ 2rd τ → τ/2 ∆r′ V′ = 4πr2 dct′= 4πrd2∆r′t′/t′dyn t′ Nshell = L0,52∆t/Γmpc2 u′ e− = εeu′int = εeεdL0,52/4πr2dcΓ2 u′γ = aT (rd)′4 A = uγ/ue−

L0,52 r0 Γ η rd τ εdL0,52 εe εb 2rd 2.3rd

S = Psync/PIC = u′B/uγ′ u′B = εbu′int

θ = kBT′/mec2

θe−

γchar′ θe− = γ_char′ /2 γ′_char = 2εeεd(mp/me)

γchar′

δtloss= E/(dE/dt) =

γ′

charmec2/(4/3)cσTγ′2charu′γ(1 + S) γ ∼ 1

tloss = δtlossγchar′

tloss/tdyn ≈ 1/((4/3)A(1 + S)γ′charτ ) = (3/4)Γ 2/3 2 (30τ )−1(0.44 + 0.48εB,−0.5εd,−1) εb= 10−6 εd= 0.1 τ≤ 1 Γ! 105 τ τ _{∼ 1} a = 2τ−2/3

100 ₁₀1 ₁₀2 ₁₀3 ₁₀4 ₁₀5 ₁₀6

Energy (keV)

10−2 10−1 100 101 102 103

EF

E

(k

eV

s

− 1

m

− 2

)

2rd [L0,52= 0.1 τ = 1 εd = 0.05 εb = 0.1 Γ = 350] [L0,52 = 10 τ = 10 εd = 0.1 εb = 10−6 Γ = 100] [L0,52= 10 τ = 20 εd = 0.2 εb = 10−6 Γ = 350] [L0,52 = 0.1 τ = 20 L0,52 = 0.1 εd = 0.2 εb = 0.1 Γ = 200] 104− 105 ∼ 30% rd 2.3rd

2.3rd 2rd

L0,52=

100 τ = 35 εd= 0.05 εb= 10−6 Γ = 250 104− 105

a b a_{∈ N} b _{∈ N} L0,52 Γ τ > 1 τ " 35 τ τ 0.1− 300L0,52 1051_{− 3 · 10}54 −1 1055 −1 εdL0,52a εd a 50 500 Γ_{≥ 100} Γ = 50 Γ Γ = 1000 εd 0.4

εe εb 0.9 10−6

−1 −1

−1

∼ 10%

Eγ ≥ 2mec2 Z

−20 0 20 40 60 Time(s) 0 2000 4000 6000 8000 10000 Co u nt s ∼ 300 ! 150 " 150 ◦

0.2 10 140

6 2

16.5_{× 10}6 2

100 −1

0 5 10 15 20 Time (s) 0 .00 0 .02 0 .04 0 .06 0 .08 Ar b . n or m . 0 5 5 10 10 20

−1 −1 −1 −2 −1 E I R(I, E) f (E) −1 −2 −1 C(I) −1 −1 f (E) = C(I)R(I, E)−1. R(I, E) C(I) = * ∞ 0 f (E)R(I, E)dE. C(I)

L(θ) = P(d|θ) d θ −2 ln(L) N P(N|S) = (ST ) N N ! e −ST_, S θ T d c P(Nd,c|Sd,c) = P Nd + i=1 Nc + i=1 P(Nd,c|Sd,c). χ2 χ2 Bd,c ˆ Bd,c σd,c Bd,c∼ N ( ˆBd,c, σd,c)

P(Nd,c|Sd,c(θ), ˆBd,c, σd,c) = L(θ) ∝ ⎧ ⎪ ⎨ ⎪ ⎩ 0Nd i=1 0Nc i=1e − ! (Sd,c(θ)+ ˆBd,c)T− σ2_{d,c T}2 2 " , N = 0. 0Nd i=1 0Nc i=1(Sd,c(θ) + ˜Bd,c)Nd,ce − ! (Sd,c(θ)+ ˜Bd,c)T + ( ˜_{Bd,c− ˆBd,c)}2 2σ2 d,c " , N > 0 . Sd,c ˜ B = 1 2 1 ˆ B_{− S(θ) − σ}2T2+ 2 (S(θ) + σ2_T2_{− ˆB)}2_{− 4(σ}2_{T S− σ}2_{N− ˆBS(θ))} 3 . d c B˜d,c χ2

T T0 T T0 T T T N T T0 p p = P(T _{≥ T}0|H0)≈ n N, n T ≥ T0 H0

P(θ_{|d) =} 4 P(θ)P(d|θ) P(θ)P(d|θ)dθ, P(θ_|d) θ d P(θ) P(d_|θ) θ 4P(θ)P(d|θ)dθ Z P(θ) θ

Z P(drep_{|d) =} * P(drep_{|θ)P(θ|d)dθ,} drep d θ p ppp pb= P (T (drep) > T (d)|d) . p P(θ_|d) T T p

ppp

ppp

92.7− 100.2 68.3 95 99.7

92.7− 100.2 X Y X Y ρX,Y = E [(X_{− µ}X)(Y − µY)] σXσY ,

72− 144

E[X] µX σX

X Y

rxy=

5n

i=1(xi− ¯x)(yi− ¯y)

#5n

i=1(xi− ¯x)2#5ni=1(yi− ¯y)2

,

xi ith n X

¯

x y

T90 α =−2/3 β = −3 E0 = 100 α = /3/2 Ep= 200 50 kBT = 50 kB EFE Epeak α β

L0,52 r0 Γ η rd τ εdL0,52 εe εb 2rd 2.3rd 2rd [L0,52= 0.1 τ = 1 εd= 0.05 εb= 0.1 Γ = 350] [L0,52= 10 τ = 10 εd= 0.1 εb= 10−6 Γ = 100] [L0,52 = 10 τ = 20 εd= 0.2 εb= 10−6 Γ = 350] [L0,52 = 0.1 τ = 20 L0,52 = 0.1 εd = 0.2 εb= 0.1 Γ = 200] 104_{− 10}5 2.3rd 2rd L0,52 = 100 τ = 35 εd = 0.05 εb = 10−6 Γ = 250 104− 105 0 11

0 5 5 10 10 20

92.7_{− 100.2} 68.3 95 99.7

γ

γ

γ

γ