D fi d p

where f$\ and j33 verify Eqs. (4.82)-(4.84) and the condition that both functions remain finite at r = 0. From Eqs. (4.117) and (4.119) one readily sees that the large-scale motion consists of a constant overall rotation of O(e1/2), a meridional flow of O(e), and a back reaction of the currents of O(e3/2).

The structure of this solution is very similar to that of a nondegenerate star. Of course, Eqs. (4.5) and (4.6) need to be modified. First, allowance must be made for a more general equation of state in the degenerate interior. Second, because energy is released throughout the star, Eq. (4.5) must be replaced by p Tu ■ grad S = div(x grad T) + pE,

where E = —T (dS/dt) is the energy released by cooling (per gram and per second) and x is the coefficient of thermal conductivity in the degenerate interior (or radiative conductivity in the nondegenerate envelope).

Given these two modifications, it is a simple matter to calculate the function u in the bulk of a cooling white dwarf. Away from the surface layers, the frictionless solution, u = uS (say), has the form

G 2m3

which is the net amount of energy crossing the spherical surface of radius r per second. (As usual, we have omitted the subscript "0" from the functions E0, x0, and T0 in the spherical model.) We have also let

The second equality defines the parameter A. Remaining symbols have their standard meanings (see Eq. [4.34]).

As one moves toward the free surface, Eq. (4.121) merely reduces to Sweet's function (4.34), since we have l = L, E = 0, and A = 1 in the nondegenerate envelope. As we know, this frictionless solution is not acceptable near the surface because it does not satisfy the kinematic boundary condition (4.38). We are thus forced to retain turbulent friction in the surface layers and, hence, to make explicit use of the sixth-order equation (4.63) for the function u. By making use of the radiative-zero boundary conditions, one can easily show that Eqs. (4.74)-(4.80) are the appropriate equations for the problem being considered. Once the function u has been calculated from r = 0 to r = R, one can solve Eqs. (4.82) and (4.83) for the functions fii and fi3.

Table 4.3 gives a detailed solution for a O.8M0 white-dwarf model. The functions u and r v are given in cm s-1. They were obtained using the formula pV = 10Nprad, with N = 2, in the nondegenerate envelope (see Eq. [4.62]). This choice of N is unimportant since u and r v depend on (pV)1/10. In Table 4.3 we also list the functions ¡3\ and p3. They were obtained using the viscosity of a degenerate electron gas in the deep interior and the above formula in the outer layers. Figure 4.7 illustrates the meridional flow, which breaks down into three regions with motions in opposite senses. This situation arises because the factor (1 — mE/1) changes its sign twice along the radius. Accordingly, this triple-circulation pattern is a mere consequence of the stratification of the spherical models that were used to obtain the function u. For a typical white dwarf, with equatorial velocity ueq that lu,

10 4. Hence, from Table 4.3 one readily sees ie| (^ er M) < 10-9 cm s-1! Moreover, since

| f3\ | and | 1 are both of order unity, e | ¡3\ | and e\fí3 \ remain in general much smaller than one, so that Eq. (4.119) provides an acceptable solution for the azimuthal motion. As regards practical applications, such as large-scale mixing and microscopic diffusion in the surface layers, we therefore conclude that the meridional currents are utterly negligible in a cooling white dwarf in a state of slow, almost uniform rotation.

Table 4.3. The velocity field in a 0.8M0 cooling white dwarf.

r/R

log(1 - m/M)

u

r v

A

0.000000

0.

0.

0.

0.

0.

0.096517

-0.00413

4.0440E-

12

1.9762E-

12

- 7.5979E-

5

-2.2525E

-4

0.172414

-0.02228

7.7119E-

12

3.6337E-

12

-2.5375E-

4

- 7.4405E

-4

0.292340

-0.09691

1.5276E-

11

6.4466E-

12

- 8.2887E-

4

-2.3373E

-3

0.374016

-0.18709

2.3923E-

11

1.0428E-

11

-1.5421E-

3

-4.1660E-

-3

0.473461

-0.34679

4.4661E-

11

1.9598E-

11

-3.2013E-

3

-7.7537E-

-3

0.583683

-0.60206

9.5232E-

11

3.7716E-

11

- 7.6818E-

3

-1.5030E-

-2

0.698923

-1.00000

2.0127E-

10

1.0222E-

11

-2.1271E-

2

-3.0718E-

-2

0.847698

-2.00000

3.6225E-

10

- 1.1825E-

09

- 8.6396E-

2

-8.2736E-

-2

0.884567

-2.50000

8.4421E-

11

-2.3871E-

09

-1.1174E-

1

-9.8335E-

-2

0.908588

-3.00000

-2.9180E-

10

-2.3595E-

12

- 9.7037E-

2

-8.5441E-

-2

0.924756

-3.50000

-4.5561E-

10

3.0146E-

09

4.6999E-

2

- 4.8949E

-2

0.936511

-4.00000

- 5.8203E-

10

9.5082E-

09

5.1515E-

2

2.0577E

-2

0.950625

-4.50000

- 5.7619E-

10

7.3009E-

09

2.4517E-

1

1.5650E-

-1

0.962850

-5.00000

-4.3912E-

10

8.3732E-

09

3.5125E-

1

2.2961E

-1

0.978424

-6.00000

- 1.4982E-

10

6.9378E-

09

3.9344E-

1

2.5786E

-1

0.987204

-7.00000

2.6621E-

11

2.2088E-

09

3.9532E-

1

2.5892E

-1

0.992409

-8.00000

1.3858E-

10

-6.4318E-

09

3.9504E-

1

2.5866E

-1

0.995597

-9.00000

2.3133E-

10

-2.2086E-

08

3.9492E-

1

2.5855E

-1

0.997503

-10.00000

3.0559E-

10

- 5.2426E-

08

3.9489E-

1

2.5853E

-1

0.998598

-11.00000

3.6040E-

10

- 1.2305E-

07

3.9488E-

1

2.5852E

-1

0.999212

-12.00000

3.7125E-

10

-2.3394E-

07

3.9488E-

1

2.5852E

-1

0.999712

-14.00000

3.7399E-

10

- 6.4598E-

07

3.9488E-

1

2.5852E

-1

0.999838

-15.00000

3.6121E-

10

-1.2407E-

06

3.9488E-

1

2.5852E

-1

0.999909

-16.00000

2.4886E-

10

-1.7597E-

06

3.9488E-

1

2.5852E

-1

0.999949

-17.00000

1.4499E-

10

- 1.8792E-

06

3.9488E-

1

2.5852E

-1

1.000000

infinite

0.

- 1.8986E-

06

3.9488E-

1

2.5852E

-1

Source: Tassoul, M., and Tassoul, J. L., Astrophys. J., 267, 334, 1983.

Source: Tassoul, M., and Tassoul, J. L., Astrophys. J., 267, 334, 1983.

4.6 Meridional circulation in a close-binary component

Consider a system of two rotating stars revolving in circular orbits about their common center of gravity. We have a chemically homogeneous, early-type star of mass M (the primary) acted on by the tidal force originating from its companion of mass M' (the secondary). We shall assume that the radii of the components are much smaller than their mutual distance d , so that the secondary may be treated as a point mass when studying the tidal distortion of the primary. Assuming that the overall rotation of the primary is synchronized with revolution, we have al = ^Ç^. (4,24)

where is the angular velocity of the primary.

4.6.1 The tidally driven currents

Since we want to fix our attention on the primary, it is convenient to choose a rotating frame of reference in which the origin is at the center of gravity of the mass M.

Fig. 4.7. Streamlines of meridional circulation in a O.8M0 cooling white dwarf. The three circulation zones are separated by the spherical surfaces r = 0.88975R and r = 0.98590R. The outer circulation pattern (0.98590R < r < R) is schematically depicted by a single curve. Source: Tassoul, M., and Tassoul, J. L., Astrophys. J., 267, 334, 1983.

The x axis points toward the center of the secondary, and the z axis is parallel to the overall angular velocity of the primary. Neglecting the inertial terms uR ■ grad uR, we thus have

2^0 x ur = -grad(V - W) - 1 gradp + 1 F(ur), (4.125)

where the three-dimensional velocity uR is measured in our rotating frame of reference, and F is the turbulent viscous force per unit volume. In spherical polar coordinates (r, d, the potential W is given by

where i = cos d and v = sin d cos The Pks are the Legendre polynomials. Since our basic assumptions are identical to those made in Section 4.3.1, Eq. (4.125) must be combined with Eqs. (4.3)-(4.6).

Following standard practice, we shall expand about hydrostatic equilibrium in powers of the nondimensional parameter e = ^ = M±ML (RI (4.127)

In particular, we shall let p = p0 + ep1 +----, etc. In the frame rotating with the angular velocity the three-dimensional velocity uR has the form ur = e u1 + e3/2u3/2 + •••, (4.128)

since the Coriolis force is of O(e3/2). By virtue of Eqs. (4.125) and (4.126), this three-dimensional velocity is the superposition of two different kinds of currents. One of them is caused by the small oblateness due to rotation around the z axis; the other one is caused by the smallprolateness due to tidal action in the direction of the x axis. Hence, the general problem can be decomposed into two subproblems. In the first subproblem, the flow is caused by the rotational distortion only; for the second, the velocity uR is due to the tidal distortion only. To evaluate the effects of the first we shall solve the equations with M = 0; for the effects of the second, at least to O(e), we shall formally let = 0 in the equations. Correct to O(e), then, the tidal flow is the superposition of these two circulation patterns.

Letting M' = 0, one readily sees that the velocity u1 becomes symmetrical with respect to both the z axis and the (z = 0)-plane. We then have, in the rotating frame,

d x and u1p = 0. To 0(e), this solution is strictly equivalent to the one obtained in Section 4.3.1, with the functions u and v being related to each other by Eq. (4.118). For the sake of completeness, one must also solve Eq. (4.125) to O(e3/2), thus expressing the balance between the Coriolis force acting on the rotationally driven currents and the turbulent friction acting on the differential rotation around the z axis. To O(e3/2), this equation is strictly equivalent to Eq. (4.57).

If we formally disregard the centrifugal and Coriolis forces in Eq. (4.125), the velocity ui becomes symmetrical with respect to the x axis (but not with respect to the (x = 0)-plane!). In order to describe this part of the solution, it is convenient to use the radial variable r, the cosine of the colatitude from the x axis v = sin d cos p, and the azi-muthal angle 0 around the x axis. Using these coordinates, one can show that the tidal contribution to the circulatory currents can be written in the form

U 1r = V) Uk(r)Pk(v), U1v = V Vk(r)(1 - v2)—^, (4.130)

and U10 = 0. Equation (4.4) implies that

(k = 2, 3, 4). This motion depends, therefore, on the three functions u2, U3, and u4.

When both the rotational and tidal terms are retained in Eq. (4.126), it is a simple matter to prove that the three-dimensional velocity u1 has the following components:

U11 = V Vk(r)(1 - l )—-+ v(r)(1 - |2)—-, (4.133)

di dx

11<p

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