Solutions to Exercises in Astrophysical Gas Dynamics

Transcription

1 1 Solutons to Exercses n Astrophyscal Gas Dynamcs 1. (a). Snce u 1, v are vectors then, under an orthogonal transformaton, u = a j u j v = a k u k Therefore, u v = a j a k u j v k = δ jk u j v k = u j v j. Hence. u v s nvarant under transformaton and s a scalar. u v j = a k u k a jl v l = a k a jl u k v l and ths s the transformaton law for a second rank tensor.. u = (a ku k ) usng the chan rule for partal dfferentaton. Now, snce x j = a jk x k x l = a kl x k = a kl x k = a kl δ kj = a jl (b). then u and s therefore a second rank tensor. u = = a k a jl u k [ ] a j u j = a j u j For the terms on the rght hand sde to equal 3 j=1 u j we requre 3 a j = 1 for each j,.e. each row of the transformaton matrx should sum to unty. Ths s an over-restrcton on the propertes of the orthogonal matrx.. The proof s smlar. Snce these quanttes are not nvarant, they are unsutable choces for the magntude of a vector. The quantty u u s nvarant and s therefore a sutable choce for the magntude of u.. Wrte The tensor s symmetrc and the tensor T j = 1 (T j + T j ) + 1 (T j T j ) S j = 1 (T j + T j ) A j = 1 (T j T j )

2 s antsymmetrc. We can further wrte S j = 1 3 S kkδ j + (S j 13 ) S kkδ j = 1 3 T kkδ j + 1 (T j + T j 3 ) T kkδ j The tensor 1 ( Tj + T j 3 T ) kkδ j s traceless. Hence, T j = 1 3 T kkδ j + 1 (T j + T j 3 ) T kkδ j + 1 (T j T j ) 3. Energy densty and energy flux n a sound wave. Mass momentum and energy flux Use the perturbaton equatons derved n lectures for a sound wave, vz, Thus Now, Thus and ρ + ρ v = v ρ + ρ c = = 1 ρ (v v ) + c ρ ρ ρ ρ ρ = ρ ρ v ρ v v = c v ρ = c v ρ c x ρ v = c (ρ v ) Snce F SW = c ρ v, then + F SW = 4. Velocty of sound n a movng medum. Consder a plane wave n a reference frame (denoted by a prme) n whch the gas s at rest. The varaton of densty n the wave s gven by: ρ = A exp [ k x ωt ] The transformaton to the movng medum s gven by: x = x + u t The equaton for the densty then transforms to: mplyng that ρ = A exp [ k x (ω + k u )t ] k = k snce ω = c s k n the statonary medum. ω = ω + k u = c s k + k u There are ways to work out the velocty of ths wave:

3 3 (a) A surface of constant phase s gven by k x (c s k + k u )t = φ Ths can be put n the form k k x = (c s + k k u )t + φ k Let n = k /k be the normal to the wave, then ths equaton mples that n x = v w t + constant where v w = c s + u n = Sound speed + Component of u n drecton of wave (b) The group velocty of waves gven by ths dsperson relaton s: c = ω k = c s k k + u so that the wave speed n the drecton of the wave s c k = c s + u k 5. Doppler Effect. Consder the frequency of a wave n a reference frame n whch the source s at rest. The medum s movng wth velocty u n ths frame. Thus, from the prevous queston, the relatonshp between the rest frequency (ω ) and the frequency n the medum n whch the source s movng (ω) s gven by: ω = ω k u = ω (1 u ) c cos θ snce n the statonary medum k = ω/c. Ths equaton then mples that ω = ω 1 u c cos θ 6. Pressure fluctuatons n a sound wave. The mean energy flux of a sound wave s: For a plane wave: F E, = p v p = c A cos(k j x j ωt) v = c ρ n A cos(k j x j ωt) where c s the sound speed n the undsturbed medum, A s the ampltude of the densty wave and n s the unt vector n the drecton of propagaton. Hence, the rms energy flux s gven by < p v > = c3 ρ A n < cos (k j x j ωt) > = c3 ρ A n F E = c3 A ρ

4 4 We can relate ths to the mean square pressure fluctuaton by: Hence p = c 4 A cos (k j x j ωt) = c4 A p F E = ρ c p = ρ c F E The background pressure can be expressed as p = ρ c /γ. Hence p 1/ p ( ) 1/ FE = γ ρ c 3 Parameters for ths problem are densty of ar, ρ 1.5 Kg m 3, c 33m s 1, γ = 1.4 and < F >= 1W/(4π 1 m ). Ths gves, < p > 1/ p Jeans mass at recombnaton. At recombnaton, the Unverse conssts manly of H and He, wth the abundance by mass of He, Y.534. The atomc masses of H and He are 1.79 and 4.6 respectvely. Hence the rato of the denstes s gven by Hence ρ He = n He 4.6 ρ H n H 1.79 = 3.97 n He =.534 n H n He n H =.638 The densty of the Unverse n terms of the number densty of atoms (n a ), can be found from Therefore, for the gven parameters. ρ n a = n Hm H (1 + Y ) n H (1 + n He /n H ) = 1.19 m H ρ m n a gm cm 3 Let us defne the Jeans mass as the mass wthn a sphere of dameter the Jeans length λ J where c λ J = π s γn a kt γkt = π 4πGρ 4πGρ = π =.1 1 cm 4πGρ 1.19m H Therefore the Jeans mass s M J = π 6 λ3 J gm = gm = M

5 8. Tmescale for gravtatonal collapse. 5 (a) When k < k J, the growth rate accordng to the Jeans theory s gven by: ω g = c s(k J k ) and the maxmum growth rate ω g = c s k J wth assocated growth tmescale τ g = 1 c s k J = 1 4πGρ (b) () For typcal ISM denstes, n 1 4 cm 3 and µ 1, the collapse tmescale s of order yrs. () Wth ρ = gm cm 3 from the prevous queston, ths gves τ g = s = yr