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Swing Amplification James Binney University of Oxford Saas Fee, - - PowerPoint PPT Presentation
Swing Amplification James Binney University of Oxford Saas Fee, - - PowerPoint PPT Presentation
Swing Amplification James Binney University of Oxford Saas Fee, January 2019 Goldreich & Lynden Bell (1965) Julian & Toomre (1966) Shearing sheet Consider large m and study small near-Cartesian patch that rotates with
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Shearing sheet
◮ Consider large m and study small near-Cartesian patch
that rotates with particle on circular orbit at its centre
◮
L = 1
2[ ˙
x2 + (R + x)2( ˙ y/R + Ω)2] − Φ(R + x) px = ˙ x py = (R + x)2 ˙ y R + Ω 1 R ≃ RΩ + 2Ωx + ˙ y
SLIDE 4
Shearing sheet
◮
H = 1
2
- p2
x +
p2
y
(1 + x/R)2
- − ΩRpy + Φ
◮ Consts of motion: H and py or
∆y ≡ py − RΩ = 2Ωx + ˙ y
◮ Relations between frequencies
A = − 1
2
∂Ω ∂ ln R B = A − Ω κ2 = 4Ω(Ω − A) = −4ΩB
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Shearing sheet
∂Φ ∂x = RΩ2 ∂Φ ∂x
- R+x
= RΩ2 + x
- Ω2 + 2RΩ∂Ω
∂R
- + O(x2)
= RΩ2 + xΩ (Ω − 4A)
- A ≡ − 1
2R∂Ω
∂R
SLIDE 6
Shearing sheet
∂Φ ∂x = RΩ2 ∂Φ ∂x
- R+x
= RΩ2 + x
- Ω2 + 2RΩ∂Ω
∂R
- + O(x2)
= RΩ2 + xΩ (Ω − 4A)
- A ≡ − 1
2R∂Ω
∂R
- ◮ Hence Φ(R + x) ≃ Φ(R) + RΩ2x + 1
2Ω(Ω − 4A)x2
H = 1
2
- p2
x + p2 y
- 1 − 2 x
R + 3 x2 R2
- − RΩpy + Φ(R)
+ RΩ2x + 1
2Ω(Ω − A)x2
≃ 1
2
- p2
x + ∆2 y − R2Ω2
+ Φ(R) − xΩ∆y + 1
2κ2x2
where ∆y ≡ py − RΩ and κ2 ≡ 4Ω(Ω − A)
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Shearing sheet
∂Φ ∂x = RΩ2 ∂Φ ∂x
- R+x
= RΩ2 + x
- Ω2 + 2RΩ∂Ω
∂R
- + O(x2)
= RΩ2 + xΩ (Ω − 4A)
- A ≡ − 1
2R∂Ω
∂R
- ◮ Hence Φ(R + x) ≃ Φ(R) + RΩ2x + 1
2Ω(Ω − 4A)x2
H = 1
2
- p2
x + p2 y
- 1 − 2 x
R + 3 x2 R2
- − RΩpy + Φ(R)
+ RΩ2x + 1
2Ω(Ω − A)x2
≃ 1
2
- p2
x + ∆2 y − R2Ω2
+ Φ(R) − xΩ∆y + 1
2κ2x2
where ∆y ≡ py − RΩ and κ2 ≡ 4Ω(Ω − A)
◮ x oscillates harmonically about x ≡ 2Ω∆y/κ2
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circular orbits
◮ Circular orbits are ones on which x = x, so
˙ y = ∆y − 2Ωx = κ2 2Ω − 2Ω
- x = −2Ax
(circular orbit)
◮ Defines shear in the sheet
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Shearing sheet
◮ Let vφ ≡ 2Ax + ˙
y be the azimuthal speed relative to the local circular orbit, then since ∆y = 2Ωx + ˙ y and x = 2Ω∆y/κ2 vφ = 2(A − Ω)x + κ2x 2Ω = 2(A − Ω)(x − x) tells us how far a star is from its guiding centre
◮ When we eliminate x from H in favour of vφ we get
H ≃ 1
2
- p2
x + ∆2 y − R2Ω2
+ Φ(R) + 1
2κ2
(x − x)2 − x2 = 1
2
- p2
x + ∆2 y
- 1 − 4Ω2
κ2
- − R2Ω2
- + Φ(R) + 1
8
κ2 (A − Ω)2 v2
φ
= 1
2
- p2
x + ∆2 y
A A − Ω − R2Ω2
- + Φ(R) + 1
2
Ω Ω − Av2
φ
= Hx(px, vφ) + Hy(∆y) (B ≡ A − Ω)
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Swing amplification
◮ We posit P = 2πGΣ/|k| is generated by a density pattern
that shears along circular orbits
◮ On account of shear, kx is a function of time:
kx(0)x + kyy(0) = kx(t)x + kyy(t) ⇒ kx(t) = kx(0) + 2kyAt
◮ Let t′ ≡ t − tc be time since tc ≡ − kx 2kyA when kx = 0
◮ kx(t′) = 2Akyt′
◮ So |k| = ky
√ 1 + 4A2t′2 goes through a minimum as kx passes through 0
◮ P = 2πGΣ1/|k| has a corresponding maximum
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Unperturbed (non-circular) orbits
x = x + X cos θr ⇒ pr = −κX sin θr vφ = 2B(x − x) = 2BX cos θr ˙ y = ∆y − 2Ωx = ∆y − 2Ω(x + X cos θr) x = 2Ω κ2 ∆y ⇒ y(t′) = y(0) + ∆y
- 1 − 4Ω2
κ2
- t′ − 2Ω
κ X sin θr = y(0) + ∆y A B t′ − 2Ω κ X sin θr Hence kxx + kyy = 2kyAt′(x + X cos θr) + ky
- y(0) + ∆y
A B t′ − 2Ω κ X sin θr
- = ky
- y(0) + 2X
- At′ cos θr − Ω
κ sin θr
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Linearizing the CBE
◮ With f = f0(H0) + f1 and H = H0 + Φ1
∂f ∂t + [f, H] = 0 ⇒ df1 dt = [Φ1, f0]
◮ On integration we have
f1 = t
t0
dt ∂Φ1 ∂x · ∂f0 ∂p where the integral is along unperturbed orbits.
◮ We take f0 = Fe−Hx/σ2 ⇒ Σ0 = 4πFσ2B/κ ◮ This DF
◮ generates biaxial Maxwellian v distribution ◮ is a function of px, and, through vφ, of both its conjugate
variable x, and the momentum ∆y
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Linearised CBE
◮ Using Hx = 1 2[p2 x + κ2(x − x)2] with Φ1(x, y) = Peik·x, we
have ∂Φ1 ∂x · ∂f0 ∂p = −iFPeik·x e−Hx/σ2 σ2 [kxpx − ky2Ω(x − x)]
◮
f1(t) = 2πGFie−Hx/σ2 σ2 t
t0
dt′ Σ1 ei[ψ(t′)+kyy(0)] × −2At′κX sin θr − 2ΩX cos θr √ 1 + 4A2t′2 , where we have introduced a phase ψ(t′) ≡ 2kyX(At′ cos θr − (Ω/κ) sin θr).
◮ As we integrate over v to get Σ1, y(0) will vary.
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Dealing with y(0)
y(0) = y(t) − ∆y A B t + 2Ω κ X sin θr = y(t) − κ2x 2ΩB A B t + 2Ω κ X sin θr = y(t) + 2At(x − X cos θr) + 2Ω κ X sin θr Note ky[y(t) + 2Atx(t)] = k · x
- t.
so kyy(0) = k · x − ψ(t)
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Introducing cpts of v
Using θr = θ0 + κt′ we write κX sin θr = κX
- sin θ0 cos κt′ + cos θ0 sin κt′
= −Ux cos κt′ + Uy sin κt′ κX cos θr = Uy cos κt′ + Ux sin κt′ Ux ≡ −κX sin θ0 = px Uy ≡ κX cos θ0 = (κ/2B)vφ. Now ψ(t′) − ψ(t) = 2(ky/κ)[Ux{A(t′S − tS) + Ω κ (C′ − C)} + Uy{A(t′C′ − tC) − Ω κ (S′ − S)}] where C(t) ≡ cos κt, S(t) ≡ sin κt
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Integrating over V
◮
Σ1(t) =
- dpxdvφ f1 = −GΣ0
σ eik·x t
t0
dt′ KΣ1(t′) √ 1 + 4A2t′2 where K(t, t′) ≡ i σ3
- d2U e−U2/2σ2ei[ψ(t′)−ψ(t)]
×
- 2At′(−UxC + UyS) + 2(Ω/κ)(UyC + UxS)
- ◮ Tickle the disc:
Σ1(t) = − κ 3.36Qeik·x t
t0
dt′ K(t, t′)[δ(t′ − t0) + Σ1(t′)] √ 1 + 4A2t′2 Q = κσ 3.36GΣ
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Evaluating K
- duy e−a2u2
y+2ibyuy
- dux e−a2u2
x+2ibxux(cxux + cyuy)
= √πe−b2
x/a2
a2
- duy e−a2u2
y+2ibyuy(cyuya + icxbx/a)
= iπ a4 e−(b2
x+b2 y)/a2(cxbx + cyby).
a2 = 1 2σ2 bx = (ky/κ)[A(t′S′ − tS) + (Ω/κ)(C′ − C)] by = (ky/κ)[A(t′C′ − tC) − (Ω/κ)(S′ − S)) cx = −At′C′ + Ω κ S′ cy = At′S′ + Ω κ C′.
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solutions
◮ Vc = const, Q = 1 ◮ Vc = const, Q = 1.1
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Swing amplification
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