Kinetic theory course of the University of Oxford (MMathPhys/MSc paper)

Normalize the symmetric

N

-particle phase-space distribution function

D_{N}

to one and write

z_{i} = (x_{i}, v_{i})

d z_{i} = d^{3} x_{i} d^{3} v_{i}

. The number-normalized one- and two-particle distribution functions are

f (z_{1}, t) = N \int D_{N} d z_{2} \dots d z_{N}, f_{2} (z_{1}, z_{2}, t) = N (N - 1) \int D_{N} d z_{3} \dots d z_{N} .

Thus

\int f d z = N

and

\int f_{2} (z, z_{⋆}) d z_{⋆} = (N - 1) f (z)

. These conventions absorb the otherwise explicit factor

N - 1

in the first BBGKY hierarchy equation.

Separate the Hamiltonian into kinetic energy, external potential energy, and pair potential terms. Their Hamilton's equations give

\dot{x}_{i} = v_{i}, \dot{v}_{i} = - \nabla U (x_{i}) - \sum_{j \neq = i} \nabla_{x_{i}} ϕ (∣ x_{i} - x_{j} ∣) .

The sign follows directly from the stated Poisson bracket:

\overset{v}{˙}_{i α} = {v_{i α}, H} = - \partial_{x_{i α}} H

. The Hamiltonian Liouville equation is consequently

\partial_{t} D_{N} + \sum_{i} [v_{i} \cdot \nabla_{x_{i}} D_{N} - (\nabla U_{i} + \sum_{j \neq = i} \nabla_{x_{i}} ϕ_{ij}) \cdot \nabla_{v_{i}} D_{N}] = 0.

Integrate over

z_{2}, \dots, z_{N}

and multiply by

N

. Assume sufficient decay in velocity space and either decay or periodic boundary conditions in position, so that integration by parts removes all transport terms with

i \geq 2

. The surviving interaction term is

N \sum_{j = 2}^{N} \int \nabla_{x_{1}} ϕ_{1 j} \cdot \nabla_{v_{1}} D_{N} d z_{2} \dots d z_{N} = \int \nabla_{x_{1}} ϕ_{12} \cdot \nabla_{v_{1}} f_{2} d z_{2},

where indistinguishability makes all

N - 1

terms equal. Relabeling

(z_{1}, z_{2})

(z, z_{⋆})

yields the exact result

\partial_{t} f + v \cdot \nabla f - \nabla U \cdot \nabla_{v} f = \int d^{3} x_{⋆} d^{3} v_{⋆} \nabla_{x} ϕ (∣ x - x_{⋆} ∣) \cdot \nabla_{v} f_{2} .

No assumption of molecular chaos for the Boltzmann equation has entered: the dependence on the two-particle distribution function is still exact.

To obtain the Boltzmann equation, take a dilute gas with a short-range pair potential. Its interaction range is small compared with the mean free path and the spatial scale on which the phase-space distribution function changes. The binary collision approximation treats encounters as isolated elastic collisions; simultaneous encounters involving three or more particles can be neglected. The encounter duration is short compared with the collision relaxation time and macroscopic evolution time, and the external force changes negligibly during an encounter. This permits local, effectively instantaneous scattering at one position

x

The essential statistical closure is molecular chaos for the Boltzmann equation: the distribution of incoming pairs factorizes as

f f_{⋆}

. Outgoing pairs need not be independent. Follow each incoming pair through its two-body scattering, and use conservation of momentum and kinetic energy, together with time reversal and the invariant scattering measure, to express both gain and loss in the same variables. This gives

C_{B} [f] (v) = \int d^{3} v_{⋆} \int_{hemisphere} d Ω B (g, θ) (f^{'} f_{⋆}^{'} - f f_{⋆}), g = ∣ v - v_{⋆} ∣.

Here

d Ω = sin θ d θ d φ

; when the measure is written as

d θ d φ

, its Jacobian is included in the Boltzmann scattering kernel. In a scattering-angle convention, that kernel contains relative speed times differential scattering cross-section. The hemisphere prevents double counting equivalent outcomes for indistinguishable particles; the precise factor is included in

B

. The primed velocities are connected to the unprimed ones by the reversible elastic collision map. This closure is appropriate for a dilute short-range gas, rather than a generic long-range interacting system with an important collective mean-field theory contribution.

The further replacement by the BGK model is a phenomenological relaxation closure. Replace the full nonlinear Boltzmann collision operator by relaxation towards a local Maxwellian, with a single, velocity-independent collision relaxation time

τ > 0

C_{BGK} [f] = - \frac{f - f ^{(0)}}{τ}, f^{(0)} = \frac{ρ}{( 2 π Θ ) ^{3/2}} exp (- \frac{∣ v - u ∣ ^{2}}{2Θ}) .

It models relaxation towards a local Maxwellian without resolving the detailed scattering. Treating

τ

as constant is an additional simplification; it is of the order of a mean time between effective collisions. Near a local Maxwellian, departures in the nonconserved velocity moments decay on this timescale.

Choose

f^{(0)}

to match the collision invariants

1

v

and

∣ v ∣^{2}

f

. For unit particle mass, this means

ρ = \int f d^{3} v, u = ρ^{- 1} \int v f d^{3} v, Θ = \frac{1}{3 ρ} \int ∣ v - u ∣^{2} f d^{3} v .

Here

ρ

is the mass density,

u

is the mean velocity, and

Θ

is the temperature in squared-velocity units,

Θ = k_{B} T / m

before setting

m = 1

. In particular,

\int (1, v, ∣ v ∣^{2}) C_{BGK} [f], d^{3} v = 0.

Thus relaxation preserves local mass, momentum and kinetic energy rather than relaxing them to externally prescribed values.

c

Solution

Use the BGK model from b. Multiply it by

v_{i}

and integrate over velocity space. Matching the collision invariants makes the collision contribution vanish. For the external force term, integration by parts gives

- \partial_{j} U \int v_{i} \partial_{v_{j}} f d^{3} v = ρ \partial_{i} U .

Define the momentum flux and force density by

Π_{ij} = \int v_{i} v_{j} f d^{3} v, F_{i} = - ρ \partial_{i} U .

The resulting momentum conservation equation is

\partial_{t} (ρ u_{i}) + \partial_{j} Π_{ij} = F_{i}, Π = ρ uu + P, F = - ρ \nabla U .

The pressure tensor is the second central velocity moment,

P_{ij} = \int (v_{i} - u_{i}) (v_{j} - u_{j}) f d^{3} v .

The zeroth velocity moment also supplies the continuity equation

\partial_{t} ρ + \partial_{j} (ρ u_{j}) = 0

, which will be needed in d.

d

Solution

Write

c_{i} = v_{i} - u_{i}

and introduce the raw third velocity moment

M_{ijk}

and the third central velocity moment

Q_{ijk}

M_{ijk} = \int v_{i} v_{j} v_{k} f d^{3} v, Q_{ijk} = \int c_{i} c_{j} c_{k} f d^{3} v .

The expansion of their product gives

M_{ijk} = ρ u_{i} u_{j} u_{k} + u_{i} P_{jk} + u_{j} P_{ik} + u_{k} P_{ij} + Q_{ijk} .

The heat flux is

q_{i} = Q_{ijj} /2

for unit mass.

Multiply the BGK model by

v_{i} v_{j}

. The external force term is

- \partial_{k} U \int v_{i} v_{j} \partial_{v_{k}} f d^{3} v = ρ (u_{j} \partial_{i} U + u_{i} \partial_{j} U) .

Therefore the raw second velocity moment satisfies

\partial_{t} Π_{ij} + \partial_{k} M_{ijk} = - ρ (u_{i} \partial_{j} U + u_{j} \partial_{i} U) - \frac{Π _{ij} - Π _{ij}^{(0)}}{τ} .

The matching local Maxwellian has

Π_{ij}^{(0)} = ρ u_{i} u_{j} + ρ Θ δ_{ij}

, so

Π_{ij} - Π_{ij}^{(0)} = P_{ij} - P_{ij}^{(0)}, P_{ij}^{(0)} = ρ Θ δ_{ij} .

To remove the bulk motion, let

D_{t} = \partial_{t} + u_{k} \partial_{k}

be the material derivative. The continuity equation and the momentum conservation equation imply

ρ D_{t} u_{i} = - \partial_{k} P_{ik} - ρ \partial_{i} U,

and hence

\partial_{t} (ρ u_{i} u_{j}) + \partial_{k} (ρ u_{i} u_{j} u_{k}) = - u_{j} \partial_{k} P_{ik} - u_{i} \partial_{k} P_{jk} - ρ (u_{i} \partial_{j} U + u_{j} \partial_{i} U) .

Subtract this identity from the raw second velocity moment equation and insert the expansion of

M_{ijk}

. The terms

u_{i} \partial_{k} P_{jk}

and

u_{j} \partial_{k} P_{ik}

cancel by the product rule. The external force terms also cancel, since a common acceleration changes the local mean velocity without directly changing the relative velocities. What remains is

\partial_{t} P_{ij} + \partial_{k} (u_{k} P_{ij} + Q_{ijk}) + P_{ik} \partial_{k} u_{j} + P_{jk} \partial_{k} u_{i} = - \frac{P _{ij} - ρ Θ δ _{ij}}{τ} .

(8)

The trace of the relaxation term is zero because

P_{ii} = 3 ρ Θ

. Thus the BGK model relaxes anisotropy of the pressure tensor while preserving the local random kinetic energy. The evolution is still a velocity-moment hierarchy, since

Q_{ijk}

requires further information about

f

2

a

Solution

Take

k > 0

initially; real solutions for

k < 0

follow by

(k, ω) \mapsto (- k, - ω)

. Put

V = v_{0} + v_{2} - v_{1}

. Approaching the real phase velocity from the upper half-plane, the Landau contour and the Sokhotski–Plemelj formula give the following Cauchy principal value decomposition

\int_{C_{L}} \frac{f _{0}^{'} ( v )}{v - u} d v = PV \int \frac{f _{0}^{'} ( v )}{v - u} d v + iπ f_{0}^{'} (u) .

The real frequency condition

ϵ = 0

therefore requires

f_{0}^{'} (u) = 0

. This is possible in the interior of the velocity plateau and outside the occupied velocity range:

u < - v_{0}, v_{1} < u < v_{2}, u > V .

These are candidate intervals, not yet solutions of the real dispersion relation. The endpoints have jumps in

f_{0}^{'}

and logarithmic singularities in the response; they are not finite-

k

undamped modes. At the cusp

u = 0

, the slope is also discontinuous and the real response diverges, so it supplies no additional mode.

The physical condition is absence of net resonant energy exchange with particles moving at the wave's phase velocity. Outside the support there are no such particles. Within a velocity plateau, the equal populations on either side of the resonance cancel Landau damping.

b

Solution

Let

A_{0} = f_{0} (0)

δ = v_{2} - v_{1}

V = v_{0} + δ

and

K = k^{2} λ_{De}^{2}

. Use the Debye length convention of the paper,

λ_{De} = v_{0} / ω_{pe}

. On the candidate intervals in a, the Landau contour contributes no imaginary part. The piecewise constant slopes give

I (u) = \int \frac{f _{0}^{'} ( v )}{v - u} d v = \frac{A _{0}}{v _{0}} [\int_{- v_{0}}^{0} \frac{d v}{v - u} - \int_{0}^{v_{1}} \frac{d v}{v - u} - \int_{v_{2}}^{V} \frac{d v}{v - u}] = \frac{A _{0}}{v _{0}} [2 ln ∣ u ∣ - ln ∣ v_{0} + u ∣ - ln ∣ v_{1} - u ∣ - ln ∣ V - u ∣ + ln ∣ v_{2} - u ∣] .

Since

n_{0} = v_{0} A_{0}

, setting the dielectric function to zero yields

ln \frac{u ^{2} ∣ v _{2} - u ∣}{∣ v _{0} + u ∣ ∣ v _{1} - u ∣ ∣ V - u ∣} = K .

The absolute values express the real response on the allowed intervals. Solving this real equation inside a sloping portion of

f_{0}

would not satisfy the imaginary part of the dielectric function.

An exact algebraic form will also make the number of undamped branches clear. Define

A = v_{0}^{2} + δ (v_{0} - v_{1}), B = v_{0} v_{1} V, q = 1 - e^{- K} .

On each allowed interval, the quantity inside the logarithm equals

u^{2} (u - v_{2}) / [(u + v_{0}) (u - v_{1}) (u - V)]

. Exponentiating and using

(u + v_{0}) (u - v_{1}) (u - V) = u^{2} (u - v_{2}) - A u + B

gives the cubic equation

q u^{2} (u - v_{2}) - A u + B = 0.

For every

k > 0

, the logarithm runs from

0

+ \infty

(- \infty, - v_{0})

, from

+ \infty

- \infty

(v_{1}, v_{2})

, and from

+ \infty

0

(V, \infty)

. The intermediate value theorem supplies a root in each interval, and the cubic equation permits at most three. Thus there is exactly one undamped branch in each interval.

c

Solution

For

K = k^{2} λ_{De}^{2} ≫ 1

, the logarithm in b must be large and positive. In each allowed interval this happens near a denominator zero, approached from the undamped side:

u_{-} = - v_{0} - O (e^{- K}), u_{a} = v_{1} + O (e^{- K}), u_{+} = V + O (e^{- K}) .

The constants hidden by

O (e^{- K})

have velocity units. Consequently,

ω_{-} \sim - k v_{0}, ω_{a} \sim k v_{1}, ω_{+} \sim k (v_{0} + v_{2} - v_{1}) .

The two outer modes have phase velocities just outside the occupied range and therefore have no resonant particles. The middle electron acoustic wave lies just inside the lower edge of the velocity plateau, where

f_{0}^{'} = 0

. All three escape Landau damping in this collisionless model. The stated neglect of ion response remains necessary for the middle mode as well.

d

Solution

For the two outer branches,

∣ u ∣ ≫ v_{0}

at small

k

. An integration by parts and a large-

u

expansion give

I (u) = \int \frac{f _{0} ( v )}{( v - u ) ^{2}} d v = \frac{n _{*}}{u ^{2}} + O (u^{- 3}), n_{*} = \int f_{0} (v) d v .

There is no boundary contribution because the phase-space distribution function vanishes at both endpoints. Direct area integration gives

n_{*} = A_{0} [v_{0} + δ (1 - \frac{v _{1}}{v _{0}})], \frac{n _{*}}{n _{0}} = 1 + \frac{δ ( v _{0} - v _{1} )}{v _{0}^{2}} .

Thus the exact small-

k

cutoff, retaining the paper's definition

n_{0} = v_{0} A_{0}

, is

ω \to \pm ω_{pe} n_{*} / n_{0}

. To leading order in the narrow-velocity plateau approximation this is

ω \approx \pm ω_{pe} .

These are Langmuir waves. The difference between

n_{*}

and the reference

n_{0}

is only first order in

δ / v_{0}

, so it does not alter the requested leading-order result.

For the finite-phase velocity branch put

u = v_{1} + δ a

, with

0 < a < 1

. To leading order in

δ / v_{0}

and

δ / v_{1}

, the real dispersion relation becomes

ln [\frac{v _{1}^{2}}{v _{0}^{2} - v _{1}^{2}} \frac{1 - a}{a}] = K .

At long wavelength set

K = 0

to leading order. Then

a = v_{1}^{2} / v_{0}^{2}

, and

ω_{a} \approx k [v_{1} + δ \frac{v _{1}^{2}}{v _{0}^{2}}] .

For comparison, the exact

k \to 0

phase velocity from the cubic equation in b is

B / A

; expanding it gives precisely

v_{1} + δ v_{1}^{2} / v_{0}^{2} + O (δ^{2} / v_{0})

at fixed

v_{1} / v_{0}

A single Maxwell-Boltzmann velocity distribution has a nonzero slope at every finite nonzero

u

and has no velocity plateau. The would-be electron acoustic wave is therefore subject to strong Landau damping, so there is no corresponding completely undamped acoustic branch in a single Maxwellian electron population with immobile ions. At its only stationary point

u = 0

, a Maxwellian has

\int f_{0}^{'} (v) / v d v < 0

, making the static dielectric function positive rather than zero. This statement does not exclude acoustic modes of suitably distinct electron populations or an ion-supported mode when ion dynamics is retained.

e

Solution

Retain

K = k^{2} λ_{De}^{2}

while making the same narrow-velocity plateau expansion as in d. With

u = v_{1} + δ a

, the factor that varies rapidly across the plateau is

(v_{2} - u) / (u - v_{1}) = (1 - a) / a

. The other factors vary slowly and may be evaluated at

u = v_{1}

, giving

\frac{v _{1}^{2}}{v _{0}^{2} - v _{1}^{2}} \frac{1 - a}{a} = e^{K} .

Therefore

ω_{a} \approx k [v_{1} + \frac{v _{2} - v _{1}}{1 + e ^{k^{2} λ_{De}^{2}} ( v _{0}^{2} - v _{1}^{2} ) / v _{1}^{2}}] .

K ≪ 1

, the denominator tends to

v_{0}^{2} / v_{1}^{2}

, recovering d. At

K ≫ 1

u_{a} - v_{1} \approx δ \frac{v _{1}^{2}}{v _{0}^{2} - v _{1}^{2}} e^{- K},

recovering the exponentially close approach to the plateau edge in c. The phase velocity remains strictly inside the velocity plateau for every finite

k

As an asymptotic expansion, the formula takes

δ \to 0

at fixed

0 < v_{1} / v_{0} < 1

. If the plateau also approaches the outer edge, so that

v_{0} - v_{1}

is comparable to

δ

, the factors treated as slowly varying need not be slowly varying. In that additional limiting regime, use the exact dispersion relation in b instead.

f

Solution

For

k > 0

, the upper Langmuir wave starts at

ω \approx + ω_{pe}

and approaches the line

ω = kV

from above. The lower Langmuir wave starts at

ω \approx - ω_{pe}

and approaches

ω = - k v_{0}

from below. The electron acoustic wave starts at the origin with slope

v_{1} + δ v_{1}^{2} / v_{0}^{2}

and approaches

ω = k v_{1}

from above; it stays inside

k v_{1} < ω < k v_{2}

. These are the only completely undamped branches, as shown by the cubic equation in b.

Figure 1.
Three undamped branches for a narrow electron velocity plateau
. Exact roots of the cubic form of the dielectric response, with $v_{0} = 1$ , $v_{1} = 0.5$ and $v_{2} = 0.56$ . Dashed lines show the three short-wavelength limits. The shaded band contains the electron acoustic branch. The exact outer cutoffs retain the small density-normalization correction explained in d. **Details**.

The small asymmetry between the outer branches reflects the extra positive-velocity population introduced by the velocity plateau. The negative-

k

sketch follows by rotating the displayed branches through the origin, since

ϵ (- ω, - k) = ϵ (ω, k)

on these undamped solutions. One should not impose symmetry about the

ω = 0

axis at fixed positive

k

3

a

Solution

For a regular integrable Hamiltonian system, action-angle variables offer two related simplifications.

Unperturbed Hamilton's equations are $\dot{J} = 0$ and $\dot{θ} = Ω (J)$ . Complicated stellar orbits become uniform motion on an invariant torus, and the unperturbed transport operator becomes $Ω \cdot \partial_{θ}$ . A Fourier series in the angles then makes orbital resonances explicit through $k \cdot Ω$ .
A phase-mixed equilibrium galactic distribution function depends on the action variables rather than the fast angles, consistently with Jeans theorem. Its slow evolution, including quasilinear diffusion in action space, can therefore be described by an angle average in two action variables rather than by resolving every stellar orbit in position and velocity.

These advantages assume regular or approximately integrable mean motion; globally defined action-angle variables need not exist in chaotic regions.

b

Solution

Use the Collisionless Boltzmann equation in action-angle variables,

\partial_{t} f + {f, H} = 0, {f, H} = \partial_{θ} f \cdot \partial_{J} H - \partial_{J} f \cdot \partial_{θ} H .

Define the orbital frequency vector by

Ω (J) = \partial_{J} H_{0} = (Ω_{φ}, Ω_{R}) .

The slow evolution of the angle average

f_{0}

starts at second order in the perturbation. At first order, the Poisson bracket therefore gives

(\partial_{t} + Ω \cdot \partial_{θ}) δ f = \partial_{J} f_{0} \cdot \partial_{θ} δ Φ.

For the

k

th term in the Fourier series, this reads

(\partial_{t} + i k \cdot Ω) δ f_{k} = i k \cdot \partial_{J} f_{0} δ ϕ_{k} .

The lowercase

δ ϕ_{k}

here denotes the Fourier coefficient of the uppercase

δ Φ

in the Hamiltonian; these are the same perturbing potential. Multiply by the integrating factor

e^{i k \cdot Ω t}

and set the initial homogeneous response to zero. One obtains

δ f_{k} (J, t) = i \int_{0}^{t} d t^{'} k \cdot \frac{\partial f _{0} ( J , t ^{'} )}{\partial J} e^{- i k \cdot Ω (t - t^{'})} δ ϕ_{k} (J, t^{'}) .

Retaining the slow time dependence of

f_{0}

inside this integral is compatible with the subsequent second-order calculation; changes in

Ω

caused by that slow evolution affect only higher orders in this approximation.

c

Solution

Let

⟨ \cdot ⟩

denote

(2 π)^{- 2} \int d^{2} θ

. Average the Collisionless Boltzmann equation to second order. All first-order angle derivatives average to zero, leaving

\partial_{t} f_{0} = - ⟨{δ f, δ Φ}⟩ .

An angle integration by parts shows that

⟨{δ f, δ Φ}⟩ = ⟨ \partial_{θ_{i}} δ f \partial_{J_{i}} δ Φ - \partial_{J_{i}} δ f \partial_{θ_{i}} δ Φ ⟩ = - \partial_{J_{i}} ⟨ δ f \partial_{θ_{i}} δ Φ ⟩ .

Thus the angle-averaged galactic distribution function obeys an action-space continuity equation,

\partial_{t} f_{0} = - \partial_{J_{i}} q_{i}, q_{i} = - ⟨ δ f \partial_{θ_{i}} δ Φ ⟩ .

This sign also follows from Hamilton's equations,

\dot{J}_{i} = - \partial_{θ_{i}} δ Φ

: the flux is

⟨ δ f \dot{J}_{i} ⟩

Reality gives

δ ϕ_{- k} = δ ϕ_{k}^{*}

and

δ f_{- k} = δ f_{k}^{*}

. Orthogonality of the Fourier series implies

⟨ δ f \partial_{θ_{i}} δ Φ ⟩ = - i \sum_{k} k_{i} δ f_{k} δ ϕ_{k}^{*} .

Insert b to obtain

q_{i} = - \sum_{k} k_{i} \int_{0}^{t} d t^{'} k \cdot \partial_{J} f_{0} (J, t^{'}) e^{- i k \cdot Ω (t - t^{'})} δ ϕ_{k} (J, t^{'}) δ ϕ_{k}^{*} (J, t) .

Although individual terms are complex, pairing

k

with

- k

makes the flux real.

Now take the marginalization specified by

F_{0} = 2 π \int_{0}^{\infty} f_{0} d J_{R}

. Assume vanishing radial-action flux at

J_{R} = 0

and

J_{R} \to \infty

: the first boundary does not permit negative radial action, and the second has no escaping flux in this calculation. Then

\partial_{t} F_{0} = - 2 π \partial_{J_{φ}} \int_{0}^{\infty} q_{φ} d J_{R} - 2 π [q_{R}]_{0}^{\infty} = - \partial_{J_{φ}} Q_{0},

with

Q_{0} = - 2 π k \sum k_{φ} \int_{0}^{\infty} d J_{R} \int_{0}^{t} d t^{'} k \cdot \frac{\partial f _{0} ( J , t ^{'} )}{\partial J} \times e^{- i k \cdot Ω (t - t^{'})} δ ϕ_{k} (J, t^{'}) δ ϕ_{k}^{*} (J, t) .

If radial-action boundary flux were present, its explicit boundary term would have to be retained.

d

Solution

The restriction

k_{φ} = \pm m

describes an

m

-fold rotating perturbation of a galactic disk, for example a transient spiral density wave or, for

m = 2

, a galactic bar. Its pattern speed is

Ω_{p}

, its amplitude peaks at

t_{peak}

, and its Gaussian function envelope has duration of order

τ

. The coefficients

u_{k} (J)

describe its strength along different stellar orbits. A real perturbing potential requires

u_{- k} = u_{k}^{*}

Write

g (t) = exp [- (t - t_{peak})^{2} / (2 τ^{2})]

. The product entering c is

δ ϕ_{k} (t^{'}) δ ϕ_{k}^{*} (t) = ∣ u_{k} ∣^{2} e^{i k_{φ} Ω_{p} (t - t^{'})} g (t^{'}) g (t) .

The phase therefore measures detuning from the rotating pattern:

ω_{k} (J) = k \cdot Ω - k_{φ} Ω_{p} = k_{φ} (Ω_{φ} - Ω_{p}) + k_{R} Ω_{R} .

Substitution gives

Q_{0} = - 2 π k_{φ} = \pm m \sum k_{R} \sum k_{φ} \int_{0}^{\infty} d J_{R} ∣ u_{k} ∣^{2} \times \int_{0}^{t} d t^{'} k \cdot \partial_{J} f_{0} (J, t^{'}) e^{- i ω_{k} (t - t^{'})} \times exp [- \frac{( t - t _{peak} ) ^{2}}{2 τ ^{2}} - \frac{( t ^{'} - t _{peak} ) ^{2}}{2 τ ^{2}}] .

The strongest long-lived response occurs near an orbital resonance,

ω_{k} = 0

, where successive force contributions do not rapidly cancel in phase.

e

Solution

The required time separation is

τ ≪ t_{sec}

, where

t_{sec}

is the evolution timescale of

f_{0}

, together with

t_{peak} ≫ τ

and an observation time well after the pulse. The first condition permits freezing

f_{0} (J, t^{'})

at its pre-pulse value; its neglected change would enter beyond the retained second order. The second makes the original initial time lie in a negligible tail of the Gaussian function, permitting replacement of the lower time limits by

- \infty

. Short duration alone would not justify that replacement for a pulse centered at the initial time.

Integrate the continuity equation in c over the whole pulse. For fixed

J

, the time factor is the ordered Gaussian response integral

I (ω) = \int_{- \infty}^{\infty} d t \int_{- \infty}^{t} d t^{'} e^{- iω (t - t^{'})} g (t) g (t^{'}) .

A common translation by

t_{peak}

removes the peak time. Pair the positive and negative azimuthal harmonics:

(- m, - k_{R})

is paired with

(m, k_{R})

. Both

k_{φ}

and

k \cdot \partial_{J} f_{0}

change sign, so their product does not;

ω_{- k} = - ω_{k}

makes the two time factors complex conjugates. Hence the pair contributes

2 Re I

For completeness, the real part of the integrand is symmetric under

t \leftrightarrow t^{'}

. The triangular time domain therefore gives half of the full plane, and the Fourier transform of a Gaussian function gives

Re I (ω) = \frac{1}{2} \int_{- \infty}^{\infty} g (t) e^{- iω t} d t^{2} = π τ^{2} e^{- ω^{2} τ^{2}} .

This derives the supplied identity rather than requiring a separate time-ordering calculation.

Define

D_{m k_{R}} = m \frac{\partial f _{0}}{\partial J _{φ}} + k_{R} \frac{\partial f _{0}}{\partial J _{R}} .

Using the paired harmonics and

Δ F_{0} = - \partial_{J_{φ}} \int Q_{0} d t

now yields

Δ F_{0} = 4 π^{2} m τ^{2} \frac{\partial}{\partial J _{φ}} \sum_{k_{R}} \int_{0}^{\infty} d J_{R} ∣ u_{m k_{R}} ∣^{2} D_{m k_{R}} e^{- ω_{m k_{R}}^{2} τ^{2}} .

By the stated assumption,

ω_{m k_{R}}

is independent of

J_{R}

, so its exponential can be taken outside the radial integral. Put

Γ = (2 τ)^{- 1}

and use

R_{Γ} (ω) = \frac{e ^{- ω^{2} / (2 Γ^{2})}}{2 π Γ}, 4 π^{2} τ^{2} e^{- ω^{2} τ^{2}} = \frac{8 π ^{5}}{Γ} R_{Γ} (ω) .

Therefore

Δ F_{0} = \frac{8 π ^{5} m}{Γ} \frac{\partial}{\partial J _{φ}} k_{R} \sum R_{Γ} (ω_{m k_{R}}) \int_{0}^{\infty} d J_{R} ∣ u_{m k_{R}} ∣^{2} (m \frac{\partial f _{0}}{\partial J _{φ}} + k_{R} \frac{\partial f _{0}}{\partial J _{R}}) .

The outer derivative acts on the Gaussian resonance broadening factor as well as on the radial integral. The result describes quasilinear diffusion in action space weighted by the squared pulse amplitude, with a detuning width

Γ

. A shorter pulse has broader frequency selectivity, while a longer pulse selects narrower orbital resonances. If the action-space boundary flux vanishes, integrating

Δ F_{0}

over

J_{φ}

gives zero, as required by conservation of mass.

f

i

Solution

For

k_{R} = 0

, the detuning is

ω_{m 0} = m (Ω_{φ} - Ω_{p})

. The Gaussian resonance broadening factor selects the corotation resonance, where the star's azimuthal frequency matches the pattern speed. The derivative driving the flux is just

m \partial f_{0} / \partial J_{φ}

. In fact this contribution alone has the form

Δ F_{0}_{k_{R} = 0} = \partial_{J_{φ}} [\frac{8 π ^{5} m ^{2}}{Γ} R_{Γ} (ω_{m 0}) \int_{0}^{\infty} ∣ u_{m 0} ∣^{2} \partial_{J_{φ}} f_{0} d J_{R}] .

A single Fourier series harmonic changes the action variables in the direction

(m, k_{R})

, because

\dot{J} = - i k δ ϕ_{k} e^{i k \cdot θ}

plus its conjugate. For this harmonic,

Δ J_{R} = 0

: stars change angular momentum and shift their epicyclic guiding center without a leading-order increase of random radial motion. This is stellar radial migration driven by the corotation resonance. The resonance condition explains the absence of leading heating:

Δ E ≃ Ω_{φ} Δ J_{φ} = Ω_{p} Δ J_{φ}

at the corotation resonance.

ii

Solution

For

k_{R} = \pm 1

, the selected orbital resonances satisfy

m (Ω_{φ} - Ω_{p}) \pm Ω_{R} = 0.

For near-circular stellar orbits,

Ω_{R}

is the radial epicyclic frequency. Thus

k_{R} = - 1

is the inner Lindblad resonance,

m (Ω_{φ} - Ω_{p}) = Ω_{R}

, and

k_{R} = + 1

is the outer Lindblad resonance,

m (Ω_{φ} - Ω_{p}) = - Ω_{R}

The flux is driven by

m \partial_{J_{φ}} f_{0} \pm \partial_{J_{R}} f_{0}

, and each resonant harmonic exchanges both angular momentum and radial action in the ratio

Δ J_{R} = \frac{k _{R}}{m} Δ J_{φ} .

These terms can therefore produce galactic disk heating as well as angular momentum transport. In the usual absorbing response, stars losing angular momentum at the inner Lindblad resonance gain radial action, and stars gaining angular momentum at the outer Lindblad resonance also gain radial action. The actual direction of flux follows from the displayed derivatives of

f_{0}

; it is not fixed by the resonance condition alone. A finite pulse replaces exact Lindblad resonance selection by a band of detunings of width of order

1/ τ

2024 (A15089W1)

Source PDF: mmathphys.physics.ox.ac.uk/sites/g/files/vzsiac2991/files/mmathphys/documents/media/a15089w1.pdf

1

a

Solution

Put

z_{i} = (x_{i}, v_{i})

and

d z_{i} = d^{3} x_{i} d^{3} v_{i}

. Normalize the symmetric

N

-particle phase-space distribution function

D_{N}

to one. Use factorial normalization for its one-particle marginal and two-particle distribution function:

f (z_{1}) = N \int D_{N} d z_{2} \dots d z_{N}, f_{2} (z_{1}, z_{2}) = N (N - 1) \int D_{N} d z_{3} \dots d z_{N} .

Thus

\int f d z = N

and

\int f_{2} (z_{1}, z_{2}) d z_{2} = (N - 1) f (z_{1})

. For unit particle mass,

f

also carries mass density. These conventions explain why no extra factor

N - 1

occurs in the result.

Separate the Hamiltonian as

H = H_{1} + H_{rest} + H_{int}

, where

H_{1} = \frac{1}{2} ∣ v_{1} ∣^{2}, H_{int} = \sum_{j = 2}^{N} ϕ (∣ x_{1} - x_{j} ∣),

and

H_{rest}

contains all terms involving only particles

2, \dots, N

. The stated Poisson bracket gives Hamilton's equations

\dot{x}_{i} = v_{i}, \dot{v}_{i} = - \sum_{j \neq = i} \nabla_{x_{i}} ϕ (∣ x_{i} - x_{j} ∣) .

Conservation of probability along the Hamiltonian flow gives the Hamiltonian Liouville equation

\partial_{t} D_{N} + \sum_{i = 1}^{N} [v_{i} \cdot \nabla_{x_{i}} D_{N} - \sum_{j \neq = i} \nabla_{x_{i}} ϕ_{ij} \cdot \nabla_{v_{i}} D_{N}] = 0.

Assume decay of the required velocity moments and either decay or periodic boundary conditions in position. Integrating over

z_{2}, \dots, z_{N}

eliminates the

H_{rest}

terms by integration by parts. The terms in

H_{int}

differentiating an integrated velocity also vanish. Indistinguishability makes the remaining

N - 1

terms equal, so

\partial_{t} f + v_{1} \cdot \nabla_{x_{1}} f = N j = 2 \sum N \int \nabla_{x_{1}} ϕ_{1 j} \cdot \nabla_{v_{1}} D_{N} d z_{2} \dots d z_{N} = \int d z_{2} \nabla_{x_{1}} ϕ_{12} \cdot \nabla_{v_{1}} f_{2} .

Relabeling the two particles gives the first BBGKY hierarchy equation:

\partial_{t} f + v \cdot \nabla_{x} f = \int d^{3} x_{⋆} d^{3} v_{⋆} \nabla_{v} f_{2} \cdot \nabla_{x} ϕ (∣ x - x_{⋆} ∣) .

This is exact for a differentiable pair potential; no independent-particle closure has been made.

b

Solution

Take the zeroth velocity moment of a. The right side is a velocity space divergence with vanishing boundary contribution, giving the continuity equation

\partial_{t} ρ + \nabla \cdot (ρ u) = 0, ρ = \int f d^{3} v, ρ u = \int v f d^{3} v .

For the first velocity moment, use integration by parts componentwise:

\int d^{3} v v_{i} \partial_{v_{j}} f_{2} = - δ_{ij} \int d^{3} v f_{2} .

Write

c = v - u

; then

\int c f d^{3} v = 0

, and the momentum flux splits into bulk transport and the pressure tensor:

\int v_{i} v_{j} f d^{3} v = ρ u_{i} u_{j} + P_{ij}, P_{ij} = \int (v_{i} - u_{i}) (v_{j} - u_{j}) f d^{3} v .

Therefore

\partial_{t} (ρ u_{i}) + \partial_{x_{j}} (ρ u_{i} u_{j} + P_{ij}) = - \int d^{3} x_{⋆} n_{2} (x, x_{⋆}, t) \partial_{x_{i}} ϕ (∣ x - x_{⋆} ∣),

where repeated Cartesian indices are summed and

n_{2} = \int f_{2} d^{3} v d^{3} v_{⋆}

. The negative sign is the attractive or repulsive force

- \nabla ϕ

, rather than an additional sign convention for the pressure tensor.

c

Solution

For fixed

R

and

t

, define

h (x, s) = n_{2} (a, b, t)

with

a = x + (s - 1) R

and

b = x + s R

. The chain rule gives

\partial_{s} h = R \cdot (\nabla_{a} + \nabla_{b}) n_{2}, \nabla_{x} h = (\nabla_{a} + \nabla_{b}) n_{2},

\partial_{s} h = R \cdot \nabla_{x} h

. The derivative with respect to

x

here translates both arguments of the two-particle number density; it is not a derivative of its first argument alone.

Take the spatial divergence of the proposed interaction pressure tensor. Its pair potential factor is independent of

x

, so the preceding identity gives

\partial_{x_{j}} P_{ij}^{ϕ} = - \frac{1}{2} \int d^{3} R \frac{R _{i} ϕ ^{'} ( R )}{R} \int_{0}^{1} \partial_{s} h d s = - \frac{1}{2} \int d^{3} R \frac{R _{i} ϕ ^{'} ( R )}{R} [n_{2} (x, x + R) - n_{2} (x - R, x)] .

Exchange symmetry of the two-particle distribution function implies

n_{2} (x - R, x) = n_{2} (x, x - R)

. In the second term, substitute

R \mapsto - R

. Since

R_{i} ϕ^{'} (R) / R

is odd, both terms become equal:

\partial_{x_{j}} P_{ij}^{ϕ} = - \int d^{3} R \frac{R _{i} ϕ ^{'} ( R )}{R} n_{2} (x, x + R) .

On the other hand, at fixed

x_{⋆} = x + R

, the first-argument derivative of the pair potential is

\nabla_{x} ϕ = - R ϕ^{'} (R) / R

. Thus the interaction force density in b is

- \partial_{x_{j}} P_{ij}^{ϕ}

. Moving it to the left yields local conservation of momentum:

\partial_{t} (ρ u) + \nabla \cdot (ρ uu + P + P^{ϕ}) = 0.

The integral over

s

distributes each pair's momentum flux along the straight segment joining the particles. The factor

1/2

removes the double counting of ordered pairs. Central forces also make

P_{ij}^{ϕ} = P_{ji}^{ϕ}

d

Solution

Spatial homogeneity makes

ρ

independent of position and makes the two-particle number density depend only on the separation. Isotropy removes its directional dependence. Consequently

n_{2} (x, x_{⋆}, t) = ρ^{2} g_{2} (∣ x_{⋆} - x ∣, t) .

Here

g_{2}

is a radial distribution function, with the finite-

N

normalization inherited from a; statistical independence is unnecessary. In particular, the translated pair in c has separation

R

for every

s

, so its

s

integral is simply

ρ^{2} g_{2} (R, t)

Isotropy of the random velocity

c = v - u

makes its second velocity moment proportional to the identity. Taking its trace fixes the proportionality constant:

θ = \frac{1}{3 ρ} \int ∣ v - u ∣^{2} f d^{3} v, P_{ij} = ρθ δ_{ij} .

Thus

θ

is the one-component random velocity variance; for unit mass in thermal equilibrium,

θ = k_{B} T

. A Maxwell-Boltzmann velocity distribution is not required for the pressure tensor to be isotropic.

For the interaction contribution, the angular integral is

\int d Ω R_{i} R_{j} = (4 π /3) δ_{ij}

, by isotropy and its trace. With

d^{3} R = R^{2} d R d Ω

, c becomes

P_{ij}^{ϕ} = - \frac{2 π}{3} ρ^{2} δ_{ij} \int_{0}^{\infty} R^{3} ϕ^{'} (R) g_{2} (R, t) d R .

Combining the two pressure tensors gives the virial pressure from pair correlations:

P + P^{ϕ} = p I, p = ρθ - \frac{2 π}{3} ρ^{2} \int_{0}^{\infty} R^{3} ϕ^{'} (R) g_{2} (R, t) d R .

For a repulsive pair potential, the outward radial force is

- ϕ^{'} (R) \geq 0

. Since

g_{2} \geq 0

, the second term in

p

is nonnegative, and positive when occupied separations feel nonzero repulsion. It expresses additional momentum transfer through interactions. The displayed integral assumes a differentiable pair potential and convergence; a hard core requires a contact formula or a limiting argument.

2

a

Solution

Use the electron electric charge

- e

, so

E = - \partial_{x} φ

gives acceleration

+ (e / m_{e}) \partial_{x} φ

. The Vlasov equation, its average, and its linearized fluctuating part are

\partial_{t} f + v \partial_{x} f + \frac{e}{m _{e}} \partial_{x} φ \partial_{v} f = 0, \partial_{t} f_{0} = - \frac{e}{m _{e}} \partial_{v} ⟨ \partial_{x} φ δ f ⟩, (\partial_{t} + ik v) δ f_{k} = - \frac{ik e}{m _{e}} φ_{k} f_{0}^{'} .

The quasilinear approximation retains the averaged quadratic term but drops products of fluctuations from the fluctuating equation. It freezes the slowly evolving

f_{0}

during the fast response calculation. A real electrostatic potential has

φ_{- k} = φ_{k}^{*}

, and the sum includes both signs of every nonzero wavenumber. Spatial averaging selects the opposite Fourier series harmonics; statistical homogeneity supplies the corresponding diagonal noise covariance.

Let

g (p) = \int_{0}^{\infty} e^{- pt} g (t) d t

. Zero initial

δ f_{k}

gives

δ f_{k} (p, v) = - \frac{ik e}{m _{e}} \frac{f _{0}^{'} ( v )}{p + ik v} φ_{k} (p) .

Insert this into the electrostatic self-consistency relation. Using

ω_{p e}^{2} = 4 π e^{2} n_{e} / m_{e}

and

p + ik v = ik (v - i p / k)

gives

ϵ (p, k) φ_{k} = χ_{k}, ϵ (p, k) = 1 - \frac{ω _{p e}^{2}}{k ^{2} n _{e}} \int_{C_{L}} \frac{f _{0}^{'} ( v ^{'} )}{v ^{'} - i p / k} d v^{'} .

This derives the stated dielectric function with the Landau contour. A noise-free growing collective response is a zero of

ϵ

The Laplace transform of the white noise covariance is particularly simple. Writing

q = p^{' *}

⟨ χ_{k} (p) χ_{k} (p^{'})^{*} ⟩ = 2 A_{k} \int_{0}^{\infty} d s \int_{0}^{\infty} d s^{'} e^{- p s - q s^{'}} δ (s - s^{'}) = \frac{2 A _{k}}{p + q} .

For clarity about contour signs, introduce the inverse-transform measures

\int_{B} \frac{d p}{2 πi}, \int_{B^{*}} \frac{d q}{- 2 πi} .

The first Bromwich inversion formula contour runs upward on

Re p = σ

; its conjugate contour runs downward on

Re q = σ

. Both measures therefore have the same positive orientation in their real frequency parameters. Combining the averaged equation with

δ f_{k}

yields

\partial_{t} f_{0} = \partial_{v} [D (v, t) \partial_{v} f_{0}], D = - \frac{e}{m _{e} f _{0}^{'}} \sum_{k} (- ik) ⟨ δ f_{k} (t, v) φ_{k} (t)^{*} ⟩ .

At a zero of

f_{0}^{'}

, the last expression means its continuous coefficient obtained before cancelling

f_{0}^{'}

. Substitution of the two inverse Laplace transforms and the transformed covariance gives

D (v, t) = \frac{e ^{2}}{m _{e}^{2}} k \sum \frac{k ^{2} A _{k}}{2 π ^{2}} \int_{B} d p \int_{B^{*}} d p^{' *} \frac{e ^{(p + p^{' *}) t}}{( p + ik v ) ( p + p ^{' *} ) ϵ ( p , k ) ϵ ( p ^{'} , k ) ^{*}} .

This is the paper's expression, with the orientation of

d p^{' *}

made explicit. Reality follows by pairing

k

and

- k

. The double integral retains switch-on transients; replacing it by a time-independent diffusion coefficient requires the additional long-time limit in b.

b

Solution

A linearly stable

f_{0}

has no zeros of the causal dielectric function in

Re p > 0

. Hence the inverse Laplace transform contours can approach

σ = 0^{+}

, retaining the causal boundary value at

p = - iω + 0^{+}

. To obtain a stationary forced response, further assume that the excited collective perturbations actually damp. This is the decay condition in the question's long-time instruction.

A transparent evaluation of the double integral uses the stationary power spectral density. The white noise covariance

2 A_{k} δ (t - t^{'})

has temporal spectrum

2 A_{k}

. Linear filtering by

1/ ϵ

therefore gives

S_{φ, k} (ω) = \frac{2 A _{k}}{∣ ϵ ( - iω + 0 ^{+} , k ) ∣ ^{2}}, ⟨ φ_{k} (t - τ) φ_{k} (t)^{*} ⟩ = \int_{- \infty}^{\infty} \frac{d ω}{2 π} S_{φ, k} (ω) e^{iω τ} .

The causal solution for

δ f_{k}

in a gives, after the initial transients have decayed,

D (v) = \frac{e ^{2}}{m _{e}^{2}} \sum_{k} k^{2} Re \int_{0}^{\infty} d τ e^{- ik vτ} ⟨ φ_{k} (t - τ) φ_{k} (t)^{*} ⟩ .

The real part can be taken because the full

k

sum pairs conjugates. Using the distribution theory identity

Re \int_{0}^{\infty} e^{ia τ} d τ = π δ (a)

, with a positive vanishing convergence factor, samples the temporal spectrum at

ω = k v

D (v) = \frac{e ^{2}}{m _{e}^{2}} k \sum \frac{k ^{2} A _{k}}{∣ ϵ ( - ik v + 0 ^{+} , k ) ∣ ^{2}} .

The factor

1/2

from the one-sided time integral cancels the

2

in the noise covariance. The result is nonnegative, as a stationary velocity-space diffusion coefficient should be.

Qualification of stability. Absence of growing modes alone does not guarantee this stationary limit. For example, the electron waterbag distribution

f_{0} (v) = n_{e} 1_{∣ v ∣ < V} / (2 V)

has

ϵ (- iω, k) = 1 - \frac{ω _{p e}^{2}}{ω ^{2} - k ^{2} V ^{2}}, ω^{2} = k^{2} V^{2} + ω_{p e}^{2}

as undamped collective roots. White noise drives their field variance to grow instead of settling. Thus the formula requires decay of excited modes, not merely absence of exponentially growing ones. For weak Landau damping, the waiting time can be much longer than

ω_{p e}^{- 1}

, and must still be short compared with evolution of

f_{0}

for the frozen-background quasilinear approximation to apply.

c

Solution

For a regular thermal

f_{0}

and noise at

k λ_{De} ≫ 1

, the susceptibility

ϵ - 1

is small at thermal phase velocities: its scale is

ω_{p e}^{2} / (k^{2} v_{the}^{2})

. To leading order the electrons cannot appreciably screen such short-scale fields, and b becomes independent of velocity:

D (v) ≃ D_{0} = \frac{e ^{2}}{m _{e}^{2}} \sum_{k} k^{2} A_{k} .

The electron kinetic energy per unit volume is

K = (m_{e} /2) \int v^{2} f_{0} d v

. Assume

v^{2} D f_{0}^{'}

and

v f_{0}

vanish at infinity. One integration by parts gives the general heating formula

\frac{d K}{d t} = - m_{e} \int_{- \infty}^{\infty} v D (v) f_{0}^{'} (v) d v .

For constant

D_{0}

, a second integration by parts and

\int f_{0} d v = n_{e}

give

\frac{d K}{d t} = m_{e} n_{e} D_{0} = \frac{e ^{2} n _{e}}{m _{e}} k \sum k^{2} A_{k} .

This is the leading short-wavelength result. With vanishing boundary flux, the diffusion equation also conserves particle number; constant

D_{0}

conserves the mean momentum while increasing the random kinetic energy.

d

Solution

Set

κ = ∣ k ∣

and first work with a representative positive

k = κ

. For

κ λ_{De} ≪ 1

, the thermal core is strongly screened. However, the causal dielectric function has weakly damped Langmuir wave roots

ω_{s} + i γ_{s}

, with

s = \pm 1

ω_{s} ≃ s ω_{p e}

and

γ_{s} < 0

. Their phase velocities

v_{s} = ω_{s} / κ

are in the tail. Both roots must be retained for each spatial Fourier series harmonic.

Write

ϵ (- iω + 0^{+}, κ) = ϵ_{R} (ω) + i ϵ_{I} (ω)

. The supplied approximation gives

ϵ_{R} ≃ 1 - \frac{ω _{p e}^{2}}{ω ^{2}}, ϵ_{I} ≃ - \frac{π ω _{p e}^{2}}{κ ^{2} n _{e}} f_{0}^{'} (ω / κ), ϵ_{R}^{'} (ω_{s}) ≃ \frac{2}{ω _{s}} .

Expanding

ϵ (ω_{s} + i γ_{s}) = 0

to first order gives the Landau damping rate

γ_{s} = - \frac{ϵ _{I} ( ω _{s} )}{ϵ _{R}^{'} ( ω _{s} )} ≃ \frac{π ω _{s} ω _{p e}^{2}}{2 κ ^{2} n _{e}} f_{0}^{'} (v_{s}) .

In particular, stable damping means

ω_{s} f_{0}^{'} (v_{s}) < 0

. Near either root,

∣ ϵ (- iκ v + 0^{+}, κ) ∣^{2} ≃ \frac{4}{ω _{p e}^{2}} [(κ v - ω_{s})^{2} + γ_{s}^{2}] .

The narrow Lorentzian resonance approximation then gives the leading resonant contribution

\frac{1}{∣ ϵ ( - iκ v + 0 ^{+} , κ ) ∣ ^{2}} ≃ \frac{π ω _{p e}^{2}}{4} \sum_{s = \pm 1} \frac{δ ( κ v - ω _{s} )}{∣ γ _{s} ∣} .

This is a formula under the integral over velocity; it is not a pointwise approximation away from the resonances. Insert it into the heating formula in c. For one spatial harmonic, the contribution from one root is

Q_{k, s} = - \frac{e ^{2}}{m _{e}} k^{2} A_{k} \frac{π ω _{p e}^{2}}{4 κ} \frac{v _{s} f _{0}^{'} ( v _{s} )}{∣ γ _{s} ∣} = \frac{e ^{2} n _{e}}{2 m _{e}} k^{2} A_{k},

because

\frac{v _{s} f _{0}^{'} ( v _{s} )}{∣ γ _{s} ∣} = - \frac{2 κ n _{e}}{π ω _{p e}^{2}} .

Thus the tiny tail slope cancels the tiny Landau damping rate: weak absorption is compensated by a large noise-driven wave intensity. The two temporal roots supply two equal halves of the answer. Negative

k

gives the same result by the conjugate response of a real electric field, without altering the original convention that the spatial sum includes both signs. Therefore

\frac{d K}{d t} = k \sum s = \pm 1 \sum Q_{k, s} ≃ \frac{e ^{2} n _{e}}{m _{e}} k \sum k^{2} A_{k} .

As in c, equality means the leading result in the stated asymptotic limit. This calculation assumes nonzero damping, observation times

t ≫ max_{s} ∣ γ_{s} ∣^{- 1}

, and forcing weak enough that

f_{0}

changes negligibly during that relaxation. Taking zero damping before the long-time limit invalidates the stationary calculation.

e

Solution

The long-wavelength limit in d is resonant heating. Most electrons screen the slowly varying spatial field. The noise nevertheless excites collective Langmuir waves, whose sharp temporal spectrum lies near

ω = \pm ω_{p e}

. Electrons with

v ≃ ω / k

remain in phase with a wave and exchange energy with it. A stable tail absorbs net wave energy through Landau damping. The heating is initially deposited in narrow tail regions of velocity space, even though the volume-integrated rate equals that in c.

The short-wavelength limit in c is stochastic heating in the terminology intended here. With

ϵ ≃ 1

, the imposed electric field is essentially unscreened. Its temporally uncorrelated kicks cause a random walk of electron velocity, so

⟨(Δ v)^{2} ⟩ = 2 D_{0} Δ t

. Electrons throughout the thermal population contribute to the resulting kinetic energy increase; no narrow collective wave peak selects a small tail population.

Both mechanisms are driven by white noise, and the formal condition

ω = k v

enters the stationary quasilinear diffusion calculation in both cases. The physical distinction is the sharp collective resonance in d versus the broad direct random forcing in c, rather than the mere presence of randomness. In collisionless theory this heating is an increase of the averaged kinetic energy; subsequent collisions can establish thermal equilibrium.

3

a

Solution

For an integrable or approximately integrable background Hamiltonian, action-angle variables turn an otherwise complicated stellar orbit into

\dot{J} = 0

and

\dot{θ} = Ω (J)

. This has two useful consequences. First, a phase-mixed equilibrium can be represented directly as

f_{0} (J)

: the action variables label the unperturbed orbits, while angle variables describe position along them. Second, the angles are periodic, so Fourier series reduce unperturbed transport to multiplication by

i n \cdot Ω

and make orbital resonances explicit. The action variables also separate rapid orbital motion from slow secular changes. These advantages apply on regular orbital regions; a globally chaotic background need not admit global action-angle variables.

The transformation is canonical, preserving the phase space measure and the stated total mass

M = \int f d^{3} θ d^{3} J

. The collisionless Vlasov equation is

\partial_{t} f + {f, H}_{θ, J} = \partial_{t} f + \frac{\partial H}{\partial J} \cdot \frac{\partial f}{\partial θ} - \frac{\partial H}{\partial θ} \cdot \frac{\partial f}{\partial J} = 0.

Here

H

is the single-star specific Hamiltonian; the normalization of

f

by mass changes neither the Poisson bracket nor the transport equation.

b

Solution

Since

H_{0}

and

f_{0}

depend only on the action variables,

{f_{0}, H_{0}} = 0

. Drop the quadratic term

{δ f, δ Φ^{tot}}

from the Vlasov equation in a. Defining the orbital frequency vector by

Ω (J) = \partial H_{0} / \partial J

, the linearized equation is

\partial_{t} δ f + Ω \cdot \partial_{θ} δ f - \frac{\partial f _{0}}{\partial J} \cdot \partial_{θ} δ Φ^{tot} = 0.

The sign of the driving term follows from the second term in the canonical Poisson bracket, because the background has no angle dependence.

c

Solution

Use the stated Fourier series normalization, so

g (θ, J, t) = \sum_{n} g_{n} (J, t) e^{i n \cdot θ}

. Then b gives

\partial_{t} δ f_{n} + i n \cdot Ω δ f_{n} - i n \cdot \frac{\partial f _{0}}{\partial J} δ Φ_{n}^{tot} = 0.

For the Laplace transform with kernel

e^{iω t}

, integration by parts yields

\int_{0}^{\infty} e^{iω t} \partial_{t} δ f_{n} d t = - δ f_{n} (0) - iω δ f_{n} .

The upper boundary vanishes for sufficiently large

Im ω > 0

, and the initial term is zero by hypothesis. Thus

i (n \cdot Ω - ω) δ f_{n} = i n \cdot \frac{\partial f _{0}}{\partial J} δ Φ_{n}^{tot},

δ f_{n} (J, ω) = - n \cdot \frac{\partial f _{0}}{\partial J} \frac{δ Φ _{n}^{tot} ( J , ω )}{ω - n \cdot Ω ( J )}, Ω = \frac{\partial H _{0}}{\partial J} .

The denominator identifies the orbital resonance between the perturbation and unperturbed orbital frequencies. The zero harmonic has no driving term and remains zero with these initial data.

d

Solution

In the periodic cube,

H_{0} = ∣ v ∣^{2} /2 = (2 π / L)^{2} ∣ J ∣^{2} /2

, so

Ω = (2 π / L) v

. The supplied action-angle variables are canonical:

d^{3} θ d^{3} J = d^{3} x d^{3} v

. For

k = 2 π n / L

n \cdot Ω = k \cdot v, n \cdot \partial_{J} f_{0} = k \cdot \partial_{v} f_{0} .

Consequently c becomes

δ f_{k} (v, ω) = \frac{k \cdot \partial _{v} f _{0}}{k \cdot v - ω} δ Φ_{k}^{tot} (ω) .

For each nonzero wavenumber, the Poisson equation for Newtonian gravity gives

- k^{2} δ Φ_{k} = 4 π G \int δ f_{k} d^{3} v .

Substitute the preceding response and add the externally imposed Newtonian gravitational potential:

δ Φ_{k}^{tot} = δ Φ_{k}^{ext} - \frac{4 π G}{k ^{2}} δ Φ_{k}^{tot} \int d^{3} v \frac{k \cdot \partial _{v} f _{0}}{k \cdot v - ω} .

Therefore

δ Φ_{k}^{tot} = \frac{δ Φ _{k}^{ext}}{ϵ _{k} ( ω )}, ϵ_{k} (ω) = 1 + \frac{4 π G}{k ^{2}} \int d^{3} v \frac{k \cdot \partial _{v} f _{0}}{k \cdot v - ω} .

This gravitational dielectric function has the opposite susceptibility sign to the electron dielectric function, reflecting attractive rather than repulsive collective interactions. The homogeneous mean field is explicitly subtracted in this fictitious periodic model: the Poisson equation for Newtonian gravity acts on density fluctuations, and the

k = 0

mode is omitted. Otherwise a nonzero uniform positive mass density would be incompatible with a periodic potential. This is the convention often called the Jeans swindle.

For analytic continuation, resolve velocity parallel to

k

: let

u = k \cdot v

and

F (u) = \int f_{0} (u, v_{⊥}) d^{2} v_{⊥}

. Then

ϵ_{k} (ω) = 1 + \frac{4 π G}{k ^{2}} \int \frac{F ^{'} ( u )}{u - ω / k} d u, k = ∣ k ∣.

Start at

Im ω > 0

and deform the Landau contour continuously below the moving pole. At real

ω

, the integral is

PV \int \frac{F ^{'} ( u )}{u - ω / k} d u + iπ F^{'} (ω / k) .

For

Im ω < 0

, this causal continuation equals the undeformed real-axis integral plus

2 πi F^{'} (ω / k)

wherever

F^{'}

has the necessary analytic extension. It must not be replaced by the undeformed integral alone.

Zeros of

ϵ_{k}

are self-consistent collective Landau modes: with no external forcing, a nonzero Newtonian gravitational potential is possible at such a root. For time dependence

e^{- iω t}

, upper-half-plane roots grow, lower-half-plane roots describe Landau damping, and real roots are neutral. Damped Landau modes are poles of the continued response; they need not be regular discrete eigenfunctions of the original collisionless operator.

Regularity qualification. Continuation throughout the complex plane presupposes suitable analytic

F

. Arbitrary smooth

F

need not allow it. For example,

F (u) = C exp [- 1/ (1 - u^{2})]

for

∣ u ∣ < 1

and zero elsewhere is smooth but not analytic at

u = \pm 1

; its Cauchy integral cannot be holomorphically continued across those endpoints. The Landau contour gives the standard causal continuation on the domains where the equilibrium permits it, with possible singularities retained.

e

Solution

Represent the perturber by the periodic Dirac delta function at

x_{p} + v_{p} t

. With spatial Fourier series coefficient normalized by

L^{- 3}

, its nonzero density coefficients are

ρ_{p, k} (t) = \frac{m _{p}}{L ^{3}} e^{- i k \cdot x_{p}} e^{- i k \cdot v_{p} t}, t \geq 0.

The mean density is subtracted as in d. The Poisson equation for Newtonian gravity gives

δ Φ_{k}^{ext} (t) = C_{k} e^{- ia t}, C_{k} = - \frac{4 π G m _{p}}{L ^{3} k ^{2}} e^{- i k \cdot x_{p}}, a = k \cdot v_{p} .

The perturbation is zero before switch-on. For

Im ω > 0

, its Laplace transform is

\int_{0}^{\infty} e^{i (ω - a) t} d t = - \frac{1}{i ( ω - a )} = \frac{1}{i ( a - ω )} .

Hence

δ Φ_{k}^{ext} (ω) = - \frac{4 π G m _{p}}{L ^{3} k ^{2}} e^{- i k \cdot x_{p}} \frac{1}{i ( k \cdot v _{p} - ω )} .

The switch-on creates a forcing pole at the real frequency

a

, approached causally from the original upper inversion contour.

f

Solution

With

C_{k}

and

a

from e, inversion gives

δ Φ_{k}^{tot} (t) = \frac{1}{2 π} \int_{Im ω = η > 0} d ω e^{- iω t} \frac{i C _{k}}{( ω - a ) ϵ _{k} ( ω )} .

For

t > 0

, deform the contour downwards. Closing below is clockwise, so the residue theorem multiplies the sum of enclosed residues by

- i

. The forcing pole has residue

i C_{k} e^{- ia t} / ϵ_{k} (a + i 0^{+})

and therefore contributes

\frac{C _{k} e ^{- ia t}}{ϵ _{k} ( a + i 0 ^{+} )} .

A simple collective root

ω_{j}

of the continued gravitational dielectric function contributes

\frac{C _{k} e ^{- i ω_{j} t}}{( ω _{j} - a ) ϵ _{k}^{'} ( ω _{j} )} .

These are the Landau mode transients excited by introducing the mass abruptly. For a strictly damped background they decay, together with any remaining phase-mixing contour contribution under the usual regularity and decay assumptions. Thus

δ Φ_{k}^{tot} (t) \sim \frac{δ Φ _{k}^{ext} ( t )}{ϵ _{k} ( k \cdot v _{p} + i 0 ^{+} )}, t \to \infty.

The causal boundary value is what the paper abbreviates by

ϵ_{k} (k \cdot v_{p})

. Physically, the moving mass polarizes the stellar population and carries a gravitational wake. Each spatial harmonic is changed in amplitude and phase by

1/ ϵ_{k}

at the forcing frequency. Self-gravity can amplify the wake. The induced part alone is

(1/ ϵ_{k} - 1) δ Φ_{k}^{ext}

; the wake's back-reaction on the perturber produces dynamical friction.

The waiting time is controlled by the slowest excited collective damping, not just an orbital or crossing time. If

ω_{j} = ω_{r j} - i Γ_{j}

with

Γ_{j} > 0

, then a modal amplitude is suppressed by

e^{- Γ_{j} t}

, and one needs

t Γ_{j} ≫ 1

for every appreciably excited mode, together with decay of the phase-mixing remainder. Large transient prefactors can require additional time. Near marginal stability, a relevant

Γ_{j}

tends to zero and the required waiting time diverges; the stationary wake approximation is then not uniform in the stability parameter.^[ref]

Qualification of stability. If stable means only non-growing, a persistent neutral mode invalidates the asserted asymptote. A concrete example is a homogeneous waterbag distribution with parallel marginal

F (u) = ρ 1_{∣ u ∣ < V} / (2 V)

. Its gravitational dielectric function is

ϵ_{k} (ω) = 1 - \frac{4 π Gρ}{k ^{2} V ^{2} - ω ^{2}} .

For

k^{2} V^{2} > 4 π Gρ

, the roots

ω = \pm k^{2} V^{2} - 4 π Gρ

are real, so the switch-on transients persist although that harmonic has no growing mode. Taking the smallest box wavenumber to satisfy the same inequality makes the whole waterbag model free of growing harmonics. The claimed late-time formula therefore requires damping of all excited modes and

ϵ_{k} (a) \neq = 0

; exact forcing at an undamped root instead produces a secular response.

2023 (A15089W1)

Source PDF: mmathphys.web.ox.ac.uk/sitefiles/kinetic-theory-2023.pdf

Solutions to all three questions of the examination held on 12 January 2023. The cover also says Hilary Term 2022; the examination date and source filename determine the year used here. The numbered headings identify source parts without reproducing the questions. Corrections to printed formulas are explained where they arise.

1

a

Solution

Write

K_{ij} = K (x_{i}, x_{j})

and

P = \sum_{i} v_{i}

. The unit particle masses make this sum the total momentum. Exchange the two indices and use the symmetry

K_{ij} = K_{ji}

\dot{P} = \sum_{i, j} K_{ij} (v_{j} - v_{i}) = \frac{1}{2} \sum_{i, j} K_{ij} [(v_{j} - v_{i}) + (v_{i} - v_{j})] = 0 .

Thus every interacting pair transfers equal and opposite momentum, even though the velocity alignment dissipates kinetic energy.

For the total kinetic energy

E_{N} = \frac{1}{2} \sum_{i} ∣ v_{i} ∣^{2}

, apply the same exchange of indices:

\dot{E}_{N} = i, j \sum K_{ij} v_{i} \cdot (v_{j} - v_{i}) = \frac{1}{2} i, j \sum K_{ij} [v_{i} \cdot (v_{j} - v_{i}) + v_{j} \cdot (v_{i} - v_{j})] = - \frac{1}{2} i, j \sum K_{ij} ∣ v_{j} - v_{i} ∣^{2} \leq 0 .

The dissipated kinetic energy is that of relative motion. Since the total momentum is constant,

E_{N} - ∣ P ∣^{2} / (2 N) = \frac{1}{2} \sum_{i} ∣ v_{i} - P / N ∣^{2}

also decreases. Equality holds when every pair with

K_{ij} > 0

has the same velocity; the hypotheses permit disconnected interacting groups, so they alone do not guarantee eventual global velocity alignment.

b

Solution

Let

A_{i} = \sum_{j \neq = i} K_{ij} (v_{j} - v_{i})

. Conservation of probability in the full phase space gives the dissipative Liouville equation

\partial_{t} ϱ + i = 1 \sum N \nabla_{x_{i}} \cdot (v_{i} ϱ) + i = 1 \sum N \nabla_{v_{i}} \cdot (A_{i} ϱ) = 0.

Normalize

\int ϱ \prod_{i} d^{3} x_{i} d^{3} v_{i} = 1

. Decay at infinity makes all boundary fluxes vanish, so the equation preserves this normalization of a probability distribution.

Unlike the Hamiltonian Liouville equation, this transport has nonzero phase-space divergence. In three dimensions,

\nabla_{v_{i}} \cdot A_{i} = - 3 \sum_{j \neq = i} K_{ij}, \partial_{t} ϱ + \sum_{i} (v_{i} \cdot \nabla_{x_{i}} ϱ + A_{i} \cdot \nabla_{v_{i}} ϱ) = 3 ϱ \sum_{i} \sum_{j \neq = i} K_{ij} .

The probability density increases along a contracting characteristic flow map; setting its material derivative to zero would lose conservation of probability. The

j = i

force is identically zero, including its derivative, which is why the last sums exclude it.

c

Solution

Put

z_{i} = (x_{i}, v_{i})

and

d z_{i} = d^{3} x_{i} d^{3} v_{i}

. Indistinguishability gives an exchangeable ensemble. Define unit-normalized marginal distributions

p_{1} (z_{1}, t) = \int ϱ d z_{2} \dots d z_{N}, p_{2} (z_{1}, z_{2}, t) = \int ϱ d z_{3} \dots d z_{N} .

Integrating the dissipative Liouville equation over

z_{2}, \dots, z_{N}

removes their boundary fluxes. The

N - 1

identical pair terms remain:

\partial_{t} p_{1} + v \cdot \nabla_{x} p_{1} + (N - 1) \nabla_{v} \cdot \int K (x, x_{⋆}) (v_{⋆} - v) p_{2} (z, z_{⋆}, t) d z_{⋆} = 0.

This is the first equation of a BBGKY hierarchy, with no independence assumed yet.

The additional physical assumption is a mean-field closure for velocity alignment: neglect pair correlations,

p_{2} (z, z_{⋆}, t) ≃ p_{1} (z, t) p_{1} (z_{⋆}, t)

. Initial independence alone does not make this exact at later times, because the particles interact. In the usual number convention,

f = N p_{1}

and the two-particle distribution function is

f_{2} = N (N - 1) p_{2}

. Consequently the closed equation at finite

N

\partial_{t} f + v \cdot \nabla_{x} f + (1 - \frac{1}{N}) \nabla_{v} \cdot (f L [f]) ≃ 0, L [f] (z) = \int K (x, x_{⋆}) (v_{⋆} - v) f (z_{⋆}, t) d z_{⋆} .

Neglecting the relative

1/ N

correction gives the intended mean-field alignment equation:

\partial_{t} f + v \cdot \nabla_{x} f + \nabla_{v} \cdot (f L [f]) = 0.

Equivalently, the closure

f_{2} ≃ f f_{⋆}

incorporates this large-

N

approximation. It cannot be an exact finite-

N

identity with

\int f = N

, since

\int f_{2} d z_{⋆} = (N - 1) f

whereas

\int f f_{⋆} d z_{⋆} = N f

There is also a consistent finite-

N

convention yielding the boxed equation directly under the independence closure: define

f = (N - 1) p_{1}

instead. Its total normalization of a probability distribution is

N - 1

, rather than the physical particle number

N

. Thus the displayed source equation needs a stated normalization of a probability distribution convention, or the large-

N

approximation, in addition to neglect of correlations. In the remaining parts we use the displayed mean-field alignment equation itself; its moment identities do not depend on the total normalization of a probability distribution.

d

Solution

Use

d z = d^{3} x d^{3} v

and set

ρ (x, t) = \int f (x, v, t) d^{3} v, E = \frac{1}{2} \int ∣ v ∣^{2} f d z, H = \int f lo g f d z .

For unit particle mass,

ρ

is both the number density and the mass density in the number convention of c. A fixed reference scale inside the logarithm changes

H

by a constant multiple of the conserved total mass, leaving its derivative unchanged.

Both prefactors in the source's part (d) need correction. Multiplication of the mean-field alignment equation by

∣ v ∣^{2} /2

and integration by parts gives

\dot{E} = \int f v \cdot L [f] d z = \iint K (x, x_{⋆}) v \cdot (v_{⋆} - v) f f_{⋆} d z d z_{⋆} .

Exchange

z

and

z_{⋆}

, use the symmetry of

K

, and average the two expressions:

\dot{E} = - \frac{1}{2} \iint K (x, x_{⋆}) ∣ v_{⋆} - v ∣^{2} f f_{⋆} d z d z_{⋆} \leq 0.

The printed right-hand side omits

1/2

. This corrected velocity-alignment energy dissipation agrees with the pair-counting calculation in a.

For velocity-alignment entropy production, let

A (x, t) = \int K (x, x_{⋆}) ρ (x_{⋆}, t) d^{3} x_{⋆}

. The velocity-space divergence is

\nabla_{v} \cdot L [f] = - 3 A

. Assuming

f > 0

and sufficient decay, integration by parts yields

\dot{H} = - \int (1 + lo g f) [v \cdot \nabla_{x} f + \nabla_{v} \cdot (f L [f])] d z = \int L [f] \cdot \nabla_{v} f d z = - \int f \nabla_{v} \cdot L [f] d z = 3 \iint K (x, x_{⋆}) ρ (x, t) ρ (x_{⋆}, t) d^{3} x d^{3} x_{⋆} \geq 0.

The printed formula omits the factor

3

from the three velocity components. In

d

spatial dimensions, that factor would be

d

. Zeros of

f

can be treated by a regularized entropy calculation when the required integrability holds.

A concrete smooth counterexample to both printed coefficients is

K \equiv 1

and

f (z, 0) = g (x) (2 π)^{- 3/2} e^{- ∣ v ∣^{2} /2}

, where

g

is a positive normalized Gaussian distribution. The total mass is one, the total momentum is zero, and

L [f] = - v

initially. The kinetic energy derivative is

- \int ∣ v ∣^{2} f d z = - 3

, whereas the printed formula gives

- 6

. The entropy functional derivative is

3

, whereas the printed formula gives

1

The signs describe cooling of relative velocities and contraction in phase space:

H

increases, so the fine-grained Gibbs entropy

- H

decreases. These dissipative particle dynamics are not an isolated Hamiltonian system; there is no contradiction with Liouville's theorem or the Boltzmann equation's usual entropy increase. The microscopic mechanism removing kinetic energy is outside this alignment model.

e

Solution

Define the local mean velocity and momentum flux by

ρ u = \int v f d^{3} v, Π_{ij} = \int v_{i} v_{j} f d^{3} v .

Here

ρ = \int f d^{3} v

and

u

is defined where

ρ > 0

. The pressure tensor separates random motion from the bulk velocity:

Π_{ij} = ρ u_{i} u_{j} + P_{ij}, P_{ij} = \int (v_{i} - u_{i}) (v_{j} - u_{j}) f d^{3} v .

Integrating the mean-field alignment equation over velocity space removes the force divergence and gives the continuity equation

\partial_{t} ρ + \nabla_{x} \cdot (ρ u) = 0.

Its first velocity moment follows by multiplying by

v_{i}

and using

\int v_{i} \partial_{v_{j}} (f L_{j}) d^{3} v = - \int f L_{i} d^{3} v

\partial_{t} (ρ u_{i}) + \partial_{x_{j}} Π_{ij} = \int f L_{i} [f] d^{3} v .

Inserting the mean-field alignment equation's force and integrating each particle velocity separately gives

\partial_{t} (ρ u) + \nabla_{x} \cdot Π = \int K (x, x_{⋆}) [u (x_{⋆}, t) - u (x, t)] ρ (x_{⋆}, t) ρ (x, t) d^{3} x_{⋆} .

The right-hand side exchanges momentum between positions and aligns their bulk velocities. Its spatial integral vanishes by exchange symmetry, recovering total momentum conservation. No closure for the pressure tensor is needed for these exact velocity moment identities.

2

a

Solution

The step in the source must be at zero. We use the Heaviside step function

H (x) = 1

for

x > 0

, zero for

x < 0

, with its value at zero immaterial here. The printed threshold

1

is inconsistent with both

H^{'} = δ (x)

and the stated number density. For example, measuring velocities in fixed units and taking

v_{F} = 2

, the literal printed cutoff fills only

v < 1

and gives one eighth of the stated number density. Its distributional derivative is

δ (x - 1)

, not

δ (x)

. The zero-threshold convention restores the stated Fermi sphere and its normalization of a probability distribution.

Let

k = ∣ k ∣ > 0

, choose

u = v \cdot k / k

, and abbreviate

V_{α} = v_{F α}

. The three-dimensional phase-space distribution function is

f_{0 α} = C_{α} 1_{v < V_{α}}

, where

C_{α} = \frac{m _{α}^{3}}{4 π ^{3} ℏ ^{3}} = \frac{3 n _{α}}{4 π V _{α}^{3}} .

Integrate over the transverse velocity disk to obtain the parallel marginal distribution

F_{α} (u) = \int f_{0 α} d^{2} v_{⊥} = \frac{3 n _{α}}{4 V _{α}^{3}} (V_{α}^{2} - u^{2}) 1_{∣ u ∣ < V_{α}}, F_{α}^{'} (u) = - \frac{3 n _{α} u}{2 V _{α}^{3}} 1_{∣ u ∣ < V_{α}} .

There are no edge Dirac delta functions in

F_{α}^{'}

because

F_{α}

itself vanishes continuously at the endpoints. In particular, the three-dimensional constant occupation does not imply a constant parallel marginal distribution.

Use time dependence

e^{pt + i k \cdot x}

, with

p = γ - iω

and

γ < 0

for damping. The given dielectric function becomes

ϵ = 1 - \sum_{α} \frac{ω _{p α}^{2}}{n _{α} k ^{2}} \int_{C_{L}} \frac{F _{α}^{'} ( u )}{u - ω / k - iγ / k} d u .

The Landau contour supplies causal continuation. For a real phase velocity

a = ω / k > V_{α}

, there is no resonant pole in the populated interval, and

\int \frac{F _{α}^{'} ( u )}{u - a} d u = - \frac{3 n _{α}}{2 V _{α}^{3}} \int_{- V_{α}}^{V_{α}} \frac{u}{u - a} d u = - \frac{3 n _{α}}{2 V _{α}^{3}} [2 V_{α} + a lo g \frac{a - V _{α}}{a + V _{α}}] .

Hence this species contributes

χ_{α} = \frac{3 ω _{p α}^{2}}{k ^{2} V _{α}^{2}} [1 + \frac{s _{α}}{2} lo g \frac{s _{α} - 1}{s _{α} + 1}], s_{α} = \frac{ω}{k V _{α}} > 1.

For comparable species number densities and fixed charge ratios,

m_{e} ≪ m_{i}

implies

V_{i} / V_{e} = O (m_{e} / m_{i})

and

ω_{p i}^{2} / ω_{p e}^{2} = O (m_{e} / m_{i})

. Since

ω > k V_{e}

, the ions have

s_{i} ≫ 1

and

χ_{i} = - ω_{p i}^{2} / ω^{2} + O (k^{2} V_{i}^{2} ω_{p i}^{2} / ω^{4})

. Their response is negligible to leading order in the mass ratio. The real dispersion relation is therefore

1 + \frac{3 ω _{p e}^{2}}{k ^{2} V _{e}^{2}} [1 + \frac{ω}{2 k V _{e}} lo g \frac{ω - k V _{e}}{ω + k V _{e}}] = 0.

This degenerate-plasma dielectric response treats a completely degenerate, nonrelativistic Fermi gas with classical particle trajectories. It does not include quantum recoil or a collision term. A finite ion response shifts the root slightly; the boxed equation is the stated leading electron approximation.

b

Solution

There is no Landau damping of this branch within the completely degenerate collisionless model. Its phase velocity obeys

ω / k > V_{e} > V_{i}

, while each parallel marginal distribution vanishes outside

∣ u ∣ \leq V_{α}

. There are no particles at the resonant velocity, and

F_{α}^{'} (ω / k) = 0

. The Landau contour consequently contributes no imaginary part to the dielectric function; a real root is allowed with

γ = 0

If a small nonzero temperature supplied a tail beyond the Fermi surface, the conclusion would change: the resonant derivative would generally be nonzero. For the convention

p = γ - iω

and a simple weakly damped root, expanding the continued dielectric function would give

γ = - ϵ_{I} (ω, k) / \partial_{ω} ϵ_{R} (ω, k)

. This conditional formula is not needed to manufacture damping for the exact cutoff. Neither collisional attenuation nor nonlinear evolution is included in the given dispersion relation.

c

Solution

Let

κ = k V_{e} / ω_{p e}

and

s = ω / (k V_{e})

. The long-wavelength electron branch has

∣ s ∣ ≫ 1

. The convergent power series

lo g \frac{s - 1}{s + 1} = - 2 (\frac{1}{s} + \frac{1}{3 s ^{3}} + \frac{1}{5 s ^{5}} + \dots)

turns the degenerate-plasma dielectric response into

0 = 1 - \frac{ω _{p e}^{2}}{ω ^{2}} - \frac{3 ω _{p e}^{2} k ^{2} V _{e}^{2}}{5 ω ^{4}} + O (\frac{ω _{p e}^{2} k ^{4} V _{e}^{4}}{ω ^{6}}) .

Substitute

ω^{2} = ω_{p e}^{2} + a k^{2} V_{e}^{2} + O (k^{4} V_{e}^{4} / ω_{p e}^{2})

. The coefficient of

k^{2}

a - 3/5

, so the degenerate-plasma Langmuir dispersion is

ω^{2} = ω_{p e}^{2} + \frac{3}{5} k^{2} V_{e}^{2} + O (\frac{k ^{4} V _{e}^{4}}{ω _{p e}^{2}}), ω_{\pm} = \pm ω_{p e} [1 + \frac{3}{10} κ^{2} + O (κ^{4})] .

The Langmuir wave therefore has a finite frequency at zero wavenumber, with the first dispersive correction set by the Fermi velocity. Indeed the occupied Fermi sphere has

⟨ v_{∥}^{2} ⟩ = V_{e}^{2} /5

, so the coefficient

3 ⟨ v_{∥}^{2} ⟩ = 3 V_{e}^{2} /5

agrees with the high-phase velocity velocity moment expansion.

Screening-length convention. The source defines

λ_{D e}^{exam} = 3 V_{e} / ω_{p e}

, so its long-wavelength condition is equivalent to

κ ≪ 1

. We retain that definition when expressing answers in its symbols. The actual electron Thomas-Fermi screening length obtained from the static response is instead

ϵ_{e} (0, k) = 1 + \frac{3 ω _{p e}^{2}}{k ^{2} V _{e}^{2}} = 1 + \frac{1}{k ^{2} λ _{TF}^{2}}, λ_{TF} = \frac{V _{e}}{3 ω _{p e}} = \frac{λ _{D e}^{exam}}{3} .

For a concrete comparison, at

k = ω_{p e} / V_{e}

, the static dielectric function is

4

; using the printed length as the physical Thomas-Fermi screening length would give

1 + 1/3 = 4/3

. Calling the printed length the Debye length thus differs from the physical degenerate-electron screening convention by a factor of three. It does not affect the boxed dimensional dispersion relation.

d

Solution

For the positive-frequency branch write

s = 1 + δ

δ > 0

, and

Q = k^{2} V_{e}^{2} / (3 ω_{p e}^{2}) = κ^{2} /3

. The degenerate-plasma dielectric response is equivalent to

1 + \frac{1 + δ}{2} lo g \frac{δ}{2 + δ} = - Q .

At large

Q

, a logarithmic divergence is needed to balance the right-hand side; hence

δ ≪ 1

. Keeping the constant as well as the large term gives

lo g \frac{δ}{2} = - 2 (1 + Q) + O (δ ∣ lo g δ ∣), δ \sim 2 exp [- 2 - \frac{2 k ^{2} V _{e}^{2}}{3 ω _{p e}^{2}}] .

More precisely the relative correction is

O ((1 + Q) e^{- 2 (1 + Q)})

. The additive constant

- 2

in the exponent fixes the leading prefactor

2 e^{- 2}

and should not be discarded.

The negative-frequency branch follows from the even real dielectric function outside the occupied interval. Thus

ω_{\pm} = \pm k V_{e} (1 + δ), δ \sim 2 exp [- 2 - \frac{2}{9} (k λ_{D e}^{exam})^{2}] .

Using the physical Thomas-Fermi screening length instead gives

δ \sim 2 e^{- 2 - 2 (k λ_{TF})^{2}}

. This is degenerate-plasma zero-point sound: the phase velocity lies exponentially close to, but strictly beyond, the edge of the Fermi sphere for every finite

k

in this approximation. The limit is a property of the supplied classical-trajectory kinetic theory; applying it at microscopic wavelengths would require checking quantum recoil, which that equation omits.

e

Solution

The two real branches start at

ω = \pm ω_{p e}

, have zero initial slope, and approach the lines

ω = \pm k V_{e}

from outside the populated parallel-velocity region. The original plot solves the degenerate-plasma dielectric response numerically; the shaded wedge shows where particles with resonant parallel velocities exist.

Figure 1.
Degenerate electron-plasma branches remain outside the resonant-velocity wedge
. The frequency is scaled by the electron plasma frequency and the wavenumber by $ω_{p e} / V_{e}$ . The dashed curves show the leading long-wavelength approximation. At large wavenumber the exact branches are exponentially close to $ω = \pm k V_{e}$ ; their visual overlap with those edges does not mean that a finite- $k$ root is inside the shaded region. Ion response, collisions and quantum recoil are omitted. **Details**.

The distinction from a nondegenerate gas with a Maxwellian Maxwell-Boltzmann velocity distribution is its tail. A Maxwellian has resonant particles at every finite phase velocity. Its long-wavelength Langmuir waves have exponentially weak Landau damping, but at shorter wavelengths that damping becomes strong. The ideal degenerate cutoff instead permits this undamped branch to persist continuously into degenerate-plasma zero-point sound. The coefficient

3 V_{e}^{2} /5

in c is also different from the classical thermal dispersive coefficient.

A natural additional branch is a degenerate ion-acoustic wave. For singly charged ions with equal electron and ion number densities, degenerate electron pressure supplies a restoring force while ion mass supplies inertia. The electron equation of state has

d P_{e} / d n_{e} = m_{e} V_{e}^{2} /3

, suggesting, at long wavelengths,

ω ≃ \pm k c_{s}, c_{s}^{2} ≃ \frac{m _{e}}{3 m _{i}} V_{e}^{2} .

Both species are degenerate here, so

V_{i} = (m_{e} / m_{i}) V_{e}

. The hierarchy

V_{i} ≪ c_{s} ≪ V_{e}

explains the damping: there are no resonant ions, but there are resonant electrons inside their parallel marginal distribution. That electron marginal distribution has a small nonzero slope at

u = c_{s}

, even though the full three-dimensional occupation is constant inside the Fermi sphere. One therefore expects weak electron Landau damping, of relative order

m_{e} / m_{i}

, rather than exact absence of damping. Ion degeneracy pressure adds a relatively small correction to

c_{s}^{2}

. This reasoning concerns an electrostatic ion-acoustic wave; transverse or magnetized branches require additional field dynamics or a background magnetic field.

(θ, J)

are canonical coordinates, with

[θ_{a}, J_{b}] = δ_{ab}

. In the convention

\overset{g}{˙} = \partial_{t} g + [g, H]

, their Poisson bracket is

[g, h] = a = 1 \sum 3 (\frac{\partial g}{\partial θ _{a}} \frac{\partial h}{\partial J _{a}} - \frac{\partial g}{\partial J _{a}} \frac{\partial h}{\partial θ _{a}}) = \partial_{θ} g \cdot \partial_{J} h - \partial_{J} g \cdot \partial_{θ} h .