Some Notes on Observing Redshifted CO with MMA/LSA bbutler, with inputs from hliszt, awootten, sradford, myun last amended - 22feb1999 reproducing the derivation of equation (1) from solomon et al. 1992 (which is the same as equation (1) in solomon et al. 1997). this equation relates the luminosity of CO to the flux density: nu_rest Lco = 1.04E-3 Sco dV dL^2 ------- (1 + z) where Lco is the CO line luminosity in Lsun (i.e., in Watts), Sco dV is the velocity integrated flux density in Jy km/s, nu_rest is the rest frequency of the transition in GHz, and dL is the luminosity distance in Mpc. according to the paper, this equation derives from conservation of energy via: nu_rest L(nu_rest) = 4 pi dL^2 nu_obs S(nu_obs) where L(nu_rest) is monochromatic luminosity (in this equation, it must be in W/Hz for units to work out, i guess) and S(nu_obs) is observed flux density. they reference weinberg's 1972 book for this, go back to weinberg to see where this comes from... on p. 453, there is the following (eqn. 14.7.10): P(nu_obs R(t0)/R(t1)) R(t1) S(nu_obs) = --------------------------- R^3(t0) r1^2 where S(nu_obs) is flux density, power per unit antenna area per unit frequency interval (W/m^2/Hz or Jy), at a fixed frequency, P is the _intrinsic power_, the power emitted per unit solid angle and per unit frequency interval (W/Hz/sr). now, the definition of z is (eqn 14.3.6): R(t0) z = ----- - 1 R(t1) so that nu_obs R(t0)/R(t1) = nu_obs (1+z), which is just nu_rest, the rest frequency of the transition. so, P(nu_rest) R(t1) S(nu_obs) = ---------------- R^3(t0) r1^2 from the definition of z, R(t1)/R(t0) = 1/(1+z), so P(nu_rest) S(nu_obs) = ------------------ (1+z) R^2(t0) r1^2 now, the definition of the angular size distance is (eqn. 14.4.17): dA = R(t1) r1 so, with some algebra (since we know that R(t0) = (1+z) R(t1)), we get: P(nu_rest) S(nu_obs) = ------------ (1+z)^3 dA^2 if you want it in terms of luminosity distance instead, the luminosity distance and the angular size distance are related via (eqn. 14.4.22): dL dA = --------- (1+z)^2 so P(nu_rest) (1+z) S(nu_obs) = ---------------- dL^2 now, how is P related to "luminosity"? P is power per unit solid angle per unit frequency interval, so: L P = ---- 4 pi where L is the power emitted per unit frequency interval (W/Hz). so, L(nu_rest) (1+z) S(nu_obs) = ---------------- 4 pi dL^2 this is equivalent to solomon et al.'s equation (1), i.e., substituting (1+z) = nu_rest/nu_obs gives: 4 pi dL^2 nu_obs S(nu_obs) = nu_rest L(nu_rest) define the total luminosity over the line as: L' = integral of L(nu_rest) over the line ~ L(nu_rest) del_nu_rest L' has units of (W). then, L' (1+z) S(nu_obs) = --------------------- 4 pi del_nu_rest dL^2 and if you substitute del_nu_rest = nu_rest del_v_rest / c, then, L' c (1+z) S(nu_obs) = ---------------------------- 4 pi nu_rest del_v_rest dL^2 or, inverting: 4 pi S(nu_obs) del_v_rest nu_rest dL^2 L' = -------------------------------------- c (1+z) which, using the same units as solomon et al. for the quantities, agrees with their equation (1), i.e., substitute: L'' = L'/Lsun = L' / 3.853d26 luminosity in Lsun S' = S * 1.0d26 flux density in Jy dv' = del_v_rest / 1.0d3 velocity width in km/s nu' = nu_rest / 1.0d9 frequency in GHz dL' = dL / 3.086d22 luminosity distance in Mpc then, 1 4 pi S' 1 L'' = -------- ---- ------ 1.0d3 dv' 1.0d9 nu' (3.086d22)^2 dL'^2 ----- 3.853d26 c 1.0d26 (1+z) 4 pi 1 L'' = 2.472d4 ---- S' dv' nu' dL'^2 ----- c (1+z) 1 L'' = 1.036d-3 S' dv' nu' dL'^2 ----- (1+z) -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- invert last equation above for S': L'' (1+z) S' = 9.653d2 -------------- dv' nu' dL'^2 use L'' = 3e5 (Wright et al. 1991), dv' = 300 km/s (Solomon et al. 1997), then: (1+z) S' = 9.653d5 --------- nu' dL'^2 and you get the following table of expected flux densities: z=.2 1 2 3 4 1-0 14.383 0.764 0.245 0.131 0.086 2-1 7.192 0.382 0.122 0.066 0.043 3-2 4.795 0.255 0.082 0.044 0.029 4-3 3.596 0.191 0.061 0.033 0.021 5-4 2.877 0.153 0.049 0.026 0.017 6-5 2.398 0.127 0.041 0.022 0.014 -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- if i believe solomon et al., then the luminosity can also be written: 1 L'co = 3.25e7 Sco dv dL^2 ---------------- nu_obs^2 (1+z)^3 where L'co is now the line luminosity in K km/s pc^2, Sco dv is still in Jy km/s, dL still in pc, and nu_obs is in GHz. invert this to get expected flux density: nu_obs^2 (1+z)^3 Sco = 3.08e-8 ---------------- L'co dv dL^2 or, using instead of nu_obs, the rest frequency nu_rest, nu_rest^2 (1+z) Sco = 3.08e-8 --------------- L'co dv dL^2 use L'co = 3.7E8 K km/s pc^2 for the Milky Way (Solomon & Rivolo 1989), and dv = 300 km/s. use the standard luminosity distance equation (see Weinberg 1972, eqn 15.3.24). ignore the difference in line intensities for the different CO lines. this might make changes of factors of possibly 2 or so at the worst (see harvey's email and note that L'co scales like Tr between lines). use H_o = 75 km/s/Mpc and q_o = 0.5 when calculating dL. then, the expected flux density, now in mJy is: nu_rest^2 (1+z) Sco = 37.9 --------------- dL^2 and you get the following table of expected flux densities: z=.2 1 2 3 4 1-0 0.865 0.046 0.015 0.008 0.005 2-1 3.460 0.184 0.059 0.032 0.021 3-2 7.784 0.413 0.132 0.071 0.046 4-3 13.837 0.735 0.235 0.126 0.083 5-4 21.617 1.148 0.367 0.197 0.129 6-5 31.125 1.653 0.529 0.284 0.186 -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- this seems partly to have come down to the question of what is the CO luminosity of the Milky Way? - Solomon & Rivolo 1989: L'co = 3.7e8 K km/s pc^2 based on measurements of CO 1-0 in quadrant 1, and extrapolation to all CO in all quadrants. - Wright et al. 1991: Lco ~ 1.0e5 W based on COBE measurements of CO 2-1, 3-2, 4-3, and 5-4. there are better COBE numbers from a better reduction (Bennett et al. 1994), but only plots are shown, no numbers. - Solomon et al. 1992: Sco(3-2) = 4.1 Jy km/s Sco(6-5) = 9.4 Jy km/s for IRAS 10214+4724. now, let's get a luminosity from: 1 L'' = 1.036d-3 S' dv' nu' dL'^2 ----- (1+z) which gives (using Ho = 75 km/s/Mpc, q0 = .5): L''(3-2) = 6.22e7 Lsun L''(6-5) = 2.85e8 Lsun and from: (1+z) L'co = 3.25e7 Sco dv dL^2 --------- nu_rest^2 which gives: L''(3-2) = 4.70e10 K km/s pc^2 L''(6-5) = 2.70e10 K km/s pc^2 now, if you used 0/10th of IRAS 10214+4724 as a Milky Way CO predictor (as Lorne Avery did), then you would get numbers like 3e9 K km/s pc^2, which is about a factor of 10 too high (if you take the solomon & rivolo number as the right one). *** turns out that IRAS 10214+4724 is lensed, with a magnification factor of about 10. so, when you get the intrinsic luminosity, you need to divide this factor of 10, which then makes the numbers agree pretty well... -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- putting it all together. take: nu_obs^2 (1+z)^3 Sco = 3.08e-8 ---------------- L'co dv dL^2 and calculate it for different lines, at different redshifts, given an excitation model. take L'co = 5e8 for the 1-0 line this is slightly higher than the solomon & rivolo number, but close to the 1/10th 10214 number (intrinsic - not lensed), and also matches pretty well with COBE results, i think. take dv = 300 km/s take Ho = 75 km/s/Mpc take q0 = 0.5 now, what to use for the relative line radiation temperatures? harvey has done a careful calculation, and gets the following: (a) T_B z 1-0 2-1 3-2 4-3 5-4 6-5 7-6 0 5.910 4.564 2.423 0.470 0.094 0.199 0.030 1 4.401 3.839 2.408 0.633 0.087 0.100 0.027 2 3.444 3.159 2.254 0.826 0.118 0.050 0.012 3 2.906 2.716 2.099 0.978 0.203 0.048 0.014 4 2.583 2.431 1.966 1.072 0.306 0.063 0.020 5 2.448 2.316 1.921 1.165 0.415 0.096 0.029 (b) z 1-0 2-1 3-2 4-3 5-4 6-5 7-6 8-7 0 16.753 15.398 10.254 4.677 1.309 0.452 0.076 0.010 1 14.965 14.522 9.980 4.694 1.342 0.443 0.079 0.010 2 13.403 13.314 9.385 4.633 1.406 0.409 0.077 0.011 3 11.929 12.055 8.652 4.479 1.477 0.410 0.076 0.011 4 10.579 10.828 7.889 4.261 1.535 0.439 0.086 0.013 5 9.361 9.684 7.152 4.011 1.570 0.480 0.104 0.016 (c) z 1-0 2-1 3-2 4-3 5-4 6-5 7-6 8-7 0 26.655 35.218 33.268 30.913 28.593 26.216 23.414 18.780 1 25.018 33.799 32.524 30.579 28.468 26.189 23.431 18.821 2 23.340 31.879 31.049 29.604 27.890 25.904 23.345 18.901 3 21.787 29.907 29.382 28.293 26.936 25.280 23.015 18.889 4 20.321 28.040 27.685 26.860 25.773 24.409 22.431 18.710 5 18.925 26.245 26.032 25.378 24.498 23.373 21.649 18.350 where (a) is a typical cold cloud (like what most of the MW emission comes from), (b) is a weak PDR, and (c) is a strong PDR. now, assume that some fraction of the emission comes from the cold material (call it fcold), and some fraction from the strong PDR's (ignore the weak PDR's). then, calculate the luminosity for a given transition and redshift as: L'co = Lco(1-0) * fCO fCO = fCO'(transition,z) / fCO'(1-0,0) fCO'(transition,z) = fcold * fCO'cold(transition,z) + (1-fcold) * fCO'hot(transition,z) where fCO'cold and fCO'hot are interpolated from harvey's calculated numbers. then, choosing fcold, you can simply calculate the expected flux density as a function of transition and z. for fcold = 0.9, this comes out as: z=.2 1 2 3 4 5 1-0 1.126 0.050 0.014 0.006 0.004 0.003 2-1 4.380 0.213 0.060 0.029 0.017 0.012 3-2 7.242 0.380 0.115 0.058 0.036 0.025 4-3 8.294 0.452 0.148 0.079 0.051 0.036 5-4 10.771 0.569 0.180 0.096 0.062 0.045 6-5 14.675 0.759 0.236 0.124 0.079 0.055 7-6 16.999 0.902 0.286 0.151 0.097 0.068 8-7 17.611 0.937 0.301 0.161 0.105 0.075 for fcold = 0.0, and L'co = 5e9 (maybe a more typical ULIRG?), you get: z=.2 1 2 3 4 5 1-0 11.560 0.583 0.174 0.087 0.053 0.036 2-1 61.353 3.152 0.952 0.478 0.294 0.200 3-2 130.864 6.823 2.086 1.058 0.652 0.446 4-3 216.661 11.404 3.535 1.810 1.125 0.773 5-4 313.494 16.587 5.203 2.693 1.686 1.166 6-5 414.122 21.970 6.957 3.639 2.299 1.601 7-6 503.514 26.750 8.533 4.508 2.876 2.018 8-7 527.544 28.059 9.022 4.832 3.132 2.234 this last table has entries which seem a bit high, so maybe i don't have the right combination of region types for the excitation model, but i don't use it below, so i press on... now, combine this with the best estimates for the noise (via the technique that i described in my previous email on this), and you get the following tables of max detectable z for the antenna size/number combinations listed (the noise here is calculated for a 50 km/s channel and 4 hour integration, and a 5-sigma detection is required): ******************** old MMA default ******************** 36 10m antennas trans max detectable z nu 1-0 0.2950 89.013 2-1 0.5800 145.910 3-2 0.6900 204.613 4-3 0.6900 272.805 5-4 0.4400 400.186 6-5 0.1000 628.612 7-6 0.2150 663.911 8-7 0.1300 815.752 ******************** MMA/LSA 10m ************************ 64 10m antennas trans max detectable z nu 1-0 0.3900 82.929 2-1 0.7850 129.153 3-2 0.8050 191.577 4-3 1.0700 222.725 5-4 1.1150 272.467 6-5 1.0150 343.163 7-6 0.2550 642.750 8-7 0.1450 805.065 ******************** MMA/LSA 12m ************************ 48 12m antennas trans max detectable z nu 1-0 0.4050 82.044 2-1 0.8050 127.722 3-2 0.8050 191.577 4-3 1.1250 216.960 5-4 1.1900 263.136 6-5 1.0500 337.304 7-6 0.2600 640.200 8-7 0.1450 805.065 ****************** with japanese? 10m ******************* 100 10m antennas trans max detectable z nu 1-0 0.4800 77.886 2-1 0.8650 123.613 3-2 1.4300 142.303 4-3 1.3300 197.872 5-4 1.6050 221.216 6-5 1.8300 244.337 7-6 1.9150 276.724 8-7 0.3850 665.559 ****************** with japanese? 12m ******************* 75 12m antennas trans max detectable z nu 1-0 0.4900 77.363 2-1 0.8650 123.613 3-2 1.5000 138.318 4-3 1.3550 195.771 5-4 1.6900 214.226 6-5 1.9550 234.001 7-6 2.0850 261.475 8-7 0.3900 663.165 ****************** pipedream? 15m ******************* 60 15m antennas trans max detectable z nu 1-0 0.5350 75.095 2-1 1.2500 102.461 3-2 1.6700 129.512 4-3 1.8750 160.362 5-4 1.8850 199.746 6-5 2.2950 209.855 7-6 2.5650 226.270 8-7 2.4500 267.188 and, as a contrast, an ULIRG (Lco = 5e9, fcold=0): 48 12m antennas trans max detectable z nu 1-0 0.7050 67.608 2-1 2.4150 67.507 3-2 4.0450 68.542 4-3 5.7300 68.505 5-4 7.4100 68.522 6-5 9.0900 68.531 7-6 >10.0000 ------ 8-7 >10.0000 ------ so, can we make a general scaling with how the maximum z (zmax) relates to the collecting area of the array (Aarray)? the following were calculated for increasing numbers of antennas: K = Nant*D^2/(No*D^2) zmax K/zmax --------------------- ---- ------ 1 .575 1.74 1.5 .730 2.05 2 .910 2.20 2.5 1.12 2.23 3 1.43 2.10 4 1.98 2.02 6 2.85 2.11 8 3.53 2.27 it's roughly linear (scaling between collecting area and max z), with a mean value of about 2.1. this was done with No = 40, and D = 8m, so the scaling looks something like: Aarray zmax ~ -------- 4200 if you change the "detection" requirements to: 75 km/s channels, 6 hour integrations, and 4-sigma detections, then this scaling becomes: Aarray zmax' ~ -------- 2200 remember that this is only for detection of a single line. if we require that 3 lines must be available for detection, then the revised requirements are: Aarray zmax ~ -------- 5000 Aarray zmax' ~ -------- 2700 note that i've only gone up to the 8-7 transition, and the result may be modified slightly by the ability to detect the higher transitions at higher z, but i think it won't change by much. better weather should improve things. remember that all of these numbers were calculated with "typical" opacities (and at an airmass of 1.3). at night in winter, it should get better, e.g...