PHYS833 Astrophysics of Compact Objects Notes Part 10

Compact objects in close binaries

A number of interesting and energetic phenomena occur in close binary systems in which one or

possibly both stars are compact objects and the orbital dimensions are such that mass transfer

occurs, either from a wind or Roche-lobe overflow.

The Roche potential

Due to tidal forces, the orbits of the stars in close binaries rapidly circularize. Consider a

coordinate frame with origin at the center of mass that co-rotates with the binary system with

angular velocity Ω. The velocity, in an inertial frame, of a particle whose position vector in the

co-rotating frame is r, is

.= + ×v r Ω r (9.1)

The inertial frame acceleration is

( )2 .= + × = + × + × ×a v Ω v r Ω r Ω Ω r (9.2)

The second term of the last expression on the right is the Coriolis acceleration and the last term is

the centrifugal acceleration.

Since

( ) ( ) ( ) ,× × = ⋅ − ⋅Ω Ω r Ω r Ω Ω Ω r (9.3)

the centrifugal acceleration can be written in terms of a potential

( ) ,φΩ× × = ∇Ω Ω r (9.4)

where

( ) ( )( )21

.2

φΩ = ⋅ − ⋅ ⋅ Ω r Ω Ω r r (9.5)

The equation of motion for a fluid in the rotating frame is

1

2 ,G

dp

dtφ φ

ρΩ+ × = − ∇ − ∇ − ∇

uΩ u (9.6)

where the gravitational potential, ϕG, satisfies Poisson’s equation

2

4 .G Gφ π ρ∇ = (9.7)

If the stars are also co-rotating with the orbit, then u = 0 and hence

1

0,pρ

∇ + ∇Φ = (9.8)

where

.Gφ φΩΦ = + (9.9)

Taking the curl of equation (9.8) gives

2

10,pρ

ρ− ∇ ×∇ = (9.10)

which means that surfaces of constant ρ and p coincide. Similarly, multiplying equation

(9.8) by ρ and then taking the curl gives that surfaces of constant ρ and ϕ coincide.

Hence the stellar surface, where the pressure and density tend to a constant value of zero, is an

equipotential surface. The problem of determining the shape of a star in a close binary is then

solved by finding the equipotential surfaces, ϕ = constant. This can be done numerically or by

using series expansions for the case in which p is a known function of ρ, e.g. for a cold white

dwarf or a polytrope. Since most stars are centrally condensed, an adequate approximation, due

to Roche, is to assume that the gravitational field from each star is that of a point mass at the

stellar center. In this Roche approximation,

( ) 1 2

1 1

,G

GM GMφ = − −

− −r

r r r r (9.11)

where M1 and M2 are the masses of the two stars, which are taken to have centers at r1 and r2.

Before writing explicit expressions for the Roche equipotentials, it is convenient to shift

the coordinate origin to the center of one of the stars, say star 1. The scale of the system is set by

the binary separation, a. In Cartesian coordinates (x, y, z) with the x-axis along the line joining

the stars and the z-axis parallel to the rotation axis, the total potential in the Roche approximation

is

( )( )

22 21 1

1 2 1 22 2 2 2 2 2

1

2

GM GMx a y

x y z x a y z

µ Φ = − − − Ω − + + + − + +

(9.12)

where

2

1 2

.M

M Mµ =

+ (9.13)

The center of mass of the system has position vector, ( ),0,0 .CM aµ=r

By Kepler’s third law,

( )1 22

3.

G M M

a

+Ω = (9.14)

Hence when a is used as the length scale,

( ) ( )

( )2 2

1 2 1 22 2 2 2 2 21

1 1.

21

ax y

G M M x y z x y z

µ µµ

Φ − Φ = = − − − − + + + + − + +

(9.15)

The topology of the potential surfaces is determined solely by the parameter, μ.

Because the potential is an even function of z, any stationary points of the potential must lie in

the plane z = 0. Hence it is instructive to consider equipotential lines in this ‘equatorial’ plane,

for which

( )( )

2 2

1 2 1 22 2 2 2

1 1.

21

x yx y x y

µ µµ

− Φ = − − − − + + − +

(9.16)

The positions of stationary points are the simultaneous solutions of

( )( )

( )3 2 3 22 2 2 2

11 0,

1

x xx

x x y x y

µ µ µ∂Φ −

= − − − − =∂ + − +

(9.17)

and

( )( )

3 2 3 22 2 2 2

1 0.

1

y yy

y x y x y

µ µ∂Φ

= − + − =∂ + − +

(9.18)

From equation (9.18), we find that either y = 0 or

( )3 2 3 2

2 2 2 2

11 0.

1x y x y

µ µ−+ − =

+ − +

(9.19)

First consider the case y = 0. From equation (9.17),

( ) 3 3

11 0.

1

x xx

x xµ µ µ

−− − − + =

− (9.20)

By considering cases the three cases i) x > 1, ii) 1 > x > 0, and iii) 0 > x separately, we can find

equations to solve for x, given μ. A simpler way is to consider μ as a function of

x,

( ) ( )

( ) ( )

31 1 1

.1 1 1 1

x x x

x x x x x x x xµ

− − −=

− − − − − − (9.21)

This is plotted below:

-1.0

-0.5

0.0

0.5

1.0

1.5

2.0

0.0 0.2 0.4 0.6 0.8 1.0

µ

x

We see that for each value of μ there are 3 stationary points on the x-axis.

Now consider the case corresponding to equation (9.19). Using this with equation (9.17),

we find two more stationary points at the intersection of the circles

2 2 1,x y+ = (9.22)

and

( )2 21 1.x y− + = (9.23)

These circles are centered on the two stars and have radii equal to the binary separation. Hence

the vectors from the two stars to each of this pair of stationary points form equilateral triangles

with side length equal to the binary separation.

The Roche potential has 5 stationary points, which are called Lagrangian points and are

labeled L1 through L5.

The figure below shows equatorial contours of the Roche potential for the case in which

the two stars have the same mass.

We can see that there are 3 X-points (saddle points) on the x-axis and 2 O-points, each at a vertex

of an equilateral triangle, with other two vertices being at the positions of the stars.

The figure below is for the case M1 = 2M2.

-1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5

-2.0

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

2.0

y

x

Note that the O points, usually referred to as the L4 and L5 points, are potential maxima and

hence are positions of unstable equilibrium (for this mass ratio).

The figure below is an enlargement of the inner parts of the above figure that shows the

location of the 5 Lagrangian points.

The equipotential surface through the L1 point encloses two cusped volumes which are called

Roche lobes.

-1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5

-2.0

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

2.0y

x

-1.0 -0.5 0.0 0.5 1.0 1.5 2.0

-1.0

-0.5

0.0

0.5

1.0

L4

L5

L3y

x

L1 L

2

The Roche lobe radius, RL, (which is defined so that the volume of the Roche lobe is 34 3LRπ ) is, for star 1, approximately

2 3

,1

2 3 1 3

0.49,

0.60 ln(1 )

LR q

a q q=

+ + (9.24)

where the mass ratio 1 2 .q M M=

The Roche lobe radii are plotted as a function of q below.

Another often used approximation is

1 3

,10.49 ,

1

LR q

a q

=

+ (9.25)

which is valid for 0.8.q ≤

When the two stars are both sufficiently small that they are inside their Roche lobes, the

binary system is said to be detached. As stars evolve they usually expand, with the initially more

massive star evolving more quickly. When one star fills its Roche lobe, the system is said to be

semi-detached. When Roche lobe overflow occurs mass is transferred to the other star. If both

stars fill their Roche lobes, a contact binary is formed. The shared envelope can co-rotate with

the binary up to the point that it fills the equipotential containing the L2 point. Further envelope

expansion will lead to mass loss from the system through the L2 point.

0.0 0.2 0.4 0.6 0.8 1.0

0.10

0.15

0.20

0.25

0.30

0.35

0.40

0.45

0.50

0.55

0.60

0.65

0.70

0.75

0.80

0.85

RL

q = M1/M

2

R1

R2

R1+R2

Roche lobe overflow

For a semi-detached system, one of the stars fills its Roche lobe. Matter on the surface of this star

is able to flow through the L1 point into the Roche lobe of the other star, which we will take to be

star 2. The surface temperatures of stars are relatively low in the sense that the thermal velocity

is much less than the escape velocity. For example, the Sun has a photospheric temperature of

6000 K, which corresponds to a thermal velocity of about 9 km s-1. The solar escape velocity is

600 km s-1. Hence the sound speed in the material at the L1 point is much less than the orbital

velocities of the stars. This means that the flow into the Roche lobe is supersonic. This simplifies

the problem of calculating the nature of the gas flow, because the internal pressure of the flowing

material can be neglected. The fluid essentially follows the path of a test particle injected at the

L1 point with a speed equal to the sound speed of material at the stellar surface.

Note that, in this picture, stellar coronae, chromospheres, winds and activity are being

neglected. These can lead to mass transfer in detached systems. A possible example of where this

might be happening is the binary V471 Tau.

In the co-rotating frame the equation of motion for a test particle is

2 .+ × = −∇Φr Ω r (9.26)

Taking the scalar product with r and then integrating with respect to time, we get

( )21,

2JE+ Φ =r r (9.27)

where EJ is a constant, known as Jacobi’s constant, which expresses the conservation of kinetic

energy and total potential energy in the co-rotating frame. Note that this result is possible

because the Coriolis ‘force’ does no work.

Because energy is conserved along the particle trajectory, particles released from rest at

L1 cannot escape from the Roche lobe. Similarly the stream of material entering the

Roche lobe of star 2 through the L1 point remains inside the Roche lobe. Because the pressure in

the stream is small, the stream remains thin and coherent, and confined mainly to the orbital

plane.

The stream is initially directed towards star 2 but due to the Coriolis effect is deflected

behind the star. If the stream does not hit star 2 directly, it will pass by, reaching some minimum

distance from star 2 before moving away again. However because the stream cannot leave the

Roche lobe it reaches a maximum distance from star 2 before falling back again. The motion of

the stream is most easily found by solving the test particle equation of motion numerically.

Using the binary separation as the length unit and 1−Ω as the time unit, the equation of

motion is

ˆ2 ,+ × = −∇Φr z r (9.28)

which, in component form, is

( ) ( )

( )

( )

( )

( )

( )

3 2 3 22 2 2 2 2 2

3 2 3 22 2 2 2 2 2

3 2 3 22 2 2 2 2 2

1 12 ,

1

12 ,

1

1.

1

x xx y x

x y z x y z

y yy x y

x y z x y z

z zz

x y z x y z

µ µµ

µ µ

µ µ

− −− = − − + −

+ + − + +

−+ = − − +

+ + − + +

−= − −

+ + − + +

(9.29)

The figure below shows the stream trajectory for equal mass stars.

Note that if the stream does not hit star 2, then it collides with itself. If this happens, since the

stream is moving supersonically, there is shock dissipation of the relative kinetic energy and

heating of the gas to high temperatures. The gas can then radiate the thermal energy gained in the

collision and cool. Note that the collision occurs close enough to star 2 that we can ignore the

gravity of star 1. This means that the collision conserves angular momentum of the gas relative

to star 2. Since the minimum energy for fixed angular momentum occurs for a circular orbit, it is

expected that after the collision the gas settles in a ring around star 2.

An estimate of the circularization radius (the radius of the ring), Rc, is obtained by

equating the angular momentum of a particle at L1 relative to star 2 to that of a particle in a

circular orbit around star 2,

2

2 2 ,cb GM RΩ = (9.30)

where b2 is the distance of the L1 point from the center of star 2. A reasonably accurate

approximation for b2 in the range 0.1 < q <10 is

( )2 0.5 0.0986ln .b a q= − (9.31)

Hence

( ) ( )4

0.5 0.0986ln 1 .cRq q

a= − + (9.32)

We can use this to compare the circularization radius to the radius of the Roche lobe of star 2,

which is given by equation (9.24) but with q replaced by q-1. We find

( ) ( )2 3 1 3

4

,2

0.60 ln(1 )0.5 0.0986ln 1 .

0.49

c

L

R q qq q

R

−+ += − + (9.33)

This is plotted below.

Hence the circulation radius is typically 30 – 50% of the Roche lobe radius. If the radius of star 2

is greater than Rc, the ring cannot form. Instead the stream hits the star directly.

However for compact objects orbiting a non-compact star, the stellar radius will be much less

than the Roche lobe radius and the ring will form.

Accretion disks

In the ring of material, dissipative processes (such as viscosity and shocks) will convert kinetic

energy of the ordered bulk motion into heat. The heat will radiate away and energy is lost from

the ring. The only way that the gas can respond to the energy loss is to sink deeper into the star’s

potential well. To do this, the gas must also lose angular momentum. The time scale for this to

occur is normally long compared to the dynamical time scale and hence gas elements slowly

move in towards the star, by following nearly circular orbits with ever decreasing radius.

-1.0 -0.8 -0.6 -0.4 -0.2 0.0 0.2 0.4 0.6 0.8 1.0

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

RC/R

L,2

log q

However, in the absence of external torques, the total angular momentum of the ring material

must be conserved. Hence some mass must also move out away from the star into higher angular

momentum orbits. Thus dissipation causes the ring to spread into a disk. Since material gets

added to the star, the disk is known as an accretion disk. In cataclysmic variable systems the

mass in the accretion disk is small compared to the mass of the stars and hence its self-gravity

can be neglected. The disk is also geometrically thin. For an excellent review of thin accretion

disks, see Pringle, J. E. 1981, ARAA, 19, 137.

The luminosity of a steady state accretion disk around a compact object can be

determined quite simply. Suppose mass joins the outer edge of the disk at a rate dM/dt.

For a steady state disk, this is also the rate of accretion by the compact object. The energy per

unit mass for a circular orbit of radius r around a mass M is

1

.2

GM

rφ = − (9.34)

For a compact object, the outer edge of the disk is at large radii compared to the size of the

compact object. Hence the luminosity of the disk i.e. energy radiated by the disk per unit time, is

1

,2

GM dML

R dt= (9.35)

where R is the inner radius of the disk. If the nature of the compact object is known, then

a measurement of the bolometric luminosity of the disk gives an estimate of the mass transfer

rate in the binary. At the inner edge of the disk, for material to be accreted by the star, excess

kinetic energy must be dissipated. If the star is not rotating and the excess kinetic energy is

dissipated in a thin boundary layer, the luminosity of the boundary layer is also given by

equation (9.35).

The specific angular momentum of the disk material is GMr. Hence for accretion to occur

most of the angular momentum must be transported outwards. There has to be torques acting in

the disk for this to occur. It is likely that tidal torques from the companion acting on the outer

edge of the disk transfers the angular momentum of the disk to the orbital motion.

The mass transfer rate

To see what determines the mass transfer rate, it is useful to consider the total orbital angular

momentum of the binary about it center of mass (CM). (Because the stars are centrally

condensed the spin angular momentum is small compared to the orbital angular momentum.) Let

a1 and a2 be the distances of star 1 and star2 from the CM, respectively.

The orbital angular momentum is

2 21 1 2 2 .J M a M a= Ω + Ω (9.36)

From the definition of the CM,

1 1 2 2.M a M a= (9.37)

Since 1 2 ,a a a= + this gives

21

1 2

12

1 2

,

.

Ma a

M M

Ma a

M M

=+

=+

(9.38)

Hence

2 2

21 2 1 2

1 2 1 2

.M M GM M a

J aM M M M

= Ω =+ +

(9.39)

Suppose some unspecified process changes the orbital angular momentum and the stellar masses

by small amounts. The resulting change in orbital separation is, from equation (9.39), given by

1 2

1 2

2 2 2 ,a J M M M

a J M M M

∆ ∆ ∆ ∆ ∆= − − + (9.40)

where M = M1 + M2.

From equation (9.24), the Roche-lobe radius of star 1 changes according to

( ),1

,1

12 ,

3

L

L

R a qg q

R a q

∆ ∆ ∆= + − (9.41)

where

( )( )

1 3 2 3 1 3

1 32 3 1 3

1 1.2 1.2.

10.6 ln 1

q q qg q

qq q

+ +=

++ + (9.42)

First consider conservative mass transfer, for which the system mass, M, and orbital angular

momentum, J, do not change. Suppose an amount of mass 2M∆ is transferred from star 1 to star

2. Since 0M∆ = and 0J∆ = for conservative mass transfer

2

1 2

1 12 ,

aM

a M M

∆= − ∆

(9.43)

and, after some algebra,

( ) ( ),1 2

,1 2

14 8 1 ,

3

L

L

R Mq q g q

R q M

∆ ∆= − + + (9.44)

From equation (9.43), we see that if star 1 is more massive than star 2, then the orbital separation

decreases. From equation (9.44), we find that the Roche lobe radius of star 1 decreases if star 1 is

greater than about 0.8 times the mass of star 2. Hence in this case, unless star 1 can adjust its

radius to the shrinking Roche lobe, the overflow of star 1 will increase and there is the potential

for runaway mass loss.

To be able to determine whether the donor (star 1) can adjust its radius, we need to know

how the stellar radius depends on the star’s mass. This can be determined from stellar models or

from measurement of the masses and radii of stars in double line eclipsing binaries. The figure

below compares result from the two approaches. The model stars are assumed to have solar

composition. The mass and radius units are solar mass and solar radius, respectively.

We see that the measured radii are greater than the theoretical radii. A plausible

explanation is that the discrepancy is due to the effects of magnetic fields on the structure of low

mass either because the presence of a strong enough magnetic field inhibits convective energy

transport or produces dark spots on the stellar surface. In either case, the radius of a model star is

found to be larger than in the absence of magnetic fields. There is strong evidence for the

presence magnetic fields in the low mass binaries. All the stars are flare stars and show light

curve variations consist with the presence of large dark spots. The magnetic fields are likely the

result of strong dynamo action in rapidly rotating stars. The binaries have orbital periods of order

1 day and because the stellar radii are relatively large compared to the orbital separations, tidal

interactions are expected to make the stars synchronously rotate with the orbit. Because the

orbital periods in cataclysmic variables are typically of order hours, the donor stars are likely to

have magnetic fields equal or larger than those in the eclipsing binaries and are likely to be larger

than non-magnetic stars.

A fit to the models of population I main sequence stars of mass 0.8 M≤

in thermal

equilibrium gives

0.85

0.82 ,R M

R M

(9.45)

and fitting the measured radii of the M dwarfs in the eclipsing binaries gives

-0.6 -0.4 -0.2 0.0

-0.7

-0.6

-0.5

-0.4

-0.3

-0.2

-0.1

0.0lo

g R

log M

ZAMS

TAMS

Eclipsing binaries

0.93

0.98 ,R M

R M

=

(9.46)

If the radius of star 1 can adjust to its equilibrium value, then

1 1 2

1 1 2

,R M M

R M q M

αα

∆ ∆ ∆= = − (9.47)

where 0.85 0.93.α ≈ − To avoid a large Roche lobe overflow, we need 1 ,1 .LR R∆ > ∆ Equations

(9.44) and (9.47) give that this can happen only if

( ) ( )3 4 8 1 .q q g qα + > − + (9.48)

which requires q < 1.2. If this condition is violated, runaway mass transfer ensues and continues

until the mass ratio is reversed. Hence in interacting binary systems, it is quite common to find

that the mass losing star is less massive than the mass gaining star.

Mass transfer in cataclysmic variables

Cataclysmic variables (CVs) are binaries containing a white dwarf that is accreting matter from a

low mass companion, which, with a few exceptions, is a low mass main sequence star. This

group of objects includes dwarf novae, nova-like variables and classical novae.

Dwarf novae are blue stars that irregularly increase in brightness by 3 to 5 magnitudes for

a few days or weeks. Classical novae are stars that increase in brightness by 7 to 20 magnitudes

over a period of 1 or 2 days and then decline more slowly. The return to pre-eruption brightness

takes decades or longer. Classical novae also eject a shell of material of mass 10-5 to 10-4 M at

velocities from 300 km s-1 up to 1500 km s-1. The ejected material is often substantially enhanced

in heavy elements, which is indicative of mixing of accreted material and white dwarf core

material. Nova-like variables, also called UX UMa stars, show irregular variations of a few

tenths of a magnitude about their mean brightness. Similar variations are seen in post novae. CV

orbital periods are typically in the range 1 to 15 hrs, with a notable absence of systems with

periods between 2 and 3 hrs.

In general, systems with periods less than 2 hrs are less luminous than systems with

periods greater than 3 hrs (Patterson, J. 1984, ApJS, 54, 443). The mass transfer rates are

estimated from the disk luminosity to range from 10-11 to 10-10 M yr-1 for dwarf novae in

quiescence up to 10-9 to 10-8 M yr-1 for dwarf novae in eruption and post-nova systems.

The binary separation is related to the orbital period by

( )2 2

31 2 1

1.

2 2

P q Pa G M M GM

qπ π

+ = + =

(9.49)

As discussed above the white dwarf star is more massive than its companion. Hence typically the

mass ratio is less than unity, q < 1. Hence, using equation (9.25),

2

3 3 3 3,1 10.49 0.49 .

1 2L

q PR a GM

q π

=

+ (9.50)

Since the companion fills its Roche lobe, this last expression relates the mean density of the

companion to the orbital period,

31 2

93 g cm ,

hrPρ −= (9.51)

where Phr is the orbital period in hours. The mean density of the eclipsing binary M dwarfs is

1.79

3.1.49 g cm .M

Mρ

−

−

(9.52)

Hence the companion mass is related to orbital period by

1.12

.8.0

hrM P

M

≈

(9.53)

This indicates that the mass losing stars in CVs have masses in the range ~0.15 to ~2.0 M. Note

that the stars in the longer period systems must be less massive than indicated to avoid runaway

mass loss and hence we conclude that they are evolved from the ZAMS.

The nuclear burning evolutionary time scale for a low mass main sequence star is roughly

3

1010 .nuc

M

Mτ

−

≈

(9.54)

Hence if expansion due to evolution drives the mass transfer, the rate is expected to be of order

4 4.48

10 1 10 11 12 10 M yr 10 M yr .

8.0

hr

nuc

M M PM

Mτ− − − −

≈ =

∼ (9.55)

A comparison with the observed rates shows that nuclear evolution is not the driving mechanism

for the shorter period systems. For example, below the period gap the nuclear time scale mass

transfer rate is < 2 10-13 M yr-1, which is an order of magnitude less than even the minimum

observed rates.

Two probable mechanisms for driving mass transfer in CVs have been identified. Both

involve angular momentum loss from the system. For systems with periods less than 2 hr, the

likely mechanism is general relativistic radiation of gravitational waves (Paczynski 1967, Acta

Astronomica, 17, 287). Orbital angular momentum is lost from a circular orbit at a rate

( )

8 35 31 2

1 35

1 2

32 2.

5

GRJ G M M

J c PM M

π = −

+

(9.56)

Using equation (9.53), we find for a short period system

( )

2 3 1.5

11 12

1 3

15.0 10 yr .

8.01

GR hrJ M P

J Mq

−

− − = − ×

+

(9.57)

Hence for a 1 M white dwarf, the angular momentum loss time scale is

( )1.5

1 39 12.5 10 1 yr .2.0

hrGR

Pqτ −

× +

∼ (9.58)

A rough estimate of the corresponding mass transfer rate is

( )0.4

1 311 112 9 10 1 M yr .

2.0

hr

GR

M PM q

τ

−−− −

× +

∼ ∼ (9.59)

which is adequate to drive mass transfer in the shorter period systems. Even so some other

process is needed to drive mass transfer at the highest observed rates.

The second angular momentum loss mechanism is magnetic braking of a stellar wind

(Verbunt & Zwaan 1981, AA, 100, L7). For main sequence G stars, the equatorial rotation

velocity is observed to be correlated with age (Skumanich 1972; Smith 1979),

14 1 2 110 cm s .ev f t− −= (9.60)

where t is the star’s age in seconds, and f is a constant of order unity. The angular momentum of

a star rigidly rotating with angular velocity ω is

2 2J k MR ω= (9.61)

where the dimensionless quantity k depends on the density profile of the star. Assuming that the

observed spin-down is due to angular momentum loss,

2 2 14 3 2 2 4 2 28 31 110 10 .

2 2

edJ dvk MR k MRf t k MR f

dt dtω− − −= = − = − (9.62)

Verbunt & Zwaan assumed that the mass losing stars in CVs and low-mass X-ray binaries also

lost angular momentum according to equation (9.62). By further assuming that the rotation of

mass losing star is tidally locked to the binary, we now have an expression for the angular

momentum lost from the system.

Taking k2 = 0.1, f = 1, and

0.93

0.98 ,R M

R M

=

(9.63)

we find

( )2.03 9 2

3 2 ,137 0.53 25.0 10 1 ,LMB RdJ M

q qdt M a

= − × +

(9.64)

where we have assumed that the donor fills its Roche lobe. Using equation (9.39),

( )0.07 5

2 ,17 0.94 1212.6 10 1 yr ,

LMB RdJ Mq q

J dt M a

− − − = − × +

(9.65)

On using equations (9.25) and (9.53),

( )2 30.82

1 39 1217.4 10 1 yr ,

8.0

MB hrdJ P Mq

J dt M

−

− − = − × +

(9.66)

The angular momentum loss time scale is now sufficiently short to drive mass loss at the highest

observed rates.

The ratio of angular momentum loss rates from the two mechanisms is

( )4 3

2 3 2.32 21.2 1 .MBhr

GR

J Mq P

J M

−

= +

(9.67)

Hence it appears that angular momentum loss from the magnetically braked wind is dominant

even at the shortest periods. However it is often argued that fully convective stars cannot

maintain a magnetic field strong enough to couple effectively to the wind. Main sequence stars

are fully convective for masses less than about 0.35 M. From equation (9.53), we find this

corresponds to a period of about 3 hrs, which is the period of the upper end of the CV period gap.

Hence one idea to explain the existence of the period gap is that the angular momentum loss

drops dramatically when the mass losing star becomes fully convective. This results in a much

reduced mass transfer rate, and so the system becomes much fainter. Radiation of gravitational

waves becomes the dominant angular momentum loss mechanism and the mass transfer rate

increases as the period decreases, so that the system brightens again. However there is no change

in coronal X-ray emission at spectral types that correspond to the transition to fully convective,

which indicates that field strength does not change significantly at the transition. An alternative

scenario is that it is the magnetic field topology that differs between the fully and partial

convective stars. A decrease in the number of open field lines will reduce the angular momentum

loss rate.

The temperature distribution in the accretion disk

The viscous stresses responsible for the outward transport of angular momentum also transport

energy. Shakura and Sunyaev (1973) showed that the rate of dissipation of energy between radii

r and r+dr is

1 2

3

31 .

4

GMM RD

r rπ

= −

(9.68)

We can now estimate the temperature at the surface of the disk if it is optically thick by equating

this to the energy radiated from a black body

42 2 2dL T rdr D rdrσ π π= = (9.69)

(The additional factor 2 comes from accounting for the two sides of the disk). Hence

1 2

4

3

31 .

8

GMM RT

r rπσ

= −

(9.70)

The maximum temperature in the disk occurs at 49 36r R= and is

1 4 3 4

4

9 8 15.6 10 K.

10 cm 10 M yrmax

M R MT

M

−

− −

= ×

(9.71)

The outer edge of the disk is usually at much lower temperature, ~ 5,000 K.

Other contributions to the radiation from the system come from the accretion stream, the

‘hotspot’ where the accretion stream hits the disk, the boundary layer between the disk and the

white dwarf, and the two stars.

Magnetic CVs

If the white dwarf has a strong enough magnetic field, the accretion disk will not extend down to

the stellar surface. Instead the magnetic field channels material from the inner edge of the disk to

a region near the magnetic polar caps of the white dwarf. If the magnetic field is large enough (>

107 G), there is no accretion disk. Matter follows magnetic field lines from near the L1 point

directly to the white dwarf. This is the case in the AM Her stars. Because the strong white dwarf

magnetic field threads the companion, the rotation rate of the white dwarf is also synchronized to

the orbital period.

The infall kinetic energy is released in a shock near the magnetic pole. If the post shock

material is dense enough it will quickly cool, so that the shock is near the stellar surface.

For a strong shock, the post shock temperature is given by

23 1 3.

8 2 8

T GMv

Rµ

ℜ= ≈ (9.72)

Hence

1

8

94 10 K,

10 cm

M RT

M

−

≈ ×

(9.73)

which indicates that much of the radiation from the shocked gas is emitted at X-ray energies of ~

30 keV. About 50% of the hard X-ray photons will be directed towards the stellar surface, where

they are re-processed into soft X-rays and EUV photons. Observations of AM Her systems show

that the ratio of hard to soft X-ray emission is about 10%, which is much less than expected from

this simple picture. One plausible solution to this difficulty is that there is other cool absorbing

material in the vicinity of the shock.

For magnetic fields of intermediate strength, ~105 - ~107 G, the disk does not extend to

the white dwarf surface. Because of the magnetic coupling between the white dwarf and the disk,

the white dwarf is expected to be rotating faster than the orbital period. This is the case for the

old nova DQ Her, which has a regular modulation of its optical light curve with period 71 s. This

period is due to illumination of the disk from two accretion columns at the magnetic poles of the

white dwarf rotating with a period of 142 s.

In AE Aqr, the magnetic CV with the shortest spin period, 33s, there is no disk. Instead

accreted matter is being centrifugally ejected due to the rapid spin of the white dwarf.

This causes the spin to slow down as is observed.