Solutions Mock Examination

Transcription

1 Solutions Mock Examination Elena Rossi December 18, The peak frequency will be given by 4 3 γ2 ν The Einstein coeffients present the rates for the different processes populating and depopulating energy levels. The spontaneous emission coefficient gives the rate of an spontaneous downward transition and has units of inverse time. The absorption coefficient is the upward transition rate given the energy density of the radiation field at the frequency of interest. It has units of volume, inverse time and inverse energy. The stimulated emission coefficient is the rate of a downward transition induced by the radiation field. It has the same units as the absorption coefficient. The differential equation for the upper level population is given by, dn u dt = n l B lu u ν n u (A ul + B ul u ν ). (1) The differential equation for the lower level is the same but with the signs in the right-hand side inverted. 3. First we need to calculate the flux from the star that reaches 51 Pegasi b. This will be, L 4πa 2 51p, (2) where we consider the luminosity of the Sun for our purpose since 51 Pegasi is a Sun-like star. The amount of energy per unit time that effectively reaches 51 Pegasi b will be this, times the area of 51 Pegasi b facing 51 Pegasi, L R51p 2, (3) 4a 2 51p 1

2 since 10% of this luminosity is reflected we have that the reflected luminosity of 51 Pegasi will be, L refl = 0.1 L 4 ( R51p a 51p ) 2. (4) Assuming that the light is perfectly reflected, its spectra will have the same shape as that of the Sun. It will be characterized by the effective temperature of the Sun, T eff = 5770 K. Using Wien s displacement law we have, which corresponds to visible light. λ max = 0.29 cm K, T eff (5) 507 nm, (6) If we consider that 51 Pegasi b behaves as a blackbody, we have that the energy it absorbs must be the same that it emits. We have then, 4πσT 4 effr 2 51p = 0.9 L 4 ( R51p a 51p ) 2, (7) Teff 4 = 0.9 L. (8) 16πσa 2 51p Inputting the values for the constants we have T eff 1200 K. Using again Wien s displacement law we have, which falls in the near-infrared. λ max 2.5 µm, (9) 4. For the pinhole camera we use the relation between flux and intensity, F ν = I ν cos θdω. (10) Ω Considering that d L, we can assume that I ν will be fairly constant through dω and we can take it out of the integral. We can consider the same for the angle θ, which will not vary much along the solid angle subtended by the circular aperture. From that we get, F ν = I ν cos θ dω. (11) Ω

3 For this we just need to integrate the solid angle. Coming from the definition, ˆr ˆndA dω =, (12) r 2 we have that r = L/ cos θ and that ˆr ˆn = cos θ. Since for our approximation all variations of θ are negligible all those values will get also out of the integral and we will have simply the integral over the area, namely πd 2 /4. Finally we have, F ν = π cos4 θ 4f 2 I ν (θ, φ). (13) For the case of the disk galaxy we consider that we are observing directly at it, F ν = I ν Ω. (14) For the solid angle we can calculate it simply from eq. 12, knowing that ˆr ˆn = cos θ. Finally we have, Ω = πr2 cos θ r 2, (15) and including this in the previous result we get the solution, ( ) 2 R F ν = πi ν cos θ. (16) r The only real difference between the two expressions is a factor cos 3 θ. This comes from the fact that in the pinhole camera there is the additional factor cos θ in the definition of the flux since the hole is not being observed directly. In the integration we will have this factor and the effect of the solid angle, while for the disk galaxy we observe it directly and the tilt is only in the disk itself. 5. The value of q will be 5/2. This can be derived from the spectral index between ν m and ν c which corresponds to (p 1)/2. We have that ν a is the absorption frequency, i.e. below this value the source is optically thick and defined by a single temperature so it behaves as a blackbody. In the case of ν m, it corresponds to the peak of the spectrum and it is given by 4γ 2 ν L /3, where ν L is the Larmor frequency. For ν c, we have the cooling frequency, corresponding to electrons that have already cooled down by time t. This can be calculated with the

4 cooling time, E e /P S, and deriving the γ corresponding to electrons already cooled. this gives us, γ c = 9m3 c 5 4e 4 B 2 t, and from this we have ν c = 4γ 2 c ν L /3. For frequencies larger than ν c the power law distribution index changes from q to q + 1 and thus we can see the spectral index will be 5/4. The electron distribution will have the power law with index q for electrons with γ < γ c and index q 1 for γ > γ c. 6. To calculate the luminosity in the optically thin regime we can use the emission per unit volume of bremsstrahlung, given by, ε ff = T 1/2 n e n i Z 2 ḡ B erg s 1 cm 3. (17) By multiplying this by the volume we can have the luminosity. Since we have a hydrogen plasma, we have Z = 1 and as usual we can use ḡ B = 1.2. We also know that n e = n i = M 0 /m p V. Getting now everything in terms of T 0, M 0 and R(t) we have, L thin = M 2 0 T 1/2 0 R 3. (18) In the optically thick case we know that the sphere will radiate as a black-body, then L = 4πR 2 σt 4 or, L thick = T 4 0 R 2. (19) We know that for large values of R we will have the optically thin case, while for small R we are in the optically thick regime. Considering this, at values of R for which both luminosities are approximately the same, will correspond roughly with the radius at which the transition occurs. Then, See Fig M 2 0 T 1/2 0 R T 4 0 R 2, (20) R M0 2 T 7/2 0, (21) R(t 1 ) M 2/5 0 T 7/10 0. (22) At t t 1 we will have just the free-free spectra. As the time approaches t 1 the absorption frequency becomes closer to the cutoff frequency, until at t = t 1 it starts having a blackbody spectrum. From then on the spectrum is the same in shape, only the luminosity will diminish.

5 Figure 1: Luminosity as a function of time for a collapsing sphere.