[mmaimcal] distant Milky Way?
Bryan Butler
bbutler at aoc.nrao.edu
Thu Feb 11 12:38:47 EST 1999
simon & all,
i like simon's treatment of CO lines, so i went through the calculation
myself, since i have no experience with this sort of thing. i went
through it fairly carefully, since i wanted to incorporate a similar
calculation in the antenna size report...
comments are not only welcome, but solicited, since i want to
incorporate a calculation of this sort in the antenna size report
before i release it, and this is about as far away from my realm of
science as it gets :)...
-bryan
here are my notes, in full glory...
i use bob brown's noise numbers which are in our antenna size report
(well, modified slightly, because there was a numerical error
in bob's straight numbers...):
nu (GHz) - 115 230 345 675
delS (mJy) - .14 .20 .34 2.4
delS' (mJy) - .044 .063 .11 0.76
delS is the noise in 1 25 km/s channel in 1 hour; delS' is the noise
averaged over 10 channels (250 km/s effective channel width). i used a
geometric area for the array of 6912 m^2, which is 48 12-m antennas,
which we derived later in the antenna size report, and is probably a
reasonable expectation for the array size.
now, calculate theoretical CO line emission flux density via
(from simon's notes - ala Solomon et al. 1992):
Sco = 3.08E-8 * L'co * (nu^2 / delv) * [(1+z)^3 / D_L^2]
this is total line flux density in Jy, where L'co is CO line luminosity
in K km/s pc^2; nu is line center frequency in GHz; delv is line full
width in km/s; and dL is luminosity distance in Mpc. use (as did simon,
basically - see Solomon & Rivolo 1989) L'co = 3.7E8 K km/s pc^2 for the
milky way, and delv = 300 km/s. use the standard luminosity distance
equation (as did simon - see Weinberg 1972). ignore, as did simon, 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
and q_o = 0.5 when calculating dL. resulting table for CO total line
flux densities in mJy in different bands:
nu (GHz) - 115 230 345 675
Sco[z=.2] - 1.24 4.97 11.2 42.8
Sco[z=1] - .183 .732 1.65 6.31
Sco[z=2] - .132 .528 1.19 4.54
Sco[z=3] - .126 .503 1.13 4.33
note that this is in the right ballpark when you compare the numbers
at z=.2 to the measured numbers in Solomon et al. 1997, and you take
into account the factor of 10 or so when comparing the milky way CO
line luminosity to the ultraluminous ones measured in Solomon et al.
now, to calculate the integration time needed to detect a line to
G-sigma:
delt = (delS * G / Sco)^2
which results in the following table detection times in hours for G=5:
nu (GHz) - 115 230 345 675
delt[z=1] - 1.4 0.19 0.11 0.36
delt[z=2] - 2.8 0.36 0.21 0.70
delt[z=3] - 3.0 0.39 0.23 0.77
if you just want to detect the line in a single 250 km/s "channel".
this is pretty easy.
but, what do we mean by "detecting" the line? is it enough to see it
in one coarse channel (the average of 10 25 km/s channels)? probably
not. i suspect that we would want to at least resolve it by a few
channels across the full width, and possibly even up to the 25 km/s
hyperresolution. in these cases, the required integration time just
scales like the ratio of the desired channel width to the 250 km/s
channel width used above. e.g., for 25 km/s resolution, the following
table for delt results:
nu (GHz) - 115 230 345 675
delt[z=.2] - .32 .04 .02 .08
delt[z=1] - 14 1.9 1.1 3.6
delt[z=2] - 28 3.6 2.1 7.0
delt[z=3] - 30 3.9 2.3 7.7
other than the 115 GHz lines, this still doesn't seem so bad to me.
----------------------------
how does this compare to simon's numbers? simon computed a quantity:
L'co 2 k nu^2 1 dL^2
X = ------ -------- -------- = ---------
delv c^2 G delS (1+z)^3
but used a different delS (and, frankly, i can't even reproduce
simon's first table of X given his delS numbers...). for the delS
numbers i'm using above in a 25 km/s channel in 1 hour, the following
table results (also using G=5):
freq 115 230 345 675 GHz
X 89 256 340 186 x 10^3 Mpc^2 in 1 min
685 1979 2635 1443 in 1 hour
1938 5597 7452 4082 in 8 hour
note that i changed the multiplier to 10^3 in order to compare more
directly with the numbers in simon's other table of X=dL^2/(1+z)^3.
here is a slightly different version of that other table:
q_o 0.05 0.25 0.5
z ----- ---- ----
0.2 | 443 425 404 X x 10^3 Mpc^2
1 | 4214 3391 2741
2 | 8237 5431 3806
3 | 11319 6307 3994
5 | 15761 6746 3730
10 | 21784 6135 2835
now, which lines might be observed? i'm no spectroscopist, so can't
give very intelligent treatment of this. the 3-2 transition requires
some 10's of K, as i recall, increasing to above 100 K for the 6-5
transition. anyway, i list below all the transitions below an
arbritrary cutoff at the 9-8 transition which are available through
decent windows at chajnantor (number in parentheses is the frequency):
z=.2 => 1-0 (96); 3-2 (288); 7-6 (672); 9-8 (864)
z=1 => 2-1 (115); 4-3 (231); 5-4 (288); 6-5 (346); 7-6 (403)
z=2 => 1-0 (38); 2-1 (77); 3-2 (115); 6-5 (231); 7-6 (269);
8-7 (301); 9-8 (346)
z=3 => 1-0 (29); 3-2 (86); 4-3 (115); 5-4 (144); 7-6 (202);
8-7 (230); 9-8 (259)
z=5 => 2-1 (38); 4-3 (77); 5-4 (96); 6-5 (115); 7-6 (134);
8-7 (154)
so, let's do some more detailed analysis involving these lines.
for each line, i calculate the visibility noise via bob's method,
using the appropriate tau0 from my default model of the atmosphere
above the site (1.5mm water vapor, -4 deg surface temp). then we
get for the required time to get to 5-sigma in a single 25 km/s
channel (Inf means the required time is > 1000 hr):
q_o=.05 q_o=.25 q_o=.5
z transition nu delS delt delt delt
.2 1-0 96 0.15 0.92 0.82 0.74
.2 3-2 288 0.25 0.030 0.028 0.025
.2 7-6 672 2.61 0.114 0.106 0.096
.2 9-8 864 4.09 0.103 0.095 0.086
1 2-1 * 115 0.24 105.3 68.7 44.8
1 4-3 * 231 0.20 4.5 2.9 1.9
1 5-4 288 0.25 2.7 1.7 1.1
1 6-5 * 346 0.35 2.6 1.7 1.1
1 7-6 403 0.55 3.6 2.3 1.5
2 1-0 38 0.16 Inf Inf Inf
2 2-1 77 0.16 903.7 391.8 191.4
2 3-2 * 115 0.24 403.6 175.0 86.3
2 6-5 * 231 0.20 17.2 7.5 3.7
2 7-6 269 0.23 11.8 5.1 2.5
2 8-7 301 0.26 9.9 4.3 2.1
2 9-8 346 0.35 9.9 4.3 2.1
3 1-0 29 0.16 Inf Inf Inf
3 3-2 86 0.15 921.8 287.8 115.2
3 4-3 * 115 0.24 763.0 238.2 95.4
3 5-4 144 0.16 136.4 42.6 17.1
3 7-6 202 0.20 54.6 16.9 6.8
3 8-7 230 0.20 33.2 10.3 4.1
3 9-8 259 0.22 24.2 7.5 3.0
5 2-1 38 0.16 Inf Inf Inf
5 4-3 77 0.16 Inf 606.9 183.9
5 5-4 96 0.15 Inf 207.2 63.4
5 6-5 115 0.24 Inf 271.1 83.0
5 7-6 134 0.16 347.7 64.1 19.6
5 8-7 154 0.17 215.6 39.3 12.0
* - on bob's list in Table 4 of report.
remember that if it is sufficient to resolve the line in only a few
channels, then you should divide the above times by a factor of 3 or
so (75 km/s channels would give you exactly a factor of 3). if you
simply want a "detection", then divide the above numbers by 10 or
so. this is why i include some numbers in the above table which seem
so ludicrously high - when you divide by 10 they aren't _so_ bad...
so, it appears to me that (depending on q_o) we may still be able to do
the higher transitions of CO even out to z=3 or so. but, as z
increases, you lose the ability to observe the lower transitions. is
this a critical loss? i don't know for sure, but if we want to do
"milky way's", then it probably is, since most of the galactic CO is
relatively cold (< 20 K or so?) i think. bob implicitly points out in
his analysis that as you go to higher z, you just pick different
molecules (CI, CII, NII, e.g.) rather than sticking to CO. however, it
seems to me that you start to probe different types of material in the
galaxies, so you don't get the same thing...
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