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 particle on circular orbit at its centre ◮ x 2 + ( R + x ) 2 ( ˙ L = 1 y/R + Ω) 2 ] − Φ( R + x ) 2 [ ˙ p x = ˙ x � ˙ � 1 y p y = ( R + x ) 2 R + Ω R ≃ R Ω + 2Ω x + ˙ y

Shearing sheet ◮ � � p 2 y H = 1 p 2 x + − Ω Rp y + Φ 2 (1 + x/R ) 2 ◮ Consts of motion: H and p y or ∆ y ≡ p y − R Ω = 2Ω x + ˙ y ◮ Relations between frequencies ∂ Ω A = − 1 B = A − Ω 2 ∂ ln R κ 2 = 4Ω(Ω − A ) = − 4Ω B

Shearing sheet ∂ Φ ∂x = R Ω 2 � ∂ Φ � Ω 2 + 2 R Ω ∂ Ω � = R Ω 2 + x + O( x 2 ) � � ∂x ∂R � R + x � 2 R∂ Ω � = R Ω 2 + x Ω (Ω − 4 A ) A ≡ − 1 ∂R

Shearing sheet ∂ Φ ∂x = R Ω 2 � ∂ Φ � Ω 2 + 2 R Ω ∂ Ω � = R Ω 2 + x + O( x 2 ) � � ∂x ∂R � R + x � 2 R∂ Ω � = R Ω 2 + x Ω (Ω − 4 A ) A ≡ − 1 ∂R ◮ Hence Φ( R + x ) ≃ Φ( R ) + R Ω 2 x + 1 2 Ω(Ω − 4 A ) x 2 R + 3 x 2 � � 1 − 2 x �� H = 1 p 2 x + p 2 − R Ω p y + Φ( R ) y 2 R 2 + R Ω 2 x + 1 2 Ω(Ω − A ) x 2 ≃ 1 p 2 x + ∆ 2 y − R 2 Ω 2 � + Φ( R ) − x Ω∆ y + 1 2 κ 2 x 2 � 2 where ∆ y ≡ p y − R Ω and κ 2 ≡ 4Ω(Ω − A )

Shearing sheet ∂ Φ ∂x = R Ω 2 � ∂ Φ � Ω 2 + 2 R Ω ∂ Ω � = R Ω 2 + x + O( x 2 ) � � ∂x ∂R � R + x � 2 R∂ Ω � = R Ω 2 + x Ω (Ω − 4 A ) A ≡ − 1 ∂R ◮ Hence Φ( R + x ) ≃ Φ( R ) + R Ω 2 x + 1 2 Ω(Ω − 4 A ) x 2 R + 3 x 2 � � 1 − 2 x �� H = 1 p 2 x + p 2 − R Ω p y + Φ( R ) y 2 R 2 + R Ω 2 x + 1 2 Ω(Ω − A ) x 2 ≃ 1 p 2 x + ∆ 2 y − R 2 Ω 2 � + Φ( R ) − x Ω∆ y + 1 2 κ 2 x 2 � 2 where ∆ y ≡ p y − R Ω and κ 2 ≡ 4Ω(Ω − A ) ◮ x oscillates harmonically about x ≡ 2Ω∆ y /κ 2

circular orbits ◮ Circular orbits are ones on which x = x , so � κ 2 � y = ∆ y − 2Ω x = ˙ 2Ω − 2Ω x = − 2 Ax (circular orbit) ◮ Defines shear in the sheet

Shearing sheet ◮ Let v φ ≡ 2 Ax + ˙ y be the azimuthal speed relative to the local circular orbit, then since ∆ y = 2Ω x + ˙ y and x = 2Ω∆ y /κ 2 v φ = 2( A − Ω) x + κ 2 x 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 v φ we get ( x − x ) 2 − x 2 � p 2 x + ∆ 2 y − R 2 Ω 2 � 2 κ 2 � H ≃ 1 + Φ( R ) + 1 � 2 1 − 4Ω 2 κ 2 � � � � = 1 p 2 x + ∆ 2 − R 2 Ω 2 + Φ( R ) + 1 ( A − Ω) 2 v 2 y φ 2 8 κ 2 � A � Ω p 2 x + ∆ 2 A − Ω − R 2 Ω 2 Ω − Av 2 = 1 + Φ( R ) + 1 y φ 2 2 = H x ( p x , v φ ) + H y (∆ y ) ( B ≡ A − Ω)

Swing amplification ◮ We posit P = 2 πG Σ / | k | is generated by a density pattern that shears along circular orbits ◮ On account of shear, k x is a function of time: k x (0) x + k y y (0) = k x ( t ) x + k y y ( t ) ⇒ k x ( t ) = k x (0) + 2 k y At ◮ Let t ′ ≡ t − t c be time since t c ≡ − k x 2 k y A when k x = 0 ◮ k x ( t ′ ) = 2 Ak y t ′ √ 1 + 4 A 2 t ′ 2 goes through a minimum as k x ◮ So | k | = k y passes through 0 ◮ P = 2 πG Σ 1 / | k | has a corresponding maximum

Unperturbed (non-circular) orbits x = x + X cos θ r ⇒ p r = − κX sin θ r v φ = 2 B ( x − x ) = 2 BX cos θ r x = 2Ω y = ∆ y − 2Ω x = ∆ y − 2Ω( x + X cos θ r ) ˙ κ 2 ∆ y 1 − 4Ω 2 � � t ′ − 2Ω y ( t ′ ) = y (0) + ∆ y ⇒ κ X sin θ r κ 2 B t ′ − 2Ω A = y (0) + ∆ y κ X sin θ r Hence k x x + k y y = 2 k y At ′ ( x + X cos θ r ) � B t ′ − 2Ω A � + k y y (0) + ∆ y κ X sin θ r � � �� At ′ cos θ r − Ω = k y y (0) + 2 X κ sin θ r

Linearizing the CBE ◮ With f = f 0 ( H 0 ) + f 1 and H = H 0 + Φ 1 ∂f d f 1 ∂t + [ f, H ] = 0 ⇒ d t = [Φ 1 , f 0 ] ◮ On integration we have � t d t ∂ Φ 1 ∂ x · ∂f 0 f 1 = ∂ p t 0 where the integral is along unperturbed orbits. ◮ We take f 0 = F e − H x /σ 2 ⇒ Σ 0 = 4 πFσ 2 B/κ ◮ This DF ◮ generates biaxial Maxwellian v distribution ◮ is a function of p x , and, through v φ , of both its conjugate variable x , and the momentum ∆ y

Linearised CBE ◮ Using H x = 1 2 [ p 2 x + κ 2 ( x − x ) 2 ] with Φ 1 ( x, y ) = P e i k · x , we have ∂ p = − i FP e i k · x e − H x /σ 2 ∂ Φ 1 ∂ x · ∂f 0 [ k x p x − k y 2Ω( x − x )] σ 2 ◮ � t f 1 ( t ) = 2 πGF ie − H x /σ 2 d t ′ Σ 1 e i[ ψ ( t ′ )+ k y y (0)] σ 2 t 0 × − 2 At ′ κX sin θ r − 2Ω X cos θ r √ , 1 + 4 A 2 t ′ 2 where we have introduced a phase ψ ( t ′ ) ≡ 2 k y X ( At ′ cos θ r − (Ω /κ ) sin θ r ) . ◮ As we integrate over v to get Σ 1 , y (0) will vary.

Dealing with y (0) B t + 2Ω A y (0) = y ( t ) − ∆ y κ X sin θ r = y ( t ) − κ 2 x B t + 2Ω A κ X sin θ r 2Ω B = y ( t ) + 2 At ( x − X cos θ r ) + 2Ω κ X sin θ r Note � k y [ y ( t ) + 2 Atx ( t )] = k · x t . � so k y y (0) = k · x − ψ ( t )

Introducing cpts of v Using θ r = θ 0 + κt ′ we write sin θ 0 cos κt ′ + cos θ 0 sin κt ′ � � κX sin θ r = κX = − U x cos κt ′ + U y sin κt ′ κX cos θ r = U y cos κt ′ + U x sin κt ′ U x ≡ − κX sin θ 0 = p x U y ≡ κX cos θ 0 = ( κ/ 2 B ) v φ . Now ψ ( t ′ ) − ψ ( t ) = 2( k y /κ )[ U x { A ( t ′ S − tS ) + Ω κ ( C ′ − C ) } + U y { A ( t ′ C ′ − tC ) − Ω κ ( S ′ − S ) } ] where C ( t ) ≡ cos κt , S ( t ) ≡ sin κt

Integrating over V ◮ � t � d p x d v φ f 1 = − G Σ 0 K Σ 1 ( t ′ ) e i k · x Σ 1 ( t ) = d t ′ √ σ 1 + 4 A 2 t ′ 2 t 0 where K ( t, t ′ ) ≡ i � d 2 U e − U 2 / 2 σ 2 e i[ ψ ( t ′ ) − ψ ( t )] σ 3 � 2 At ′ ( − U x C + U y S ) + 2(Ω /κ )( U y C + U x S ) � × ◮ Tickle the disc: � t d t ′ K ( t, t ′ )[ δ ( t ′ − t 0 ) + Σ 1 ( t ′ )] κ 3 . 36 Q e i k · x √ Σ 1 ( t ) = − 1 + 4 A 2 t ′ 2 t 0 κσ Q = 3 . 36 G Σ

Evaluating K � � d u y e − a 2 u 2 y +2i b y u y d u x e − a 2 u 2 x +2i b x u x ( c x u x + c y u y ) √ π e − b 2 x /a 2 � d u y e − a 2 u 2 y +2i b y u y ( c y u y a + i c x b x /a ) = a 2 = i π a 4 e − ( b 2 x + b 2 y ) /a 2 ( c x b x + c y b y ) . 1 a 2 = 2 σ 2 b x = ( k y /κ )[ A ( t ′ S ′ − tS ) + (Ω /κ )( C ′ − C )] b y = ( k y /κ )[ A ( t ′ C ′ − tC ) − (Ω /κ )( S ′ − S )) c x = − At ′ C ′ + Ω κ S ′ c y = At ′ S ′ + Ω κ C ′ .

solutions ◮ V c = const , Q = 1 ◮ V c = const , Q = 1 . 1

Swing amplification

Conclusion ◮ WKB theory is misleading in putting an exclusion zone around CR ◮ CR is where the key action takes place ◮ illustrated by modes in N-body models ◮ Key process: amplification of waves as leading → trailing