Fluids in astrophysics
S.M.Lea
March 1998 revised 2000
1 The equations of -uid dynamics
Fluid dynamics describes the behavior of liquids and gases when the mean free path for
particles is very small compared with the other length scales of interest in the problem. In
astrophysics, densities are often low and mfp correspondingly large, but the relevant length
scales are also very large. In addition, interactions between the particles of gas and the
magnetic ®eld serve to reduce the mfp. Thus in most astrophysical situations we may safely
use the -uid equations to describe the motion of matter.
1.1
The continuity equation
The density in a region is due to -ow of matter into or out of that region. Thus:
Z
Z
@
½dV = ¡ ½~v ¢ d~s
@t
Using the divergence theorem, we may write this relation as:
Z µ
¶
@½ ~
+ r ¢ (½~v ) dV = 0
@t
Since this relation must be true for any volume V ; we may conclude:
@½ ~
+ r ¢ (½~v ) = 0
(1)
@t
Expanding the divergence, we get:
@½
~ + ½r
~ ¢ ~v = 0
+ ~v ¢ r½
(2)
@t
D½
~ ¢ ~v = 0
+½r
Dt
where
D
@
~
=
+ ~v ¢ r
(3)
Dt
@t
is the Lagrangian derivative, the rate of change as seen by an observer moving with the -uid.
This is often the best derivative to use since it refers to ®xed particles, as Newtonzs law do.
1
1.2
The equation of motion
We apply Newtonzs 2nd law, F~ = m~a to a -uid }particle} of mass ½dV : The forces acting
on the particle are pressure forces and gravity. Thus:
D~v
~ + ½~g
½
= ¡ rP
(4)
Dt
Here we have ignored the effect of magnetic ®elds, which can be very important in
astrophysics. (These effects are studied in Physics 712.) We can expand out the Lagrangian
derivative to get:
@~v
~ v = ¡ rP
~ + ½~g
½
+ ½~v ¢ r~
(5)
@t
1.3
The energy equation
So far we have two equations (one vector, one scalar) for the varaiables ~v ; ½; and P: So we
need another equation. (We also need ~g : This may be speci®ed independent of the -uid
under consideration, or it may be derived from Poissonzs equation in the case of self gravity.
) The third -uid equation is the energy equation. Sometimes we can use a short cut by using
an equation of state:
P = nkT
(6)
for an ideal gas at constant temperature, or
P = K½°
(7)
for an adiabatic process. When neither idealization is valid we need to use the ®rst law of
thermodynamics:
dU = Q ¡ W
We apply this to the speci®c internal energy (energy per particle) Uspecif ic = 32 kT with ¡
and ¤ representing the heating and cooling rates (per unit volume) for the -uid. Then the
®rst law of thermodynamics may be written:
µ
¶
µ ¶
D 3
¡¡¤
D 1
kT
=
¡ nkT
Dt 2
n
Dt n
¡¡¤
kT Dn
=
+
(8)
n
n Dt
or
µ
¶
D 3
Dn
n
kT ¡ kT
= ¡ ¡¤
Dt 2
Dt
µ
¶
D 3
5
Dn
nkT ¡ kT
= ¡ ¡¤
(9)
Dt 2
2
Dt
Setting ¡ ¡ ¤ = 0; we retrieve the adiabatic equation of state with ° = 5=3 for standard
astrophysical situations where the -uid is a monatomic gas. (The situation could be different
in molecular clouds, however.)
Note that we have neglected thermal conductivity in this derivation.
2 Applications of the -uid equations
2
2.1
The virial theorem
The virial theorem applies to a system enclosed within a surface S: Take the momentum
equation and dot with the position vector ~r :
³
´
³
´
@~v
~ ~v = ¡~r ¢ rP
~ + ½~r ¢ ¡r©
~
½~r ¢
+ ½~r ¢ ~v ¢ r
@t
Integrate over the volume V containing the system. Consider ®rst the steady state case, so
that the ®rst term with the time derivative is zero. Then the orther terms are:
Z
Z
³
´
~
½~r ¢ ~v ¢ r ~v dV =
½ri vj @j v idV
V
ZV
=
[@j (½riv j v i) ¡ vi@j (½ri vj )] dV
ZV
Z
=
(½riv j v i) nj dS ¡
[viri @j (½vj ) + ½vi vj @j ri] dV
S
V
Z
Z
=
½ (~r ¢ ~v ) ~v ¢ ^n dS ¡ (0) ¡
½vi vj ±ij dV
S
V
Z
= ¡2T +
½ (~r ¢ ~v )~v ¢ ^n dS
S
where
T =
Z
V
1 2
½v dV
2
is the kinetic energy of the system.
On the other side of the equation we have:
Z
Z h
i
~
~ (~rP ) ¡ P r
~ ¢ ~r dV
¡
~r ¢ rP dV = ¡
r
V
ZV
Z
= ¡
P~
r ¢n
^ dS + 3
P dV
S
V
Z
= 3¦ ¡
P ~r ¢ n
^ dS
S
where
¦=
and
Z
P dV =
V
Z
nkT dV =
V
¡
Z
V
2
3
Z
V
3
2
nkT dV = U
2
3
~ ´W
½~r ¢ r©
So the volume-integrated equation of motion is:
2T + 3¦ + W + surface intregrals = 0
If the system is ®nite, then we can put the surface outside the system and thus ensure that the
surface integrals are all zero. Then we have the steady state virial theorem:
2T + 3¦ + W = 0 = 2T + 2U + W
(10)
Note that T is always positive, as is ¦: Thus W is negative and is necessary for the system
3
to be in a steady state. This is a gravity-bounded system. Writing the kinetic plus internal
energy of the system as E = T + U; equation 10 becomes:
2E + W = 0
or
W
E=¡
2
which is often the most useful form of the virial theorem.
R
W is related to the gravitational energy - = 12 ½©dV for the system:
¶
Z µ 2
g
©
~r ¢ ~g
~
W =-¡
~r +
~g ¡
~g ¢ dS
8¼G
8¼G
4¼G
S
To see this, we start with the answer and work backwards:
¶
Z µ 2
1
g
©
~
~r + ~g ¡ (~r ¢ ~g )~g ¢ dS
4¼G S 2
2
µ
¶
Z
1
gj gj
©
= ¡
@i r j g j g i ¡
ri ¡ gi dV
4¼G V
2
2
¶
Z µ
1
gjgj
@i ©
©
= ¡
rj gj @igi + ±ij gj gi + rj g i@i gj ¡
@iri ¡ gj ri@i gj ¡
g i ¡ @ig i dV
4¼G V
2
2
2
Now we use Poissonzs equation, @ igi = ¡4¼G½; and simplify:
Z µ
¶
Z µ
¶
©
1
3
¡gi
surface integral =
rj gj ¡
(¡½) dV ¡
g 2 ¡ rj g i@i @j © ¡ g 2 + gj ri@i @j © ¡
g i dV
2
4¼G V
2
2
V
Z
Z
1
= -+
rj @j ©½dV ¡
(¡rj g i + g j ri ) @i @j ©dV
4¼G V
V
= -¡ W
The remaining volume integral is zero because the integrand is an antisymmetric tensor
times a symmetric tensor, and is thus identically zero.
The surface integral terms are often zero. For a bounded system, we know that © » 1=R
at large distances, and similarly g » 1=R2 : Thus g 2 R » 1=R 3 ; which goes to zero faster
than the surface area goes to in®nity (as R2 ). Thus the whole integral goes to zero as
R ! 1: In fact for any spherically symmetric system, the surface terms sum to zero. For
then the surface S is a sphere of radius a, ~g = ¡g^
r ;© (a) = ¡GM (a) =a = g (a) a; and so:
Z µ 2
¶
µ 2
¶
g
©
g
ag
2
~
~r + ~g ¡ (~r ¢ ~g) ~g ¢ dS=4¼a
a+
g ¡ g2a = 0
2
2
2
2
S
In such cases W = -: For a system of mass M and radius R; - » ¡M 2 =R: For example,
for a spherical system with ½ = ½0 R=r; we have
Z r
Z r
M (r) =
½ (r) 4¼r2 dr = 4¼½0 R
rdr = 2¼½0 Rr 2
0
0
M (r)
© (r) = ¡G
= ¡2G¼½0 Rr
r
4
and
=
1
2
Z
1
½©dV = ½0 R (¡2G¼½0 R)
2
3
GM2
2 R
= ¡G (2¼½0 R)
=¡
3
3 R
- is negative, as expected.
-
Z
0
R
1
r4¼r2 dr
r
2.1.1 Example: a slowly contracting star
If the star constracts very slowly, then T ¼ 0;and so the virial theorem reduces to:
- = ¡2U
As the star contracts, j-j _ M =R increases, so the internal energy U = 12 j-j increases
also. Thus the star heats up. Notice that to remain in equilibrium, only half of the
gravitational energy released is converted to thermal energy: the other half must be radiated
away.
2
2.1.2 Example: a spherical interstellar cloud
The cloud is embedded in the surrounding medium, so we cannot ignore all the surface
terms. However, if the cloud is spherical, the gravitational surface terms sum to zero, and if
~v = 0 at the surface, then only the pressure term remains. So:
Z
2 (T + U ) + - =
P ~r ¢ n
^ dS = 4¼a3 P (a)
(11)
S
3
This system can
R be pressure bounded even if ¡- is small. Notice that 4¼a = 3V; and
2U = 3¦ = 3 P dV ; so equation 11 becomes:
Z
2T + - + 3 (P ¡ P (a)) dV = 0
If the kinetic energy T ¼ 0; and - ¼ 0; the pressure at the surface must be:
Z
1
P (a) =
P dV =< P >
V
the average pressure inside the cloud. If gravity is negligible, there can be no pressure
gradients: pressure is constant inside the cloud.
Magnetic ®elds can be included in this analysis. Magnetic ®eld inside the cloud always
tends to expand the cloud, but external ®elds can help to con®ne the cloud.
2.1.3 Example: stellar systems
The virial theorem is traditionally applied to stellar systems such as clusters of galaxies or
globular clusters. For such systems, which are }pressureless}, U = 0 but T 6= 0: The virial
theorem becomes:
2T + - = 0
Thus by measuring the systemzs kinetic energy T ; we can measure its mass (through -).
5
2.2
The linearized equations
³
´
~ ~v terms in
The -uid equations are, in general, non-linear (see, for example, the ½ ~v ¢ r
equation 5). This means that the solutions are interesting and complex, but also dif®cult to
®nd. In many cases the -uid has a steady equilibrium state (@=@t ´ 0) and we can look for
small perturbations to that state. We shall label the steady state variables with a substcript 0
and the perturbed quantities with the subscript 1. Then we have:
Equation 2:
@ (½0 + ½1 )
~ (½0 + ½1 ) + (½0 + ½1 ) r
~ ¢ (~v 0 + ~v1 ) = 0
+ (~v 0 + ~v1 ) ¢ r
@t
Use the fact that @½0 =@t ´ 0 and ignore products of two perturbed quantities:
@½1
~ 0 + ~v0 ¢ r½
~ 1 + ~v1 ¢ r½
~ 0 + ½0 r
~ ¢ ~v0 + ½0 r
~ ¢ ~v1 + ½1 r
~ ¢ ~v0 = 0
+ ~v0 ¢ r½
@t
The underlined terms cancel: this is the equation describing the original steady state. Thus
we are left with:
@½1
~
~
~ v1 + ½ r
~ ¢ ~v0 = 0
+ ~v0 ¢ r½
v1 ¢ r½
(12)
1 +~
0 + ½0 r ¢ ~
1
@t
Now we do the same thing with the momentum equation (5):
@ (~v0 + ~v1 )
~ (~v0 + ~v1 ) = ¡r
~ (P0 + P 1 )+(½ + ½ ) (~g0 + ~g 1 )
(½0 + ½1 )
+(½0 + ½1 ) (~v0 + ~v 1 )¢r
0
1
@t
Linearize:
³
´
³
´
³
´
@~v1
~ ~v1 + ½ ~v1 ¢ r
~ ~v 0 + ½ ~v 0 ¢ r
~ ~v0 = ¡rP
~ 1 + ½ ~g 1 + ½ ~g 0 (13)
½0
+ ½0 ~v 0 ¢ r
0
1
0
1
@t
Before we perturb the energy equation, letzs look at some examples.
2.2.1 Special case: uniform, static equilibrium state with ~g = 0
With ~v 0 = 0 and gradients of all equilibrium quantities zero, equations 12 and 13 reduce
to:
@½1
~ ¢ ~v 1 = 0
+ ½0 r
(14)
@t
and
@~v 1
~ 1
½0
= ¡rP
(15)
@t
Now³we look for ´a solution in which all the perturbed quantities are proportional to
exp i~k ¢ ~x ¡ i!t Equations 14 and 15 become:
¡i!½1 + i½0 ~k ¢ ~v 1 = 0
(16)
and
¡i!½0~v1 = ¡i~kP 1
Now we add the linearized version of the adiabatic equation of state 7:
dP
P0
P0
°¡1
°¡1
P1 =
½1 = °K½1 ½0 = ° ° ½1 ½0 = ° ½1
d½
½0
½0
6
(17)
(18)
Combining equations 16, 17 and 18, we get:
¡i!½1 + i½0 ~k ¢
¡!½1 + ~k ¢
Ã
Ã
~ 1
¡ikP
¡i!½0
~k° P 0 ½1
½
0
!
!
!
= 0
= 0
Now this equation can be satisifed with a non-zero perturbation ½1 only if:
P0
! 2 ¡ k2 °
=0
½0
or
s
!
P0
= °
k
½0
q
0
So we have sound waves with speed ° P
:
½
0
2.2.2 The Jeans instability.
Now we take ~g 1 6= 0; but keep ½0 constant and ~v 0 = 0: Thus we still have equation 14, but
the momentum equation becomes:
@~v1
~ 1 + ½0 ~g 1
½0
= ¡ rP
@t
We ®nd ~g 1 from Poissonzs equation:
~ ¢ ~g1 = 4¼G½
r 2© 1 = ¡r
(19)
1
Again we assume the wave form for all the perturbed quantities.
¡i~k ¢ ~g1 = 4¼G½1
(20)
and
¡i!½0 ~v 1 = ¡i~kP1 + ½0 ~g1
~
We dot equation 21 with k; and use equations 20, 16 and 18:
¡i!½0~k ¢ ~v1 = ¡ik2 P 1 + ½0~k ¢ ~g1
=
=
P0
4¼G½1
¡ik2 ° ½1 + ½0
½0
¡i
µ
¶ ~
P0
½0 k ¢ ~v1
i ¡k2 °
+ ½04¼G
½0
!
Again we want a solution with a non-zero ~k ¢ ~v 1 ; and this is only possible if
P0
! 2 = k2 °
¡ ½0 4¼G
½0
¢
P0 ¡ 2
= °
k ¡ k2J
½0
7
(21)
(22)
where
4¼G½0
(23)
°P0 =½0
Thus the frequency is real (the perturbation oscillates) if k > kJ ; but ! is imaginary (the
perturbation grows) if k < kJ;or equivalently, the wavelength of the perturbation ¸ is greater
than the Jeans wavelength
s
s
2¼
°P 0 =½0
¼°P 0 =½0
¸J =
= 2¼
=
kJ
4¼G½0
G½0
k2J =
We can also express this in terms of the minimum mass of the perturbation:
s
µ
¶3
¼°P 0
1
3
M J » ½0 ¸ J =
½0 G
½0
When the wavelength is short, the increased pressure in the compressed regions is able to
overcome the increased gravitational attraction, and the system oscillates. But for large
wavelengths, the increased density causes suf®cient gravitational attraction to overcome the
pressure forces, and the system collapses.
This result is generally interpreted to mean that an interstellar cloud needs to have a mass
greater than MJ in order to collapse. The limiting mass may be expressed in terms of the
cloud temperature, density, and mean molecular weight ¹ as:
s
µ
¶3
¼°kT
1
MJ »
¹nG
n¹
s
µ
¶3
¼ 53 (1:38 £ 10 ¡14 erg/K) (100 K) T 100
1
=
m (1:66 £ 10¡ 24 g) n (6:7 £ 10¡ 8 cm3 =g ¢ s2 )
nm (1:66 £ 10¡ 24 g)
3=2
T 100
g/cm6
m 2 n2
3=2
3=2
4: 1 £ 10 41 T 100
6
8 T100
=
M
/cm
=
2
£
10
M ¯ /cm6
¯
2 £ 10 33 m 2 n2
m 2 n2
which is large! Cold, dense clouds collapse the most easily. For a temperature of 20 K, a
mean molecular weight of 2 (molecular hydrogen) and a density of 1000 cm ¡3 ; we ®nd the
minimum mass for collapse to be
=
4: 1 £ 1041
MJ » 2 £ 108
0:2 3=2
2 M ¯ = 4: 5 M ¯
22 (1000)
However these parameters are not typical of a cloud before it collapses.
Notice that Jeanzs original analysis, as presented here, has a serious -aw: the initial
equilibrium is not an equilibrium! We cannot have a cloud with uniform ½0 and P 0 since
it will not have ~g0 = 0; and without pressure gradients it will collapse! However, the
calculation does give the salient features
p of a more accurate (and complicated!) calculation,
which leads to a critcal mass about 2 times larger than we found here. (See Spitzer for
details.)
8
2.2.3 Thermal instability
This instability arises because, in general, heating processes are proportional to n while
cooling processes go like n2 : When a system is perturned from equilibrium, heating and
colling no longer balance. In particular, if ±n > 0; cooling increases faster than heating, T
and hence P are reduced, and the density increases yet more.
To investigate this instability, we consider the energy equation in the form (equation 8):
µ
¶
D 3
Dn
n
kT = ¡ ¡ ¤ + kT
Dt 2
Dt
In equilibrium, ¡ = ¤: Let
¡ ¡ ¤ = ¡n$ (n; T )
where £ is the net energy loss rate, and
$ (n0 ; T 0 ) = 0
(24)
Now we perturb, to get
µ
¶µ
¶
µ
¶
@
3
@
~
~ (n0 + n1 )
(n0 + n1 )
+ ~v ¢ r
k (T 0 + T 1 ) ¡ k (T 0 + T 1 )
+ ~v ¢ r
@t
2
@t
·
¸
@$
@$
= ¡ (n0 + n1 ) $ (n0 ; T0 ) +
n1 +
T1
@n
@T
³
´
Setting ~v0 = 0; and assuming all perturbed quantities go like exp i~k ¢ ~x ¡ i!t ; we have
µ
¶µ
¶
µ
¶
·
¸
@
3
@
~
~ n1 = ¡ (n0 + n1 ) $ (n0 ; T 0 ) + @$ n1 + @$ T1
n0
+ ~v1 ¢ r
kT1 ¡kT 0
+ ~v1 ¢ r
@t
2
@t
@n
@T
Now linearize, and use equation 24 to get:
µ
¶
·
¸
3
@$
@$
n0 (¡i!)
kT 1 ¡ kT 0 (¡i!) n1 = ¡n0
n1 +
T1
2
@n
@T
Next we use the perturbed ideal gas equation of state to relate P 1 to n1 and T 1 :
P 1 = n1 kB T 0 + n0 kB T 1
Substituting this into equation 22, we have:
~ ¢ ~v 1 = ¡ik 2 (n1 kB T 0 + n0 kB T 1 ) + ½ ~k ¢ ~g1
¡i!½0k
0
Again we have dotted with ~k so we can use equation 20
~k ¢ ~g1 = 4¼G½1 = i4¼G½
1
¡i
So
¡i!½0~k ¢ ~v1 = ¡ik2 (n1 kB T 0 + n0 kB T 1 ) + i4¼G½0 ½1
and, as before, the continuity equation 10 gives:
½ ~k ¢ ~v1
½1 = 0
!
9
(25)
So
Ã
!
~k ¢ ~v 1 kB T 0
½
½ ~k ¢ ~v1
0
¡i!½0~k ¢ ~v1 = ¡ik2
+ n0 kB T 1 + i4¼G½0 0
(26)
!
m
!
and ®nally the energy equation gives
µ
¶
3
@$
@$
¡in0 !
kB T 1 + n0
T 1 = ¡ikB T0 !n1 ¡ n0
n1
2
@T
@n
So
k B T0
¡ @$
@n ¡ i! n0
T 1 = n1 @$ 3
¡ 2 i!kB
@T
Now write
2 @$ 1
1
kT =
¼
3 @T kB cs
cooling length
where the isothermal sound speed is
r
kB T 0
cs =
m
and the }cooling length} = sound speed times (cooling time at constant density). Similarly,
we de®ne
2 n0 @$=@n
k½ =
3 kT0 cs
and then
µ
¶
T1
n1 ¡k½ cs ¡ 23 i!
=
T0
n0
kT cs ¡ i!
µ
¶
~
k¢~
v1 ¡k½ cs ¡ 23 i!
=
!
kT cs ¡ i!
Now putting this into the momentum equation 26, and cancelling the factor ~
k¢v
~1 that
appears in each term, we have:
µ
µ
¶¶
½0 kB T 0
1 ¡k½ cs ¡ 23 i!
4¼G½20
¡i!½0 = ¡ik 2
+ n0 kB T 0
+i
! m
!
kT cs ¡ i!
!
µ
µ
¶¶
2
kB T 0
kB T0 ¡k½ cs ¡ 3 i!
!2 = k 2
+
¡ 4¼G½0
m
m
kT cs ¡ i!
Then we introduce kJ (equation 23 with ° = 1);to get
µ
¶
¡k½ cs ¡ 23 i!
2
2 2
! = k cs 1 +
¡ kJ2 c2s
kT cs ¡ i!
10
which is, in general, a messy cubic equation for !:
¡ 2 ¡ 2
¢ ¢
! ¡ k ¡ k2J c2s (kT cs ¡ i!)
µ
¶
2
= k 2 c2s ¡k½ cs ¡ i!
3
2
= ¡c3s k 2 k½ ¡ ic2s k2 !
3
! 2 kT cs ¡ i! 3 ¡ c3s k2 kT + ic2s k2 ! + c3s k2J kT ¡ ic2s kJ2 !
µ
¶
¡
¢
5 2
¡i!3 + !2 kT cs + ic2s!
k ¡ kJ2 ¡ c3s k2 (kT ¡ k½ ) ¡ k2J kT
= 0
3
First note that setting kT and k½ = 0 gives back our previous result for the Jeans
instability.
¡
¢¢
1 ¡
¡ i 3! 2 ¡ c2s 5k2 ¡ 3k2J ! = 0
3
which has solutions ! = 0 (which is a spurious root here) and
r
r
5
3
!=§
cs k2 ¡ k2J
3
5
(remember that our new de®nition of kJ differs from the old one by a factor of ° = 5=3):
To get an idea of the thermal effects, let both kT and k½ ¿ k; and expand ! in a series
! = !0 + !1 + ¢ ¢ ¢
where ! 1 is 1st order in kT ; k½ and so on. The to ®rst order, equation 27 becomes:
µ
¶
¡
¢
5 2
2
2
2
2
¡i3!0 ! 1 + ! 0 kT cs + ics ! 1
k ¡ kJ ¡ c3s k 2 (kT ¡ k½ ) ¡ kJ2 kT = 0
3
So
¡
¢
!20 kT cs ¡ c3s k 2 (kT ¡ k½ ) ¡ k2J kT
¡
¢
i!1 =
3! 20 ¡ 53 k2 ¡ kJ2 c2s
The new result has !0 = 0; and then:
i!1
=
=
¡
¢
¡c3s k2 (kT ¡ k½ ) ¡ k2J kT
¡
¢
¡ 53 k2 ¡ kJ2 c2s
¡ ¡
¢
¢
cs kT k2 ¡ k2J ¡ k 2 k½
¡5
¢
k2 ¡ k2J
3
(28)
Remembering that our functions
q have time dependence exp (¡i!t) ; this shows that we will
3
k ;
5 J
have growth (i! < 0) if k <
kT
¡
as before, and
¢
k 2 ¡ k2J ¡ k2 k½ >
So we would need
kJ
2
k (kT ¡ k½ ) >
s
kT
<k<
kT ¡ k½
11
r
0
kT k2J
3
kJ
5
(27)
or if k >
p
3
5 kJ
and
s
3
kT
kJ < k < kJ
5
kT ¡ k½
Since the frequency has no real part, this mode has pure exponential growth.
The previous Jeans result is modi®ed. With:
q
q
!0 = § 53 cs k2 ¡ 35 k2J
we ®nd:
i!1
r
=
5 2
3 cs
= cs
¡
¢
¡
¢
k 2 ¡ 35 kJ2 kT cs ¡ c3s k2 (kT ¡ k½ ) ¡ k2J kT
¡
¢ ¡
¢
5c2s k2 ¡ 35 k2J ¡ 53 k 2 ¡ k2J c2s
2 2
k kT
3
10 2
k
3
+ k2 k½
¡ 2k2J
= cs
1 2
k (2kT + 3k½ )
3
¡
¢
10
k 2 ¡ 35 kJ2
3
Again we have growth for k < kJ if kT and k½ are positive, but also if k > kJ and kT or
k½ is suf®ciently negative. In both these cases we have a growing wave, sometimes called
overstability.
To see whatzs going on in these modes, look at the ®rst solution (pure instability) in the
case kJ = 0 (i.e. gravity is not important). Then
¡
¢
c s kT k2 ¡ k2 k½
3
i! 1 =
= cs (kT ¡ k½ )
5 2
5
3k
independent of k: Putting back the de®nitions of kT and k½ ; we have:
µ
¶
3
2 @$ 1
2 n0 @$=@n
i! 1 =
cs
¡
5
3 @T kB cs
3 kT 0 cs
·
¸
° ¡ 1 @$
n0 @$
=
¡
°kB @T
T 0 @n
Now
° ¡1 m
1
=
° kB
cP
where cP is the speci®c heat at constant pressure (see, e.g, Lea and Burke equation 19.18),
and
¯
¯
¯
@$ ¯¯
@$ ¯¯
@½ ¯¯ @$
=
+
@T ¯ P
@T ¯ ½
@T ¯ P @½
¯
µ
¶
@$ ¯¯
½0 @$
=
+ ¡
@T ¯ ½
T 0 @½
Thus
¯
1 @$ ¯¯
cP @T ¯ P
Since !0 = 0; accelerations are zero to 1st order, so this is a constant pressure perturbation.
But if @$=@T at constant pressure is < 0; the the perurbation is unstable. The temperature
and density perturbations are out of phase (see equation 25 with P1 = 0), so if ±n > 0; then
i! 1 =
12
±T < 0: Then cold regions have
±$ =
¯
@$ ¯¯
±T > 0
@T ¯P
(both terms negative) and so cool faster. Hot regions expand and get hotter while cool
regions contract and get colder.
What about the other mode? Again with kJ = 0; we have:
µ
¶
1 2
k (2kT + 3k½ )
1
3
i! 1 = cs 3
=
c
k
+
k
s
T
½
10 2
k
5
10
3
which is also independent of k: Now
¯
¯
¯
@$ ¯¯
@$ ¯¯
@½ ¯¯ @$
=
+
@T ¯S
@T ¯½
@T ¯S @½
where S is the entropy. Now if entropy is constant (no heat -ow) then P _ ½° ; T _ ½°¡ 1 ;
and so ½ _ T 1=(° ¡1) ; and thus
¯
@½ ¯¯
1 ½0
=
¯
@T S
° ¡ 1 T0
So
¯
¯
@$ ¯¯
@$ ¯¯
1 ½0 @$
=
+
¯
@T S
@T ¯½ 2=3 T 0 @½
and so
µ
¶
cs 2 @$ 1
3 2 n0 @$=@n
i!1 =
+
5 3 @T kB cs
2 3 kB T 0cs
µ
¶
2
@$
3 n0 @$
=
+
15kB @T
2 T 0 @n
¯
2 @$ ¯¯
=
15k @T ¯
B
S
This mode is a sound wave, and we get instability if @$
along an adiabat is negative. Then
@T
the dense parts of the wave heat up, and the wave grows.
2.3
Discontinuities
Sharp boundaries often exist where two distinct -uid phases exist in pressure equilibrium
with each other. (Processes such as thermal conductivity broaden the boundary to a ®nite
width.) There is no velocity of material across such a boundary. Pressure must balance
across the boundary or the boundary will accelerate. To see this mathematically, integrate
the momentum equation across the boundary. Let x be a coordinate perpendicular to the
13
boundary. Then:
Z xb +"
xb ¡"
D~
v
½
dx
Dt
=
=
Z
Z
xb+ " ³
xb ¡"
xb+ " ³
xb ¡"
´
~ ¡ ½r©
~ g dx
¡ rP
´
~ ¡r
~ (½©g ) + © g r½
~
¡ rP
dx
= ¡ (P + ½© g )jxxbb +"
+ ©g
¡"
x +"
= ¡ (P + ½© g )jxbb ¡" +
Z
xb +"
xb ¡"
x +"
©g ½jxbb ¡"
~
r½dx
= P (x b ¡ ") ¡ P (xb + ")
where we have used the fact that ©g changes negligibly across the thin boundary. Thus to
make the left hand side zero (no acceleration) we need P to be the same on both sides of the
boundary.
2.3.1 Stability of contact discontinuities
Ref: p 222 Spitzer
Imagine a cold cloud in a hot surrounding medium in the galaxy. Gravity pulls the
cloud toward the galactic plane, and the motion is opposed by pressure gradients in the
surrounding inter-cloud medium. Letzs examine the stability of this situation.
We begin by writing linearized -uid equations for the material on both sides of the
boundary. For simplicity, we consider a plane interface. As usual we choose ~v0 = 0; and in
the unperturbed state we have:
~ = ½~g
rP
. Now Izm going to use the labels 1 and 2 to denote the two sides of the boundary, so Izll use
a prime to denote the small perurbation. Here g is not due to self gravity, so it does not
change when we perturb the -uid.
Continuity equation:
¡i!½01 + i½1 ~k ¢ ~v 01 = 0
(29)
Momentum equation:
¡i!½1~v10 = ¡i~kP 10 + ½01~g
(30)
with similar equations for medium 2. Now we add the boundary conditions. Letzs choose
a coordinate system with x running along the boundary and z upward, perpendicular to
the boundary. The boundary is at z = 0: Then ~g = ¡g^z; and for each -uid vz0 = v b ; the
boundary velocity. We can expect to ®nd a solution with no dependence on y:
Next we look for incompressible perturbations in which there is no density change,
½0 ´ 0: Then equation 29 becomes:
~k ¢ ~v 0 = 0
(31)
1
0
Solving equation 30 for P ; we have:
!½1 v b
P10 =
(32)
kz
14
Next we dot the momentum equation with ~k , using 31, to get:
k2 P 10 = 0
and since we expect P 10 6= 0; then k 2 = 0 , or
kx2 + kz2 = 0
and so either kx or kz must be imaginary. We expect the perturbations to get smaller as
we go away from the boundary, so we take kz = i·; with solutions of the form e¡ ·z for
z > 0 and e+·z for z < 0: Then kx = §i·: Then we have pressure balance at the perturbed
boundary:
DP
~ = ¡i!P 0 + v b (¡½g)
= ¡i!P 0 + ~vb ¢ rP
Dt
So the pressure change at the perturbed boundary is
vb ½g
±P = P 0 +
i!
!½1 vb
v b ½1 g
!½2 vb
v b ½2 g
=
+
=
+
i·
i!
¡i·
i!
Thus
µ
¶
½1 + ½2
!2
= g (½2 ¡ ½1 )
·
½ ¡ ½1
! 2 = g· 2
(33)
½2 + ½1
Thus ! 2 is real (boundary is stable) if ½2 > ½1 (denser -uid on the bottom) but is unstable
(! 2 < 0) if ½2 < ½1 (denser -uid
p on the top). This is the Rayleigh-Taylor instability. Note
that in the linear regime, j!j _ gk; and so the shortest wavelengths have the fastest growth
rates.
Thus we conclude that cold interstellar clouds should }drip} towards the galactic plane
with a timescale
s
1
½2 + ½1 1
t»
»
j!j
½2 ¡ ½1 gk
For a disk, we ®nd g » G¾;where ¾ is the surface density in the disk.
¡
¢
1012 2 £ 10 31 Kg
1012 M ¯
2
¾»
2 =
2 = 71 kg/m
¼ (10 kpc)
¼ (10 £ 3 £ 10 19 m)
If ±½=½ » 10; for a cloud of radius 1 pc this time is less than or of the order of
s
3 £ 1016 m
t »
¡
¢
= 3 £ 10 11 s
¡
11
10 (6:7 £ 10
m3 =kg ¢ s2 ) 71 kg/m2 2¼
» 10 4 y
which is rather short by astronomical standards.
Additional applications include extragalactic radio sources in which the clouds of
radio-emitting plasma are decelerated by the surrounding, denser, intergalactic medium. The
instability may be important as a mechanism for driving turbulence, and hence acceleration
15
of particles in the radio source. In accreting magnetic neutron stars, the instability is
necessary to allow the incoming material to reach the neutron star surface.
2.4
Discontinuities with ~
v normal to the boundary
2.4.1 Shock waves
Reference: Spitzer page 218
Consider a sound wave propagating through a -uid. From equation 17 we have:
~kP1
~v1 =
!½0
and from equation 16
~k ¢ ~v1
k 2 P1
P1
½1 = ½0
=
= 2
2
!
!
cs
Thus all the -uid quantities oscillate in phase. However, the sound speed is
s
°P
cs =
½
and as the wave grows, the sound speed does not remain constant:
s
s
° (P + ±P )
°P (1 + ±P =P )
cs + ±cs =
=
½ + ±½
½ (1 + ±½=½)
µ
¶µ
¶
1 ±P
1 ±½
= cs 1 +
1¡
2 P
2 ½
µ
¶
1
±P
±½
= cs + cs
¡
2
P
½
µ 2
¶
1
cs ±½ ±½
= cs + cs
¡
2
P
½
µ
¶
1
°P ±½ ±½
= cs + cs
¡
2
P½
½
1
±½
= cs + cs (° ¡ 1)
2
½
Since ° > 1 always, the sound speed in the compressed part of the wave is higher than in
the rari®ed part. The wave starts to steepen, and ultimately }breaks}. This is a shock wave.
16
After a time t; the crest of the wave has advanced by a distance
° ¡ 1 ±½
±x = t±cs = cs t
2 ½
t ° ¡ 1 ±½
= ¸
T 2 ½
The steepening is noticeable when ±x » ¸=2; or when
±½
T
½
where T is the wave period. This time can become quite short as ±½=½ approaches 1.
Viscosity begins to get important when the velocity shear becomes large. Viscosity acts
to stabilize the steepened wave front into a zone whose thickness is of order the collisional
mean free path. Usually this is small compared to other scales of interest in the problem, so
we shall ignore this thickness and regard the shock as a sharp discontinuity.
The shock usually travels through the -uid in the }lab} frame. The analysis is easiest if
we work in the frame moving with the shock. In this frame, the -uid properties are
t » (° ¡ 1)
upstream
P1 ; ½1
¡! ~v1
shock
j
j
downstream
P 2 ; ½2
¡! ~v2
We also assume a steady state, and take ~g ´ 0: Then the -uid equations for a plane shock
are:
Continuity (equation 1):
~ ¢ (½~v ) = 0 = d (½v)
r
(34)
dx
and momentum (equation 5)
³
´
~ ~v = ¡rP
~
½ ~v ¢ r
½v
dv
dP
+
dx
dx
17
=
0
(35)
and the energy equation (8)
µ
¶
D 3
¡ ¡¤
kT Dn
P Dn
kT =
+
= ¡$ + 2
Dt 2
n
n Dt
n Dt
(36)
We can also write an equation for the speci®c entropy of the gas, since dS = Q=T and thus
DS
dS
1 dQ
1
=v
=
=¡ $
(37)
Dt
dx
T dt
T
¡ 3=2 ¢
The expression for the speci®c entropy S (entropy per particle)is S = (k=m) ln T =½
The goal is to write all these equations in conservative form, that is in the form
divergence (something) = 0
Equation 34 is already in this form. We can put the momentum equation 35 into the right
form by adding a term that is identically zero:
d
dv
dP
v
(½v) + ½v
+
= 0
dx
dx
dx
¢
d ¡ 2
½v + P
= 0
(38)
dx
This is Bernoullizs equation. Now we can integrate equations 34 and 38 across the shock, to
get:
Z 1
d
(½v) dx = 0
2 dx
or
[½v] = ½1 v1 ¡ ½2 v 2 = 0
(39)
and similarly:
£ 2
¤
½v + P = 0
(40)
Digression on the energy equation
Letzs look at the rate of change of total energy:
µ
¶ µ
¶
µ
¶
@ 1 2 3 kT
1 2
@½
@~v
@U
½v + ½
=
v +U
+ ½ ~v ¢
+
(41)
@t 2
2 m
2
@t
@t
@t
where U =
3 kT
2 m
is the speci®c internal energy. Then from equation 36
@U
DU ³ ~ ´
$
P D½ ³ ~ ´
=
¡ ~v ¢ r U = ¡ + 2
¡ ~v ¢ r U
@t
Dt
m
½ Dt
³
´
$
P~
~ U
= ¡ ¡ r
¢ ~v ¡ ~v ¢ r
m
½
Now we use equations 1, 5 to simplify the RHS of equation 41:
µ
¶
µ µ ³
¶
´
³
´ ¶
1 2
1~
$
P~
~
~
~
¡
v + U r (½~v ) + ½ ~v ¢ ¡ ~v ¢ r ~v ¡ rP ¡
¡ r ¢ ~v ¡ ~v ¢ r U
2
½
m
½
µ
¶
µ µ 2¶
¶
µ
¶
1 2
v
1
$
P~
~
~
~
~
= ¡
v + U r (½~v ) ¡ ½~v ¢ r
+ rP + rU ¡ ½
+ r ¢ ~v
2
2
½
m
½
Now
1~
P~
~ + Pr
~ ¢ ~v = r
~ ¢ (~v P )
½~v ¢ rP
+½ r
¢ ~v = ~v ¢ rP
½
½
18
So
¶¶
µ ¶
1 2
P
$
RHS
½~v
v +U +
¡½
2
½
m
µ µ
¶¶
µ ¶
1
$
~ ½~v
= ¡r
v2 + H
¡½
2
m
where H is the enthalpy.
¡ ¢ This is the conservative form that we want. We can write the
~ ¢ F~ where F~ is the radiative -ux vector. Thus in steady state
radiative loss term ½ $
=r
m
where the total energy does not change in time (@=@t = 0) we can integrate across the shock
to get:
· µ
¶¸
1 2
½v
v +H
= [Fx ]
2
which reduces to
·
¸
1 2
v + H = [Fx ]
(42)
2
since [½v] = 0: Also note that
P
° P
H=U +
=
½
° ¡1 ½
Let
mU
¿ cool ¼
$
be the cooling time behind the shock front. Then the distance material travels before cooling
substantially is
¸ cool = v2 ¿ cool
(43)
~
= ¡r
µ
µ
Case in which cooling is unimportant in the shock itself.
This case is described by
¸cool À shock thickness and other scales of interest
In this case we ignore cooling (and the jump in Rx ) while analysing the shock. De®ne the
Mach number
v1
v1
M ´
= p
c1
°kT 1 =m
Then from equation 39:
½
½
v2 = 1 v 1 = 1 M c1
½2
½2
and from equation 40
µ
¶
µ
¶
P2
P1
½2 v22 +
= ½1 v12 +
½2
½1
19
Now eliminate the vzs:
õ
!
¶2
½1
P 2 ½1 c21
½2
M c1
+
½2
P 1 ½2 °
õ
!
¶2
½2
½1
P 2 ½1 1
M
+
½1
½2
P 1 ½2 °
½1 2 P2
M +
½2
P1
µ
¶
½
°M 2 1 ¡ 1
½2
°
µ
M 2 c21
c2
+ 1
°
=
½1
=
M2 +
=
°M 2 + 1
=
P2
¡1
P1
=
1 2
° P1
v1 +
2
° ¡ 1 ½1
¶
1
°
(44)
Next we work on equation 42:
1
2
µ
1 2
° P2
v2 +
2
° ¡ 1 ½2
¶2
½1
° P2 ½1 c21
M c1
+
½2
° ¡ 1 P1 ½2 °
µ
¶2
½1
2 P2 ½1
M
+
½2
° ¡ 1 P1 ½2
Ã
µ ¶2 !
½1
M2 1 ¡
½2
=
1 2 2
° c21
M c1 +
2
° ¡1 °
=
M 2 c21 +
=
2
° ¡1
µ
2
° ¡1
P2 ½1
¡1
P1 ½2
¶
(45)
Now divide equation 45 by equation 44, and write ½2 =½1 = e
½; and P2 =P1 = Pe: Then
³
´
³
´
¡2
2
e =e½ ¡ 1
M 2 1 ¡ e½
P
°¡ 1
³
´ =
¡
1
Pe ¡ 1
°M 2 1 ¡ e
½
³
´³
´
´
2° ³ e
Pe ¡ 1 1 + e
½¡1
=
P =e½ ¡ 1
° ¡1
So
µ³
¶ ³
´
´
2° 1
2°
Pe
1 + e½¡ 1 ¡
= 1+e
½¡1 ¡
° ¡1e
½
°¡1
and so
°+ 1
1 ¡ (°¡
½
e
1)
Pe =
(46)
°+1
½ ¡ (°¡1)
e
20
Now substitute back into equation 44 to get:
µ
¶
°+1
1 ¡ (°¡1)
½
e
1
°M 2 1 ¡
=
¡1
°
+1
½
e
½ ¡ (°¡ 1)
e
³
´
°+1
°+ 1
µ
¶
1
¡
½
e
¡
½
e
¡
e½ ¡ 1
(°¡1)
(°¡ 1)
°M 2
=
°+1
½
e
e½ ¡ ( °¡1)
³
´
°+1
°+ 1
1 + (°¡1)
¡ (°¡
+
1
e½
1)
=
°+1
e½ ¡ ( °¡1)
2 ° °¡1 (1 ¡ e
½)
=
and so
°M
and therefore
e½
2
µ
=
=
½¡
e
°+1
(°¡1)
¶
°+1
°
¡e
½ =2
½
e
°¡1
°¡1
M2
³
2
°¡1
°+1
°¡1
´
+ M2
° +1
1
³
´
° ¡1 1+ 2 2
M ( °¡1)
(47)
So for large values of M (M 2 À 2= (° ¡ 1) = 3)
v1
½
°+1
= 2 !
=4
v2
½1
°¡1
where the numerical value is for ° = 5=3: Finally, from equation 46:
³
´
°+1
2
2
2 ° +1
+
M
¡
M
°¡1
(°¡1)
° ¡1
³
´
³
´
Pe =
°
+1
°+1
2
M 2 ° ¡1 ¡ ( °¡1) °¡1 + M 2
h
i
2
2
2 (° ¡ 1) + M 2 (° ¡ 1) ¡ (° + 1)
=
¡2 (° + 1)
2 (° ¡ 1) ¡ 4°M 2
=
¡2 (° + 1)
2°M 2 ¡ (° ¡ 1)
(48)
°+1
Thus while the increase in density across a shock remains bounded, the pressure increase
can be large if M is large. Equations 47 and 48 are the Rankine-Hugoniot relations for the
=
21
shock. From these relations we can also obtain an equation for the temperature jump:
0
1
2
2
T2
P 2 ½1
2°M 2 ¡ (° ¡ 1)
°¡1 + M
@
³
´A
=
=
T1
P 1 ½2
° +1
M 2 °° +1
¡1
¡
¢¡
¢
2
2
2°M ¡ (° ¡ 1) 2 + (° ¡ 1) M
=
M 2 (° + 1)2
¡
¡
¢¢
+2°M 4 (° ¡ 1) ¡ 2° + 2 ¡ M 2 °2 + 1 ¡ 6°
=
M 2 (° + 1)2
¡
¢
¡
¢
4
2 °M ¡ 1 (° ¡ 1) ¡ M 2 ° 2 ¡ 6° + 1
=
M 2 (° + 1)2
which becomes for large M
T2
2° (° ¡ 1) 2
5
!
M2
2 M =
T1
16
(° + 1)
(49)
(50)
The value of M
By the
¡ second¢ law of thermodynamics, the entropy must increase across
the shock. Since S _ ln T 3=2 =½ ; then T 3=2 =½ _ P 3=2 =½5=2 must increase, and so
µ ¶°
P2
½2
>
P1
½1
Thus
0
2
2°M ¡ (° ¡ 1) @
>
°+1
Set ° = 5=3: Then we need
10M 2 ¡ 2
8
5M 2 ¡ 1
4
Ã
M2
2
°¡1
³
° +1
°¡ 1
´ 1°
+M2
A
¡ ¢ ! 5=3
M 2 82
>
3 +M2
µ
¶ 5=3
4M 2
>
M2 + 3
22
2.5
2
1.5
1
0.5
0
0.2
0.4
0.6
M0.8
1
1.2
1.4
Solid = LHS, Dashed = RHS. Both sides = 1 at M = 1:
Thus we ®nd that we need M > 1 for a valid solution.
A shock wave can only occur when material approaches the shock at greater than the
sound speed. In the shock, collisions between particles convert directed kinetic energy of
-uid -ow into random kinetic energy of thermal motion.
Cooling shocks
When ¸ cool is not very large, the structure of the shock looks like:
We can compute the jump conditions across the entire shock plus cooling region, if we
pick x 2 where the temperature has returned to its pre-shock value. The previous energy
equation is replaced by the ideal gas law at constant temperature T : The sound speed is
23
c2s = kT =m = P=½ and is the same in state 1 and state 2. Then equation 40 becomes:
¡
¢
¡
¢
½1 v 21 + c2s = ½2 v 22 + c2s
So:
¢
½1 ¡ 2
v 1 + c2s =
½2
So
Ã
v 21
µ
M2 + 1
½
e
¡
¢
e2 ¡ M 2 + 1 e
½
½ +M2
and the solution is:
M2 + 1 §
½1
½2
¶2
=
+
c2s
M2
e½2
!
+1
= 0
p
M 4 + 2M 2 + 1 ¡ 4M 2
2 ¢
¡
M2 + 1 § M 2 ¡ 1
=
2
With the minus sign we would get e½ = 1; (no shock) so the correct solution is:
e½
=
e½ = M 2
Thus the density increase can be very large when cooling occurs.
2.4.2 Supernovae
References: Spitzer, pg 255
Review article by Chevalier, ARAA 15, 175, 1977
Details of the initial supernova explosion are not well known, since we only see them
after the explosion, and usually not immediately after. We expect a blast of radiation and
ejection of some material at high energy. Since this material is moving at v À sound speed
of surrounding material, a shock is formed.
Phase I.
Shock velocity is approximately constant. The temperature behind the shock
can be found from the Rankine-Hogoniot relations, speci®cally equation 49 or 50:
5
5 v 21
M 2 T1 =
16
16 °k=m
3 2m
=
v
16 1 k
Thus by measuring T 2 ; we can determine v1 : The temperature is usually determined from
x-ray measurements. Typical kinetic energy of a supernova is estimated to be approximately
4 £ 10 50 erg, and the mass of the ejecta Me j » 0:25 M ¯ for a Type I supernova (the numbers
are rather uncertain). Thus we ®nd
s
r
1
2E
2 (4 £ 1050 erg)
2
E = Mv ) v »
=
= 10 9 cm/s = 104 km/s
2
M
0:25 £ 2 £ 10 33 g
T2
=
24
and thus
3 18
2
T2 »
10 (cm/s)
16
µ
1:7 £ 10 ¡24 g
1:4 £ 10¡ 16
qerg/K
For a Type II, E » 10 51 erg, Mej » 5 M ¯ ; so v »
¶
= 2 £ 109 K
2(1051 erg)
5£ 2£1033 g
= 4: 5 £ 103 km/s and
T » 4 £ 108 K.
This phase lasts until the matter swept up by the outward moving shock is of the same
order as the ejected mass. At this point the shock begins to decelerate. This happens when
4 3
¼r ½ = M ej
3 s 1
So for a Type I, and for surrounding ISM density of 1 cm¡3 ;
Ã
! 1=3
3 £ 0:25 £ 2 £ 1033 g
rs »
¡
¢
= 4 £ 1018 cm » 1 pc
4¼ 1:7 £ 10¡ 24 g/cm 3
The time to reach this radius at v » 109 cm/s is
t » 4 £ 109 s » 130 y
which is very short!
The Sedov phase
The following phase is called the Sedov phase. We shall neglect
radiation losses, which we can show a posteriori are small. Then the energy behind the
shock wave equals the initial supernova energy, and thus the energy density is
3
u=
E
4¼r3s
The mean pressure inside is
2
E
P = nkT = u =
3
2¼rs3
The pressure right behind the shock is some numerical factor of order 1 times this mean
pressure. (Numerical simulations typically ®nd this numerical factor to be about 2.) Thus,
from equation 48 for large M :
P2 =
5 2
5
v12
kT 1
3
M P1 =
½
= ½1 v12
4
4 °kT1 =m 1 m
4
So
3
E
½ v2 =
4 1 1
2¼r3s
and thus
µ
¶1=2
drs
2 E
v1 =
=
rs¡3=2
dt
3 ¼½1
We can integrate this equation to get:
µ
¶ 1=2
2 5=2
2 E
r
=
t
5 s
3 ¼½1
25
(51)
or
rs
or, expressing t in years:
¶ 1=5 µ ¶ 2=5
5
t2=5
2
¡
¢
= 8 £ 10 14 cm t2=5
=
µ
2 E
3 ¼½1
(52)
(53)
¡
¢
2=5
rs = 8 £ 10 17 cm t2=5
yr = (0: 27 pc) t yr
where we used the numbers for a type I supernova, and an ISM density of 1 cm¡ 3 : Notice
that rs has a very small dependence (1/5 power) on both E and ½1 : The temperature behind
the shock decreases with time:
¡
¢2
drs
v1 =
= 8 £ 10 14 cm t¡ 3=5
dt
5
and thus
¢2 4 ¡6=5
3 mv 2
3 m¡
T2 =
=
8 £ 10 14 cm
t
16 k
16 k
25
= 2: 3 £ 10 11 t¡6=5
yr
Remember this is valid for t & 130 yr. Equivalently, we can ®nd T as a function of the
shock radius:
µ
¶ ¡3
³
´¡3
rs
11
T2 = 2: 3 £ 10 11 t2=5
=
2:
3
£
10
yr
0:27 pc
µ
¶¡ 3
rs
= 4: 5 £ 10 9
1 pc
The temperature increases toward the interior because the Mach number of the shock was
higher as it passed through that material. Numerical calculations give T » r¡4:3 ; in fair
agreement with this estimate. The same simple arguments suggest a constant density in the
interior since ½2 = 4½1 for strong shock. But that would lead to strong pressure gradients,
which would change the density distribution. If the pressure in the interior were constant,
then we would have to have ½ _ r3 (highest density right behind the shock). In fact the
density increases more steeply than this, because the pressure is actually greatest right
behind the shock.
We have observational evidence supporting this picture: both x-ray and radio maps tend
to show SNR as shells, with the emission concentrated in a thin circular region on the sky. If
the emission comes from a spherical shell of radius r and thickness 4r; then we see emission
from a region
of length ` » 2r
cos µ » (r ¡ 4r) =r = 1 ¡ 4r=r = 1 ¡ µ2 =2;
p
p sin µ where p
so µ » 24r=r and ` » 2r 24r=r = 2 2r4r _ r1=2 : Thus the emission measure
a very strong dependence on r:
n2 ` » r6 r1=2 » r6:5
Isothermal phase.
As expansion continues and T drops, line emission becomes more
important as the ionization state decreases. The amount of energy radiated away becomes
26
signi®cant. For T < 10 6 K, colling dominates the SNR evolution.
µ
¶ ¡3
¡
¢1=3
rs
9
4: 5 £ 10
< 106 for rs > 4:5 £ 103
pc = 17 pc
1 pc
In this phase we can consider the shock to be isothermal. The compression is very high,
and the velocity behind the shock is very low. Thus a thin, dense, shell moves outward.
The thickness of the shell is ¸ cool (equation 43). The shock is no longer driven by internal
pressure, but merely conserves momentum. Thus this phase is also called the snowplow
phase. By now the mass inside the shell is mostly swept-up interstellar medium:
4
M » ¼rs3 ½1
3
and so momentum conservation gives:
4 3 drs
¼r ½
= M t vt = constant
3 s 1 dt
and thus:
4
r4
¼½1 s = M t vt t
3
4
µ
¶1=4
3M t vt
rs =
t1=4
(54)
¼½1
In the ®nal phase, the interior cools adiabatically and
T V ° ¡1 = constant
so
¡ ¢¡ 2=3
T _ r3
= r¡ 2
27
When the heated gas interior to the shell pushes the shell outward, (a light -uid pushing
on a heavier one) the shell becomes Rayleigh-Taylor unstable (section 2.3.1). In addition
the shell is thermally unstable (section 2.2.3). Both effects cause the shell to break up into
clumps.
Effects of supernovae
1. Heating of the ISM. As the shock wave sweeps up the ISM it also heats it.
² Shocks may overlap to form }tunnels} of hot gas that may occupy as much as 30% of the
volume of the spiral arms in a galaxy (See eg Smith, Ap J 211 p 404)
² Conduction may evaporate cold clouds into the hot, }tunnel} region. (McKee and
Ostriker, 1975)
² X-rays and cosmic rays from supernovae can heat and ionize gas at large distances from
the supernova itself. (Salpeter, 1976, Lea and Silk 1974)
² Heating by Supernovae may lead to the formation of galactic winds that blow ISM into
the intergalactic space (Mathews and Baker 1971 Ap J 170 241)
² Supernovae lead to fast-moving clouds whose energy is ultimately dissipated as heat in
the ISM.
2Relativistic particles: Particles generated in the SN itself are trapped in the remnant,
because streaming particles generate plasma waves that scatter the particles. Particles
thus suffer adiabatic expansion losses. Acceleration in or behind the shock wave has
been discussed but remains controversial.
Radio properties:
Radio observations show 3 classes of SN remnants:
1. Shells with tangential magnetic ®eld (compressed interstellar ®eld?)
2. Shells with radial magnetic ®eld (stretched out stellar ®eld?)
3. Plerions- ®lled remnants. These are usually assumed to be powered by a pulsar that
ejects relativistic particles into the remnant. The proptypical plerion is the Crab Nebula.
2.4.3 Ionization fronts
Reference Spitzer page 246. Also Shu
Regions of ionized gas (HII regions) form around a source of ionizing radiation. The
classical HII region is powered by a bright, young star (O or B type) that is hot enough to
produce a substantial amount of its radiation in the UV. Such stars have short lifetimes
(107 years or so), so an HII region is a transient event.
Formation of an HII region
Ref Stromgren Ap J 89, 526, 1939
28
A photon with hº > 13:6 eV (¸ < 912 Å) can travel a distance 1=n¾ in neutral gas,
where ¾ is the cross section for ionization of a neutral H atom.
µ
¶3
¡
¢
¸
¾ = 6 £ 10 ¡18 cm 2
(55)
912 Å
Therefore the mean free path is
µ
¶3
1
1018 912 Å
=
cm
n¾
6n
¸
µ
¶3
1
912 Å
=
pc
18n
¸
As radiation streams out from the star it ionizes gas in a small region. The following photons
can travel almost unimpeded through the }hole} created by their predecessors, and then
penetrate only a small distance into the surrounding material. In this way the photons }eat}
their way out into the ISM, forming an HII region. The region is fully formed when all of
the -ux from the star is used up reionizing those atoms that recombine within the HII region.
Heating goes along with ionization, since the photon energy is hº = 13:6 eV +4E ;
and the extra energy 4E of the liberated electron ends up as thermal kinetic energy in the
HII region. Thus the pressure in the HII is greater than in the HI, and the hot gas expands
outward. The expansion speed (approximately the sound speed in the hot HII) is greater
than the sound speed in the surrounding cold HI and so a shock wave forms.
In a steady state we can ®nd the radius of the HII region (the }Stromgren sphere}) by
noting that all the photons emitted by the star are used to reionize the atoms that recombine
within the sphere.
Notation:
n( HII )
x =
where n = n (HI) + n (HII) is assumed uniform in the initial undisturbed medium
n
¡13
® =
recombination coef®cient
cm3 s¡1 at 10 4 K
R to levels above R1: ® = 2:5 £ 10
¾ = mean ionization coef®cient = ¾ (º) I (º) dºR= I (º) dº where I (º ) is the stellar spectrum
r
¿ =
optical depth = 0 n¾ (1 ¡ x) dr
Then the rate of change of ionization is due to net ionizations minus recombinations:
dx
© ¡¿
=
e ¾ (1 ¡ x) ¡ ®nx2
(56)
dt
4¼r2
where
Z 1
¾ (º ) L (º )
©=
dº
¾
hº
13:6 e V
is the photon -ux above the ionization edge. Thus in a steady state we have:
1¡ x
4¼®nr 2
=
(57)
x2
¾©e¡ ¿
which is a horrible equation for x: (Remember that ¿ also contains x:)
Now de®ne rS by the condition that the sphere be fully ionized, and ionizations equal
recombinations in the sphere:
4 3 2
¼r n ® = ©
(58)
3 S
29
Then equation 57 becomes:
Then if r ¿ rS ; ¿ ¿ 1 and
1¡x
3 (r=rS )2
=
x2
¾nrs e¡ ¿
¼ 1: Since
Z r
¿ = n¾
(1 ¡ x) dr
e¡¿
(59)
0
then, using equation 59,
d¿
3 (r=rS )2 2
= n¾ (1 ¡ x) =
x
dr
rs e¡¿
and so
¡¿
e
d¿
= 3
µ
= x2 d
Now let r=rS = z: Then
So for r ¿ rS ; x ¼ 1 and
r
rS
µ
¶2
r
rS
x2
¶3
dr
rS
de¡ ¿
= ¡x 2
dz
de¡ ¿
¼ ¡1 ) e¡ ¿ ¼ 1 ¡ z
dz
Thus 1 ¡ x remains small until e ¡¿ becomes small, i.e. z ¼ 1: In fact we need
¾ nrse¡ ¿ ¼ 1
or
1¡z ¼
So the thickness of the ionization front is
1
¾nrs
1
1
=
¿S
¾nrs
and thus 4r = rS 4z = 1=n¾; as predicted above.
Typical numbers for an O star are © » 10 48 s¡ 1 , which gives
µ
¶ 1=3 µ
¶ 1=3
3©
3 £ 1048 s¡1
30 pc
rS =
=
= 2=3
4¼n2 ®
4¼n2 (2:5 £ 10¡ 13 cm 3 s¡1 )
n
cm 2
4z »
(60)
Dynamical evolution
The HII region is not in equilibrium dynamically since the
pressure inside exceeds the pressure outside. Letzs investigate its evolution.
First, the size of the region is changing as the ionization front eats its way out. Let ri be
the radius of the ionization front at time t: Then
µ
¶
d 4 3
4
¼r n
= © ¡ ¼ ®n2 r3i
dt 3 i
3
rate of increase in # of ions = ionization rate - recombination rate
30
Using equation 58, we have:
1 dz
= 1¡z
®n dt
which has the solution
µ ¶3
ri
= z = 1 ¡ e ¡®nt
rs
The volume grows linearly for t ¿ 1=®n
¡ and¢asymptotically approaches the steady state
solution. The time scale ¿ = 1=®n » 104 y =n is independent of the ionization rate ©:
The timescale is rather short. Then
dri
d
rs
dz
®nrs 1 ¡ z
= rs z 1=3 = z ¡2=3
=
dt
dt
3
dt
3 z 2=3
¡t=¿
rs
e
=
(61)
¡
3¿ 1 ¡ e¡t=¿ ¢2=3
which can exceed the speed of sound in the HII region if rs (and thus ©) is large. The gas is
essentially motionless as the front expands through it. With T I I ¼ 10 4 K, cI I ¼ 10 km/s.
Then with n = 10 cm ¡3 and rs ¼7 pc (equation 60),
vi =
for
dri
7 £ 3 £ 10 18 cm 1 ¡ z
=
= 10 £ 10 5 cm/s
dt
3 £ 10 3 £ ¼ £ 107 s z 2=3
(1 ¡ z)
z
ri
= z 2=3 4: 5 £ 10¡ 3
= 0: 99547
= 0: 9985rs
i.e. the HII region is almost formed.
Note that the pressure inside the HII region greatly exceeds the presure outside, so the
gas expands outward into the neutral surroundings. We shall neglect radiation pressure here,
because it is small compared with the difference in gas pressures. (You should verify this.)
Jump conditions across the ionization front
Again we work in the frame moving with
the front. In deriving the jump conditions across a planar shock, nothing we did depended
on the fact that the discontinuity was a shock per se, so we get the same jump conditions
here.
HII
½2 ; v2 ; T 2
ionization front
j
HI
½1 ; v1 ; T 1
½1 v1 = ½2 v2
(62)
and
¡
¢
¡
¢
½1 v 21 + c21 = ½2 v 22 + c22
p
Here c1 is the isothermal sound speed kT 1 =m: With T1 ¼ 100 K
s
(1:38 £ 10 ¡16 erg/K) (100 K)
c1 =
= 9: £ 10 4 cm/s = 0:9 km/s
1:67 £ 10¡ 24 g
31
(63)
and in the HII, T2 ¼ 104 K, so
s
(1:38 £ 10 ¡16 erg/K) (10 4 K)
c2 =
= 1: 3 £ 10 6 cm/s = 13 km/s
0:5 £ 1:67 £ 10 ¡24 g
These temperatures are enforced by the radiative terms in the energy equation.
Now use equation 62 to eliminate ½2 from equation 63:
¡
¢
¢
v1 ¡ 2
½1 v21 + c21 = ½1
v + c22
v2 2
then solve for v2 :
¡ 2
¢
v1 + c21
2
v2 ¡
v 2 + c22 = 0
v1
s
¡ 2
¢
µ
¶2
v1 + c21
1
c2
v2 =
§
v1 + 1
¡ 4c22
2v 1
2
v1
Real solutions exist only if the quantity inside the square root is positive:
c21
v1
2
v1 ¡ 2c2 v1 + c21
v1 +
With an equals sign, the quadratic
p
and v D = c2 ¡ (c22 ¡ c21 )
v21
¡ 2c2 v1 + c21
(64)
> 2c2
> 0
= 0 has solutions: vR = c2 +
(65)
p
(c22 ¡ c21 )
(In the plot I have multiplied c1 by 4 for clarity.)
Thus our problem has solutions only for v1 < vD or v 1 > vR : Since (c1 =c2 )2 =
32
T1 =T 2 ¿ 1;
vD
vÃ
u
µ ¶2 !
u
c1
t
= c2 ¡ c2
1¡
c2
Ã
Ã
µ ¶2 !!
µ ¶
1 c1
1 c1
¼ c2 1 ¡ 1 ¡
=
c1 ¿ c1
2 c2
2 c2
In fact for our numbers,
vD »
1 0:9
0:9 km/s = 3 £ 10¡ 2 km/s
2 13
(66)
and
v R ¼ 2c2 = 26 km/s
(67)
Since ½v = constant, small v corresponds to large ½: Therefore D stands for }dense} while
R stands for }rare}.
The forbidded region arises because the sound speeds are imposed by external conditions,
unlike shocks, where jump conditions determine c2 and all values of v 1 > c1 are allowed.
Also note that
vR vD = c21
so that v R=c1 > 1 while v D =c1 < 1:
All R¡type ionization fronts have v1 > vR > c1 and thus the front moves supersonically.
Conversely, D¡type fronts have v 1 < vD < c1 ; and so the front moves subsonically.
Using equation 64 to solve for v2 ; we get
õ
!
¡ 2
¢
¶2
v 1 + c21
1
c21
2
v2R v 2D =
¡
v1 +
¡ 4c2
4v1
4
v1
= c22
(68)
So again one solution is subsonic in the HII region, and the other is supersonic. The solution
for v2 is:
8
9
vÃ
>
µ ¶2 u
µ ¶ 2 !2
µ ¶2 >
<
u
v2
½
1
c1
c1
c2 =
= 1 =
1+
§t 1+
¡4
v1
½2
2>
v1
v1
v1 >
:
;
D-type front
v 1 < c1 :
With the plus sign, we ®nd that ½1 =½2 is large since c1 =v1 is and the square root is > 0:
This is a strong D-type ionization front. Then
8
9
vÃ
!2
>
µ ¶2 u
µ
¶
µ
¶
>
2
2
<
u
v2
1 v1
c1
c1
c2 =
=
1+
+t 1+
¡4
c2
2 c2 >
v1
v1
v1 >
:
;
Since v1 +
c2
1
v1
> 2c2 (equation 65)
v2
>1
c2
33
So with the strong D¡type front, the -ow is supersonic in the HII.
With the minus sign, ½1 =½2 is small. This is a weak D-type ionization front. By equation
68, the -ow is subsonic in the HII.
R-type v1 > c1
Thus
Equation 67 shows that in fact we have v 1 À c1 ; and also v 1 > c2 :
v2
v1
=
=
8
9
s
µ ¶2 =
½1
1<
c2
¼
1§ 1¡ 4
½2
2:
v1 ;
Ã
Ã
µ ¶ 2 !!
1
c2
1§ 1 ¡2
2
v1
With the plus sign, we have
v2
½
= 1 =1¡
v1
½2
µ
c2
v1
¶2
¼1
This is a weak R-type front. We also have
v2
v1
c2
=
¡
>1
c2
c2
v1
so the -ow is supersonic in the HII.
With the minus sign,
This is the strong R-type.
v2
½
= 1 =
v1
½2
µ
c2
v1
¶2
<1
v2
c2
=
<1
c2
v1
so in this case the -ow is subsonic in the HII.
Summarizing, we have:
Type
sign v 1 =c1 v 2 =c2
Weak R
+
>1
>1
Strong R ¡
>1
<1
Weak D ¡
<1
<1
Strong D +
<1
>1
So we trace the evolution as follows.
Initially v 1 > c1 : The front is R¡type, and we have already noted that gas will not be
accelerated behind the front since it passes by rapidly. So it is a weak R: But v1 = dri=dt
decreases with time ( equation 61), and so approaches vR : But v1 cannot decrease below v R:
Note that if v 1 = v R; then v 2 = c2 : Signals can begin to propagate up to the front (v2 was
previously > c2 ) bringing news of the high pressure behind. A pressure wave catches up to
and passes through the ionization front, where it steepens into a shock.
Now the structure looks like:
34
j
dense, hot HI
j
HI
j
Ã
j
Ã
j
v1
j
v0
ionization front
shock wave
The front becomes weak D¡type. This happens when formation of the Stromgren
sphere is essentially complete.
So how is it that material still -ows through the front? Notice that the number of
particles inside the sphere is
4
©
N = ¼r3s n =
3
®n
(equation 58). Thus N increases as n decreases, i.e. as the region expands. With a lower
particle density, there are fewer recombinations. As expansion continues, the pressure in
the interior falls and the shock weakens. The expansion slows.
HII
Ã
v2
The ®nal phase
To analyze this phase, assume
1. Uniform density ½I I in the HII region
2. Constant temperature T II in the HII region
Then as the region expands, let r be the coordinate of a particle within the region. The
total mass interior to r remains constant as the region expands, so
½II r 3 = const
Thus
Right behind the front, r = ri;
1 d½II
3 dr
vI I
=¡
= ¡3
½I I dt
r dt
r
1 d½I I
v II
= ¡3
½II dt
ri
Now from the de®nition of rs ;(equation 58),
(69)
n2II r3i = constant
So
2 d½II
3 dri
vi
=¡
= ¡3
(70)
½II dt
ri dt
ri
Then comparing equations 69 and 70, we ®nd
1
vI I = v i
2
Thus the speed with which matter -ows through the front is
1
v 2 = v i ¡ vII = vi
2
Now look at the jump conditions across the whole structure (shock plus ionization front).
Neglect the small difference between v s and v i: Then v0 = vs:
35
Ã
v2
½II
j
j
j
ionization front
The jump conditions are:
But v s À cI ; so
j
j
j
shock
Ã
v0
½I
¡
¢
¡
¢
½I vs2 + c2I = ½I I v 22 + c2I I
½
c2I I I I
½I
vs2
¼
vs2
= c2I I
Ã
1
1+
4
µ
vs
cII
¶2 !
½I I
1
½I (1 ¡ ½II =4½I )
(71)
3=2
Now since ½2I I r3i =const, and initially ½II = ½I ; we set ½I I =½I = (ri0 =ri)
where ri0 is
a constant that is close to rs : Then since v s ¼ v0 ¼ v i; and ½I I ¿ ½I ; equation 71 gives:
³ r ´3=4
dri
i0
¼ cI I
dt
ri
and integrating gives:
4 7=4
3=4
r
= cII ri0 t + constant
7 i
or
µ
¶ 4=7
ri
7 cI I t
= 1+
ri0
4 ri0
in the ®nal phase, where t is measured from the time at which the shock forms. The
³
´¡6=7
expansion stops when PI I = PI : But ½II =½I ¼ 1 + 74 crIIi0t
» 0:1 for t = an O starzs
lifetime, and so PI I =P I = ½II T II =½I T I ¼ 10: Thus the expansion never stops. Instead,
ionization ceases and recombination ends the HII regionzs life.
Summary
1. Dynamical evolution of the HII region. Rapid expansion of ionization front (weak
R-type) to ri » 0:999rs:
2. Formation of shock ahead of ionization front and slow expansion into surrounding
region. Front has become weak D-type.
3. Recombination
Notice that some of the UV energy (a few percent) ends up as kinetic energy of the ISMthe star is a heat engine!
2.5
Flow under the in-uence of gravity: accretion and winds
36
2.5.1 Supersonic -ow past a gravitating object.
Recall the steady state version of the momentum equation (5):
~
~ v = ¡ rP + ~
v
~ ¢ r~
g
½
These terms are of order:
v2
c2
» s +~
g
L
L
2
2
So if v À cs ; we can ignore the pressure term to ®rst order. Then we have:
µ 2
¶
v
~
r
+ Ág = 0
2
or total energy (kinetic plus potential) is conserved in the -ow. The -uid -ows like a set of
non-interacting particles. Each -uid |particle} follows an orbit around the object. Assume
a straight line orbit to get a rough order of magnitude estimate. The particle is captured by
the body if
v2
M
¡G
<0
2
r
or
2GM
r<
= rA
v2
the accretion radius. Then the accretion rate is of order ¼r2A ½v:
More accurately, each -uid particle orbits around the gravitating body. On the back
side, the orbits intersect (particles collide) and our non-interacting particle model fails. This
implies that in the -uid a shock forms behind the mass M:
Digression - review of orbit theory.
Use polar coordinates with origin at the mass M ;
and assume that the orbiting particle has much smaller mass.
Conservation of angular momentum:
dµ
r2
= ` = constant = angular momentum per unit mass
(72)
dt
Conservation of energy:
1
E =
mv2 + V (r) = constant
2
1
GM m
=
mv2 ¡
(73)
2
r
where
dr
dµ
v = ^r+r µ^
~
dt
dt
and so
µ ¶2 µ
¶2 µ ¶2
dr
dµ
dr
`2
v2 =
+ r
=
+ 2
dt
dt
dt
r
37
So equation 73 becomes:
1
2
õ
dr
dt
¶2
`2
+ 2
r
!
¡
GM
E
=
="
r
m
Now make a change of variable to u = 1=r: Then r = 1=u and so
1 du
dr
¡ 2
=
u dt
dt
Equation 74 becomes:
µ ¶2
1 du
+ `2 u 2 ¡ 2GM u = 2"
u 4 dt
Also from equation 72,
d
` d
d
= 2
= `u 2
dt
r dµ
dµ
Thus
µ
¶2
1
2 du
`u
+ `2 u 2 ¡ 2GM u = 2"
u4
dµ
µ ¶2
du
2GM
"
+ u2 ¡
u = 2 2
2
dµ
`
`
Now let y = u ¡
GM
:
`2
The differential equation becomes:
µ ¶2
µ
¶2
dy
"
GM
+ y2 = 2 2 +
=K
dµ
`
`2
So
p
dy
= K ¡ y2
dµ
Z
dy
p
=µ
K ¡ y2
³ y ´
µ = µ 0 + sin ¡1
K 1=2
Integrating, we get:
Thus
and
y = K 1=2 sin (µ ¡ µ 0 )
Returning to our original variables:
s
µ
¶2
1
GM
"
GM
= 2 + 2 2+
sin (µ ¡ µ0 )
r
`
`
`2
With a change of constant µ a = µ 0 ¡ ¼=2; we get
r=
`2 =GM
D
=
1 + ² cos (µ ¡ µ a )
1 + ² cos (µ ¡ µ a )
38
(74)
where D = `2 =GM and
s
²=
E
2 2
m`
µ
`2
GM
¶2
+1=
s
E
2
m
µ
`
GM
¶2
+1
The accretion radius
Returning to our original problem, the angular momentum per
unit mass for our -uid |particle} at 1 is ` = pV ; where p is the impact parameter, and
E = 12 mV 2 . Thus
p2 V 2
D=
GM s
sµ
¶2
µ ¶2
2
pV
D
²=
+1=
+1
GM
p
The particle starts out at 1 where µ = ¼; so
1 + ² cos (¼ ¡ µ a )
1 ¡ ² cos µa
= 0
= 0 ) ² cos µ a = 1
The particle reaches the axis (µ = 0) on the downstream side at a distance d from the mass
M; where
D
D
d=
=
1 + ² cos µ a
2
At this point it collides with particles orbiting in the opposite direction, and the µ component
of velocity is converted to thermal energy in the resulting shock. Only the radial component
remains. The |particle} is captured if
1
GM m
mv 2r ¡
<0
2
d
But from energy conservation:
¢ GM m
1 ¡ 2
1
m vr + vµ2 ¡
=
mV 2
2
d
2
¢
1
GM m
1 ¡ 2
mvr2 ¡
=
m V ¡ v 2µ
2
d
2
and from angular momentum conservation:
dvµ = pV
Thus the capture condition becomes:
µ
¶
p2
V 2 1¡ 2 < 0 ) d< p
d
or
p2 V 2
<p
2GM
2GM
p<
= RA
(75)
V2
All -uid with impact parameter less than R A is ultimately accreted by the mass M : The
39
accretion rate is:
2
dM
4 (GM )
(GM )
= ¼R2A ½1 V = ¼½1 V
= 4¼½1
dt
V4
V3
Notice that the accretion rate decreases as V increases.
2
(76)
2.5.2 Subsonic accretion
Ref: Bondi MNRAS 112, 195, 1952
We study this situation by ®rst looking at the case in which the -uid is stationary at
in®nity. We also use the adiabatic equation of state, P = k½° : Then the governing equations
are:
Conservation of mass:
dM
= 4¼½r2 v = accretion rate
(77)
dt
and the steady state momentum equation (5) in spherical coordinates:
µ ¶
d v2
1 dP
=¡
+ gr
dr 2
½ dr
where
The sound speed is:
gr = ¡
d©
dr
dP
= °k½°¡1
(78)
d½
so
µ
¶
µ 2 ¶
1 dP
°k½°¡1 d½
d
°
d
cs
°¡1
=
=
k½
=
½ dr
½
dr
dr ° ¡ 1
dr ° ¡ 1
µ
¶ °¡1
2
c1 d
½
=
° ¡ 1 dr ½1
where the subscript 1 denotes values at in®nity. Then integrating the momentum equation
gives Bernoullizs equation:
µ
¶ °¡ 1
v2
c21
½
+
= ¡© (r) + constant
2
° ¡ 1 ½1
At in®nity, v = 0 and © = 0; so:
"µ
#
¶ °¡1
v2
c21
½
GM
+
¡ 1 = ¡© (r) =
(79)
2
° ¡1
½1
r
c2s =
Now write the mach number ¹ = v=cs and use the dimensionless radius variable
40
¡
¢
» = r= GM=c21 : Then equation 77 becomes:
dM
dt
where
=
4¼½r 2 v = 4¼
=
4¼¸½1
½ 2 cs
» ¹
½ c1
½1
c1 1
(GM )2
c31
µ
GM
c21
¶2
(80)
cs ½
¹
(81)
c 1 ½1
is the dimensionless accretion rate. Now use equation 78 to eliminate the sound speed:
µ
¶ °+1
2
½
¸ = »2
¹
½1
¸ = »2
Now we use this relation to eliminate ½ from Bernoullizs equation (79:
"µ
#
µ
¶2
¶° ¡1
1 v2
c1
1
½
1 c21
+
¡
1
=
2 c2s
cs
° ¡1
½1
» c2s
"
#
µ
¶ ¡(°¡ 1)
µ
¶ ¡ (°¡1)
1 2
1
½
1
½
¹ +
1¡
=
2
°¡1
½1
» ½1
"
#
µ
¶
µ
¶ ¡2(°¡ 1)=(°+1)
¡2(°¡1)=(°+ 1)
¹2
1
¸
1
¸
+
1¡
=
2
° ¡1
» »2 ¹
» 2¹
To simplify the notation, let ® = 2 (° ¡ 1) = (° + 1) : Then
"
µ
¶ ¡® #
¹2
1
¸
+
1¡
= » 2®¡1 ¹® ¸¡ ®
2
°¡1
» 2¹
Now multiply by ¹¡ ® and collect the terms in ¹ on one side:
µ
¶
¹2¡®
¹¡®
»2®
+
= ¸ ¡® »2®¡ 1 +
2
° ¡1
° ¡1
(82)
This equation for ¹ could be solved numerically in a speci®c case, but letzs see what we can
learn about the general case. First notice that
° ¡1
2
4
2¡®= 2¡2
=2
=
° +1
° +1
°+1
and
° ¡1
° +1
3° ¡ 5
2® ¡ 1 = 4
¡
=
° +1
° +1
°+1
Since ° · 5=3 always, then 2® ¡ 1 is always · 0; while 2 ¡ ® is always positive. So each
function
¹2¡®
¹¡®
F (¹) =
+
(83)
2
° ¡1
41
and
»2®
(84)
° ¡1
has one positive power and one negative power. Thus each function has a maximum or
minimum. At in®nity, ¹ = 0 and F (¹) ! 1: Also » ! 1 and G (») ! 1 . Therefore
each function has a minimum at ®nite »:
Consider F (¹) ®rst.
dF
2 ¡ ® 1¡®
®
=
¹
¡
¹¡®¡1
d¹
2
°¡1
¢
2 ¡ 1¡ ®
=
¹
¡ ¹¡®¡1
°+1
which is zero at ¹ = 1: Thus F is minimum at ¹ = 1 and
1
1
° +1
F min = +
=
2
° ¡1
2 (° ¡ 1)
Now letzs look at G :
dG
» 2®¡1
= (2® ¡ 1) » 2(®¡1) + 2®
d»
°¡1
which equals zero for
µ
¶
(2® ¡ 1) (° ¡ 1)
3° ¡ 5 (° + 1) (° ¡ 1)
5 ¡ 3°
» = »m = ¡
=¡
=
(85)
2®
° +1
4 (° ¡ 1)
4
and then
µ
¶2® µ
¶
5 ¡ 3°
4
1
Gmin =
+
4
5 ¡ 3°
° ¡1
µ
¶2®
5 ¡ 3°
° +1
=
4
(5 ¡ 3°) (° ¡ 1)
µ
¶2®¡ 1
5 ¡ 3°
° +1
=
4
4 (° ¡ 1)
G (») = » 2®¡1 +
Topology of the solutions.
Bernoullizs equation has been reduced to:
¸ ® F (¹) = G (» )
Letzs look at the functions F and G: (The plots are drawn for ° = 3=2:)
42
(86)
8
3
7
2.5
6
2
5
4
1.5
3
1
2
0.5
1
0
0
1
2
3
4
0.2
0.4
0.6
F
G
®
Now de®ne ¸ c = G min =F min : Then
(a) If ¸ > ¸ c ;then Gmin < ¸ ® Fmin : There is a region of » for which the RHS of equation
86 is less than F min ; which is impossible.
(b) If ¸ < ¸ c ; then ¸ ¡® G > F min for all »: The solutions are always supersonic or always
subsonic.
(c) If ¸ = ¸ c ; then G = Gmin coincides with F = F min : Thus the -ow goes through a
sonic point (¹ = 1) at » = »m = 14 (5 ¡ 3°) : A solution is possible which passes smoothly
from subsonic to supersonic or vice-versa. For the accretion problem, the boundary
condition is ¹ ! 0 as » ! 1: Case (b) which is always subsonic requires large pressure
gradients, all the way out to in®nity, to oppose gravity and slow the -ow. This is a settling
solution. Thus solution (c) is the usual accretion solution.
43
0.8
1
¸c
µ
¶1=®
G min
F min
"µ
# 1=®
¶ 2®¡1
5 ¡ 3°
° + 1 2 (° ¡ 1)
4
4 (° ¡ 1) (° + 1)
"µ
¶ 2®¡1 # 1=®
5 ¡ 3°
1
4
2
µ
¶ 2¡1=®
5 ¡ 3°
1
4
21=®
=
=
=
=
where
2¡
Thus
¸c
=
=
1
° +1
3° ¡ 5
=2¡
=
®
2 (° ¡ 1)
2 (° ¡ 1)
µ
¶( 3°¡5)=2( °¡1)
5 ¡ 3°
2 ¡( °+1)=2(°¡ 1)
4
"µ
# 1=2(°¡1)
¶ (3°¡5)
5 ¡ 3°
2 ¡(° +1)
4
Then the accretion rate is:
(87)
2
dM
(GM )
= 4¼½1
¸c
(88)
dt
c31
The value of ¸ c varies from 0.87 at ° = 1:2 to 0.5 at ° = 1:5: Notice that the two values
° = 5=3 and ° = 1 cause trouble. Letzs look at them.
Special case: ° = 1:
We have to go back to the original derivation. In this case the
sound speed is constant, and the equations are simpler:
1 dP
1 d½
d
= c2s
= c2s
ln ½
½ dr
½ dr
dr
so that Bernoullizs equation becomes:
v2
GM
+ c2s ln ½ ¡
= c2s ln ½1
2
r
and
¸ = »2
Thus
¹2
+ ln
2
µ
¸
¹»2
¶
¹2
¡ ln ¹
2
½
¹
½1
=
1
»
=
1
+ 2 ln » ¡ ln ¸
»
44
In this case
F (¹) =
has a minimum when
¹2
¡ ln ¹
2
1
=0
¹
or ¹ = 1; as before. The minimum value is F min = 1=2:
1
G (») = + 2 ln »
»
has a minimum for
¡1
2
+ =0
»
»2
or » = 1=2: The minimum value is
¹¡
Gmin = 2 ¡ 2 ln 2
So ¸ c is given by:
ln ¸ c
=
¸c
=
=
Gm in ¡ Fmin = ¡ ln 4 + 3=2
1 3=2
e
for ° = 1
4
1: 1204
Special case: ° = 5=3
According to equation 85, »m = 0 when ° = 5=3: The function
has its minimum at the accreting object. This means that the -ow never becomes supersonic.
Then ° ¡ 1 = 2=3, ° + 1 = 8=3; and with 5 ¡ 3° = "; equation 87 becomes:
³ " ´¡3"=4
0
1
¸ c = lim
2 ¡2 = lim
"0¡3"
0 !0
"!0 4
"
4
0
where " 0 = "=4: Now let z = " 0¡3" : Then
ln z = ¡3"0 ln "0 ! 0 as "0 ! 0
since the lograithm ! 1 slower than any algebraic power. Then z ! 1 and ¸ c = 1=4 for
° = 5=3:
Transitions from one branch of the solutions to another can occur. For example, when
the -ow goes supersonic, information about the solid object upstream cannot be transmited
back through the -uid. A shock wave forms and -ow makes a transition to the subsonic,
settling solution. Flow cannot become supersonic again behind the shock because pressure
gradients slow the -ow.
2.5.3 Transonic accretion
Ref: Hunt, MNRAS,154, 141,1971
Letzs summarize our results so far. For supersonic accretion, we have (equation 76)
2
dM
(GM )
= 4¼½1
dt
V3
45
while in the subsonic case (equation 88)
2
dM
(GM )
1
= 4¼½1
¸c ;
< ¸c < 1:12
dt
c31
4
Thus we might guess that the transonic case must be of the form:
2
dM
(GM )
= 4¼ ½1
¸
3=2
2
dt
(V + c21 )
where ¸ is a constant of order 1. Huntzs numerical calculations con®rm this guess. See his
paper for details.
2.5.4 Applications
Neutron stars in binaries with stellar winds
The neutron star moves through the wind
as it orbits its stellar companion. The semi-major axis of the binary is not much larger than
the primary starzs radius, about 10 12 cm. The orbital period is of the order of a few days,
and thus the orbital speed is about
2¼r
2¼ £ 10 12
7 £ 10 7
=
cm/s =
cm/s
T
T da ys £ 24 £ 60 £ 60
T da ys
The wind speed is about 600-1000 km/s for an O-B type star. The temperature is regulated
by the x-ray source itself via Compton/inverse Compton cooling/heating, so we expect
T » T X » a few £10 7 K. Then the accretion radius is:
2GM X
M X;¯
RA » 2
= 2:7 £ 1010 2
cm
¡2
vrel (1 + ¹ )
V 8 (1 + ¹¡2 )
v or b »
46
where
2
2
2
V rel
= V wind
+ V orb
dM X
= ¼R2A ½wind V rel
dt
is the mass accretion rate onto the x-ray source,
dM ¤
= 4¼R2x ½wind V wind
dt
is the mass loss rate by the star, Rx is the radius of the neutron star orbit, and
¡
¢2 2
µ
¶2
2:7 £ 10 ¡2 M X;¯
MX
1 RA
Vr el
V rel
=
»
4
M¤
4 Rx
V wind
4
V 8 V wind (1 + ¹¡2 )2 R2x;12
»
1: 8 £ 10 ¡4
2
MX;¯
V rel
4
V 8 Vwind (1 + ¹¡2 )2 R 2x;12
Note the very strong dependence on the wind speed. If the stellar mass loss rate is about
3£10 ¡6 M¯ /y; then the mass accretion rate onto the neutron star is
dM X
5:4 £ 10 ¡10
=
M ¯ /y
dt
V 84
and the x-ray luminosity is then
Lx
»
=
GMx
dM X
dM X
» 0:1c2
dt
dt
Rne utron star
3 £ 10 36
erg/s
V 84
which is consistent with observed values.
2.5.5 The solar wind
Ref: Parker: Interplanetary Dynamical Processes
In our previous analysis, the -uid velocity appears only as v 2 (or ¹2 ), so vr can have
either sign - i.e. the solutions work for winds as well as for accretion. We have to modify
the analysis to allow for a ®nite wind speed at in®nity (or at some reference point). So we
rewrite equation 79 using new variables
v
u=
ca
and
» = r=a
where r = a is the reference point and ca is the sound speed at that point. The equation
becomes:
µ ¶ °¡1
v2
1
½
GM
va2
1
GM
+
¡
=
+
¡ 2
2
2
2
2ca
° ¡ 1 ½a
car
2ca
° ¡1
caa
µ ¶ °¡1
2
2
u
1
½
GM a
ua
1
GM
+
¡ 2
=
+
¡ 2
2
° ¡ 1 ½a
ca a r
2
° ¡1
caa
47
and from the continuity equation, we have:
r2 ½v
½
½a
a2 ½ a v a
1 ua
»2 u
=
=
Next write GM=c2a a = H; and u21 ´ u2a + 2= (° ¡ 1) ¡ 2H: Then Bernoullizs equation
becomes:
2
u +
° ¡1
µ
µ
µ
¶
°¡1 2
2H
¼
u1 1 + 2
2
»u 1
2
Solution at large »
ua
u»2
¶° ¡1
¡
2H
= u21
»
(89)
If u ¿ 1 at large » ; then we may ignore the term in u 2 to get:
ua
u» 2
¶ °¡1
Taking the root:
u»2
ua
=
µ
2
(° ¡ 1) u 21
¶ 1=(°¡1)
³
1
´1=( °¡1)
2H
1 + »u
2
1
¶1=(°¡ 1) µ
1¡
2
µ
¶
ua
2
2H
u ¼
(° ¡ 1) » u21
»2 (° ¡ 1) u 1
Whereas if u remains & 1; we have
Ã
!
µ
¶ °¡1
2H
2
ua
1
2
2
u
= u1 1 + 2 ¡
» u1
° ¡ 1 u» 2
u 21
Ã
!
µ
¶ °¡1
H
1
ua
1
u ¼ u1 1 + 2 ¡
» u1
° ¡ 1 u» 2
u 21
(90)
(91)
These two solutions correspond to the upper and lower branches that we found in the
accretion problem. (See Figure on page
). Notice that on the upper branch, u ! u1
as » ! 1 (for ° > 1): The wind reaches a terminal velocity. Because energy is conserved,
thermal energy in the inner regions is converted to kinetic energy in the outer regions. The
-ow becomes cold and supersonic at in®nity. (See below.)
Solutions at small » :
48
Lower branch
If u ! 0 as » ! 0; then equation 89 becomes:
µ
¶ °¡1
µ
¶
2
ua
2H
2H
u21 »
2
=
+ u1 =
1+
° ¡ 1 u»2
»
»
2H
µ
¶1=(°¡ 1) µ
¶ ¡1=(°¡1)
ua
»
u 21 »
u =
1+
2H
» 2 (° ¡ 1) H
µ
¶
ua
1 u21 »
1=(°¡ 1)¡2
¼ »
1¡
° ¡ 1 2H
[(° ¡ 1) H]1=(° ¡1)
This is self consistent (i.e. u ! 0) only if
1
¡2 > 0
° ¡1
1
°¡1 <
2
3
° <
2
(92)
(93)
Upper branch
On the other branch, (u not small), we would have:
µ
¶ °¡ 1
2H
2
ua
u2 =
+ u21 ¡
»
° ¡ 1 u» 2
µ ¶1=2 Ã
µ
¶°¡ 1 !1=2
p
H
u 21 »
»
ua
u =
2
1+
¡
»
2H
(° ¡ 1) H u»2
µ ¶1=2 µ
¶
p
H
u 21 »
» 1¡2° +2 ³ u a ´°¡1
¼
2
1+
¡
»
4H
2 (° ¡ 1) H u
µ ¶1=2 µ
³ u ´°¡1 ¶
p
H
u2 »
»3¡2°
a
=
2
1+ 1 ¡
»
4H
2 (° ¡ 1) H u
p ³ ´1=2
We can simplify the last term by using the zeroth order term, u ¼ 2 H»
:
!
µ ¶ 1=2 Ã
µ
¶ ° ¡1
2
p
H
u 21»
» 3¡2°+(°¡ 1)=2 u 2a
u ¼
2
1+
¡
»
4H
2 (° ¡ 1) H
2H
!
µ ¶ 1=2 Ã
5¡ 3°
µ
¶ ° ¡1
2
p
H
u 21»
»2 2
u 2a
=
2
1+
¡
»
4H
2 (° ¡ 1) H 2H
Then:
(94)
which shows that u » » ¡1=2 as » ! 0: This analysis is valid only if the power of » in
the last term inside the parentheses is greater than zero, or ° < 5=3: Otherwise our
series expansion does not converge.
The solar breeze.
The case u 1 ´ 0 and ° = 5=3 is a special case. For these values of
49
the parameters, equation 89 becomes:
2
u +3
µ
ua
u» 2
¶ 2=3
¡
2H
=0
»
and the condtion u1 ´ 0 is
u2a
u2
3
+ 1= (° ¡ 1) ¡ H = 0 ) H = a +
2
2
2
Thus we have:
µ
¶ 2=3
ua
u 2a 3
2
u +3
¡
¡ =0
»
»
u»2
This equation has a solution
ua
u = 1=2
»
which is called the solar breeze.
(95)
Boundary conditions at in®nity
So which solution describes the actual solar wind?
We want a solution with P ! 0 as » ! 1: From the continuity equation (77),
½ ua
½= a 2
u»
On the upper branch, u ! u 1 ; and so
½ ua
½! a 2 !0
u1 »
and therefore P ! 0 as well.
On the lower branch
½a ua
½ =
³
´1=(° ¡1)
2
»2 u»a2 ( °¡1)u
2
1
µ
¶1=°¡1
(° ¡ 1) u 21
= ½a
2
and therefore
µ ¶°
µ
¶ °=°¡1
½
(° ¡ 1) u 21
P = Pa
= Pa
½a
2
is comparable to Pa :
For the solar breeze,
½a
½ = 3=2
!0
»
Thus the upper branch and the breeze are the only allowable solutions. But the breeze is a
singular case, requiring very special values of the parameters.
Mach number
Now letzs look at the Mach number.
µ ¶ (°¡1)=2
v
ca
½a
¹=
=u
=u
cs
cs
½
Then for our various solutions we have:
50
1. Large »
² Upper branch:
¶ (°¡ 1)=2
u1 » 2
! 1 as » ! 1
ua
Thus this solution corresponds to a cold -ow.
¹ ! u1
² Lower branch:
¹
¼
ua
»2
=
ua
»2
µ
µ
for all values of °:
² Solar breeze:
µ
¹=
2
(° ¡ 1) u 21
2
(° ¡ 1) u 21
ua
»1=2
¶ 1=(°¡1) µ
¶ 12 1+°
° ¡1
2
(° ¡ 1) u 21
¶ 1=2
! 0 as » ! 1
» 3=2 = u a » ! 1 as » ! 1
(96)
2Small »
In general:
So
¹=u
µ
½a
½
¶(° ¡1)=2
½
ua
= 2
½a
u»
=u
µ
u» 2
ua
¶ (°¡ 1)=2
= ua
µ
u
ua
¶ 1+ (°¡1)=2
» °¡1 = ua
µ
u
ua
¶ ° +1
2
»°¡ 1
(97)
² Lower branch:
Here we use equation equation (92) to get:
Ã
! °+1
2
» (3¡ 2°)=(°¡1)
¹ = ua
» °¡1
[(° ¡ 1) H]1=(°¡1)
(3¡2°)(°+1)
ua
=
» 2(° ¡1) » °¡1
(°+ 1)=2(°¡1)
[(° ¡ 1) H]
(3¡2°)(°+1)+2(°¡1)2
ua
2(°¡1)
=
»
(°+ 1)=2(°¡1)
[(° ¡ 1) H]
The power of » is:
(3 ¡ 2°) (° + 1) + 2 (° ¡ 1)2
5 ¡ 3°
=
2 (° ¡ 1)
2 (° ¡ 1)
which is always positive if ° < 5=3: So ¹ ! 0 as » ! 0 unless ° = 5=3; in which case ¹
approaches a constant value. Notice that there is a problem with ° = 1; but we have
51
already noted similar problems above.
² Upper branch:
This branch exists only for ° < 1:5:
Here we use equation (94) in equation (97) to get:
à p µ ¶ ! ° +1
Ãp
! °+1
2
2
1=2
° +1
2 H
2H
¹ = ua
»°¡1 = u a
» °¡1¡ 4
ua
»
ua
Ãp
! °+1
2
2H
= ua
»(3°¡5)=4 ! 1 as » ! 0 for ° < 5=3
ua
We have previously noted that this branch exists only for ° < 5=3:
² Solar breeze:
From equation 96, ¹ ! 0 as » ! 0: For the breeze, we have the interesting situation
that the lower branch in u is the upper branch in ¹; and vice versa.
Thus we have the following situation:
quantity
at in®nity
as » ! 0
upper branch
lower branch
upper branch
lower branch
1
³
´ ° ¡1
³ ´1=2
3¡2°
ua
2
2H
speed
u1
» °¡1 [(°¡ 1)Hua]1=(°¡1)
»2
(°¡1)u2
»
1
°
³
´ À¡1
½ a ua
(°¡1) u2
1
pressure
Pa
2
u 1» 2
³ 2 ´ °¡1
³
´ 1+°
³ p ´ °+1
5¡3°
3° ¡5
2
2
2(° ¡1)
ua
2
2H
ua
2(° ¡1)
Mach number
u 1 uu1»a
u
» 4
2
2
a
°+1 »
»
(°¡1)u
ua
1
[( °¡1)H ] 2(°¡1)
Parker argues that the solution through the sonic point is always set up. His argument
goes like this: Assume ®rst that there is a high pressure at in®nity, so that we are on the
lower branch of the solution. Now slowly decrease P 1 : The wind will accelerate and u
will increase. But P 1 is independent of u a ; so the -ow will continue to accelerate until it
switches to the upper branch through the sonic point. And then we have the critical solution.
52
so
What is the signi®cance of the constraint ° < 1:5 for the existance of the lower branch
solution u ! 0 as » ! 0? Physically, it means that the gas must be heated as it expands.
The critical solution is analagous to the expansion of a gas through a Laval nozzle. Gravity
acts like the throat of the nozzle.
For the upper branch to exist we need u21 > 0; i.e.
2
2GM
u2a +
¡ 2
>0
° ¡1
ca a
or
GM
u 2a
1
<
+
(98)
2
caa
2
° ¡1
The gravitational ®eld must not be too strong or the gas will form a static atmosphere rather
than a wind.
To ®nd another constraint, we differentiate equation 89:
µ ¶ °¡ 1
³ u ´° ¡1
du
ua
du
2H
a
¡ 2°+1
2u
¡4
»
¡2
u¡ °
+ 2 = 0
2
d»
u
d»
»
»
Ã
!
µ ¶ °¡1
³ u ´°¡ 1
du
ua
H
a
¡°
u¡
u
=
2
»¡ 2°+1 ¡ 2
2
d»
u
»
»
We want the velocity to be increasing outward at the reference point » = 1: At » = 1 the
term in parentheses on the LHS is
1
ua ¡
ua
which is < 0 for u a < 1: Then the RHS must also be < 0; which means we need H > 2:
This is the nozzle condition. From equation (98) with ua ¿ 1; we ®nd:
1
2< H<
° ¡1
which gives the same condition (93) that we have previously found for °:
Thermal conductivity is important in the solar wind. Heat is conducted up from the
corona and keeps ° close to 1 in the inner regions, easily satisfying this constraint.
2.5.6 Radiation pressure driven winds
Now we add to our theory the interaction of the -uid with the radiation from the star (or the
accreting object). For winds, this modi®cation is important when discussing luminous stars
such as O and B stars. To get the basic idea, we make some simplifying assumptions (eg
Castor, Abbot and Klein, Ap.J. 195, 160, 1975)
Radiation pressure is greater than electron scattering pressure because of absorption in
spectral lines due to atomic and molecular transitions. For electron scattering only, we
would have a radiation pressure force
f lux
¾ T L½
Fp = ¾ T ne
=
c
4¼r2 cm
The mass m is one half the proton mass if the -uid is ionized hydrogen. So we write the
53
total radiation pressure force as:
¾ T L½
(1 + M (t))
4¼r 2 cm
where M (t) _ t¡ ® is the effect of the lines, and t is the optical depth. (CAK take
M » t¡ 0:7 =30): M ! 0 as t ! 1 because all the lines become optically thick and hence
less ef®cient at absorbing radiation.
When this extra force is included in the momentum equation, we get:
dv
1 dP
GM ¤
¾T L
v
=¡
¡
+
(1 + M )
2
dr
½ dr
r
4¼r 2 cm
The temperature at each point in the wind is assumed to be given by the local energy balance
at that point which is dominated by the radiation (heating and cooling). The timescale to
reach thermal equilibrium is short compared with the -ow time scale. Thus we take
Fp =
P = ½c2s
where c2s is the local, isothermal sound speed, and is presumed known. Then:
1 dP
c2 d½ dc2s
= s
+
½ dr
½ dr
dr
and as usual
½=
Then
and so
dM=dt
4¼r2 v
(99)
1 d½
2
1 dv
=¡ ¡
½ dr
r
v dr
1 dP
2
c2 dv
dc2
= ¡ c2s ¡ s
+ s
½ dr
r
v dr
dr
Now write
¡=
¾T L
4¼cGM ¤ m
so that the momentum equation becomes:
µ
¶
c2s dv
2
dc2
GM ¤
v¡
= c2s ¡ s ¡
(1 ¡ ¡ ¡ ¡M )
v dr
r
dr
r2
If M = 0; we have the previous equation with a reduced mass M eff = M ¤ (1 ¡ ¡) : If
¡ = 1; Meff = 0 and there is no effective gravitational force!
The optical depth depends on the wind speed gradient, since Doppler shift moves
material into or out of the optically thick region near the line center.
Z r
¾L
t (º 0 ) =
½· (º) dr
· L; 0 0
where ¾ L is the total cross section for absorption in the line, ½ is given by equation 99 and
v th
· (º ) = ·L ;0
jdv=drj
In this expression · L;0 is the absorption coef®cient at the line center, vth is the thermal
54
velocity width of the line and dv=dr is the speed gradient in the wind. Thus vth = jdv=drj is
the distance before the frequency is Doppler shifted out of the line. There is one such term
for each relevant line. Then:
½¾ L vth
t ¼
jdv=drj
dM =dt ¾ L v th
=
4¼ r2 v jdv=drj
and so
µ
¶¡ ® µ
¶®
dM=dt
dv
M = kt¡ ® = k
¾ L v th
r2v
4¼
dr
Now de®ne the variables
1
w = v2
2
1
u=¡
r
2
2
2 dc s
h (u) = ¡GM¤ (1 ¡ ¡) + 2cs r ¡ r
= a known function of r
dr
µ
¶®
4¼
C = ¡GM ¤ k
¾ L vth dM =dt
Then Bernoullizs equation takes the form:
µ
¶
c2s
®
0
F (u; w; w ) ´ 1 ¡
w 0 ¡ h (u) ¡ C (w 0)
2w
where w 0 ´ dw=du: As usual we look for a solution that goes from 0 at u = ¡1 (r = 0) to
a large value at u = 0 (r ! 1) : It turns out that there are two critical points. One is the
usual sonic point, and one where
©
ª
c2
®
w= s ¡
h messy function of c2s and h
2
1¡®
As a ! 0; this point approaches the sonic point. For h < 0 (which is always true) the Mach
number is greater than 1 at this 2nd critical point.
The wind density must be below a maximum value. If the density is too high, the wind
gets optically thick too soon and there is not enough pressure at high altitude to drive the
wind.
Accretion -ows are similarly affected by radiation pressure. With ¡ = 1; the accretion
is stopped. This is the Eddington limit.
4¼cGM¤ m
LEdd =
¾
¡ T
¢¡
¢ ¡
¢
¢ M¤
4¼ 3 £ 10 10 cm/s 6:67 £ 10¡ 8 cm3 =g ¢ s2 12 1:67 £ 10¡ 24 g ¡
=
2 £ 10 33 g
¡25
2
6:6 £ 10
cm
M¯
M
¤
= 6: 4 £ 10 37
erg/s
M¯
Steady state spherical accretion is not possible above the Eddington limit.
55
3 Accretion disks
Reference: Pringle, Annual Reviews of Astronomy and Astrophysics, 19, 137, 1981
3.1
Introduction
When the accreting material has high angular momentum, we get an accretion disk. Material
must orbit the gravitating object, and cannot move in arbitrarily close without losing angular
momentum. If the angular momentum per unit mass is `; then the tangential velocity at
distance r from the central object is
`
vÁ =
r
and the force per unit mass required to keep the -uid in circular motion is:
v2
`2
= 3
r
r
The inward gravitational force equals this centripetal force at a distance r` where
FC =
GM
`2
`2
= 3 ) r` =
(100)
2
r`
r`
GM
Gas cannot move in any further since all the inward force is needed to maintain the orbit and
hence there is no dr=dt:
In the absence of forces preventing it, each parcel ofqgas moves inwards to the r`
corresponding to its `; and its speed there is v Á =
`
r
=
GM
r`
: Thus the -uid parcel is in
a Keplerian orbit. Since v = v (r) ; we have velocity shear and viscosity and instabilities
act to smooth out the shear. Angular momentum is transported outward and material is
transported inward through the disk.
Disks are common phenomena.
1. Cataclysmic variables and dwarf novae. The disk contributes most of the light
from the system. Observational evidence for the existence of disks includes Doppler
broadening of spectral lines and observed hot spots where the accretion stream from the
companion meets the disk.
2. X-ray binaries. Disks are mostly conjectured rather than directly observed. But in
certain sources such as HER X-1 the warped, processing disk occasionally occults the
x-ray emitter.
3. Gas disks observed in galactic nuclei (in 21-cm radio emission and molecular lines.
Also HST images.) May be disks around central black holes.
4. Disks around young stars may be proto-planetary systems. These are observed
primarily in the infra-red.
5. Symmetry of extra-galactic radio sources may be explained if there is a disk around the
central engine that acts as a collimator for the jets.
6. Disks are conjectured to be present in many stages of binary star evolution- e.g x-ray
56
binaries. SS433 may have one.
3.2
Basic theory
As usual we shall simplify the discussion by assuming the system to be in a steady state. We
also assume that the mass M of the central object remains constant and that density in the
disk is high enough for cooling to occur. Thus the -ow is supersonic.
3.2.1 The momentum equation
a) Radial component
2
vÁ
@v r
1 @P
GM
¡
+
+ 2 =0
(101)
@r
r
½ @r
r
As
p usual we work with the mach number ¹ = vÁ =cs; where cs is the isothermal sound speed
P =½: Then equation 101 becomes:
µ ¶
v 2Á
¢
@ v 2r
GM
@ ¡ 2
¡
+ 2 +
cs ln ½ = 0
@r 2
r
r
@r
Ã
!
µ 2¶
2
vÁ
v 2Á
@ vr
GM
@
¡
+ 2 +
ln ½
= 0
@r 2
r
r
@r ¹2
vr
If ¹ À 1; then the last term may be ignored with respect to the second. Then if we also
have vr ¿ 1; we ®nd
r
GM
vÁ =
r
as we concluded in the introduction.
b) z-direction (perpendicular to the disk)
We assume that v z = 0 so that the
momentum equation becomes the equation of hydrostatic equilibrium:
1 @P
GM z
=¡ 2
½ @z
r r
where z=r = cos µ for z ¿ r: Then we can estimate the disk thickness:
GM z
r3
vÁ2 ³ z ´2
GM z 2
ln ½ = C ¡ 3
=
C
¡
r 2c2s
2c2s r
(C is an integration constant), and thus
Ã
!
µ
¶
v 2Á ³ z ´2
1 ³ z ´2
½ = ½0 exp ¡ 2
= ½0 exp ¡
2cs r
2 H
c2s
@
ln ½
@z
= ¡
where the scale height of the disk is
H=
57
r
¹
(102)
and is much less than r; as required for self-consistency. Note that the disc }-ares} - H
increases as r increases. As material -ows inward through the disc, its z¡component of
velocity is of order:
H
r vÁ
vÁ
vz »
=
=
¿ vÁ
T
¹ r
¹
consistent with our previous approximation vz ¼ 0:
3.2.2 Mass -ow inward
Mass -ows inward through the disk at a rate
dM
= 2¼r§ (¡vr )
dt
where § (r) is the surface mass density in the disk:
Z +1
§ (r) =
½ (r; z) dz » 2H½ (r)
(103)
¡1
To make further progress, we need v r : Recall that it is viscous drag that allows material
to fall inward, so we expect v r to depend on viscosity. Since we found v Á from the
radialp
momentum equation, we look at the Á component to get vr : Write vÁ = r- where
- = GM =r 3 is the keplerian angular speed of the disk at radius r: Then the angular
momentum contained in an annular ring of thickness dr is dL = dM r2 - = 2¼r§r 2 -dr:
This angular momentum is carried outward by viscous torques. The net change of angular
momentum in the annulus is:
¯
¯
-ux in - -ux out = 2¼r§r 2-v r¯
¡ 2¼r§r2 -vr ¯
r+dr
r
¢
d ¡
=
2¼§r3 -vr dr
dr
Now the viscous force is given in terms of the kinematic viscosity º :
F (r) = 2¼rº§ (rate of velocity shear) = 2¼rº§r
ddr
and thus the viscous torque is
G (r) = rF (r) = 2¼r 3º §
The net torque on an annulus of thickness dr is
ddr
µ
¶
d
d3
dG = G (r + dr) ¡ G (r) =
2¼r º §
dr
dr
dr
Then setting torque equal to rate of change of angular momentum, we have:
µ
¶
¢
d ¡
d
d3
3
2¼§r -vr
=
2¼r º§
dr
dr
dr
º dk
vr =
+ 3
(104)
- dr
r §where k is an integration constant that must be determined by the boundary conditions.
58
For a disk around a star with radius R¤ and angular speed
p -¤ ; there must be a boundary
layer where - changes rapidly from the Keplerian value GM =R¤3 to - ¤ : Since we must
have
GM
> - 2¤ R¤
R2¤
for the star to hold together, then - ¤ < -Kep (R¤ ) : The angular speed as a function of
radius looks like:
0.6
0.5
0.4
0.3
0.2
0.1
0
1
2
x
3
4
5
Thus d-=dr must be zero at some r0 not much larger than R¤ ; and - (r0 ) ¼
Then from equation 104,
k
k
v r (R¤ ) ¼
p
=
p
R 3¤ § GM =R3¤
§ (R¤ ) (GM R3¤ )
and hence
p
GM =R3¤ :
p
vr (R¤ ) § (R¤ ) (GM R3¤ )
º dvr =
+
- dr
r 3 §µ ¶ ¡ 3=2
º dr
dM=dt
=
¡
(105)
- dr
R¤
2¼§R¤
where we used equation 103 evaluated at R¤ to simplify the last term. Then using this
equation again, we ®nd:
Ã
!
µ ¶ ¡3=2
dM
º dr
dM =dt
= ¡2¼r§
¡
dt
- dr
R¤
2¼§R¤
Ã
µ ¶ ¡1=2 !
dM
r
º dº
1¡
= ¡2¼r§
= 3¼r§ = 3¼§º
(106)
dt
R¤
- dr
r
So §º must vary with r: Note also that dM =dt ! 0 as º ! 0; as expected.
From equation 105, we ®nd that for r À R ¤ ;
3º
vr ¼ ¡
2r
59
which is negative (i.e. the -ow is inward) and depends directly on the viscosity, as expected.
The viscosity also dissipates energy. The power expended by the vsicous forces on our
annulus of thickness dr is:
d~ ¢~
dP = F
v =2¼r2 º§
((r + dr) - (r + dr) ¡ r- (r))
dr
µ ¶2
d= 2¼r3 º §
dr
dr
and thus the power per unit area is
¡ ¢2
µ
¶2
2¼r3 º§ ddr
dP
ddr
=
= º§ r
dA
2¼rdr
dr
This energy is dissipated as heat. using the Keplerian expression for -; and using equation
106 for º§; we get:
Ã
µ
¶¡ 1=2 ! µ
¶2
1 dM
r
3
GM
D (r) =
1¡
¡
3¼ dt
R¤
2
r3
Ã
µ
¶¡ 1=2 !
3 dM GM
r
=
1¡
(107)
4¼ dt r3
R¤
In this expression, the viscosity º has disappeared! The total luminosity of the disk is
then:This is one half of the total accretion energy.
1. Solid Star: the other half of the energy must be released in a boundary layer at the
stellar surface.
2. Neutron star with magnetosphere: Replace R¤ with RM : Energy can be carried
through the magnetosphere and released at the star itself.
3. Black hole: R¤ must be the radius of the last stable orbit. The other half of the energy
disappears down the hole!
Now letzs look at the energy radiated between radii R1 and R2 ; both of which are À R¤ :
µ
¶
3 dM
1
1
L (R1 ¡ R2 ) =
GM
¡
2 dt
R1
R2
This is 1.5 times the energy released in the region between R 1 and R2 : Thus energy must be
transported out from the inner regions of the disk.
3.2.3 Local disk structure
We already found the density pro®le (equation 102). The density falls off rapidly above the
central plane of the disk. For supersonic -ow, as we have assumed, H ¿ r; i.e. the disk
is thin. The temperature gradient perpendicular to the disk is much greater than the radial
gradient, thus radiation escapes primarily perpendicular to the disk. In equilibrium, the disk
reaches a temperature T (r) such that the radiated energy 2¾T 4 equals the heat input D (r) :
60
(¾ is the Stefan-Boltzmann constant.) Thus:
Ã
3 dM GM
2¾T =
4¼ dt r3
4
and thus:
T =
Ã
3 dM GM
8¾¼ dt r3
For r À R¤ ;
T = T1
where
Ã
r
R¤
1¡
r
R¤
r
¶ ¡1=2 !
R¤
r
!!1=4
¶ ¡3=4
¶1=4
3 dM GM
8¾¼ dt R3¤
Thus the temperature is greatest in the inner regions of the disk.
The total radiated spectrum is:
Z Rmax
Fv =
Bº (T (r)) 2¼rdr
T1 =
µ
µ
1¡
µ
R¤
At high frequencies, we get an exponential drop off. At low frequencies, º ¿ kT out =h;
radiation comes from the outer regions of the disk, and the spectrum _ º 2 . To investigate
the intermediate region, let x = hº =kT : Then
µ
¶
hº dT
hº
3
3x
dx = ¡ 2
dr = ¡ 2 ¡ T dr =
dr
kT dr
kT
4r
4r
2hº 3
ex ¡ 1
µ
¶4=3
kT1
= R¤
x
hº
Bº (T ) =
and
r = R¤
So
Fv
=
=
_
µ
Z
T1
T
¶ 4=3
xmax
1
4r
r dx
x ¡ 1 3x
e
x¤
µ
¶ 8=3 Z xmax
16
kT1
1
¼hº 3 R2¤
x 5=3 dx
x
3
hº
e ¡1
x¤
Z xmax 5=3
x
º 1=3
dx
x¡ 1
e
x¤
4¼hº 3
So the spectrum looks like:
61
3.2.4 Viscosity
The usual molecular viscosity proves inadequate to explain disk structure.
viscosity is usually assumed to be due to magnetic ®elds or turbulence.
Therefore,
Turbulence
The -ow is strongly sheared and supersonic. Thus the Reynolds number
R = Lv=º is high. This leads us to believe that turbulence will develop in the disk, but
this has not been proven.
Magnetic viscosity
The -ow in the disk winds up the magnetic ®eld lines, and
reconnection will occur.
The ®eld may be chaotic.
The magnetic ®eld energy is
constrained to be
B2
1
/ ½c2s
8¼
2
since reconnection will generate heat in the disk.
For both of these processes, we can parametrize our ignorance with a parameter called
the ®¡parameter as follows
1. Turbulent viscosity:
º ¼ `v
where ` is the scale length and v is the speed of the largest turbulent eddies.
` < H; the disk scale height, and probably v < cs ; so
Obviously
º / Hcs = ®Hcs
where the parameter ® 6 1:
But H=r ¼ cs =v Á (equation102), so
º¼®
c2s r
vÁ
(108)
1
2. Magnetic
The magnetic stress is of order 8¼
(BrB Á ) » B2 =8¼ »
¡ 2 viscosity:
¢
2
2
½cs v A =cs where vA is the Alfven speed. Now we derived the viscous stress =
62
force/length as º §r-0 ; so
¯ µZ
¯ Z
¶
¯
¯
¡
¢
0¯
¯º
½dz
r½c2s v 2A =c2s dz
¯
¯=
¶ 0 for a keplerian disk,
so, evaluating ½-
where in this case ® ¼
2
3
¡
º ¼ c2s
2
vA
r
= ®c2s
2
cs (3=2) vÁ
¢
2 =c2 / 1
vA
s
again.
So the usual method is to use the so-called ®¡viscosity given by equation 108. Then
we have a radial velocity component
¶0
¶0
cs
cs
vr ¼ º
= ®c2s 2 = ®cs
=® ¿1
(109)
-r
¹
Note that the inward velocity is directly proportional to the value of ®: Then the Reynolds
number is
rvÁ2
Lv
¹2
R»
=
=
À1
º
®c2s r
®
which is self-consistent.
3.2.5
Radiation from the disk:
If the disk is optically thick, then the spectrum does not depend on º (or ®); as we
have seen. But if energy is dissipated in optically thin regions, we do not have complete
thermalization of the radiation and hence we can have a higher effective temperature T eff :
If the surface of the disk gets hot, there can be a disk corona above and below the central
plane. For high luminosity systems we can even have a disk wind. Radiation from the
inner regions can be reabsorbed farther out and cause heating and evaporation.
3.2.6 Timescales
1. Rotational timescale is
tÁ »
2. z-structure.
tz »
r
1
»
vÁ
-
H
H r vÁ
H
»
» tÁ ¹ » tÁ
cs
r v Á cs
r
3. surface density is governed by dM=dt and º :
2
tº »
r
r
(-r) 1
1
»
-r =
= ¹2 t Á À t Á
2
2
vr
®cs
®cs ®
4. Thermal timescale
tth »
§c2s
§c2s
c2s v Á
tÁ
=
=
=
& tÁ
2
2
2
D (r)
º §®cs r®
63
3.2.7 Instabilities
² Thermal. Since tth ¿ tº ; we may assume § = constant. In equilibrium, heating
= cooling. But if the heating rate increases with temperature faster than the cooling
rate, a small perturbation will grow. Since the scale height H _ cs ; increased T
)increased H ) decreased ½ and therefore decreased cooling on the timescale tz » tÁ :
Thus the disk should be thermally unstable.
² Viscous. Since tº À tth ; thermal equilibrium always holds, heating balances cooling
and the temperature T (r) is ®xed. Then º = º (§; r) : De®ne ¸ = º §; and consider a
perturbation § ! § + ±§; ¸ ! ¸ + ±¸; where
@¸
±¸ =
±§
@§
The continuity equation gives us:
@§
1 @
+
(r§vr ) = 0
(110)
@t
r @r
while the Á¡component of the momentum equation gives:
¢ 1 @ ¡
¢
@ ¡ 2 ¢ 1 @ ¡
§r - +
r§vr r2 - =
º §r 3- 0
@t
r @r
r @r
where the LHS equals the rate of change of angular momentum and the LHS equals the
net viscous torque. Expanding the second term, we get:
¢
@ ¡ 2 ¢ 1 @
r§v r @ ¡ 2 ¢ 1 @ ¡
§r - +
(r§vr ) r2 - +
r - =
º§r3 - 0
@t
r @r
r @r
r @r
Now - does not depend on time, and we can use equation 110 to simplify the second
term:
¢
@
@§ 2
r§vr @ ¡ 2 ¢
1 @ ¡
r2- § ¡
r -+
r =
º§r3 - 0
@t
@t
r @r
r @r
¢
@ ¡ 2 ¢
@ ¡
r§vr
r =
º §r3 - 0
@r
@r
¢
1
@ ¡
r§v r =
º §r 3 -0
0
2
(r -) @r
Now we substitute back into equation 110:
µ
¶
¢
@§
1 @
1
@ ¡
3 0
+
º§r =0
@t
r @r (r2 -)0 @r
p
p
p
¡
¢0
Now for a keplerian disk, - = GM =r 3 ; so r2 - = GM r and r2 - = 12 GM=r:
Also, - 0 = ¡ 32 -=r: Thus:
Ã
Ã
!!
r
@§
1 @
2
@
3
GM
2
p
= ¡
¡ º§r
@t
r @r
2
r3
GM =r @r
µ
¶
3 @ p @ ¡ p ¢
=
r
º§ r
r @r
@r
64
Then
@
@r
@
@¸ @±§
@¸ 3 @
±¸ =
=
@t
@§ @t
@§ r @r
Letzs look at
µ
¶
p @ ¡ p ¢
r
±¸ r
@r
=
=
=
µ
p @ ¡ p ¢
r
±¸ r
@r
¶
(111)
1 ¡1=2 @ ¡ p ¢ p @ 2 ¡ p ¢
r
±¸ r + r 2 ±¸ r
2
@r
@r
µ
¶
µ
¶
p @ p @
1 ¡1=2 p @
1
1
r
r
(±¸) + p ±¸ + r
r
(±¸) + p ±¸
2
@r
2 r
@r
@r
2 r
µ
¶
p
p @2
1 @
1
1 @
1
1 @
p
(±¸) +
±¸ + r
(±¸) + r 2 (±¸) ¡ 3=2 ±¸ + p
(±¸)
2 @r
4r
2 r @r
@r
4r
2 r @r
1 @
1
1 @
@2
1
1 @
(±¸) +
±¸ +
(±¸) + r 2 (±¸) ¡ ±¸ +
(±¸)
2 @r
4r
2 @r
@r
4r
2 @r
µ
¶
3 @
@2
1 @
@
@
=
(±¸) + r 2 (±¸) =
(±¸) +
r
(±¸)
2 @r
@r
2 @r
@r
@r
1 @
=
(±¸) + rr2 ±¸
2 @r
Thus equation 111 is of the form:
@
@¸ 2
@¸ 3 @
±¸ =
3r ±¸ +
(±¸)
@t
@§
@§ 2r @r
which is a diffusion equation for ±¸:
=
The standard form of the diffusion equation is:
@F
= Dr2 F
@t
To solve it, separate variables: F = X (x) T (t) ; to get
1 dT
D d2 X
=
= constant = k
T dt
X dt2
The solution is:
T = T0 ekt
and
p
X = X0 e k=Dx
For stability,
µ qwe need
¶ k < 0 (perturbation decays in time) in which case X =
jkj
X0 exp §i D x which is an oscillating disturbance in space if D > 0. For the disk,
spatial growth is as bad as temporal growth, since the disk is disrupted either way. So we
need D > 0:
Digression: solving the equation for ±¸
Our equation is a little more complicated
because of the extra term, but the same general ideas apply.
µ
¶
@
@¸ 3 @ p @ ¡ p ¢
±¸ =
r
±¸ r
@t
@§ r @r
@r
65
Let ±¸ = R (r) T (t) : Then
1 dT
1 @¸ 3 @
=
T dt
R @§ r @r
Then T = T0 e kt and
µ
¶
p @ ¡ p ¢
r
R r
= constant = k
@r
µ
¶
@ p @ ¡ p ¢
kRr
r
R r
=
@r
@r
3D
@¸
where D = @§
: Expand the LHS:
@ 2 R 3 @R krR
+
¡
=0
@r2
2 @r
3D
³p
´
Look for a solution of the form R = g (r) exp
k=3Dr : Then:
³p
´ r k
³p
´
@R
dg
=
exp
k=3Dr +
g exp
k=3Dr
@r
dr
3D
and
Ã
!
r
³p
´
@2R
d2 g
k dg
k
=
+2
+
g exp
k=3Dr
2
2
@r
dr
3D dr
3D
Stuf®ng into the differential equation gives:
Ã
!
Ã
!
r
r
d2 g
k dg
k
3 dg
k
kr
r
+2
+
g +
+
g ¡
g = 0
dr2
3D dr
3D
2 dr
3D
3D
à r
!
r
d2 g
k
3 dg
3
k
r 2 + r2
+
+
g = 0
dr
3D
2 dr
2 3D
P
which is ugly. Look for a series solution: g = n an rn+p
à r
!
r
X
X
k
3
3
k X
n+p¡ 2
n+p¡1
r
an (n + p) (n + p ¡ 1) r
+ 2r
+
an (n + p) r
+
an r n+p
3D
2
2 3D
Ã
Ã
!
!
r
r
X
k
3 n+p¡ 1
3
k n+p
n+ p¡1
n+p
an (n + p) (n + p ¡ 1) r
+ (n + p) 2r
+ r
+
r
3D
2
2 3D
r
Looking at the lowest power of r; which is r p¡1 ; we get:
3p
a0 p (p ¡ 1) +
a0 = 0
2
which has roots p = 0 pr p = 1 ¡ 32 = ¡ 12 : Then the recursion relation is found by looking
at the power r p+m :
r
r
k
3
3
k
am+1 (p + m + 1) (p + m) + am (m + p) 2
+ am+1 (p + m + 1) +
am = 0
3D
2
2 3D
66
= 0
= 0
am+1
r
k
3=2 + 2 (m + p)
3D (p + m + 1) (p + m + 3=2)
r
am
3=2 + 2 (m + p)
k
¡
(p + m + 1) (p + m + 3=2)
3D
=
¡am
=
So for p = 0 we get:
am
=
am¡ 1 3=2 + 2 (m ¡ 1)
¡
m
m + 1=2
=
(¡1)m
giving the solution
R = R0 exp
k
3D
a0 2m ¡ 1=2 2m ¡ 5=2
3=2
¢ ¢¢
m! m + 1=2 m ¡ 1=2
3=2
µ
k
3D
¶m=2
µ
¶ m=2
1
m
´X
(¡1) 2m ¡ 1=2 2m ¡ 5=2
3=2 kr 2
k=3Dr
¢¢¢
m! m + 1=2 m ¡ 1=2
3=2 3D
m=0
³p
while for p = ¡1=2
am
=
=
giving
r
am¡1 2m ¡ 1
¡
m ¡ 1=2 m
m
(¡1)
r
k
3D
a0 2m ¡ 1 2m ¡ 3
1
¢¢¢
m! m ¡ 1=2 m ¡ 3=2
1=2
µ
k
3D
¶ m=2
µ 2 ¶ m=2
1
³p
´X
R0
(¡1)m 2m ¡ 1 2m ¡ 3
1
kr
R = p exp
k=3Dr
¢¢¢
r
m!
m
¡
1=2
m
¡
3=2
1=2
3D
m=0
Both series converge, so our conclusion about stability is not changed. We need D > 0 for
stability.
3.2.8 The two-temperature accretion disk
Reference: Shapiro, Lightman and Eardley. Ap.J. 204, 187, 1976
Consider an optically thin disk, where cooling is primarily by Bremsstrahlung. Then
¤ = 5 £ 10 20 ½2 T 1=2 erg ¢ s¡1 ¢ cm¡3
and therefore the heat removed from each unit area of the disk is
Q¡ = 2¤H = 10 21 ½2 HT 1=2 erg ¢ s¡1 ¢ cm¡2
Equate this cooling rate to heating by viscous dissipation (equation 107):
Ã
µ ¶ ¡1=2 !
3 dM GM
r
1¡
= 10 21 ½2 HT 1=2 erg ¢ s ¡1 ¢ cm ¡2
4¼ dt r 3
R¤
67
(112)
We previously found
r
cs
cs
=
r=
¹
vÁ
µ
¶ 1=2
r
¡
¢
Ti + T e
r
8
=
4:
0
£
10
cm/s
r
9
10 K
GM
Now letzs express the radius in terms of the Schwarzschild radius of a mass M :
2GM
rs =
c2
c2 r
r¤ = 2r=rs =
GM
Then:
r
µ
¶ 1=2
¡
¢ r¤ GM
Ti + Te
8
H =
4: 0 £ 10 cm/s
r¤ 2
10 9 K
c2
c
¡
¢
¡11
33
¢ 3=2 6:7 £ 10
2 £ 10
M
1=2 ¡
8
= T9
4 £ 10 r¤
3
10
M¯
(3 £ 10 )
¡
¢
M
1=2 3=2
=
2 £ 10 3 cm T9 r¤
M¯
From equations 103 and 109, we ®nd § :
dM §=
dt 2¼®c2s
and then
§
dM
dM - 2
½ =
=
=
2H
dt 4¼®c2s H
dt 4¼®c3s
¡ 17
¢
GM
= 10 g/s M17
3 3=2
3
r 4¼® (4: 0 £ 108 cm/s) T 9
¡ 17
¢
µ 2 ¶3
10 g/s
M 17 GM
c
=
3
3=2 r3
8
GM
4¼ (4: 0 £ 10 cm/s) ®T 9
¤
¡ 17
¢
10 g/s
M 17 1
c6
=
3
2
3=2
3
4¼ (4: 0 £ 10 8 cm/s) ®T 9 r¤ (GM )
¡ 17
¢
¡
¢6
2
10 g/s
3 £ 1010 cm/s
M 17 1
(M ¯ =M )
=
4¼ (4: 0 £ 10 8 cm/s)3 ®T 93=2 r¤3 (6:7 £ 10 ¡8 cm3 =gs2 )2 (2 £ 10 33 g)2
µ
¶
¡
¢ M¯ 2 M 17 1
= 5: 0 g/cm 3
3=2 r 3
M
®T
¤
H
=
9
68
Then we can solve equation 112 for the temperature:
T
1=2
=
'
1=2
10 4:5 T 9
=
1
3 dM GM
21
2
¡1
¡
2
10 ½ Herg ¢ s ¢ cm 4¼ dt r 3
µ
³
´2
¯
1021 5: 0 M
M
5=2 9=2 M ¯
®2 T 9 r¤
and thus
This disk is very hot!
1
¶2 ³
M17 1
3=2 r 3
¤
®T9
1
M M 17 r¤3
T92
=
T9
=
Ã
1¡
M M 17
M ¯ ® 2 r3=2
¤
µ
¶ 1=2 1=2
M
M 17
40
3=4
M¯
®r¤
69
r
R¤
¶ ¡1=2 !
1=2 3=2 M
r¤ M ¯
2 £ 103 T9
19: 385
1: 6 £ 10 3
µ
3
GM
17
´ 4¼ 10 M 17 r¤3
µ
c2
GM
¶3