Prof. dr. A. Achterberg, Astronomical Dept., IMAPP, Radboud Universiteit

Transcription

1 Prof. dr. A. Achterberg, Astronomical Dept., IMAPP, Radboud Universiteit

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4 Central concepts: Phase velocity: velocity with which surfaces of constant phase move Group velocity: velocity with which slow modulations of the wave amplitude move

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6 Definition phase S

7 Definition phase S Definition phase-velocity

8 Definition phase S Definition phase-velocity

9 + dk ξ( x, t) = ( k) exp( ikx - iωt) 2π Fourier phase factor amplitude Fourier integral

10 Narrow packet with k k : 0 ω ω( k) ω k k k ( ) + ( ) 0 0 k k 0

11 Same for wave phase S = kx ωt: S Sk ( ) S k k k ( ) + ( ) 0 0 k k 0 ω = kx 0 ω( k0) t+ ( k k0) x t k k 0 This should vanish for constructive interference!

12 Wave-packet, Fourier Integral

13 Wave-packet, Fourier Integral Phase factor x effective amplitude

14 Wave-packet, Fourier Integral Phase factor x effective amplitude Constructive interference in integral when

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22 1. Incompressible, constant density fluid (like water!) ρ = constant, V =0 2. Constant gravitational acceleration in z-direction; dv dt P = ρ g zˆ 3. Fluid at rest without waves

23 Pz ( ) = P + ρg H z atm ( ) V = 0, P= ρgz= ˆ constant (!)

24 atm ( ) Pz ( ) = P + ρg H z V = 0, P= ρgz= ˆ constant (!) δv = V =, δv = = 0 = 0 t t

25 δv = V = t =0 dv = gzˆ = 2 dt ρ t 2 P δ P ρ SAME as for SOUND WAVES!

26 δv = V = t =0 δp δp δp t ρ t ρ ρ = = = = 0 2 2

27 δv = V = t =0 P 0, try: P( x, z, t) P ( z)exp( ikx i t) cc. 2 δ = δ = ω + d dz 2 P 2 kz 2 kp = Pz = P + P plane wave part 0 ( ) e + e kz

28 δp( x, z, t) = P ( z)exp( ikx iωt) + cc. kz Pz ( ) = P e + P e + plane wave part kz 2 t δ P = x z t = a z ikx iωt + cc ρ & (,, ) ( )exp( ) 2

29 δp( x, z, t) = P ( z)exp( ikx iωt) + cc. plane wave part kz Pz ( ) = P e + P e + kz 2 ρω a = ik P P x ( kz kz e + e ) + 2 dp ρω az = = k P P dz ( kz kz e e ) +

30 δp( x, z, t) = P ( z)exp( ikx iωt) + cc. plane wave part kz Pz ( ) = P e + P e + kz 2 ρω a = ik P P x z ( kz kz e + e ) + 2 ρω a = k P P ( kz kz e e ) + 1. At bottom (z=0) we must have a z = 0: P = P +

31 δp( x, z, t) = P ( z)exp( ikx iωt) + cc. plane wave part Pz ( ) = P e + e + ( kz kz ) 2 ρω a x = ikp + ( kz kz e + e ) 2. At water s surface we must have P = P atm : 2 ρω a z = kp + ( kz kz e e ) P P= δp+ ξz = dz ( ) d 0

32 δp( x, z, t) = P ( z)exp( ikx iωt) + cc. plane wave part Pz ( ) = P e + e + ( kz kz ) 2 ρω a x = ikp + ( kz kz e + e ) 2. At water s surface we must have P = P atm : 2 ρω a z = kp + ( kz kz e e ) δpz ( = ) = ρg

33 δp( x, z, t) = P ( z)exp( ikx iωt) + cc. plane wave part Pz ( ) = P e + e + ( kz kz ) kgp+ ρgaz ( ) = e e = P e + e 2 ω ( k k ) ( k k ) +

34 kgp+ ρgaz ( ) = e e = P e + e 2 ω δp( x, z, t) = P ( z)exp( ikx iωt) + cc. ( k k ) ( k k ) + k k 2 e e ω = kg = kg tanh( k ) k k e + e plane wave part Pz ( ) = P e + e + =ξ z ( kz kz ) ( )

35 2 ω = kg tanh( k ) Shallow lake: k << 1: tanh k k, ω k g ( ) Deep lake: k >> 1: tanh k 1, ω kg ( )

36 ω = = g / 1/2 ν κ tanh κ κ = k ν κ deep lake shallow lake ν κ

37 Situation in rest frame ship: quasi-stationary

38 wave frequency: ω = kg wave vector: k = kcosϑ xˆ + ksinϑ yˆ w w Ship moves in x-direction with velocity U 1: Wave frequency should vanish in ship s rest frame: Doppler: ω = ω k U= kg ku cosϑ = 0 w

39 wave frequency: ω = kg wave vector: k = kcosϑ xˆ + ksinϑ yˆ w w Ship moves in x-direction with velocity U 2: Wave phase should be stationary for different wavelengths in ship s rest frame: Sx ( ', yk, ) k ships rest frame = 0

40 ω = ω k U = ω ku cosϑ = 0 w Ship moves in x-direction with velocity U ( ) Sxyt (,,) = kx+ ky ωt= k x' + Ut+ ky ωt x y x y ( ) = kcos ϑwx' + ksinϑwy ω kcosϑwu t vanishes due to "Doppler condition" ω= 0

41 ω = ω k U = ω ku cosϑ = 0 w Ship moves in x-direction with velocity U Wave phase in ship s frame: Wavenumber: Sx ( ', yk, ) = kcos ϑ x' + ksinϑ y w g ω = kg = ku cos ϑw k = K ϑ 2 2 w U cos ϑ w w ( )

42 Ship moves in x-direction with velocity U ω = ω k U k = kg ku cosϑ = 0 = g K 2 2 U cos ϑ w w ( ϑ ) w Stationary phase condition for Sx ( ', yk, ) = kcos ϑ x' + ksin ϑ y: w w S d g = 0 ( x'cosϑ + ysinϑ ) = w w k dϑw U cos ϑw

43 Situation in rest frame ship: quasi-stationary

44 Shocks occur whenever a flow hits an obstacle at a speed larger than the sound speed

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46 1. Shocks are sudden transitions in flow properties such as density, velocity and pressure; 2. In shocks the kinetic energy of the flow is converted into heat, (pressure); 3. Shocks are inevitable if sound waves propagate over long distances; 4. Shocks always occur when a flow hits an obstacle supersonically 5. In shocks, the flow speed along the shock normal changes from supersonic to subsonic

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48 t = coll L V D Time between two `collisions V sh D D = = V t L D coll `Shock speed = growth velocity of the stack.

49 1 2 Go to frame where the `shock is stationary: Incoming marbles: L = = V1 V Vsh V L D Marbles in stack: D = = V2 Vsh V L D

50 1 2 Flux = density x velocity Incoming flux: 1 L V = n V = V = L L D L D Outgoing flux: 1 D V = n V = V = D L D L D 2 2 2

51 Conclusions: 1. The density increases across the shock 2. The flux of incoming marbles equals the flux of outgoing marbles in the shock rest frame: = 1 2

52 Cs T ρ ( γ 1)/2

53 Generic conservation law: + = 0 t x

54 t = x Change of the amount of in layer of width 2e: t + ε + ε ε d x = d x = ( ε) ( + ε) x ε = in out flux in - flux out

55 What goes in must come out : in = out

56 What goes in must come out : in = out Formal proof: use a limiting process for e 0 + ε d x = in out t ε + ε lim d x ( x) 0 ε 0 = ε in = out

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58 Starting point: 1D ideal fluid equations in conservative form; x is the coordinate along shock normal, velocity V along x-axis! ρ + ( ρv ) = t x 0 Mass conservation ( ρv ) t ( 2 + ρv + P) = x 0 Momentum conservation 2 2 V P V γ P ρ + + ρv + = t 2 ( γ 1) ρ x 2 ( γ 1) ρ 0 Energy conservation

59 Three conservation laws means three fluxes for flux in = flux out! ( ρv) ( ρv) = 1 2 Mass flux ( 2 ) ( 2 ρv P ρv P) + = Momentum flux 2 2 V γp V γp ρv + = ρv + 2 ( γ 1) ρ 2 ( γ 1) ρ 1 2 Energy flux Three equations for three unknowns: post-shock state (2) is uniquely determined by pre-shock state (1)!

60 1D case: Mach Number s pre-shock flow speed V = = pre-shock sound speed C s 1 Shocks can only exist if s >1! Weak shocks: s =1+e with e<< 1; Strong shocks: s >> 1.

61 V, ρ and P are all small!

62 ρ mass conservation: ( ρ) V1+ ρ1 V = 0 = ρ ( ρ) momentum conservation: V + 2ρV V + P= V V 1 1 P= V ρ 2 1 Energy conservation: V V + 1 γ P P ρ = ( γ 1) ρ P1 ρ1 0 γp ρ γp V = 0 V = = C s1 ρ1 ( γ 1) ρ ρ1

63 P= V ρ = C ρ s1 Sound waves: ( ) ρ = ρ γ P 2 P= ρ = Cs ρ ρ P= γ P ( )

64 Approximate jump conditions: put P 1 = 0! ρv = ρ V (1) ρv = ρ V + P (2) 2 2 V1 V2 γ P2 = ( γ 1) ρ 2 (3)

65 ρv = ρ V (1) ρv = ρ V + P (2) 2 2 V1 V2 γ P2 = ( γ 1) ρ 2 (3) (1) + (2) V V = 1 2 P 2 2γ PV (3) V V = ( V V )( V + V ) = γ V + V = 1 2 2γV2 γ 1

66 ρv = ρ V (1) ρv = ρ V + P V1 V2 γ P2 = ( γ 1) ρ 2 (2) (3) V + V = 1 2 2γV2 γ 1 γ 1 ρ 1 2 V V, V γ + = ρ = = ρ, P = ρv V V = ρv ( ) γ + 1 V2 γ 1 γ + 1