Homework 3 Solutions

ASTR/PHYS 4080: Introduction to Cosmology Spring 2019

1. Suppose the energy density of the cosmological constant is comparable to the −3 present critical density εΛ = εcrit,0 ∼ 5 GeV m . What is the total energy of the cosmological constant within a sphere of 1 AU radius? Compare this value to the rest energy of the Sun. Should the cosmological constant have a significant effect on the motion of planets within the solar system given the energies at play? Given the rest mass of the Milky Way galaxy (a hundred billion stars, although 90% of the total mass is in the form of dark matter), how far away would another galaxy need to be for the influence of the cosmological constant and the Milky Way’s gravity to be comparable?

Solution

−3 If the energy density of the cosmological constant was εΛ = εc,0 = 5 GeV m then a sphere of radius 1 AU would have a total energy of:

3 −3 4π 1.5×1011m 34 EΛ = εΛV = 5 GeV m × 3 × 1 AU × 1 AU = 7 × 10 GeV

Furthermore, the rest energy of the sun is:

2 30 8 2 47 57 E = M c = (2 × 10 kg) (3 × 10 m/s) = 1.8 × 10 J ≈ 10 GeV

Therefore, the cosmological constant, even if it accounted for the entire energy density of the universe, is insigniﬁcant in our solar system (the local gravity of our sun dominates).

Using the same logic for the Milky Way, now we want to ﬁnd the volume in 11 total mass which the energy density εMW = 10 E × (10 baryonic mass )/V is equal to EΛ. Solving for the radius R of volume V , we have

1/3 3 12 1/3 3 69 m3 1 Mpc R = 4π × 10 E /εΛ = 4π × 10 GeV × 5 GeV × 3.086×1022 m

R ≈ 1 Mpc.

Dark energy, at the current epoch, becomes important for isolated galaxies on Mpc scales assuming its energy density is comparable to the critical density (it is: ∼0.7Ωcrit,0). Massive galaxies like the Milky Way, without nearby massive neighbors like M31, are destined to die alone and friendless.

1 ASTR/PHYS 4080: Introduction to Cosmology Spring 2019

2. Consider Einsteins static universe, in which the attractive force of the matter density is exactly balanced by the repulsive force of the cosmological constant (Λ = 4πGρ). Suppose that some matter is converted into radiation (by stars for instance, since that’s precisely what they do!). Will the universe start to expand or contract? Explain your answer.

Solution

The acceleration of the universe is governed by Ryden Eq. 4.68: a¨ 4πG Λ = − (ε + 3P ) + (1) a 3c2 3 Initially only a density ρ of non-relativistic matter is present. Non-relativistic matter has P = 0, so a cosmological constant with Λ = 4πGρ produces zero acceleration. If some of the non-relativistic matter is converted to photons by, for exmaple, stars or decaying particles, then by energy conservation the energy density in non-relativistic matter, εnr, and relativistic matter, εr, must still equal ρc2. From the equation of state for relativistic matter, e.g. Ryden 1 equation 4.61, Pr = 3 εr. Therefore, equation 1 becomes: a¨ 4πG ε Λ 4πG 4πGρ = − [ε − ε ] + ε + 3 r + = − ρc2 + ε + (2) a 3c2 nr r r 3 3 3c2 r 3

a¨ 4πGεr =⇒ a = − 3c2

Here [εnr − εr] is the remnant energy density of non-relativistic matter at the moment we’re considering because of energy conservation. Therefore, the universe would start to collapse because of the pressure associated with the radiation. The fact that increased pressure initiates collapse is counter-intuitive, but is consistent with pressure contributing to gravitational acceleration in General Relativity.

2 ASTR/PHYS 4080: Introduction to Cosmology Spring 2019

3. Again consider Einstein’s static universe. If ρ = 2.7 × 10−27 kg m−3, what is the radius of curvature R0? How long would it take such a photon to circumnavigate such a universe?

Solution

Following Eq. 1 witha ¨ = 0 to have a static universe and ignoring the con- tribution of radiation (so ε = ρc2 and P = 0), Λ = 4πGρ. The Friedmann equation (Ryden Eq. 4.66) with κ = +1 (sincea ˙ = 0) can be solved for R0 (Ryden Eq. 4.73):

8 1 Gpc c (3 × 10 m/s) 3.086×1025 m R0 = √ = q (3) 4πGρ 4π(6.67 × 10−11 kg−1 m3 s−2)(2.7 × 10−27 kg m−3)

The curvature of the universe R0 is then 6.5 Gpc. To ﬁnd the time it would take a photon to circumnavigate this universe, we can use the metric (Ry- den Eq. 3.31) and read the following text to see that C = 2πR0. The travel time is then just

C 3.086 × 1025 m 1 1 yr 1 Gyr t = = 2π(6.5 Gpc) × × × , c 1 Gpc 3 × 108 m s−1 π × 107 s 109 yr (4) which plug-and-chugs to ≈130 Gyr.

3 ASTR/PHYS 4080: Introduction to Cosmology Spring 2019

4. The principle of wave-particle duality tells us that a particle with momentum p has an associated de Broglie wavelength of λ = h/p; just like with light, this wavelength is tied to the expansion of space: λ ∝ a. The total energy density of a gas of particles can be written as ε = nE, where n is the number density of particles and E is the energy per particle. For simplicity, lets assume that all the gas particles have the same mass m and momentum p. The energy per particle then simply follows the relation:

E2 = (m2c4 + p2c2) = (m2c4 + h2c2/λ2) (5)

Compute the equation-of-state parameter w for this gas as a function of the scale factor a. Show that w = 1/3 in the highly relativistic limit (a → 0, p → ∞) and that w = 0 in the highly non-relativistic limit (a → ∞, p → 0).

Solution

The intended, more straightforward solution (but not the only possible one) starts from the ﬂuid equation (Ryden Eq. 4.44), which allows us to solve ε in terms of a, and ε is a function of E and a and P = wε, so we can then solve −3 −3 for w. The particle number density n = N/V ∝ a , or n = n0a , and since −1 −1 p ∝ a or similarly p = p0a and ε = nE, it is possible with Eq. 5 to ﬁnd ε˙. (Recall that the subscript 0 indicates the value at the current time t0, when a ≡ 1.) 2 4 2 −2 2 1/2 −1 3 E = (m c + p0a c ) = E(a) = ε/n = εn0 a (6) So taking the time derivative,

−4 −3 ˙ ε˙ = n0a˙(−3a E(a) + a E(a)) , (7)

˙ 1 1 2 2 −3 ˙ where E(a) = 2 E(a) p0c (−2a ). Substitution of E(a) into Eq. 7 yields

ε p2c2a−3 a˙ p2c2 ε˙ = a˙ −3a−1E(a) − 0 = −ε 3 + 0 (8) E(a) E(a) a a2E2(a)

a The ﬂuid equation says P = −ε˙ 3a ˙ − ε = wε, so our expression for w in terms of a is 2 2 2 2 ε˙ a p0c p0c w = − + 1 = − − 1 + 2 2 + 1 = 2 2 4 2 −2 2 . ε 3˙a 3a E (a) 3a (m c + p0a c ) (9) 2 2 p0c 1 As a → 0, w → 2 2 2 −2 = , returning the relativistic equation of state. 3a p0c a 3 2 2 p0c For a → ∞, w → 3a2m2c4 → 0, returning the expected value of w for the non-relativistic case.