1. Lecture 3 By Tom Wilson

2. Summary of Lecture 1 Noise in a Receiver time on source Receiver itself, atmosphere,ground and source Analying bandwidth (for lines, need 3 resolution elements on the line above the ½ power point) Temperatures from thermal hot and cold load measurements using the receiver.

3. Hot-cold load measurements Absorber at a given temperature Input to receiver (to determine receiver noise contribution) OK for heterodyne receivers, but for Bolometers OK for heterodyne receivers, but not for Bolometers

4. Current Receiver Noise Temperatures T min =h /k for coherent receivers T SYS =T RX e  Atmospheric optical depth

5. Receivers Heterodyne for spectral lines High velocity resolution High velocity resolution flexibility, but not multi-pixel receivers in the mm/sub-mm flexibility, but not multi-pixel receivers in the mm/sub-mm Bolometers for continuum Multi-pixel cameras Multi-pixel cameras NoiseEquivalentpower (about 1)

6. Types of Receivers Fractional Resolution

7. Lecture 3 page 7 BOLOMETERS VS COHERENT RECEIVERS = 345 GHz JCMT: 15 m sub mm dish,  A = 0.5 at = 0.87 mm, = 345 GHz With SCUBA, can detect a source with 0.16 Jy in 1 second. So RMS is ¼ of this peak value or  S  0.04 Jy in 1 sec. Compare to a coherent receiver: T SYS = 50 K,  = 2 GHz, integration time T SYS = 50 K,  = 2 GHz, integration time= 1 sec In antenna temperature. From Lecture 2: For JCMT, Or So comparable for 1 beam, but SCUBA has 37 beams & MAMBO has 117 beams.

8.  (arc sec)=  k  mm) /D(m) of order 1.2 for single dishes S =3520 T A /(  A D 2 )

9. Rayleigh-Jeans Or Gaussian beams: Summary of Lecture 2 (Show that these are consistent) In Jy True source size and temperature apparent source size and temperature

10. Can make a relation for flux density similar to that for Main Beam Brightness temperature: S(total)=S(peak). (  S 2 +  B 2 )/  B 2 Example: Orion A is an HII region with a total flux density of 380 Jy at 1.3 cm. The size is 2.5’ (FWHP). If the radio telescope beam size is 40” (FWHP), what is the peak flux density? Use the R-J relation to find the peak main beam brightness temperature. Solution: peak Jy/beam=9.5; T A =8.8K, T B =24 K

11. Far Field Diffraction and Fourier Transforms (Exact calculations require programs such as GRASP) (radiation passing through an opening) Lecture 3 page 1

12. Grading Across the Aperture and Far E Field Lecture 3 page 2

13. ALMA Front-End Digitizer Clock Local Oscillator ANTENNA Data Encoder 12*10Gb/s 12 Optical Transmitters 12->1 DWD Optical Mux Digitizer 8* 4Gs/s -3bit ADC 8* 250 MHz, 48bit out IF-Processing (8 * 2-4GHz sub-bands) Fiber Patch-Panel From 270 stations to 64 DTS Inputs Optical De-Mux & Amplifier Digital De-Formatter Correlator Technical Building Tunable Filter Bank Fiber Lecture 3, page 3

14. Sketch of 2 element interferometer

16. (u,v) plane and image plane These are related by Fourier transforms The distance between antennas varies, so we sample different source structures On the next overheads, we indicate how structures are sampled. Following tradition, u represents one dimension distributions, with x as the separtion in wavelenghts u=2  x  and  v=2  y  Earth Rotation Aperture Synthesis

17. Above: the 2 antennas on the earth’s surface have a different orientation as a function of time. Below: the ordering of correlated data in (u,v) plane.

18. Gridding and sampling in (u,v) plane Sensitivity: http://www.eso.org/projects/alma/science/bin/sensitivity.html

19. VLA uv plane response

20. Data as taken Data with MEM with MEM and Self- Calibration The radio galaxy Cygnus A as measured with all configurations of the VLA

21. From W. D. Cotton (in ‘ The Role of VLBI in Astrophysics, Astrometry, And Geodesy, ed Mantovani & Kus, Kluwer 2004)

22. Lecture 3 page 16 BROADBAND RADIATION Black body (Moon, planets, 3K background) Dust thermal emission Bremsstrahlung (free-free) Synchrotron (relativistic electrons in magnetic fields) Inverse Compton Scattering (S-Z) Dust: Mostly carbon, silicon with ice mantles “ground up planets” From Hildebrand (1983)

23. Lecture 3 page 17 For warm grains Use R-J get EXAMPLE: Dust emission from Orion KL The Orion “hot core” has the following properties:

24. Lecture 3 page 18, if z = z Sun, b = 1.9 Calculate the column density N(H 2 ) = n L and the 1.3 mm dust flux density, S, if z = z Sun, b = 1.9 If the value of L = diameter, use L (diameter in cm) x (size in radians) = = 7.5 10 16 cm   Them N(cm -2 ) = 7.5  10 16 cm x 10 7 cm -3 = 7.5  10 23 cm -2 for H 2 N(H 2 ) = 1.5  N(H) = 2 N(H 2 ) = 1.5  10 24 cm -2 So S is 81 times smaller or 120 mJy. At 0.39 mm, S is 81 times larger or 810 Jy At 4 mm, S is 81 times smaller or 120 mJy. At 0.39 mm, S is 81 times larger or 810 Jy

25. Lecture 3 page 19 BREMSSTRAHLUNG (FREE-FREE) Hydrogen is ionized by O, B, electron and protons interact, electrons radiate. Classically: Power radiated during encounter: Find From the Kirchhoff relation, get velocity ‘ p ’ is impact parameter frequency

26. Lecture 3 page 20    = 1: When   = 1: 0 For Orion A,  0 = 1 GHz, or 30 cm. But - What is the Relation for T B ?

27. Lecture 3 page 21 Orion A HII Region Te = 8500 K,  = 2.5’ (FWHP), so use = 23 GHz so T = T e  is much less than 1, at = 23 GHz so T = T e  = From 100-m, T M B (main beam) = 24 K in a 40” beam, so T MB (main beam) = T(true)  (8.235 10 -2 )  (8500) -1.35  (23) -2.1  EM 24 = (8500)  (8.235 10 -2 )  (8500) -1.35  (23) -2.1  EM so EM = 4  10 6 cm -6 pc = N e  N i  L If L = 25’ = 0.33 pc converted to radians @ 500 pc get N e = N i = 3.5  10 3 cm -3 This is the RMS density. Calculate the mass of ionized gas. = 10 4 cm -3, so L = 0.03 pc. Then M = 0.6 M Sun in ionized gas Rough number since know Orion A is not spherical. From spectral lines know N e = 10 4 cm -3, so L = 0.03 pc. Then M = 0.6 M Sun in ionized gas

28. Free-Free Intensity and Flux Density as function of Frequency (Problem: Use the example of Orion to check these Curves) Lecture 3 page 22

29. Lecture 3 page 4 Free-Free Emission from Planetary Nebulae S = 5.4 Jy at 1.3 cm. What is the T MB (main beam brightness temperature) if the 100-m FWHP beam size is 43”? NGC7027 (a PNe) has S = 5.4 Jy at 1.3 cm. What is the T MB (main beam brightness temperature) if the 100-m FWHP beam size is 43”?Use Where Where  0 is the telescope beam size in are min. Suppose the “true” gaussian source size is 10”, what is T B (true brightness temperature). Could use (Problem: Repeat for the 30-m, with beam 27’’, wavelength 3.5 mm, flux density 4.7 Jy)

30. Lecture 3 page 5 Or And get We know that the electron temperature of NGC7027 is T e = 14000 K. Use equation of radiative transfer: To get This is a source which is thermal, so the radiation is free-free or Bremsstrahlung

31. Lecture 3 page 23 SYNCHROTON RADIATION (NON-THERMAL) Highly relativistic electrons spiraling in a B field with a frequency P: Power radiated by electron (lab) P’: Power radiated by electron (rest frame) so P = P’ Transformation of acceleration So

32. Radiation patterns of an electron in a B field B Field, V about 0 Velocity B Field, V about 0.2 c Lecture 3 page 24

33. Lecture 3 page 25 Then E: Particle energy Is difficult to separate energy of electron from B field strength To get spectral distribution, use  : Find a synchrotron spectrum  : Synchrotron radiation is found to be linearly polarized (power law distribution of Cosmic Rays)

34. Lecture 3 page 26 SINGLE ELECTRON SYNCHROTRON EMISSION For relativistic electrons, the emitted pulse is 1/  shorter due to relativistic beaming while the Doppler effect gives rise to a factor 1/  2  B : Frequency of rotation  G, So for B = 10  G,  B is even lower when  <1 Thus in frequency reach a critical value  G, C = 10 GHz, 10 4 So if B = 10  G,  G = 176 Hz, to reach C = 10 GHz,  = 1.6 10 4 In Synchrotron emission, we measure only the most relativistic particles

35. Lecture 3 page 27 SYNCHROTRON ENERGY CONSIDERATIONS Allow one to determine the minimum or equipartion energy Inverse Compton effect When the radiation density is equal to magnetic energy density there can be energy losses in addition to synchrotron energy losses. R & W don’t do much, but Kellermann & Owen give: This is the basis of the statement: “10 12 K is the highest source temperature possible”

36. Lecture 3 page 6 Non-thermal sources S = 3 10 4 Jy,  4’ (source size), Cas A: at 100 MHz, S = 3 10 4 Jy,  s =4’ (source size),  = 3 m = 300 cm  10 4 K Thermal sources have limit T = 2  10 4 K Assume that for Cas A, T=7.5 10 8 (  m  3) -2.8 What is the source temperature at 3 mm? SourceSpecralIndex

37. Lecture 3 page 29 SUNYAEV-ZELDOVICH EFFECT Clusters of galaxies are filled with hot diffuse gas. Photons from the 3 K background are scattered in this cluster gas. More photons are given energy than lose energy on the low frequency side of the Planck curve. On the high frequency side, some photons are shifted to lower energies, but still a reduction in the 3 K background. At 160 GHz, have a cross over from absorption at longer wavelengths to emission at shorter wavelengths, so have zero absorption. The absorption is: L, can solve for source distance. Given velocity of source, get HUBBLE CONSTANT. However there can be systematic effects such as clumping. When combined with X ray luminosity, which is Bremsstrahlung (free- free), proportional to N e 2 L, can solve for source distance. Given velocity of source, get HUBBLE CONSTANT. However there can be systematic effects such as clumping.

38. Lecture 3 page 30 EXAMPLE OF S-Z EFFECT  K at 1 cm wavelength The cluster CL0016 +16 shows on S-Z absorption of –700  K at 1 cm wavelength Z = redshift of CL0016 +16 is 0.541 X ray data: T e = 1.6 10 8 K Cluster size = 30“ to 19” RMS N e = 1.2 10 -2 cm -3 So