Re: [guide-user] Redshift to distance conversion
Bill J Gray Nov 15, 2005
Hi Pat,
(Pause to do some on-line research... I'm not all that knowledgeable
about cosmology; does that mean I'm not a cosmopolitan kind of guy?)
Unfortunately, Guide can't tell you that. I ought to add in some
code to enable it to compute new numbers using those found in a
catalog... but at present, it can just take the redshift and spit that
out at you. However, there's another way to do it, as follows.
I eventually found some formulae (more on that below) which, with a
modest bit of effort, could be turned into the following table. You
can interpolate within the table as needed. For example, an object
with z=.4 is receding from us at about one-third of the speed of light,
and is about 1400 million parsecs or 4.5 billion light-years away.
(Please ignore the meaningless precision! If the Hubble Constant really
is known to within about one part in twenty, then these distances
should also be good to within one part in twenty.)
z vel distance
(% of c) Mpc Mlightyears
0.1 9.5 401 1308
0.2 18.0 761 2482
0.3 25.7 1083 3530
0.4 32.4 1369 4464
0.5 38.5 1624 5294
0.6 43.8 1850 6031
0.7 48.6 2051 6687
0.8 52.8 2230 7272
0.9 56.6 2390 7793
1.0 60.0 2533 8259
1.5 72.4 3057 9967
2.0 80.0 3377 11012
3.0 88.2 3725 12145
4.0 92.3 3897 12706
5.0 94.6 3994 13021
6.0 96.0 4053 13214
7.0 96.9 4092 13341
8.0 97.6 4119 13429
9.0 98.0 4138 13492
10.0 98.4 4153 13539
Anyway. My on-line research turned up a relationship between the
velocity v of the galaxy and its redshift z:
1 + z = sqrt( (1+v/c) / (1-v/c))
which, after a bit of algebra, can be solved for v:
k = (z+1)^2
v = c (k-1) / (k+1)
Then you can solve for the distance to the object using Hubble's Law:
D = v / H
where D = distance in megaparsecs, v = velocity in km/second, and
H = Hubble's constant, currently thought by at least some to be about
71+/-3.5 km/sec/Mpc.
-- Bill