Particle Physics And Cosmology Problems and Solutions 4

 Problem#1

A Ξ^–1 particle at rest decays to a Λ⁰ and a π^–1 . (a) Find the total kinetic energy of the decay products. (b) What fraction of the energy is carried off by each particle? (Use relativistic expressions for momentum and energy.)

Answer:
Given: m_Ξc² = 1321 MeV, m_Λc² = 1116 MeV and m_πc² = 139.6 MeV.

does the decay Ξ^–1 → Λ⁰ + π^–1, p_Λ = p_π = p; E_Ξ = E_Λ + E_π;

(a) the total kinetic energy of the decay products is

K = m_Ξc² – m_πc² – m_Λc² = 1321 MeV – 139.6 MeV – 1116 MeV = 65.4 MeV

(b) we use

E_Ξ = E_Λ + E_π

E_Ξ = (E_Λ² + p²c²)^1/2 + (E_π² + p²c²)^1/2

E_Ξ – (E_Λ² + p²c²)^1/2 = (E_π² + p²c²)^1/2

E_Ξ² – 2E_ΞE_Λ + p²c² + E_Λ² = E_π² + p²c²

m_Ξ²c⁴ – 2m_Ξc²E_Λ + p²c² + m_Λ²c⁴ = m_π²c⁴ + p²c²

E_Λ = (m_Ξ²c⁴ + m_Λ²c⁴ – m_π²c⁴)/2m_Ξc²

K_Λ = [(m_Ξ²c⁴ + m_Λ²c⁴ – m_π²c⁴)/2m_Ξc²] – m_Λc²

K_Λ = [(1321 MeV)² + (1116 MeV)² – (139.6 MeV)²)/(2 x 1321 MeV] – 1116 MeV

K_Λ = 8.5 MeV

So fraction of the energy is carried off by each particle is

8.5 MeV/65.4 MeV = 13.0%

and
E_π = E_Ξ – E_Λ

E_π = (m_Ξ²c⁴ – m_Λ²c⁴ + m_π²c⁴)/2m_Ξc²

So that,

K_π = [(m_Ξ²c⁴ – m_Λ²c⁴ + m_π²c⁴)/2m_Ξc²] – m_πc²

K_π = [(1321 MeV)² – (1116 MeV)² + (139.6 MeV)²)/(2 x 1321 MeV] – 139.6 MeV

K_π = 56.9 MeV

So fraction of the energy is carried off by each particle is

56.9 MeV/65.4 MeV = 87.0%

Problem#57
Consider the spherical balloon model of a twodimensional expanding universe (see Fig. 44.17 in Section 44.6). The shortest distance between two points on the surface, measured along the surface, is the arc length r, where r = Rθ. As the balloon expands, its radius R increases, but the angle θ between the two points remains constant. (a) Explain why, at any given time (dR/dt)/R, is the same for all points on the balloon. (b) Show that v = dr/dt is directly proportional to r at any instant. (c) From your answer to part (b), what is the expression for the Hubble constant H₀ in terms of R and dR/dt. (d) The expression for H₀ you found in part (c) is constant in space. How would R have to depend on time for H₀ to be constant in time? (e) Is your answer to part (d) consistent with the observed rate of expansion of the universe?

Answer:
(a) For this model, dR/dt = HR, so

(dR/dt)/R = H

Presumed to be the same for all points on the surface.

(b) For constant θ, dr/dt = θdR/dt = HRθ = Hr

(c) see part (a) H₀ = (dR/dt)/R

(d) The equation (dR/dt) = H₀R is a differential equation, the solution to which, for constant H₀, is
R(t) = R₀e^Hot

where R₀ is the value of R at 0. t = This equation may be solved by separation of variables, as
(dR/dt)/R = dln(R)/dt = H₀ and integrating both sides with respect to time.

Problem#3
Suppose all the conditions are the same as in Problem 2, except that v = dr/dt is constant for a given θ, rather than H₀ being constant in time. Show that the Hubble constant is H₀ = 1/t and, hence, that the current value is 1/T where T is the age of the universe.

Answer:
We know that

r = Rθ or R = r/θ, then

dR/dt = θ^-1dr/dt – rθ^-2dθ/dt ≈ θ^-1dr/dt. So,

(dR/dt)/R = (Rθ)^-1dr/dt = r^-1dr/dt

v = dr/dt = r(dR/dt)/R = H₀R

now, dv/dθ = 0 = d[r(dR/dt)/R]/dθ

dv/dθ = 0 = d[θ(dR/dt)]/dθ

θ(dR/dt) = K where K is a constant.

dR/dt = K/θ

R = Kt/θ

Since dθ/dt = 0

H₀ = (dR/dt)/R = (θ/Kt)(K/θ)

H₀ = 1/t

So the current value of the Hubble constant is 1/T where T is the present age of the universe.  

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