ASTR240: Radio Astronomy
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1 ASTR240: Radio Astronomy HW#3 Due Feb 27, 2013 Problem 1 (4 points) (Courtesy J. J. Condon & S. M. Ransom) The GBT (Green Bank Telescope, a steerable radio telescope roughly the size of a football field see photo below!) has sensitive receivers covering the frequency ranges 18 to 26.5 GHz and 26 to 40 GHz. Show that this frequency coverage is sufficient to detect CO emission from galaxies at any redshift from z 1.9 to at least z 10 (the estimated redshift at which stars started producing significant amounts of interstellar CO). You can look up the rest frequency of CO lines using splatalogue (note that the low-j lines are most likely to be excited in the cold ISM): The most commonly observed CO emission lines arise either from the J=1 0 or J=2 1 transition, with rest frequencies 115 and 230 GHz. For the upper limit on the redshift, z For the lower limit on the redshift, z
2 Problem 2 (17 points) You go to Prof. Novick s lab and measure the rest frequency of the J=3 2 transition of the CO molecule to be GHz. A) Estimate the distance between the C and O atoms in the molecule. (2 points) Recall that the frequency of the rotational line of a small molecule is given by ν = hj 4π 2 µre 2 (1) In this case, J=3 and µ = m1m2 m 1+m 2 = m H. Solving for r e gives cm Now you go to a telescope and observe the following CO line emission from a giant molecular cloud in the Milky Way galaxy (rest frame is LSR, position on the sky is α=18 h 30 m, δ=-10 6, in B1950 coordinates): B) What is the cloud s velocity relative to the sun? Is it moving towards or away from the sun? Remember that the conversion between LSR and heliocentric velocity is: V LSR = V heliocentric + V [cos(α α 0 ) cos δ 0 cos δ + sin δ 0 sin δ], where V = 19.7 km/s, and the direction of motion is towards (α 0 = 271, δ 0 = +30 ) in B1950 coordinates. (3 points) First, calculate LSR velocity: ν/ν = v/c. Solving for v gives 50 km/s. Note that the emission is redshifted (lower frequency) and has an amplitude larger than the maximum LSR-heliocentric velocity difference, so it must be moving away from the sun. To get the object s velocity relative to the sun, we must convert to the heliocentric rest frame. Using the values provided in the problem (with V LSR =50 km/s), the heliocentric conversion gives V hel = 35 km/s C) At what galactic longitude l do you observe this cloud? (One good calculator to help you out is the NED Coordinate and Extinction calculator at ). At what distance(s) from the earth might it be located? Assume that the galactic rotation curve exhibits a constant velocity of 220 km/s as a function of radius (i.e., ΩR = constant = 220 km/s). (4 points) 2
3 Plugging the coordinates into the calculator (converting from equatorial to galactic in epoch B1950) gives l = 21.9 Now comes the exciting part... Solve for the galactocentric distance R of the source based on its recessional velocity, using the assumption that ΩR = constant = 220 km/s: v = R 0 (Ω Ω 0 ) sin l (2) ΩR = constant = 220km/s (3) v = (R 0 Ω Ω 0 R 0 ) sin l (4) = 220km/s( R 0 1) sin l R (5) R = R 0 v 220km/s sin l + 1 (6) Plugging in the appropriate numbers (R 0 = 8.5 kpc, v = 35 km/s) gives R = 6 kpc. Once we have calculated R, we can use geometry to show that the distance of the cloud from earth must be x ± y, where x = R 2 (R 0 sin l) 2 and y = R 2 0 (R 0 sin l) 2. Here is a drawing of how that relationship is derived: (7) Plugging in appropriate values for x and y indicates that the cloud must be 8.4±5.9 kpc from earth, or either 2.5 or 14.4 kpc from earth D) Assuming the line is thermally broadened, estimate the temperature of the cloud. Check to make sure the cloud is hot enough to excite the CO J=3-2 line. (2 points) 3
4 8kT ln 2 Recall: ν D (F W HM) = ν 0 mc. From the drawing, ν 2 D (F W HM) = MHz, implying that the temperature is roughly 100 K. The estimated minimum temperature at which a line will be excited is given by T min = νh(j+1) 2k =33 K. Yes, the kinetic temperature indicated by the FWHM is high enough to excite the CO(3 2) line. E) Assume that the source fills the beam. Is the cloud optically thick, or optically thin? (2 points) Spectral line is characterized by thermal emission at temperature T s. When the source fills the beam, the antenna temperature T A is equal to the brightness temperature T B. If the source were optically thin, T A = T B τ ν, with τ ν < 1. However, in this case we see that T A = T B = T K (the kinetic temperature measured in part B), so the line has achieved its maximum brightness and is most likely optically thick. F) Can you measure the total column density of material through the cloud? If so, do it! If not, why not? (2 points) Can t measure column density for an optically thick line, because the object is opaque and so you can t see all the material along the line of sight. Instead you see surface temperature. G) The Einstein A coefficient for the CO J=3-2 line is s 1. Estimate the critical density of the line. Do you think the cloud is collisionally or radiatively excited? Why? Does this tell you anything about the density of material in the cloud? (2 points) n crit A21 σ calculate n crit cm 3 m kt k. With A 21 given in the problem, σ cm 2, T K 100 K, and m = (6 + 12)m H, we Since we know the kinetic temperature T K 100 K from the width of the line, and the flux of the line indicates optically thick emission at T S =100 K, it appears that the line is collisionally excited. This tells us that the density of material in the cloud must be higher than the critical density calculated above. Problem 3 (8 points) (Courtesy M. Faison) Qualitatively sketch the beam pattern of this aperture: 4
5 There are effectively four elements to this pattern: The beam pattern will be the combination of the Fourier transformations of these four functions: Note: (1) is a narrow peak with Airy rings, (2) is a broad trough with Airy rings, and for (3) and (4) the long axis FTs to a narrow sinc function, while the short axis FTs to a broad sinc function. The total should look something like this: 5
6 And in case you don t believe me, here is the actual FT, done in Mathematica: Problem 4 (6 points) Derive the expressions for α ν and j ν in terms of the Einstein coefficients that we used in class: Some helpful hints: α ν = g 2 + g 2 nc 2 h 8πkT s ν f(ν)a 21 (8) j ν = g 2 + g 2 nhν 4π f(ν)a 21 (9) I ν α ν is the number of net absorptions per unit time (i.e., absorption minus stimulated emission), times the energy hν, per unit solid angle (4π). Similarly, j ν is the number of spontaneous emissions per unit time, times the energy, per unit solid angle. Use the Rayleigh-Jeans approximation liberally. 6
7 n is defined as n 1 + n 2, and can also be written as n 1 (1 + n 2 /n 1 ). Don t give up: j ν is much simpler to derive than α ν! Number of net absorptions per unit time = number of absorptions - number of stimulated emissions: = (n 1 B 12 n 2 B 21 )f(ν)i ν Note that B 21 = g1 g 2 B 12 and n2 n 1 = g2 exp ( hν kt ) Then, = n 1 B 12 I ν f(ν)[1 exp ( hν kt )] Now use R-J approximation: hν kt << 1: = n 1 B 12 I ν f(ν) hν kt Now, I ν α ν = net absorptions per unit time hν 4π, implying: α ν = h2 ν 2 4πkT n 1B 12 f(ν) Recall: B 12 = g2 c 2 n 2 n 1 = g2 exp ( hν kt ) g2 Then, n = n 1 (1 + g2 2hν A 3 21 and n = n 1 + n 2 = n 1 (1 + n2 when hν << kt ) and n 1 = n 1+ g 2 = g1 g 1+g 2 n Substituting for B 12 and n 1 in the expression for α ν above gives: α ν = h2 ν 2 g 2 nc 2 4πkT +g 2 2hν A 3 21 = Now for j ν... g2 +g 2 nc 2 h 8πkT ν A 21f(ν) j ν = number of stimulated emissions per unit time hν 4π : j ν = n 2 hν 4π f(ν)a 21 Once again, express n 2 in terms of n and statistical weights: n 2 = Substitute in: j ν = g2 nhν +g 2 4π f(ν)a 21 n 1 ) g2 +g 2 n 7
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