1Stellar ExplosionsThree assumptions make modeling tractable analytically. First, the pres-sure of the expanding remnant is dominated by radiation pressure. Second,the energy radiated from the surface and by gamma emission from radioac-tivity in the interior are small compared to the total energy. Third, sphericalsymmetry is valid to zeroth order.The first law o f thermodynamics dE + P dV = T dS = dQ is˙E + P˙V = −∂L∂M+ ˙ǫ (1)where E = aT4V and P = aT4/3 are the energy per gram and the pressure.Dots represent time derivatives. The volume per gram is V = 1/ρ, ˙ǫ is theenergy input per gram per second from radioactivity, L(r, t) is the luminos-ity, a nd M(r, t) is the mass. Define x = r(M, t)/R(t) as a dimensionlessLagrangian radius containing mass M at time t and radius r. R(t) is thesurface (defined below). The density at a given enclosed mass scales in timewith R(t)−3sinceM (x) = 4πR ( t)3Zx0ρ (r, t) x2dx (2)is independent of time. We define a dimensionless density η(x) byρ (r, t) ≡ 1/V = ρoη (x)RoR (t)3, (3)where ρo≡ ρ(0, 0) and Ro≡ R(0). Thus˙V /V = 3˙R/R.If the right-hand side of Eq. (1) was neglected, adiabaticity applies and˙T /T = −˙R/R. This is the major time dependence of T . Now defineT4(r, t) = Ψ (x) φ (t) T4oR4o/R4(t) (4)so˙T /T = −˙R/R +˙φ/4φ. Here, To≡ T (0, 0). The luminosity becomesL (r, t) = −4πr23acκρ∂T4∂r= −4πx23acκρoηφT4oRodΨdx, (5)where κ is the opacity. The equation of energy, Eq. (1) becomes−3˙ǫρ2oR2oacT4oηφΨ+3R3oρoc˙φφR=1Ψx2ddxx2ηκdΨdx. (6)2For the moment, ignore the radioactive contributions. The opacity isdominated by electron scattering, for which κ=constant, except very nearthe surface where the temperature is low. There, a Kramer’s opacity ;withκ ∝ ρ; T−3.5∝ η(Ψφ)−7/8R1/2would be appropriate. For simplicity, wetake κ = κo. The differential equation (6) is now separable. It becomes3ρoκoR3oc˙φφR= −α =1Ψx2ddxx2ηdΨdx. (7)The time dependence is easily solved for:φ (t) = exp−αc6ρoκoR3oZt0R (t) dt= exp−12RoτoZt0R (t) dt(8)where the usual diffusion timescale isτo=3R2oρoκoαc. (9)The explicit time behavior depends upon R(t) (see below).Spatial SolutionsWe next examine the spatial solution, which permits us to evaluate theeigenvalue α. The boundary conditions on Ψ ar e easily given. First, atthe origin, Ψ(0) = 1 and Ψ′(0) = 0. At the surface we use the Eddingtonboundary conditionTTe4=ΨΨe=34τ +23(10)where τ(r) = −R∞rκρdr is the optical depth and the subscript e refers tothe effective radiating surface where τ = 2/3 and r = R or x = 1. FromEq. (10) we have thatΨ (1) ≡ Ψ (τ = 0) =12Ψe(11)and by differentiation, using Ψ′= dΨ/dx,Ψ′(1) = −34Ψe(κρ)x=1R. (12)3Combining these two, and setting κ = κo, we haveΨ (1) = −23Ψ′κoρRx=1. (13)With ηκ constant, the solution of Eq. (7) is a polytrope of index 1:Ψ (x) =sin√αx√αx. (14)The boundary condition Eq. (13) im pliessin√α√α= −231κoρRcos√α −sin√α√α, (15)or√α ≃ π1 −231κoρR. (16)Note that κoρR is the total optical depth in the case of unifo r m density.For κoρR → ∞, we have Ψ(1) = 0 and α = π2. Only if the total opticaldepth is less than about 10 is there significant deviation f r om this result,and this generally occurs only after several months. The leading correctionis Ψ(1) ≃ 2/(3κoρR). In what f ollows, we will simply impose the outerboundary condition a s Ψ(1) = 0.Suppose the density η is not constant. For η = Ψmthe solution is relatedto the polytropic Lane-Emden solution θnfor the index n :Ψ = θnn, η = θn−1n, x = ξpn/α, n = 1/ (1 − m) . (17)With the b oundary condition Ψ(1) = 0, some cases are shown in Ta ble1. Note that m = 1/3(2/3)[4/5] is the n = 3/2(3)[5] polytrope, and theeigenvalue α is α = nξ21. The case m = 4/5 has α = ∞; a density spike inthe center with a zero density mantle – not physically very relevant.The po lytropic solutions all have η (density) decreasing with distancefrom the center. A shell-like behavior can also be modelled:η =11 − βx2; Ψ = 1 + γx2+ δx4. (18)The b oundary condition on the outside becomes γ + δ = −1. Physicalsolutions ar e possible for two cases:a) α = (140/9)(1 −p8/35) ≃ 8.119, β = α/28, γ = −α/6, δ = −1 + α/6.b) α = 6, β = 3/5, γ = −1, δ = 0.The ratios of densities of the surface and center are η(1)/η(0) = 1/(1 − β)= 2.5 (1.41).4Table 1: Spatial solutions to Eq. (7)η α IMIKITαIMIM/IKη = θ23142.7 0.0132 0.0024 0.00615 1.882 5.42η = θ1/23/220.03 0.151 0.071 0.0556 3.028 2.15η = ηoπ21/3 1/5 0.101 3.290 5/3a: η = (1 − βx2)−18.119 0.409 0.260 0.113 3.303 1.572b: η = (1 − βx2)−16 0.553 0.366 0.133 3.319 1.510Temporal EvolutionThe temporal evolution will yield the light curves. When the shockemerges at the surface Roat time t = 0, the energy is nearly evenly dividedbetween thermal and kinetic energies:ET(t) =ZR0aT44πr2dr = 4πR3oaT4oRoRφ (t) IT= ET(0) φ (t)RoR, (19)EK(t) =12ZR0ρv24πr2dr = 2πρoR3o˙R2IK= EK(0)˙R2o˙R2, (20)where IT=R10Ψx2dx, IK=R10ηx4dx, v = dr/dt = x˙R and˙Rois its initialvalue. The total energy ESNis a constant o f the motion if the energy lostin radiation or gained from r adioactivity is negligible, and is given by itsinitial value:ESN= ET(0) + EK(0) ≃ 2ET(0) . (21)Reference to Eq. (5) allows us to now write the luminosity asL (t) = −4π3acκoρoRoT4oΨ′ηx=1φ (t) = −4πcIMRoET(0)3κoMITΨ′ηx=1.(22)The total ejected mass is M =RR04πρr2dr = 4πρoR3oIM, where IM=R10ηx2dx. From Eq. (7) we have the identity−αIT=x2Ψ′ηx=1=Ψ′ηx=1, (23)5so, with Eq. (22), we findL (t) =2πc3κoESNMαIMRoφ (t) =ESNφ (t)2τo. (24)Conservation of energy ESN= ET(t) + EK(t) yields˙R2=˙R2o(2 − φRo/R) . (25)At t = 0, φ(t)Ro/R(t) ≃ 1, R increases linearly with time: R ≃ Ro+˙Rot.The expansion timescale initially isτh=Ro˙Ro= 2.5 × 105Ro1014cm4 × 108cm s−1˙Ros. (26)However, after a time of several τh, i.e., at most days, the term φRo/R <<1. The expansion is still linear in time, but√2 times faster than it wasinitially: the thermal energy has been converted into kinetic energy of ex-pansion. A reasonable a pproximation isR (t) = Ro+√2˙Rot. (27)We are finally able to solve for the function φ(t): using Eq. (8) we haveφ (t) =