where j = Q.m2 is the angular momentum per unit mass. Then, under what conditions is this configuration stable with respect to small isentropic disturbances? Although no definitive answer can be given at the present time, some interesting results can be obtained for axially symmetric motions (i.e., motions for which the specific angular momentum of each fluid particle is preserved along its path). Departures from axial symmetry will be discussed briefly in Section 3.4.3.

Two types of description can be used to analyze the oscillations of a star about a known state of equilibrium: Either we specify the Eulerian change noted by an external observer who, at every instant t, views a given volume of fluid at a fixed location in space, or we describe the Lagrangian change within a given mass element, which is followed along its path in the course of time. Let Q(r, t) and Q0(r, t) be the values of any physical quantity in the perturbed and unperturbed flows, respectively. Consider also the Lagrangian displacement £(r, t), which describes any departure from the state of equilibrium. Given these definitions, one finds that

is the Eulerian variation of the quantity Q, whereas

is its Lagrangian variation. In the linear approximation, we can thus write

One also has


where dm = pdv is the mass element and V is the total volume. Very much for the same reason, we can also write d f Q dm = ( —dm. (3.53)

3.4.1 An energy principle

As was originally shown by Fj0rtoft (1946), one can derive the appropriate stability criterion on the basis of the energy equation. His analysis relies upon the fact that the total energy - which is extremal for configurations satisfying Eq. (3.48) - is a minimum for dynamically stable equilibria and fails to be a minimum for dynamically unstable ones. This can be seen as follows. Take the scalar product of Eq. (3.1) with the velocity v, and integrate over the volume V. After performing an integration by parts and using the fact that the pressure vanishes at the free surface, we obtain d r 1 2 r p r

—- - |v | dm = — div vdm — v ■ grad Vdm. (3.54)

By making use of Eqs. (3.9) and (3.53), one finds that f P ,• f DU , d f TT 7 dUj

Jv P Jv Dt dt Jv dt where UT is the total thermal energy. Similarly, we can write f dW

Jv dt where W is the gravitational potential energy. Using Eqs. (3.55) and (3.56), we can integrate Eq. (3.54) to obtain

Suppose now that an axially symmetric motion is superimposed upon the state of equilibrium. Equation (3.57) then becomes f - |vp|2 dm + E = constant, (3.58)

Jv 2

where vp is the velocity field of the axially symmetric pulsation, and E is the total energy

Obviously, any increase of the kinetic energy of the axially symmetric motion must be supplied from the total energy E. Accordingly, this energy must be a minimum for stable, isentropic motions.

Let us now compute the first and second variations of the total energy E by keeping constant the total mass M and the total angular momentum J. Dynamically stable equilibria correspond to the conditions

In the case of axially symmetric motions, the specific angular momentum is preserved for each fluid particle. We thus have Dj/Dt = 0, so that we can write

Similarly, for isentropic motions one has DS/Dt = 0. Equation (3.9) thus implies that

S f U dm = i AU dm = — f — div £ dm = f — £ ■ grad pdm. (3.63) Jv JV JV P JV P

Finally, the first variation of the potential energy is

From Eqs. (3.61), (3.63), and (3.64), it thus follows that

By making use of Eq. (3.48), one readily sees that the condition SE = 0 defines a state of mechanical equilibrium. Similarly, it is a simple matter to prove that

= grad S V + - grad Sp - gradp + £ • grad (^2m) lm. (3.67)

P P2

Hence, we can rewrite Eq. (3.66) in the compact form

Jv where

L£ = -gradS V — gradSp + gradp--- $ ■ grad(Q2m4) 1m. (3.69)

p p2 m3

Equations (3.3) and (3.10) imply that

Sp = — r1 p div £ — £ ■ grad p. (3.71) Similarly, by assuming that the density vanishes at the free surface, we have

where Sp is given in terms of £ by Eq. (3.70). Thus, if there exists a displacement £ such that i £ ■ L£ dm > 0, (3.73) Jv we have S2E < 0, and the system is dynamically unstable because its total energy fails to be an absolute minimum.

For the sake of completeness, let us briefly consider the small oscillations about the state of mechanical equilibrium defined in Eq. (3.48). A rigorous discussion was made by Lebovitz (1970), who derived the following equation:

where L is the time-dependent operator defined in Eq. (3.69) and £ is a two-dimensional vector with components and . This operator is symmetric in the sense that


where £ and n are two arbitrary vectors. As was shown by Lebovitz, the symmetry of L implies that the configuration is unstable if, for any trial function £, condition (3.73) is satisfied. The strength of this result is that it avoids any assumptions about the existence of normal modes or about their properties when they do exist. It can be put into a more familiar form if we consider the normal-mode solution

for which

By virtue of Eq. (3.75), this equation provides a variational basis for the determination of the allowed values of a2, with the smallest eigenvalue being the minimum of the expression on the right-hand side of Eq. (3.78). Accordingly, if condition (3.73) is satisfied for some vector £ = £o, the right-hand side of Eq. (3.78) is negative for such a choice. This implies a negative value for the least eigenvalue a2, so that the mechanical equilibrium is dynamically unstable. The equivalence of the two methods is therefore demonstrated.

To proceed any further, we must now insert Eqs. (3.69)-(3.72) into Eq. (3.73). Integrating by parts and rearranging the various terms in this equation, we eventually obtain

jv jv jv where the tensor M has the form

Ti pp

M = — gradp ( — grad p--gradp ) +--- grad m grad(Œ2m4). (3.80)

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