1
Stellar Explosions
Three assumptions make modeling tractable analytically. First, the pressure of the expanding remnant is dominated by radiation pressure. Second,
the energy radiated from the surface and by gamma emission from radioactivity in the interior are small compared to the total energy. Third, spherical
symmetry is valid to zeroth order.
The first law of thermodynamics dE + P dV = T dS = dQ is
Ė + P V̇ = −
∂L
+ ǫ̇
∂M
(1)
where E = aT 4 V and P = aT 4/3 are the energy per gram and the pressure.
Dots represent time derivatives. The volume per gram is V = 1/ρ, ǫ̇ is the
energy input per gram per second from radioactivity, L(r, t) is the luminosity, and M (r, t) is the mass. Define x = r(M, t)/R(t) as a dimensionless
Lagrangian radius containing mass M at time t and radius r. R(t) is the
surface (defined below). The density at a given enclosed mass scales in time
with R(t)−3 since
Z x
3
ρ (r, t) x2 dx
(2)
M (x) = 4πR (t)
0
is independent of time. We define a dimensionless density η(x) by
Ro 3
ρ (r, t) ≡ 1/V = ρo η (x)
,
R (t)
(3)
where ρo ≡ ρ(0, 0) and Ro ≡ R(0). Thus V̇ /V = 3Ṙ/R.
If the right-hand side of Eq. (1) was neglected, adiabaticity applies and
Ṫ /T = −Ṙ/R. This is the major time dependence of T . Now define
T 4 (r, t) = Ψ (x) φ (t) To4Ro4 /R4 (t)
(4)
so Ṫ /T = −Ṙ/R + φ̇/4φ. Here, To ≡ T (0, 0). The luminosity becomes
4πr2 ac ∂T 4
dΨ
4πx2 ac
L (r, t) = −
=−
φTo4 Ro ,
3 κρ ∂r
3 κρo η
dx
where κ is the opacity. The equation of energy, Eq. (1) becomes
η
1 d x2 dΨ
3Ro3 ρo φ̇
3ǫ̇ρ2o Ro2
=
+
.
−
φΨ
c φR
acTo4
Ψx2 dx ηκ dx
(5)
(6)
2
For the moment, ignore the radioactive contributions. The opacity is
dominated by electron scattering, for which κ=constant, except very near
the surface where the temperature is low. There, a Kramer’s opacity ;with
κ ∝ ρ; T −3.5 ∝ η(Ψφ)−7/8 R1/2 would be appropriate. For simplicity, we
take κ = κo . The differential equation (6) is now separable. It becomes
3ρo κo Ro3 φ̇
1 d x2 dΨ
= −α =
.
(7)
c
φR
Ψx2 dx η dx
The time dependence is easily solved for:
Z t
Z t
1
αc
R (t) dt = exp −
R (t) dt
φ (t) = exp −
2Ro τo 0
6ρo κo Ro3 0
(8)
where the usual diffusion timescale is
3Ro2ρo κo
.
τo =
αc
(9)
The explicit time behavior depends upon R(t) (see below).
Spatial Solutions
We next examine the spatial solution, which permits us to evaluate the
eigenvalue α. The boundary conditions on Ψ are easily given. First, at
the origin, Ψ(0) = 1 and Ψ′ (0) = 0. At the surface we use the Eddington
boundary condition
4
T
Ψ
3
2
=
=
τ+
(10)
Te
Ψe
4
3
R
where τ (r) = − r∞ κρdr is the optical depth and the subscript e refers to
the effective radiating surface where τ = 2/3 and r = R or x = 1. From
Eq. (10) we have that
1
Ψ (1) ≡ Ψ (τ = 0) = Ψe
2
(11)
and by differentiation, using Ψ′ = dΨ/dx,
3
Ψ′ (1) = − Ψe (κρ)x=1 R.
4
(12)
3
Combining these two, and setting κ = κo , we have
2 Ψ′ .
(13)
Ψ (1) = −
3 κo ρR x=1
With ηκ constant, the solution of Eq. (7) is a polytrope of index 1:
√
sin αx
Ψ (x) = √
.
(14)
αx
The boundary condition Eq. (13) implies
√
√ √
sin α
2 1
sin α
√
=−
cos α − √
,
(15)
α
3 κo ρR
α
or
√
2 1
α≃π 1−
.
(16)
3 κo ρR
Note that κo ρR is the total optical depth in the case of uniform density.
For κo ρR → ∞, we have Ψ(1) = 0 and α = π 2 . Only if the total optical
depth is less than about 10 is there significant deviation from this result,
and this generally occurs only after several months. The leading correction
is Ψ(1) ≃ 2/(3κo ρR). In what follows, we will simply impose the outer
boundary condition as Ψ(1) = 0.
Suppose the density η is not constant. For η = Ψm the solution is related
to the polytropic Lane-Emden solution θn for the index n:
p
(17)
Ψ = θnn , η = θnn−1 , x = ξ n/α, n = 1/ (1 − m) .
With the boundary condition Ψ(1) = 0, some cases are shown in Table
1. Note that m = 1/3(2/3)[4/5] is the n = 3/2(3)[5] polytrope, and the
eigenvalue α is α = nξ12 . The case m = 4/5 has α = ∞; a density spike in
the center with a zero density mantle – not physically very relevant.
The polytropic solutions all have η (density) decreasing with distance
from the center. A shell-like behavior can also be modelled:
1
η=
;
Ψ = 1 + γx2 + δx4 .
(18)
2
1 − βx
The boundary condition on the outside becomes γ + δ = −1. Physical
solutions are possible for
p two cases:
a) α = (140/9)(1 − 8/35) ≃ 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).
4
Table 1: Spatial solutions to Eq. (7)
η
α
η = θ32
1/2
η = θ3/2
IM
IK
IT
αIM IM /IK
142.7 0.0132 0.0024 0.00615 1.882
5.42
20.03 0.151
0.071
0.0556 3.028
2.15
η = ηo
π2
1/3
a: η = (1 − βx2 )−1 8.119 0.409
6
0.553
b: η = (1 − βx2 )−1
1/5
0.260
0.366
0.101
0.113
0.133
3.290
5/3
3.303 1.572
3.319 1.510
Temporal Evolution
The temporal evolution will yield the light curves. When the shock
emerges at the surface Ro at time t = 0, the energy is nearly evenly divided
between thermal and kinetic energies:
Z R
Ro
Ro
aT 4 4πr2 dr = 4πRo3 aTo4 φ (t) IT = ET (0) φ (t)
ET (t) =
, (19)
R
R
0
Z
Ṙo2
1 R 2
2
3
2
ρv 4πr dr = 2πρo Ro Ṙ IK = EK (0)
,
(20)
EK (t) =
2 0
Ṙ2
R
R
where IT = 01 Ψx2 dx, IK = 01 ηx4 dx, v = dr/dt = xṘ and Ṙo is its initial
value. The total energy ESN is a constant of the motion if the energy lost
in radiation or gained from radioactivity is negligible, and is given by its
initial value:
ESN = ET (0) + EK (0) ≃ 2ET (0) .
(21)
Reference to Eq. (5) allows us to now write the luminosity as
′
4πcIM Ro ET (0) Ψ′
4π ac
4 Ψ
φ (t) = −
.
L (t) = −
R o To
3 κo ρ o
η x=1
3κo M IT
η x=1
(22)
RR
The total ejected mass is M = 0 4πρr2 dr = 4πρo Ro3 IM , where IM =
R1 2
0 ηx dx. From Eq. (7) we have the identity
2 ′
′
x Ψ
Ψ
−αIT =
=
,
η
η
x=1
x=1
(23)
5
so, with Eq. (22), we find
L (t) =
2πc ESN
E φ (t)
αIM Ro φ (t) = SN
.
3κo M
2τo
(24)
Conservation of energy ESN = ET (t) + EK (t) yields
Ṙ2 = Ṙo2 (2 − φRo /R) .
(25)
At t = 0, φ(t)Ro /R(t) ≃ 1, R increases linearly with time: R ≃ Ro + Ṙo t.
The expansion timescale initially is
8 cm s−1 4
×
10
Ro
R
o
= 2.5 × 105
s.
(26)
τh =
1014 cm
Ṙo
Ṙo
However, after a time of several τ h , i.e., at most
√ days, the term φRo /R <<
1. The expansion is still linear in time, but 2 times faster than it was
initially: the thermal energy has been converted into kinetic energy of expansion. A reasonable approximation is
√
R (t) = Ro + 2Ṙo t.
(27)
We are finally able to solve for the function φ(t): using Eq. (8) we have
t2
t
.
(28)
φ (t) = exp − − √
τo
2τo τh
Since τo >> τh , φ initially decreases exponentially, but before much decay
has occurred, the decrease steepens into a Gaussian with time constant
q√
τdecay =
2τo τh .
(29)
It is useful to now express things in terms of the quantities M, Ro and
ESN which we can hope to extract from a supernova’s light curve. For
example, the diffusion time becomes, using Eqs. (9),
14
κo π 2
M
10 cm
7
τo = 1.6 × 10
s.
(30)
Ro
αIM cm2g−1 M⊙
The maximum observed velocities in the ejecta’s spectra are, using Eqs. (20)
and (21),
s
s
√
2ESN IM
ESN M⊙ IM
≃ 109
cm s−1 . (31)
vmax ≃ 2Ṙo =
51
M IK
10 ergs M IK
6
Thus, the decay time, given by Eq. (29), becomes
s
1/4 κo π 2
M 3/4 1051ergs
IK 1/4
6
τdecay = 1.7 × 10
s.
ESN
IM
αIM cm2g−1 M⊙
(32)
Additionally, we may write the luminosity, Eq. (24), as
2 −1 E
SN M⊙ Ro
43 αIM cm g
φ erg s−1 ,
L = 3.1 × 10
2
51
14
κo π
10 ergs M 10 cm
(33)
and the effective temperature, Te = (L/4πσR2 )1/4 , as
1/4 1/4 1/2
ESN M⊙ 1/4 1014 cm
αIM cm2 g−1
Ro
4
Te = 4.6×10
φ1/4 K
2
51
Ro
R
κo π
M 10 ergs
(34)
The largest time dependence in the effective temperature is due to the expansion. The density and opacity dependences in the ejecta have a relatively
√
small effect on observables, through αIM and (IM /IK )1/4 , which are insensitive to assumptions about η (Table I). For the last two cases shown,
√
√
IM = β −3/2 (tanh−1 ( β) − β) and IK = (IM − 1/3)/β. For a Population
I composition, κo ≃ 0.33 cm2 g−1 , and the combination κo π 2 /αIM ≃ 1.
Similarly, IM /IK ≃ 1.7.
Summarizing: measurement of the largest velocities in the spectrum
gives the ratio ESN /M from Eq. (31). The effective temperature gives the
initial radius, by Eq. (34). The peak luminosity, through Eq. (33) then
can give the distance to the supernova. Finally, measurement of the decay
timescale provide an estimate of the ejected mass, through Eq. (32).
Radioactive Heating
Typical Type 2 light curves do not show the effects of radioactive energy
input until a year or more after peak light, but in Type 1s, it dominates
the light curve from the first few days. Much more radioactive nickel is
produced, and the ejecta mass and initial radius are much smaller.
Choose a simple parametrization of the radioactivity:
ǫ̇ = ǫ̇o ξ (x) e−t/τr
(35)
where ǫo is the energy released per gram of radioactive nuclei per second, ξ
is the density distribution of the radioactive nuclei, and τ r is the radioactive
7
decay timescale. For 56 Ni→56Co, ǫo = 4.78 × 1010 erg g−1 s−1 and τr =
7.6 × 105 s. For 56 Co→56Fe, ǫo = 7.97 × 109 erg g−1 s−1 , and τr = 9.82 ×
106 s. In Type 1 events, however, the exploding envelope is transparent
to much of the cobalt decay gamma rays and positrons, making a simple
characterization of ǫ̇ difficult. This might have contributed to a mistaken
identification with the decay time of Cf in some supernovae. For Ni-Co-Fe
decay,
h
i
Ni
Ni
Co
N
i
−t/τ
Co
−t/τ
−t/τ
r + ǫ̇
r
r
ǫ̇ = ξ (x) ǫ̇o e
1−e
e
.
(36)
o
The energy input from Co exceeds that from Ni when
N
i
Co
ln ǫ̇o /ǫ̇o
+ e−t/τN i
t
≃
≃ 2.08
τN i
1 − τN i /τCo
(37)
or about 18 days. After 4τN i , the Ni decay is only producing 1/8 of the
total energy, and Eq. (35) becomes a good approximation. The decay
(56 Ni→56Fe) liberates 0.12 MeV/nucleon or about 2×1050 ergs per solar
mass of radioactive material, which does not destroy the homologous expansion, since it is such a small fraction of ESN .
Using
ǫ̇ = ǫ̇o ξ (x) f (t) ,
(38)
one notes that if the function ǫ̇η/φΨ is approximately independent of x, we
can separate Eq. (6) as before. Taking ξ(x)η(x)/Ψ(x) ≡ b, a constant, is
tantamount to concentrating radioactive nuclei in the center, not an obviously bad assumption. The initial mass of radioactive nuclei becomes
Z 1
I
3
Mr = 4πRo ρo
ξ (x) η (x) x2 dx = b T M.
(39)
IM
0
We may now rewrite Eq. (6) as
φ̇ + φ
R (t) 2ǫ̇oMr
R (t)
.
= f (t)
τo Ro
Ro ESN
Define
(40)
2
u = t/τo + t2 /τdecay
.
u̇ = R (t) / (Ro τo ) ,
The second follows from R(t) (Eq. (27)). The solution of Eq. (40) is
Z u
−u
eu f du + e−u
(41)
φ = ǭe
0
8
where
2ǫ̇o Mr τo
Mr ǫ̇o
Mr M 1014 cm 1051 ergs
ǭ =
=
,
= 3.1 2
ESN
L (0)
Ro
ESN
M⊙
(42)
evaluated for nickel decay. This quantity measures the importance of radioactivity to the light curve, especially at early times. For times t < τN i ,
we have f ∝ e−t/τN i , and for t > 3τN i , we have f ∝ e−t/τCo . For extremely
early times, u << 1, we may use the fact that τN i , τdecay << τo to find
φ = 1 + (ǭ − 1) u
u << 1.
(43)
For Type 1 supernovae, the initial radius is very small. For the envelope of
the presupernova to be in hydrostatic equilibrium, the presupernova luminosity must be less than the Eddington limit 4πcGMcore/κ. This implies
that the envelope radius is limited to
r
1/2 cGMcore 1
6000K 2
13 Mcore
= 2.1 × 10
Ro <
cm.
(44)
σκ Te2
1.4M⊙
Te
Eq. (42) thus shows that ǭ >> 1, and we expect the bolometric luminosity
to initially increase after the shock reaches the photosphere. For Type 2
supernovae with a red supergiant progenitor (we note the important exception of SN1987A!), Ro ∼ 1014 cm, Mr < 0.1M⊙, and so ǭ ≤ 1 and the
luminosity will begin to fall almost immediately.
Over very long times, Eq. (41) becomes
u >> 1
(45)
φ ≃ e−u + ǭ f − e−u .
Even for ǭ < 1 the radioactive term is important. For t >> τdecay , u ∝ t2
and f ∝ exp(−t/τCo ). The f ǭ term will eventually dominate.
The luminosity function Eq. (41) has the property that it reaches a
maximum when φ = f ǭ, i.e., the light curve peaks when it has the same
value as the radioactive term, at least when ǭ > 1. If ǭ < 1, the maximum
is at t = 0. The width of the maximum in the light curve, which may of
course be measured, can be estimated from
2 −1/2
r
d φ
τCo
∆peak ≃
=
τ
(46)
decay
2φmax tmax
dt2 tmax
where tmax is the time when the luminosity is maximum. Thus, the larger
the decay time, the broader the peak of the light curve.
9
In summary, the luminosity, when radioactive input is considered, is
Z u
M
E
R
⊙
o
eu f du e−u ergs
1 + ǭ
L = 3.1 × 1043 51SN
14
10 ergs M 10 cm
0
Z u
3.1 Mr
ESN M⊙ Ro
43
u
= 3.1 × 10
e−u ergs.
e f du + 51
14
.5 M⊙ 0
10 ergs M 10 cm
(47)
The upper (lower) number in the second equation corresponds to Ni (Co)
decay. We have used representative values for the opacity and the spatial
part of the solution. The more massive the ejecta, and the more compact
the envelope, the more important radioactivity is.P For consistency, the
maximum amount of energy input by radioactivity ( ǫ̇ro τr ) should be small
compared to ESN . In the case of the Type 2 supernovae, with massive and
extended envelopes, radioactivity becomes important at late enough times.
From Eq. (45) this occurs when e−u ≈ ǭf or
!2
t
t
ESN M⊙ 0.1M⊙ Ro
−
≃ ln 1 + 40 51
,
(48)
τdecay
τCo
10 erg M Mr 1014 cm
which is generally t ≃ (1 − 2)τdecay ≃ 100 − 200 days.
10
The figure shows three typical cases of supernova light curves. The left
panel is for the Type 2 supernova SN 1987A. A more typical ’red giant’
Type 2 supernova is in the middle panel. The right-hand panel is for a
Type 1 supernova. The solid lines are the full solutions, the dashed lines
assume the radioactivity is due only to Co decay, and the dash-dot lines
assume no radioactive energy input.
Application to SN 1987A
SN 1987A appeared to be subluminous for a Type 2. Eq. (33) immediately suggests that the presupernova envelope must have been smaller than
usual; in fact, from photographic identification, the progenitor was a blue,
not a red, supergiant. Neutrinos being observed less than 4 hours before
the optical display is corroborative evidence: the time necessary for the
shock wave with a velocity equal to the largest velocities initially observed
in the spectrum, about 5×108 cm s−1 , to traverse the envelope is equal to
two hours (it takes 2 hours after shock breakout for the optical emission to
occur) if Ro ≃ 3 × 1012 cm. The bolometric luminosity remained nearly
constant for the first few weeks and then increased. The shape of the initial
light curve thus cannot completely determine the envelope properties.
The initial luminosity, based on the assumed distance to the Large
Magellanic Cloud of 50 kpc, was about 1.8×1041 ergs s−1 . Coupled with
Eq. (47), evaluated at t = 0, Ro ≃ 2.5 × 1012 cm gives an energy to mass
ratio of ESN /M ≃ .18×1051 ergs/M⊙. Eq. (30) says that the diffusion
p time
was about τo ≃ 20.2(M/M⊙ ) yrs, and the decay time τdecay ≃ 26.6 M/M⊙
days, for nominal values of the opacity, α, IM and IK .
After several months, the light curve became exponential decay with a
halflife equal to that of Co decay. The total mass of radioactive nuclei may
be found from the absolute position of the light curve’s tail:
Mr
43
(49)
L (t → ∞) ≃ 1.55 × 10
e−t/τCo ,
M⊙
which gives Mr ≃ 0.075M⊙. Therefore, for cobalt decay, one finds
ǫ̇ Mr
2ǫ̇ Mr τo
= Co
ǭ = Co
≃ 8.5,
ESN
L (0)
(50)
and for nickel,√ǭ ≃ 40. However, Eq. (43) demonstrates that until ǭu ≃ 1,
or t ≃ τdecay / ǭ, the radioactive heating will have little effect. In the case
p
of SN1987A, this occurs (for nickel) 32 M/41M⊙ days after the explosion.
11
The fact that the light curve began to increase due to radioactive heating
after about 1 1/2 weeks is already an indication that the ejecta mass is in
the range of 5–10 M⊙.
The intermediate times are more complicated to describe analytically,
but if we have simple radioactive decay of one species, so that f = exp(−t/τCo ),
one can describe the peak conditions according to:
upeak + y ≈ 1.8 − 2.0;
φpeak ≈ 0.17ǭ/y.
(51)
This approximation is valid when ǭ >> 1 and 0.3 < y < 1.5 with
1/4
τdecay
M 3/4 1051ergs
y=
= 0.15
.
2τCo
M⊙
ESN
(52)
√
Thus, t/τCo ≃ 2y u. From the observation that the light curve peaked
around tpeak = 90 days after the explosion, and the relations
upeak = tpeak /τdecay
2
≃ 12.3 (M⊙/M ) ;
p
y = 0.23 M/M⊙ ,
we deduce that M ≈ 9−11M⊙. We used the energy to mass ratio described
above. The mass scales as the square root of the energy to mass ratio, and is
inversely proportional to the opacity, whose value was taken to be equal to
that for a solar composition gas (=0.33). For metal-rich ejecta the opacity
is greater than for a solar gas, making our estimate of the mass too high.
It is worth noting that the light curve for SN 1987A, although agreeing
with the general behavior of the solutions we have discussed, differs from
them in some important respects. First, the observed peak value of the
luminosity was about 5 times the initial luminosity (excluding transient
effects due to the ultraviolet burst) compared to about φpeak ≃ 0.17ǭ/y ≃ 2
times in the analytic solution. Second, the peak of the observed light curve
does not lie exactly on the asymptotic radioactive decay line, and it lacks the
symmetry and breadth of the analytic solutions. This behavior implies the
existence of additional energy sources or uneven distribution of radioactive
components. One additional energy source is due to recombination in to
the cooling wave that eats inward into the ejecta. This contribution cannot
be easily included in the analytic solution.
12
Application to Type 1 Supernovae
In the case of Type 1 supernovae, involving the explosion of white dwarfs,
the radius is of order 1000 km. The explosion energy is again of order 2×1051
ergs. Using Eq. (42), we find for Ni that ǭ ≈ 4 × 105 if the ejecta mass is 1
M⊙ and 1/2 is Ni. We also note that the parameter
1/4
M 3/4 1051 erg s−1
y = τdecay /2τN i = 1.3
≈ 1.1
M⊙
ESN
(53)
for Ni decay. Eq. (51) then suggests that the peak in the light curve is
reached when upeak ≈ 0.6 and φpeak ≈ 0.16ǭ. The maximum luminosity
is therefore about 0.16Mr ǫ̇N i = 8 × 1042 ergs/s, which is completely independent of R and is also insensitive to the explosion energy and the ejected
mass. This is what makes Type 1 SN so valuable as distance indicators.
Note that the light curve peak is noticeably shifted from the explosion time
by a few weeks:
p
tpeak = upeak τdecay ≈ 17 days.