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ASTR 3740 Relativity & Cosmology Spring 2000. Problem Set 7. Due Fri 5 May

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1. Wm from the Coma Cluster

Click on the picture for a PostScript version This is the famous ``Slice of the Universe'' picture (de Lapparent, Geller & Huchra 1986, ApJ Letters 302, L1). It is a redshift map of galaxies brighter than apparent magnitude 15.5 in a slice 6 thick passing through the north galactic pole. The Milky Way is at the vertex of the wedge. The prominent `finger of god' at 13 hours Right Ascension centered about 7,000 km/s away is the Coma cluster.

Clicking on the picture will give you a PostScript version of it; but be warned, the resolution is lower than the image handed out in class. You will be better off using the image handed out, or else photocopying it from the Astrophysical Journal.

(a) Velocity dispersion, radius

From the diagram measure approximately (i) the velocity dispersion v of the Coma cluster in km/s, (ii) the radius (not distance!) H0 r of the Coma cluster in km/s, and hence determine (iii) the ratio v / ( H0 r ). Explain what you did. [Hint: The velocity dispersion is the root mean square orbital velocity of galaxies in Coma, relative to the center of Coma. You can estimate the velocity dispersion from the length of the finger-of-god. You can estimate the radius from the transverse size of the finger-of-god. Question: How far out should I measure? Answer: We are interested in the collapsed part of the Coma cluster, so that we can apply the virial theorem. Clearly the inner parts of the Coma cluster have a higher velocity dispersion than the outer parts, so your answer depends on where exactly you choose to do your measurement. However, it turns out that the final answer we are looking for, Wm in part (d), is (or should be) insensitive to where you choose to cut the cluster, as long as you are consistent all the way through.]

(b) Density of Coma relative to the mean mass density

(i) Estimate the number density nComa, in galaxies (km/s)-3, of galaxies in the Coma cluster

nComa =

N
4
3
p (H0 r)3
(1.1)
and hence determine (ii) the ratio nComa/n of the densities of Coma to the mean density n of galaxies in the slice. [Hint: Start by making an approximate count of the number N of galaxies in the Coma cluster (the number of galaxies within the radius you chose in part (a)). Then use the radius H0r from part (a) to determine the density nComa. Notice that the formula for nComa quoted above assumes that the Coma cluster is spherical. The effective volume of the survey is about 1.2×1011 (km/s)3 (this comes from the fact that the slice is 117.5 wide by 6 thick, and it has an effective depth of 12,000 km/s), so the mean density of galaxies in the slice is
n = 1061   galaxies
1.2 ×1011  (km/s)3
 .
(1.2)

(c) Density of Coma relative to the critical density

The velocity dispersion v of Coma is related to its mass M and radius r by (this is the virial theorem)

v2 = G M
r
 .
(1.3)
The density rComa is related to its mass and radius by the usual formula for a spherical object
rComa =

M
4
3
p r3
 .
(1.4)
Use these formulae, along with the definition of critical density rc
H02 = 8
3
p G rc
(1.5)
to derive an expression for the ratio rComa/rc of the density of Coma to the critical density in terms of the ratio v/(H0r) that you measured in part (a).

(d) Omega in matter

Assume that mass densities are in proportion to galaxy number densities, rComa/r=nComa/n. Use your results from parts (a)-(c) to infer Wm from the defining formula

Wm = r
rc
 .
(1.6)

2. Relation between Horizon and Flatness Problems

Show that Friedmann's equation can be written in the form (compare Problem Set 5, Question 3a)

W - 1 = k xH2
(2.1)
where xHc/(aH) is the comoving Hubble distance. Use this equation to argue in your own words how the horizon and flatness problems are related. [The main part of this question is not the math but the explanation. You should convince the grader that you understand physically what is going on.]


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On 3 May 2000, 13:54.