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# Download ASTR 5340: Radio Astronomy Problem Set 1 Due: 13 September

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ASTR 5340: Radio Astronomy Problem Set 1 Due: 13 September, 2010 1. (6 points) If the Sun were a black body with a T = 5800 K photosphere, it would be undetectably faint at ν = 1 GHz at the distance of the nearest stars. However, that doesn’t mean it is impossible to detect blackbody photospheric emission from any star at any radio frequency. The best candidate would be a very large and luminous star, and the best frequency might not be 1 GHz. The star Betelgeuse is an excellent candidate. Betelgeuse (aka α Orionis because it is the brightest star visible in the Orion constellation) is the red star that defines the left shoulder of Orion. It is an extremely luminous (bolometric luminosity L ≈ 105 L⊙ , where the ⊙ subscript refers to the Sun) supergiant star with photospheric radius R ≈ 103 R⊙ and photospheric (color) temperature T ≈ 3600 K. For such a cool black body to appear so bright, the Planck law implies that its angular diameter must be about θ = 0.045 arcsec at its known distance of ∼130 pc. In a ten-minute observation, the VLA can detect a source whose flux density S is larger than the limit Smin . In the six main VLA frequency bands, the detection limits are: ν (GHz) 1.4 4.8 8.4 15 23 43 Smin (mJy) 0.30 0.30 0.25 1.60 0.55 1.35 Assume that Betelgeuse is a black body. Which observing frequency would you choose to obtain the highest ratio S/Smin ? What is the expected flux density S from the photosphere of Betelgeuse at this frequency? 2. (5 points) The blue star Rigel (β Orionis, the right foot of Orion) is another supergiant star that is about as luminous as Betelgeuse (both have bolometric luminosities L ≈ 105 L⊙ ) and lies at about the same distance, but Rigel is much hotter, TR ≈ 11, 000 K, versus TB ≈ 3, 600 K for Betelgeuse. Again using the black-body approximation, compare its expected flux density at any radio frequency ν with that of Betelgeuse (HINT: Use scaling relations!). 3. (13 points) The brightness Bν of blackbody radiation falls off exponentially at frequencies higher than ν ∼ kT /h. This problem illustrates why (1) Kirchoff’s law is not intuitively obvious to us because even a black body at environmental temperatures T ≈ 300 K emits almost zero visible photons and (2) thermal emission from cold astronomical sources such as the cosmic microwave background (a nearly perfect T ≈ 2.73 K blackbody) and interstellar dust and molecular gas (typically T < 100 K) can be detected only at radio and infrared wavelengths. (a) (4 points) Show that the spectral power per unit area emitted by a black body is Pν = πBν , where Bν = 1 2hν 3 hν 2 c exp( kT ) − 1 is the surface brightness of a blackbody at temperature T . (b) (3 points) The human eye is sensitive only to radiation at wavelengths shorter than λ0 ≈ 7500 Å. 0 Calculate the corresponding frequency ν0 and show that exp( hν kT ) ≫ 1 for objects of ambient temperature T = 300 K. (c) (4 points) Show that a very good approximation to the number Ṅ of photons with frequencies greater than ν0 emitted per second per unit area by a black body is Z hν 2π ∞ 2 dν . ν exp − Ṅ = 2 c ν0 kT (d) (2 points) The integral above can be integrated by parts twice (optional excercise for the student) to yield Z ∞ hν0 kT ν02 2k 3 T 3 2k 2 T 2 ν0 hν dν = . exp − + + ν 2 exp − kT h h2 h3 kT ν0 Given this result, show that Ṅ ≈ 10−6 cm−2 s−1 ; that is, 1 cm2 of a T = 300 K blackbody only emits one visible photon per million seconds! 4. (6 points) At what opacities τ do absorbers allow 99%, 50%, 36.8%, 1% and 1×10−5 of the input brightness through? If the absorber is in thermodynamic equilibrium at those same τ s, what fraction of the absorber’s kinetic temperature T is seen in thermal emission? Page 2