Problem#1
Cosmic Jerk. The densities of ordinary matter and dark matter have decreased as the universe has expanded, since the same amount of mass occupies an ever-increasing volume. Yet observations suggest that the density of dark energy has remained constant over the entire history of the universe. (a) Explain why the expansion of the universe actually slowed down in its early history but is speeding up today. “Jerk” is the term for a change in acceleration, so the change in cosmic expansion from slowing down to speeding up is called cosmic jerk. (b) Calculations show that the change in acceleration took place when the combined density of matter of all kinds was equal to twice the density of dark energy. Compared to today’s value of the scale factor, what was the scale factor at that time? (c) We see the galaxy clusters in Figs. 1a and 1b as they were 300 million years ago and 10.2 billion years ago. Was the expansion of the universe slowing down or speeding up at these times? (Hint: See the caption for Fig. 1b.)
Answer:
(a) When the matter density was large enough compared to the dark energy density, the slowing due to gravitational attraction would have dominated over the cosmic repulsion due to dark energy.
(b) Matter density is proportional to 1/R3, so R ∝ 1/ρ1/3.
Therefore
R/R0 = (ρnow/ρpast)1/3
If ρm and ρDE are the present-day densities of matter of all kinds and of dark energy, we have ρDE = 0.726ρcrit and ρm = 0.274ρcrit at the present time. Putting this into the above equation for R/R0 gives
R/R0 = {0.274ρDE/(2 x 0.726ρDE)}1/3 = 0.574
Problem#2
The K0 meson has rest energy 497.7 MeV. A K0 meson moving in the +x-direction with kinetic energy 225 MeV decays into a π+ and a π–1, which move off at equal angles above and below the +x-axis. Calculate the kinetic energy of the π+ and the angle it makes with the +x-axis. Use relativistic expressions for energy and momentum.
Answer:
Given: │pπy+│ = │pπy-│; pπ+ sinθ = pπ- sinθ and pπ+ = pπ- = pπ; mKc2 = 497.7 MeV; mπc2 = 139.6 MeV.
Conservation of momentum for the decay gives pK = 2pπx and pK2 = 4pπx2.
EK = 497.7 MeV + 225 MeV = 722.7 MeV, so
(pKc)2 = EK2 – mK2c4 = (722.7 MeV)2 – (497.7 MeV)2 = 2.746 x 105 (MeV)2 and
(pπxc)2 = 2.746 x 105 (MeV)2/4 = 6.865 x 104 (MeV)2.
Conservation of energy says EK = 2Eπ, then
Eπ = EK/2 = 722.7 MeV/2 = 361.4 MeV.
Kπ = Eπ – mπc2 = 361.4 MeV – 139.6 MeV = 222 MeV
(pπc)2 = Eπ2 – (mπc2)2 = (361.4 MeV)2 – (139.6 MeV)2 = 1.11 x 105 (MeV)2
The angle θ that the velocity of the π+ particle makes with the -axis +x is given by
cosθ = {(pπxc)2/(pπc)2}1/2 = {6.865 x 104/1.11 x 105}1/2
cosθ = 0.78643
θ = cos-1(0.78643) = 38.150
Problem#3
A ∑–1 particle moving in the +x-direction with kinetic energy 180 MeV decays into a π–1 and a neutron. The π–1 moves in the +y-direction. What is the kinetic energy of the neutron, and what is the direction of its velocity? Use relativistic expressions for energy and momentum.
Answer:
Given: K∑ = 180 MeV; m∑c2 = 1197 MeV; mnc2 = 939.6 MeV; mπc2 = 139.6 MeV.
E∑ = K∑ + m∑c2 = 180 MeV + 1197 MeV = 1377 MeV
Conservation of the x-component of momentum gives
p∑ = pnx
then,
(pnxc)2 = (p∑c)2 = E∑2 – (m∑c2)2 = (1377 MeV)2 – (1197 MeV)2 = 4.633 x 105 (MeV)2
E∑ – (mn2c4 + pn2c2)1/2 = (mπ2c4 + pπ2c2)1/2
Square both sides:
E∑2 + mn2c4 + pnx2c2 + pny2c2 – 2E∑En = mπ2c4 + pπ2c2, where pπ = pny, so
E∑2 + mn2c4 + pnx2c2 – 2E∑En = mπ2c4
En = {E∑2 + mn2c4 + pnx2c2 – mπ2c4}/2E∑
En = {(1377 MeV)2 + (939.6 MeV)2 + 4.633 x 105 (MeV)2 – (139.6 MeV)2}/[2(1377 MeV)]
En = 1170 MeV
Kn = En – mnc2 = 1170 MeV – 939.6 MeV = 230 MeV
Eπ = E∑ – En = 1377 MeV – 1170 MeV = 207 MeV
Kπ = Eπ – mπc2 = 207 MeV – 139.6 MeV = 67 MeV
(pnc)2 = En2 – mnc2 = (1170 MeV)2 – (939.6 MeV)2 = 4.861 x 105 (MeV)2
The angle θ the velocity of the neutron makes with the -axis +x is given by
cosθ = pnx/pn = {4.633 x 105/4.861 x 105}1/2 = 0.9763
θ = cos-1(0.9763) = 12.50 below the +x-axis.
Problem#4
Consider a collision in which a stationary particle with mass is bombarded by a particle with mass m speed v0 and total energy (including rest energy) Em. (a) Use the Lorentz transformation to write the velocities vm and vM of particles m and M in terms of the speed vcm of the center of momentum. (b) Use the fact that the total momentum in the center-of-momentum frame is zero to obtain an expression for vcm in terms of and m, M and v0. (c) Combine the results of parts (a) and (b) to obtain Eq. (44.9) for the total energy in the center-of-momentum frame.
Answer:
(a) For mass m, u = –vm, v’ = v0, and vm = (v0 – vm)/(1 – v0vcm/c2). For mass M, u = –vcm, v’ = 0, so vM = –vcm.
(b) The condition for no net momentum in the center of mass frame is
mγmvm + MγMvM = 0
where γm and γM correspond to the velocities found in part (a). The algebra reduces to
βmγm = (β0 – β’)γ0γM
where β0 = v0/c; β’ = vcm/c and the condition for no net momentum becomes
m(β0 – β’)γ0γM = Mβ’γM or
β’ = β0/[1 + M/mγ0] = β0mvcm/[m + M(1 – β02)1/2]
β’ = mv0/{m + M[1 – (v0/c)2]1/2}
(c) Substitution of the above expression into the expressions for the velocities found in part (a) gives the relatively simple forms
vm = v0γ0M/(m + Mγ0) and vM = –v0γ0m/(M + mγ0)
After some more algebra,
γm = (m + Mγ)/(m2 + M2 + 2mMγ0)1/2 and
γM = (M + mγ)/(m2 + M2 + 2mMγ0)1/2 from which
mγm + MγM = (m2 + M2 + 2mMγ0)1/2
This last expression, multiplied by c2, is the available energy Ea in the center of mass frame, so that
Ea2 = (m2 + M2 + 2mMγ0)c4
Ea2 = m2c4+ M2c4+ 2mMγ0c4
Ea2 = (mc2)2+ (Mc2)2+ 2Mc2Em
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