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13.1 Lagrangian perturbation theory

Following [44Jump To The Next Citation Point45Jump To The Next Citation Point], we work with Lagrangian variations. We have already seen that the Lagrangian perturbation ΔQ of a quantity Q is related to the Eulerian variation δQ by
ΔQ = δQ + ℒ Q, (227 ) ξ
where (as before) ℒξ is the Lie derivative that was introduced in Section 2. The Lagrangian change in the fluid velocity follows from the Newtonian limit of Equation (135View Equation):
i i Δv = ∂tξ , (228 )
where ξi is the Lagrangian displacement. Given this, and
Δgij = ∇i ξj + ∇jξi, (229 )
we have
Δvi = ∂tξi + vj∇i ξj + vj∇jξi. (230 )

Let us consider the simplest case, namely a barotropic ordinary fluid for which ℰ = ℰ (n). Then we want to perturb the continuity and Euler equations from the previous Section 12. The conservation of mass for the perturbations follows immediately from the Newtonian limits of Equations (134View Equation) and (138View Equation) (which as we recall automatically satisfy the continuity equation):

i i Δn = − n∇i ξ, δn = − ∇i (nξ ). (231 )
Consequently, the perturbed gravitational potential follows from
∇2 δΦ = 4πGm δn = − 4πGm ∇i (nξi). (232 )
In order to perturb the Euler equations we first rewrite Equation (218View Equation) as
( ) (∂t + ℒv )vi + ∇i ˜μ + Φ − 1-v2 = 0, (233 ) 2
where ˜μ = μ∕m. This form is particularly useful since the Lagrangian variation commutes with the operator ∂t + ℒv. Perturbing Equation (233View Equation) we thus have
( 1 ) (∂t + ℒv)Δvi + ∇i Δ ˜μ + Δ Φ − --Δ (v2) = 0. (234 ) 2

We want to rewrite this equation in terms of the displacement vector ξ. After some algebra one finds

∂2ξ + 2vj∇ ∂ξ + (vj∇ )2ξ + ∇ δΦ + ξj∇ ∇ Φ − (∇ ξj)∇ ˜μ + ∇ Δ ˜μ = 0. (235 ) t i j ti j i i i j i j i
Finally, we need
( ) ( ) Δ ˜μ = δ˜μ + ξi∇ ˜μ = ∂μ˜ δn + ξi∇ μ˜= − ∂˜μ- ∇ (nξi) + ξi∇ ˜μ. (236 ) i ∂n i ∂n i i
Given this, we have arrived at the following form for the perturbed Euler equation:
[( ) ] 2 j j 2 j ∂˜μ j ∂tξi + 2v ∇j ∂tξi + (v ∇j ) ξi + ∇i δΦ + ξ ∇i ∇j (Φ + ˜μ) − ∇i ∂n- ∇j(n ξ ) = 0. (237 )
This equation should be compared to Equation (15) of [44Jump To The Next Citation Point].

Having derived the perturbed Euler equations, we are interested in constructing conserved quantities that can be used to assess the stability of the system. To do this, we first multiply Equation (237View Equation) by the number density n, and then write the result (schematically) as

2 A ∂tξ + B ∂tξ + Cξ = 0, (238 )
omitting the indices since there is little risk of confusion. Defining the inner product
∫ ⟨ i ⟩ i∗ η ,ξi = η ξidV, (239 )
where η and ξ both solve the perturbed Euler equation, and the asterisk denotes complex conjugation, one can now show that
∗ ∗ ⟨η,A ξ⟩ = ⟨ξ,A η⟩ and ⟨η,B ξ⟩ = − ⟨ξ,B η⟩ . (240 )
The latter requires the background relation ∇i (nvi) = 0, and holds provided that n → 0 at the surface of the star. A slightly more involved calculation leads to
∗ ⟨η,C ξ⟩ = ⟨ξ, Cη ⟩ . (241 )
Inspired by the fact that the momentum conjugate to ξi is ρ(∂ + vj∇ )ξi t j, we now consider the symplectic structure
⟨ ⟩ ⟨ ⟩ W (η,ξ ) = η,A ∂tξ + 1B ξ − A∂tη + 1-Bη, ξ . (242 ) 2 2
Given this, it is straightforward to show that W (η,ξ) is conserved, i.e. ∂tW = 0. This leads us to define the canonical energy of the system as
Ec = m-W (∂tξ,ξ) = m- {⟨∂tξ,A ∂tξ ⟩ + ⟨ξ,C ξ⟩}. (243 ) 2 2
After some manipulations, we arrive at the following explicit expression:
1 ∫ { (∂ μ) 1 } Ec = -- ρ|∂tξ |2 − ρ|vj∇jξi|2 + ρ ξiξj∗∇i ∇j(˜μ + Φ ) + --- |δn|2 − ----|∇iδΦ |2 dV (244 ) 2 ∂n 4πG
which can be compared to Equation (45) of [44Jump To The Next Citation Point]. In the case of an axisymmetric system, e.g. a rotating star, we can also define a canonical angular momentum as
m ⟨ 1 ⟩ Jc = − --W (∂ϕξ, ξ) = − Re ∂ϕξ,A ∂tξ + -B ξ . (245 ) 2 2
The proof that this quantity is conserved relies on the fact that (i) W (η,ξ) is conserved for any two solutions to the perturbed Euler equations, and (ii) ∂ϕ commutes with ρvj∇j in axisymmetry, which means that if ξ solves the Euler equations then so does ∂ϕ ξ.

As discussed in [44Jump To The Next Citation Point45Jump To The Next Citation Point], the stability analysis is complicated by the presence of so-called “trivial” displacements. These trivials can be thought of as representing a relabeling of the physical fluid elements. A trivial displacement i ζ leaves the physical quantities unchanged, i.e. is such that i δn = δv = 0. This means that we must have

∇i (ρζi) = 0, (246 ) i (∂t + ℒv )ζ = 0. (247 )
The solution to the first of these equations can be written as
ρζi = εijk∇j χk (248 )
where, in order to satisfy the second equations, the vector χk must have time-dependence such that
(∂t + ℒv)χk = 0. (249 )
This means that the trivial displacement will remain constant along the background fluid trajectories. Or, as Friedman and Schutz [44Jump To The Next Citation Point] put it, the “initial relabeling is carried along with the unperturbed motion”.

The trivials have the potential to cause trouble because they affect the canonical energy. Before one can use the canonical energy to assess the stability of a rotating configuration one must deal with this “gauge problem”. To do this one should ensure that the displacement vector ξ is orthogonal to all trivials. A prescription for this is provided by Friedman and Schutz [44Jump To The Next Citation Point]. In particular, they show that the required canonical perturbations preserve the vorticity of the individual fluid elements. Most importantly, one can also prove that a normal mode solution is orthogonal to the trivials. Thus, normal mode solutions can serve as canonical initial data, and be used to assess stability.

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