ASTR240: Radio Astronomy

Transcription

1 AST24: adio Astronomy HW#1 Due Feb 6, 213 Problem 1 (6 points) (Adapted from Kraus Ch 8) A radio source has flux densities of S Jy and S Jy at frequencies of ν 1 6 MHz and ν MHz, respectively. A) Show that its spectral index α [log (S 1 /S 2 )/[log (ν 2 /ν 1 ) (2 points) Spectral index α; S ν ν α S 1 S 2 ( ν1 ν 2 ) α log (S 1/S 2) α log (ν 2/ν 1) ( ) α ν2 ν 1 α log (S 1/S 2)/ log (ν 2/ν 1) B) Calculate its spectral index. (2 points) α log (12.1/8.3)/ log (1415/6).44 C) Is the spectrum thermal or nonthermal? (2 points) Thermal spectra have α-2, so this spectrum is nonthermal. Problem 2 (8 points) (ybicki & Lightman Problem 1.5) A supernova remnant has an angular diameter θ4.3 arc minutes and a flux at 1 MHz of F cm 2 s 1 Hz 1. Assume that the emission is thermal. A) What is the brightness temperature T b? What eny regime of the blackbody curve does this correspond to? (2 points) Definition of brightness temperature: T b Iνc2 2kν 2 1

2 To calculate I ν, need to know the flux density and angular size: I ν Fν Ω Ω π 4 θ2 π 4 I ν ( ) π Sr cm 2 s Hz Sr cm 2 s Hz Sr Now, plug and chug: T b Iνc (3 cm 2 s Hz Sr 11 cm/s) 2 2kν K K (18 Hz) 2 T b K What eny regime does this correspond to? Check hν kt : hν kt s 1 8 Hz /K K hν kt << 1 -J regime (big surprise!) B) The emitting region is actually more compact than indicated by the observed angular diameter. What effect does this have on the value of T b? (2 points) If θ is smaller than assumed, then Ω decreases, I ν increases, T b increases. C) At what frequency will this object s radiation be maximum, if the emission is blackbody? (2 points) Wien displacement law: hν max 2.82 k T ν max 2.82 k K h Hz D) What can you say about the temperature of the material from the above results? (2 points) Graybodies: Planck function gives max emission for temp T. Therefore we know only that the temp of the object is T b calculated above. 2

3 Problem 3 (8 points) (ybicki & Lightman 1.9) A spherical, opaque object emits as a blackbody at temperature T c. Surrounding this central object is a spherical shell of material, thermally emitting at a temperature T s (T s < T c This shell absorbs in a narrow spectral line; that is, its absorption coefficient becomes large at the frequency ν and is negligibly small at other frequencies, such as ν 1 : α ν >> α ν1 (see Fig The object is observed at frequencies ν and ν 1 and along two rays A and B shown above. Assume that the Planck function does not vary appreciably from ν to ν 1. A) At which frequency will the observed brightness be larger when observed along ray A? Along ray B? (4 points) Along ay A: Intensity at surface of BB: I ν B ν (T c ) Since no (or very little) absorption at ν 1, intensity remains constant along the ray at frequency ν 1. Now consider ν. Source function in shell (thermal radiation) is J ν jν α ν B ν (T s Since T s < T c, B ν (T s ) < B ν (T c ), and intensity drops as ray passes through shell (I A ν < I A ν 1 3

4 Along ay B: I B ν 1 I B (no incident radiation Since T s >, B ν (T s ) > B ν (), and intensity increases as ray passes through shell (I B ν > I B ν 1 B) Answer the preceding questions if T s > T c. (4 points) Along ay A: Since T s > T c, B ν (T s ) > B ν (T c ), and intensity increases as ray passes through shell (I A ν > I A ν 1 Along ay B: Since T s > (still), B ν (T s ) > B ν (), and intensity still increases as ray passes through shell (I B ν > I B ν 1 Illustration of all four cases: Problem 4 (8 points) (Courtesy J. Moran) Calculate the quietest (i.e., darkest) place in the radio spectrum. Neglect noise from the earth s atmosphere. At low frequencies the sky noise is dominated by synchrotron emission from relativistic electrons rattling around all over the galaxy. Away from the galaxy plane, the brightness temperature is ( ν T B 18K 18MHz which is more or less independent of direction. At high frequencies, the CMB dominates with T B 2.7 K in all directions ) 2.6 4

5 A) What is the frequency of minimum background brightness temperature? Of minimum background flux density? (4 points) Background brightness temperature: T B (ν) 18 K ( 18 MHz) K. No minimum TB ; asymptotically approaches 2.7 K. ν Minimum flux density: ν 2 S ν I νω 2kT B c Ω 2 S ν ν (small angles) 2kΩ [ 18 K(18 MHz) 2.6 ν K ν 2 at the minimum 2kΩ [ 18 K(18 MHz) 2.6 (.6)ν 1.6 min K ν min ν min 57 MHz B) What is the incident power on the earth from 1 MHz 1 THz ( Hz)? If we intercept this power could we reduce our reliance on fossil fuels? (4 points) I ν 2k [ 18 K(18 MHz) 2.6 ν K ν 2 S ν 2π π/2 I ν cos θ sin θdθdφ 5

6 where S ν is integrated over the total amount of sky visible to a patch of earth s surface: Assume that I ν is uniform and isotropic... F [ [ 1 12 Hz 2π I νdν 1 7 Hz π/2 cos θ sin θdθdφ [ 1 12 Hz [ 2k 1 7 Hz c [18K( Hz) 2.6 ν Kν 2 dν 2π 1 2 (sin2 θ π/2 ) π Sr.28π cm 2 s Sr So flux per unit area is.28π cm 2 s Sr, and multiplying by the surface area of the earth (4π2 ) gives the total power: P (4π 2 )(.28π) /s, or W. Compare this with the power of the sun, P 1 17 W, and the eny used to generate electricity from fossil fuels worldwide, currently around, P ff 1 13 W. Problem 5 (8 points) (Courtesy J. Moran) The full moon at millimeter wavelengths has a brightness temperature distribution that might be approximated as (see figure): T B (θ) T + T 1 cos θ. A) What is the flux density at the earth? Hint: First calculate the brightness temperature as a function of polar angle in the earth coordinate system. (4 points) T B (θ) T + T 1 cos θ Geometry: cos θ 2 x 2, and φ x/d x Dφ S ν I ν cos θdω (In earth-based coordinates, θ φ I ν 2kT Bν 2 Earth-based coordinates: T B (φ) T + T 1 2 D 2 φ 2. Then: 6

7 S ν 2kν2 /D 2 D (T + T 2 φ 2 1 ) cos φ sin φdφ 2π cos φ 1 and sin φ φ (small angles) [ /D /D T φdφ + 4πkν2 4πkν2 2πkν2 ( 1 2 T( D )2 1 2 T1 D D φt 1 (/D)2 φ 2 dφ ) u 1/2 du ( T D 2 ( D )2 T 1 3 ( ( D )3 ) ) Note : u ( D )2 φ 2 S ν 2πkν2 ( D )2 (T T 1) B) What is the mean brightness temperature, i.e., disk temperature? (4 points) Mean T B of the moon: T mean 1 /D Ω moon (T + T 2 D 2 φ 2 1 )2πφdφ where Ω moon π( [ D )2, and we have already evaluated the integral: T mean 1 π ( D )2 2 D (T T 1)π T T 1 7