Part I) Consider a spherical cluster of galaxies with a dark matter (DM) mass of

Question

Part I) Consider a spherical cluster of galaxies with a dark matter (DM) mass of 1015 M, and a radius R of 2 Mpc.

i) What is the average mass density of DM, DM, in gm/cm3?
ii) Assume that the DM consists of weakly interacting elementary particles with a mass of mDMc2 = 100 GeV. (1 GeV = 109 eV.) What is mDM in gm?
iii) What is the average number density of DM particles in the cluster nDM (in cm-3)?

Part II) Imagine that two such identical clusters collide head-on at a relative speed of 3000 km/s. They will travel through one another, with each particle moving a total distance of twice the cluster radius, or 2R, within the other cluster. Let be the cross-sectional area for two DM particles to interact with (e.g., scatter) one another as they pass through.

iv) Give an expression for the probability that a given DM particle will interact with a particle from the other cluster as they pass through one another.

v) Observations indicate that the two DM halos passed through one another without any interaction between the two halos. Let’s say that this indicates that the probability of any individual DM particle interacting is less than 10%. Use this to derive an upper limit to the interaction cross-section (in cm2).

vi) A useful quantity to describe the DM interaction is the cross-section per unit mass, or /m. Give the upper limit on /m in cm2/gm.

vii) Show that the limit of /m is actually independent of the rest mass of the DM particle.

Explanation / Answer

i. mass, M = 10^15 Mo
Mo = 1.989*10^30 kg
R = 2 Mpc = 6.1714*10^22 m
hence
average mass density = 10^15*1.989*10^30*1000 g/(4*pi*(6.1714*10^24)^3/3) = 2.0202*10^-27 g/cm^3

ii. Md*c^2 = 100*10^9 eV = 100*10^9*1.6*10^-19 J
Md = 1.777*10^-22 g

iii. average numerb density = rho/Md = 0.000011363625 particels per cm^3

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.