 Solution

Let $M_h,R_h$ describe the halo and let $v=v_{\mathrm{vir}}$. The cosmic baryon fraction is $f_b={\Omega }_b/{\Omega }_m=0.20$, so the specified settled fraction gives $M_d=0.10M_h$. The specific angular momentum constraint at the disk edge isTo infer an actual circular speed one must specify the disk's radial mass distribution: enclosed mass does not determine a disc rotation curve. Introduce a finite geometry coefficient $\kappa$ by $v_d^2=\kappa G{M}_d/R_d$. A rounded, centrally concentrated disk permits a monopole estimate $\kappa \simeq 1$ at its outer edge; it is an explicit approximation, not the spherical shell theorem applied exactly to a razor-thin disk. Ignore the force of the unsettled baryons as well as the dark matter within the disk. Using $v^2=G{M}_h/R_h$, the angular-momentum estimate of a self-gravitating galactic disk givesAn exact disk answer cannot be fixed by total disk mass alone; the dependence on $\kappa$ records that missing input.

For the numerical virial radius of a dark-matter halo, adopt mean density ${\Delta }_c=200$ times the critical density at the formation epoch. This conventional definition givesThe supplied expansion law gives $H(3)=286.49\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$, and hence $R_h=v/(10H)=97.74\mathrm{kpc}$. Taking $\kappa =1$ yieldsThe corresponding virial mass of a dark-matter halo is $M_h\simeq 1.78\times 10^{12}{M}_{\odot }$, and $M_d\simeq 1.78\times 10^{11}{M}_{\odot }$. Other overdensity conventions give $R_d\propto {\Delta }_c^{-1/2}$ at fixed $H,v,\kappa$.

Interpret the wavelength separation as an observed-frame local mean near the redshift in question. Since Lyman-alpha absorption appears at ${\lambda }_{\mathrm{obs}}={\lambda }_{\alpha }(1+z)$, the incidence isAssume one counted absorption system per intercepted disk, no unrelated forest systems or missed absorbers, unity neutral covering fraction, and a locally slowly varying population. Let ${\sigma }_{\mathrm{abs}}$ be the proper interception cross-section and $n_c$ the comoving number density. The proper density is $n_p=n_c(1+z{)}^3$, and the proper line element along the light path is $d{l}_p=cdz/[H(z)(1+z)]$. Thus the absorber incidence and comoving number density relation isFor a thin circular disk, the projected area is $\pi R_d^2|\cos i|$. Isotropically oriented normals have $⟨|\cos i|⟩=1/2$, so the random-orientation absorbing-disk cross-section is ${\sigma }_{\mathrm{abs}}=\pi R_d^2/2$. ConsequentlyIf all disks are taken face-on instead, $n_c\simeq 0.398{\mathrm{Mpc}}^{-3}$. Generally the random-orientation result scales as ${\kappa }^2({\Delta }_c/200)$ and is divided by the neutral covering fraction. The finite mean redshift spacing is sizable, so a precision inference would integrate the incidence over the actual redshift interval rather than identify it with one local value.

There is also a halo abundance mass-budget bound on this formal result. The present mean matter density for these parameters is ${\bar{\rho }}_{m,0}={\Omega }_{m}3H_0^2/(8\pi G)\simeq 3.40\times 10^{10}{M}_{\odot }{\mathrm{Mpc}}^{-3}$. Distinct haloes of this mass cannot have $n_c{M}_h>{\bar{\rho }}_{m,0}$, even if every matter particle belonged to them. Yet the inferred random-orientation population hasEven the face-on estimate exceeds this bound by about twenty-one. The numerical incidence result is conditional; the supplied population assumptions are not cosmologically consistent under this standard virial and compact-disk estimate. A larger neutral-gas absorption radius, a different absorber population, or different physical inputs are needed. For example, random orientations would require an absorbing radius at least about $63\mathrm{kpc}$ merely to reach the all-matter upper bound, much larger than the calculated centrifugal radius. This check does not change the algebraic answer, but prevents treating it as a realizable population.