Last class: Adiabatic and Isothermal shocks

adiabatic and cooling regions in discontinuity

1

In both cases R-H equations conserving mass and momentum apply to material on either side of shock

R-H energy conservation equation applies to adiabatic case For isothermal, condition becomes T1 = T2

Adiabatic: Isothermal: maximum compression factor ≤ 4 ρ2/ρ1 = u1

2/cs2 = M1

2

compression can be large cs isothermal sound speed

dx infinitely thin → “jump”

Shock wave theory applied: Supernovae Clarke & Carswell Chapter 8; Draine Chapter 39

Supernovae explosions from gravitational collapse of stellar core + envelope ejection

Type II SN – end point of massive stars M > 8M.

“instantly” ~ 1 – 5 M of ejecta velocities ~ 104 km/s

energy released ~ 1051 ergs

Gas sweeps up, accelerates, compresses, heats ISM → blast wave

Typical ISM: nHI ~ 1 cm-3, ρcloud ~ 1.6 x 10-24 g/cm3, Tcloud ~ 104 K

2

Phases of a Supernova Explosion (interaction between ejecta & ambient medium) 1. Mejecta >> Mswept up ISM

Free expansion of expelled gas at constant vE

2. Mswept up ISM > Mejecta Expansion slows somewhat, shocked gas has no time to cool

→ adiabatic shock 3. Expansion slows and radiative losses become important

→ isothermal shock 4. SN remnant slows to point where pressure balance with

ambient medium is achieved → merges

Idealized case: homogeneous gas explosion energy E instantaneously from point surrounded

by region of uniform density ρ 3

Phase I –free expansion Free expansion, vE const; radius increases with time;

assume uniform density ρ in front of ejecta Phase I terminates when mass swept up = mass ejected

,energy emitted ε = 4 x 1050 ergs, vE

2 = 2ε/ME and vE ~ 104 km/s

∴ ME = (8x1050) /1018x 2x 1033) = 0.4 M when mass swept up = mass ejected, 4/3πρR3 ~ 0.4 M

R3 = 0.4 x 2x 1033/4 x 1.6 x 10-24 ~ 1057/8 ∴R ~ 0.5 x 1019 cm ~ 1.7 pc

∴duration = R/vE = 1.7 x 3 x 1018/109 x3x107 ~ 170 yrs in practice, phase I lasts about 100 years

“bubble” continues to expand at high velocity → shock

energy conserved → shock adiabatic

4

Phase II Blast Wave [Adiabatic or Sedov Phase] Explosion propagates outwards → shock Assume all swept up medium in narrow shell ∴ for small D, Adiabatic shock:

for γ = 5/3, D/R = 1/12 << 1 i.e. OK to assume shell thin

diatomic gas, γ = 7/5, ρ0/ρ1 =6 – even thinner ∴ assume all gas in thin shell moves at same

average velocity at any instant

In frame of shock, diagram as seen earlier

Words to fill an cover

1

0

11

ρ γρ γ

+=

−

3 20 1

4 43

R R Dπ ρ πρ=

1 13 1

D Rγγ

−∴ = +

5

R-H (1) from continuity equation: ρ0u0 = ρ1u1, so that velocity of shocked to unshocked gas = U As shell grows, its radial momentum increases as (momentum due to pressure on inside of shell, Pin outside ρ, T are 0)

Let pressure within shell. P1 scale as pressure Pin Pin = αP1 For strong shock And rate of change of radial momentum = force over area ∴

01 0 0

1

11

u u uρ γρ γ

−= = +

00 1

21

uu uγ

= − = +

3 00

243 1

ud Rdt

π ργ

+

20 0

21isP uρ

γ

= +

3 2 200 0 0

24 243 1 1

ud R R udt

π ρ πα ργ γ

= + +

6

1

From Obtain i.e. shock advances on undisturbed gas at u0 = dR/dt,

d/dt[R3Ŕ] = 3αR2Ŕ2

Substituting R ∝ tb, here → equn w. solution b = 1/(4-3α) And power law scalings for R and u0 that describe

how radius of shell increases with time in terms of α

[Clarke & Carswell p 93]

How to determine α?

3 2 200 0 0

24 243 1 1

ud R R udt

π ρ πα ργ γ

= + +

3 2 20 03d R u R u

dtα =

14 3

3 33 34 3

o

R t

u t R

α

ααα

−

−−−

∝

∝ ∝

7

Energetics of explosion define α Pin = αP1 relates pressure in shocked gas to pressure on shell from explosion

Adiabatic case: energy (= KE + ε) is conserved Internal energy /unit volume = P/(γ-1)

Most energy is in material in cavity (little mass)

∴ Internal energy = 4πR3Pin/(γ-1) = (4/3) πR3αP1/(γ-1) and KE = ½ (4/3) πρ0 R3U2

E = KE + ε = ∴ Since energy must be conserved, 6α - 3 = 0, and α = ½

2 23 0 0 0

02 24 1

3 2 1 1 1u uaR ρπ ρ

γ γ γ

+ + − +

→ R ∝ t2/5, u0 ∝ t-3/5 P1 ∝ t-6/5

6 33 2 4 3

0E R u tα

α−

−∝ =

8

∝

with increasing R, u0 and P1 decreasing C& C §8.1, Kwok §16.2, Tielens §12.3.2

More rigorous derivation: Similarity/Sedov Solutions C&C §8.1.2

Let λ = scale parameter → size of blast wave at time t after explosion Assume λ variation with t depends on E, ρ0

∗

Dimensionally, E ≡ ML2T-2 and ρ ≡ ML-3

To keep E/ρ const, ML2T-2/ ML-3 ~ L5T-2 = const ∴(Et2/ρ)1/5 has dimensions of length → λ ≡ (Et2/ρ)1/5 Sedov-Taylor solution Replace radius of shell r by ξ = r/λ and assume solution self-similar; i.e. since no natural scale length all variables X(r,t) take form X = X(t)X(ξ); substitution → scaling relation for Rshock from ξ = r/λ = r/ (E/ρ0)1/5 t2/5

SIMILARITY SOLUTION - All physical properties scale wrt radius of shock front

Rshock = ξ0(E/ρ0)1/5t2/5 ∝ t2/5

(ξ0 ≡ value of similarity variable that labels location of shock front ~1) ∗ E and ρ are key parameters; ejecta expand without losing energy; density important since initial energy redistributed to larger and larger amount of gas.

X = X(t)X(ξ) has same shape But scaled by time-dependent X(t)

X

ξ ξ

9

Sedov: solved R-H equations analytically

(C and C §8.3 adapt, using ξ = r/λ = r (Et2/ρ0)1/5 and dimensionless variables ρ′(ξ), u′(ξ), P′(ξ),

so that ρ(r,t) = ρ1ρ′(ξ) = ρ0(γ+1)/γ-1) ρ′(ξ) etc. → set of equations independent of r and t

(integration “elementary but laborious”)

Exact solution traces u, ρ, P, T, behind shock (assumes γ = 7/5)

At r=o t= 0, instantaneous energy release and shock front values us, ρs, Ps, Ts

Plots show variation of u/us , ρ/ρs, P/ Ps, T/ Ts, with ξ

Similarity solutions more or less validated closer to shock, density , pressure high

inner regions : gas hotter , more tenuous, but pressure maintained velocity over-simplified?

1

0 ξ0

10

“Experiments” also validate self-similar solutions R ∝ t2/5 and R(t) = ξ0(Et2/ρ0)1/5

e.g. blast wave from atomic bomb Snapshots of fireball (C&C §8.2) confirm scaling of R5/2 with t (log r plotted v logt) Relation enables determination of explosion energy R(t) = ξ0(Et2/ρ0)1/5

For SN, expansion velocity U = dR/dt = 2/5ξ0(E/t3ρ0)1/5 = 2R/5t Energy E ~ (R/t)2

SN ejecta ~ 1 M expands with velocity ~ 104 km/s Energy ~ mv2 ~ 1033 x 1018 – 1051 ergs

Rewrite R(t) in terms of typical SN energy and density 1 gm/cm3

→ R(t) ~ 1013t2/5 and u = 4 x 1012t-3/5 . R ~ 0.3t2/5 pc, u ~ 105t-3/5 km/s

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Alternatively can rewrite R(t) = ξ0(Et2/ρ0)1/5 as R(t) ~ 10-0,7(E50/ρ24)1/5t2/5 with t in yrs, R in pc

Again leads to R ~ 0.3t2/5 pc, and u ~ 105t-3/5 km/s

For Crab nebula, with R ~ 2 pc, E = 4 x 1050, derive age ~ 103 yrs

What next?

Radiation losses increase when vs < 250 km/s ( T<106 K)

→ no longer adiabatic shock → isothermal at that time t3/5 = 1/250, t~ 4 x 104 yrs, R ~ 25 pc

i.e. Sedov phase ends after ~ a few x 104 yrs (Tielens)

Note: Approximations do not hold for t < 100 yrs (the free expansion stage with U = 104 km/s)

12

SN 1987A

In LMC, at 50kpc. Star (Sanduleak) SK-69, mass 20 M. Strong stellar wind 106 years ago → cavity in ISM

1987 explosion → blast wave (Sedov phase) . Shock moves at 4.5 x 103 km/s.

X-ray emission as it meets dense ring of gas deposited by wind 13

Why transition from adiabatic conditions at 106 K?

Here Λ = cooling rate (ergs cm-1 s-1) Below 106 K Λ increases as T decreases Time scale for Sedov phase: Λ= φne

2T-1, where φ= 8x1017 (cgs) For sphere radius R, cooling ~ 4/3(πR3) φne

2T-1, i.e. fastest cooling for larger R, low T

Since kT ~ ½ m(3/4)v2, T =3/32(mv2/k)

At T = 106 K, v2 ~ 32/3 x10-16/10-24x 106 ~ 1015, V ~ 3 x 107 ~ 300 km/s

From Sedov tables, v =300 km/s and t = 104 yrs Adiabatic phase lasts ~ 104 yrs

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More formally: recall R-H conditions P0 negligible, u1 much decreased behind shock ∴

Isothermal shock: ρ1/ρ0 ~ M0

2 = u02/cs

2, Suppose blast wave at 100 km/s moves into gas with cs = 1 km/s,

M02 = 104 = ρ1/ρ0

→ density increase of 104 ?

→ In practice, radiation inside SNR ionizes and heats gas ahead cs ~ 10 km/s more typical, and ρ1/ρ0 ~ 100

Note: since u1 is small, material remains near shock → thin, highly compressed shell

1 1 0 02 2

1 1 1 0 0 22

1 0 0~

u uu P u P

P u

ρ ρ

ρ ρ

ρ

=

+ = +

15

SN Evolution (Sedov-like tables for γ = 5/3) t(yrs) 1 10 100 1000 10,000 100,000 R(pc) 0.315 0.791 0.791 4.91 12.5 31.5 V(km/s) 124,000 81,000 7820 1970 4.94 1.24

As shock speed decreases, Tshock decreases - after about 104 yrs, T ~ 4 x 106 K and still decreasing Recombination of C, N, O ions → subsequent line cooling i.e. radiative losses and shock becomes isothermal temperature determined by conservation of momentum

- snowplow phase mass swept up slows front to sub-sonic

Sedov onset radiative cooling

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Phase III Snowplow or Momentum-Conserving Phase (Kwok 16.2.1)

Begins at t= 0, when momentum of shell p0 = Msvs = const ∴equn of motion Integrating → with Solutions:

4 310 0 04c R R Rt

•

= −

( )( )

340 03

3403

4 014

0

34

o

o

p R R

p R R

pR t c

π ρ

π ρ

πρ

•

•

=

=

= +

( ) ( )

1 34 4

0 00 000

0 0

4 41 1

R t t R t tR R and R R

R R

−

• •• •

− −

= + = +

R ∝ t1/4 dR/dt ∝ t-3/4

Adiabatic phase: R ∝ t2/5, dR/dt ∝ t-3/5

Snowplow phase: expansion slowing down much more

17

Supernova remnant evolution – schematic, realistic?

Remaining questions: - effect of magnetic field on SNR growth? -interaction of shock front with ISM inhomogeneities?

-how does the real ISM (irregular clouds - e.g. SN 1987A) affect Sedov phase?

-role of dynamical instabilities in structure of SNR - Rayleigh-Taylor instabilities in Crab nebula filaments

-energy input into ISM - Kwok 16.2.1

Draine & McKee 1993 AnnRevAstrAstrophys 31, 373 18

Cygnus Loop - 7.5 kpc - 2 x 105 yrs - radiative phase

19

20

Expansion of HII Regions - dynamics

So far: expansion of static HII regions → Strömgren sphere rate of input of ionizing photons ≡ recombinations

Assumed density in HII region same as before gas ionized But: gas much hotter inside HII region 10,000K versus 100K

∴pressure inside ~ 100 x pressure outside → expansion and drop in density

recombination rate ∝ ne2 - more ionized gas than in static

Stages of expansion: 1) Initial Strömgren sphere Produced by ionizing photons from star, very little motion α = recombn coefft = 2.6 x 10-11/T1/2.

For O5 star, Rinit ~ 5.6 pc

13

*2

34 ( )init

H

SRn Tπ α

=

21