The first step is to rewrite the dynamical equation

dR / dt = [ 8 G R ^{2}/ 3 ]^{1/2}= [ 8 G_{0}/ 3 R ]^{1/2}and thus

R ^{1/2}dR = [ 8 G_{0}/ 3 ]^{1/2}dtis the relation we want. Note, however, that in a flat universe with the critical density that

H _{0}^{2}= 8 G_{0}/ 3so we can write our differential simply as

R ^{1/2}dR = H_{0}dt.The integrand becomes

c dt / R = ( c / H _{0}) R^{-1/2}dRwhich integrates to

(R) = ( 2 c / H _{0}) [ 1 - R^{1/2}]which you can check by taking the derivative. Substitution of

R = ( 1 + *z*)^{-1}gives us the expression

(R) = ( 2 c / H _{0}) [ 1 - ( 1 +*z*)^{-1/2}]we wanted.

As

*z*goes to infinity, we get the current*horizon distance*_{}= 2 c / H_{0}= 6000*h*^{-1}Mpcwhich gives 10000 Mpc for

*h*=0.6 (H_{0}= 60 km/s/Mpc).Simple substitution gives

= ( D H _{0}/ 2 c ) [ 1 - (1+*z*)^{-1/2}]^{-1}=_{}(1+*z*)^{1/2}/ [ (1+*z*)^{1/2}- 1 ]which tends toward the limiting angular diameter

_{}= ( D H_{0}/ 2 c ) = D / 6000*h*^{-1}Mpc

= 1.667 × 10^{-4}( D / 1*h*^{-1}Mpc ) rad

= 34.4" ( D / 1*h*^{-1}Mpc )and our fiducial cluster of D = 1.8

*h*^{-1}Mpc tends to the limiting diameter of 61.9 arcseconds (approximately 1 arcminute) at infinite redshift.Upon substitution we find

= ( D H _{0}/ 2 c ) (1+*z*) / [ 1 - (1+*z*)^{-1/2}] =_{}(1+*z*)^{3/2}/ [ (1+*z*)^{1/2}- 1 ]which now diverges at

*z*= 0 and infinity (try it). Thus, it must be a minimum somewhere in between. If we make the substitution a = (1+*z*)= _{}a^{3/2}/ [ a^{1/2}- 1 ]and the derivative is

d/da = _{}3/2 a^{1/2}/ [ a^{1/2}- 1 ] - 1/2 a / [ a^{1/2}- 1 ]^{2}which when set to zero, and dividing out common factors, gives

3 a ^{1/2}= a / [ a^{1/2}- 1 ]or

3 [ a ^{1/2}- 1 ] = a^{1/2}.Our solution is then

a ^{1/2}= 3/2 ---> 1 +*z*= 9/4 --->*z*= 5/4thus the angular size reaches a minimum at

*z*= 1.25.Note: you could also have computed the redshift for which the angular diameter distance d

_{A}is a maximum, which is actually more straightforward. Again, with our variable changed _{A}= ( 2 c / H_{0}) [ ( 1 +*z*)^{-1}- ( 1 +*z*)^{-3/2}] =_{}( a^{-1}- a^{-3/2})we set the derivative

d/da ( d _{A}) =_{}( -a^{-2}+ 1.5 a^{-5/2}) = a^{-5/2}_{}( 1.5 - a^{1/2})equal to zero gives

1.5 - a ^{1/2}= 0 ---> a^{1/2}= 3/2 ---> a = 9/4& nbsp; --->*z*= 5/4as before.

At the minimum a = 9/4 and thus

_{min}=_{}27/8 / [ 3/2 - 1 ] = 6.75_{}versus the

*comoving*angular diamter at infinte redshift (calculated above).Our fiducial galaxy has a size of

D = 0.1 *h*^{-1}Mpc --->_{}= 3.44"so its minimum apparent angular size is 23.2".

Using

H _{0}= 100*h*km/s/Mpc / 3.086 × 10^{19}km/Mpc = 3.240 × 10^{-18}*h*s^{-1}the current critical density is

= ( 3 H _{0}^{2}) / ( 8 G ) = 1.88 × 10^{-26}*h*^{2}kg m^{-3}or 2.76 × 10

^{11}*h*^{2}M_{sun}Mpc^{-3}. The critical energy density is thereforeu _{cr,0}= 1.88 × 10^{-26}*h*^{2}c^{2}kg m^{-3}= 1.69 × 10^{-9}*h*^{2}J m^{-3}.The radiation constant a (not the scale factor a in the last problem) has the value

a = 4 / c = 7.56 × 10 ^{-16}J m^{-3}K^{-4}which means for T

_{0}= 2.73 Ku _{rad,0}= 7.56 × 10^{-16}× 2.73^{4}= 4.20 × 10^{-14}J m^{-3}for the radiation energy density. Thus,

_{rad,0}= ( 4.20 × 10^{-14}J m^{-3}) / ( 1.69 × 10^{-9}*h*^{2}J m^{-3}) = 2.48 × 10^{-5}*h*^{-2}is the fraction of the critical density currently contributed by radiation.

Since

u _{rad}/ u_{cr}( 1 +*z*)^{4}/ ( 1 +*z*)^{3}1 +*z*then we will have matter-radiation equality at

1 + *z*~ 1 /_{eq}_{rad}= 4.03 × 10^{4}*h*^{2}and thus the redshift of equality

*z*~ 14500 for_{eq}*h*=0.6.The radial velocity dispersion reprents a 1-D projection of the true 3-D velocities of the galaxies, so

_{v}^{2}= 3_{r}^{2}as the squares of the velocities add in quadrature. For Coma

_{v}= 3^{1/2}_{r}= 1692 km/s.The virial relation 2K = -U gives

2 · (1/2) M _{v}^{2}= (3/5) G M^{2}/ R_{vir}for mass M and virial radius R

_{vir}. Thus,M _{vir}= (5/3) R_{vir}_{v}^{2}/ G = 5 R_{vir}_{r}^{2}/ Gwhich for radial velocity dispersion 977 km/s and

R _{vir}= 1.8*h*^{-1}Mpc = 5.55 × 10^{22}*h*^{-1}mgives us

M _{vir}= 3.97 × 10^{45}*h*^{-1}kg = 1.99 × 10^{15}*h*^{-1}M_{sun}for the virial mass.

We use the usual relation between luminosity and absolute magnitude

log( L/L _{sun}) = [ +4.83 - (-26.92) ]/2.5 = 12.7for the Sun's absolute visual magnitude of +4.83. Technically, though it wasn't stated in the problem, that since the absolute magnitude is determined from the apparent magnitude and luminosity distance d

_{L}(in*h*^{-1}Mpc), then the luminosity will be proportional to h^{2}, which givesL = 5.01 × 10 ^{12}*h*^{2}L_{sun}for Coma. Thus,

M / L = 397 *h*M_{sun}/L_{sun}for Coma cluster, which is significantly larger than that for the centers of spiral and elliptical galaxies ( ~10 h) and the extended halos of giant elliptical galaxies ( ~100 h). (Note: if you missed the factor of h^2 from the luminosity, thats OK, because I missed it also the first time.)

If we count up the mass associated with the light which we see, which comes from the galaxies in the cluster, then we would estimate

M _{gal}~ 10*h*M_{sun}/L_{sun}· 5.01 × 10^{12}*h*^{2}L_{sun}= 5.01 × 10^{13}*h*^{-1}M_{sun}which represents a fraction of the total virial (gravitational) mass

M _{gal}/ M_{vir}~ 10 / 397 = 0.052and thus 94.8% of the gravitational mass is unaccounted for by the galaxies, and must be

*dark matter*in the extended cluster halo!The density of the Coma cluster within the virial radius is

= ( 3 M _{vir}) / ( 4 R_{vir}^{3}) = 8.15 × 10^{13}*h*^{2}M_{sun}Mpc^{-3}= 5.54 × 10^{-24}*h*^{2}kg m^{-3}which when compared with

= 1.88 × 10 ^{-26}*h*^{2}kg m^{-3}gives

/ = 295 and indeed the Coma cluster is extremely overdense compared to the critical density, and thus should behave like a part of a

~ 295 >> 1 closed universe. Since the density

R ^{-3}---> R^{-1/3}---> R_{L}^{1/3}295^{1/3}= 6.66which for R = 1.8

*h*^{-1}Mpc gives R_{L}= 12.0*h*^{-1}Mpc. Thus, since the early Universe had a density close to the critical density (as all FRW models do), the Coma cluster has collected up all the matter from a region equivalent to a sphere of radius 12 h^-1 Mpc today!

*smyers@nrao.edu*
*Steven T. Myers*