ASTR/PHYS 4080: Introduction to Cosmology Spring 2018 Homework 4 Solutions 1. Consider a flat, single component universe. (a) For a light source at redshift z that is observed at time t0, show that z changes at a rate dz 3(1+w)=2 = H0(1 + z) − H0(1 + z) : (1) dt0 Solution Starting from the relationship between z and a, 1 + z = a(t0) , and the a(te) 1 da(t) definition of the Hubble parameter H(t) = a(t) dt , we can solve for dz=dt0 in terms of H0. First, note that as the time of observation t0 changes to t0 + dt0, we observe light that was also emitted at a later time te + dte, so both a(t0) and a(te) are changing with t0. Taking the derivative then yields dz da(t0) 1 a(t0) da(te) = − 2 (2) dt0 dt0 a(te) a (te) dt0 1 da(t ) a(t ) a(t ) 1 da(t ) dt = 0 0 − 0 e e (3) a(t0) dt0 a(te) a(te) a(te) dte dt0 a(t0) dte = H0(1 + z) − H(te) : (4) a(te) dt0 Looking at the right hand side (RHS) of Eq. 4, we see we need an ex- pression for dte=dt0. If you recall way back when we derived the equation for a(z), we used the fact that the comoving distance at te and t0 are the same, which yields Ryden Eq. 3.58. Noting that λ/c is a small interval of time, we can rewrite it as Z te+dte dt Z t0+dt0 dt = ; (5) te a(t) t0 a(t) which implies that dte=a(te) = dt0=a(t0) and the RHS becomes simply H(te). For a flat, single component universe (i.e., Ω0 = 1 and κ = 0), the Friedmann equation is 2 2 −3(1+w) 2 3(1+w) H = H0 a = H0 (1 + z) : (6) Substituting that in to Eq. 4, we arrive at Eq. 1. 1 ASTR/PHYS 4080: Introduction to Cosmology Spring 2018 (b) For what values of w does the observed redshift increase with time? Solution Due to the minus sign in Eq. 1, the sign of dz=dt0 depends on which term is larger. For positive values of dz=dt0, 3(1+w)=2 3(1+w)=2−1 H0(1 + z) > H0(1 + z) ! 1 > (1 + z) : (7) Since 1 + z > 1, the exponent must be < 0 for dz=dt0 > 0, so w > −1=3. Similarly, for the redshift to increase with time, we need w < −1=3. (c) Assuming the single component is matter and H0 = 68 km/s/Mpc, you observe a galaxy at z = 1. Using Equation 1, determine how long you will have to keep observing the galaxy in order to see its redshift change by 1 part in 106. Solution Rearranging the equation to solve for dt0 and then plugging in numbers (w = 0 for matter and dz = −10−6 because w > −1=3): dz dt0 = 3(1+w)=2−1 (8) H0(1 + z)[1 − (1 + z) ] −10−6 s Mpc 3:086 × 1019 km kyr = × × × (9) 2(1 − 21=2) 68 km Mpc π × 1010 s = 17 kyr (10) Note that this is several times longer than the lifespan of the longest continuous human civilization (China's, and that's for a relatively loose definition). Good luck keeping systematic measurement uncertainties un- der control between the fall and rise of entire civilizations! 2. Although current data are consistent with dark energy being due to a cos- mological constant w = −1, other values of w are acceptable depending on the exact nature of dark energy. One possibility is called \phantom energy," where w < −1. Suppose that the universe is spatially flat and contains matter with a density parameter Ωm;0, and phantom energy with a density parameter Ωp;0 = 1 − Ωm;0 and equation of state parameter wp < −1. (a) At what scale factor amp are the energy densities of phantom energy and matter equal? Solution 2 ASTR/PHYS 4080: Introduction to Cosmology Spring 2018 The energy densities (or equivalently density parameters) of the compo- nents will equal each other at amp: Ωm = Ωp (11) −3 −3(1+wp) Ωm;0amp = Ωp;0amp (12) −1=3wp −1=3wp Ωm;0 Ωm;0 ! amp = = (13) Ωp;0 1 − Ωm;0 (b) Write down the Friedmann equation for this universe in the limit that a amp. Integrate this equation to show that the scale factor a goes to infinity at a finite cosmic time trip, given by the relation 2 −1=2 H0(trip − t0) ≈ (1 − Ωm;0) : (14) 3 j 1 + wp j Solution The universe in the limit a amp will be dominated by phantom energy, so we can ignore other components. The Friedmann equation is then 8πG H2 = " (15) 3c2 p and dividing both sides by the critical density yields 2 2 2 da =a −3(1+wp) −3(1+wp) (H=H0) = 2 2 = Ωp;0a = (1 − Ωm;0)a : (16) H0 dt Rearranging to separate a and t terms for integration, we have Z trip Z 1 −1=2 (1+3wp)=2 ±H0 dt = (1 − Ωm;0) a da (17) t0 a(t0) with the ± resulting from the square root of both sides. Since a(t0) = 1, the right integral evaluates to 1 3(1+wp)=2 1 (1+w )=2+1 2(1 − 1) a p = (18) (1 + 3wp)=2 + 1 1 3(1 + wp) For phantom energy, w < −1 by definition, so the exponent on infinity is negative and thus that term is 0, yielding 2 −1=2 H0(trip − t0) = ± (1 − Ωm;0) : (19) 3(1 + wp) We know the rip happens in the future, so the LHS is postitve, Ωm;0 is positive and less than 1 (so the square root term is fine), and wp < −1 so (1 + wp) < 0. To complete the equation, one can either change the ± sign to a negative sign, or one can remove it and take the absolute value of the 1 + wp term (or the entire equation, but that's unnecessary). 3 ASTR/PHYS 4080: Introduction to Cosmology Spring 2018 (c) Current observations of our own universe are consistent with H0 = 68 km/s/Mpc, Ωm;0 = 0:3, and wp = −1:1. If these numbers are correct, how long do we have remaining until the Big Rip? Solution 2 s Mpc 3:086 × 1019 km kyr trip − t0 ≈ p × × × (20) 3 × 0:1 0:7 68 km Mpc π × 1016 s ≈ 115 Gyr (21) So plenty of time yet to party. 3. In the Benchmark Model, what is the total mass of all the matter within our horizon? What is the total energy of all the photons within our horizon? How many baryons are within the horizon? Solution We can estimate the amount of stuff within our horizon by considering a sphere of radius dhor(t0) = 14 Gpc (Ryden Eq. 5.115). Because the Benchmark model is flat, the volume is just 4π(14 Gpc)3 3:086 × 1025 m3 V = = 3:4 × 1080 m3 (22) 3 Gpc The values given in Ryden Table 5.2 are relative to the critical density, which is 4.87 GeV m−3 or 8:7 × 10−27 kg m−3 (Ryden Eq. 4.31-4.32), so the total 53 mass is 0:31ρc;0V = 9:2 × 10 kg . The total energy in photons can be ap- 76 proximated by the total energy in the CMB: Ωγ;0"c;0V = 8:9 × 10 GeV = 67 Ωbary;0 1:4 × 10 J . The mass of baryons is similarly calculated as ρm;0 = Ωm;0 53 0:048 53 (9:2 × 10 kg) 0:31 = 1:4 × 10 kg , but the number depends on mass. Bary- onic mass is tied up in protons and neutrons, which have similar masses, and protons outnumber neutrons (since the number of hydrogen atoms exceeds the number of helium atoms), so it's not unreasonable to assume all baryons have the mass of a proton. Therefore, the number of baryons within the horizon is 1:4 × 1053 kg=1:67 × 10−27 kg = 8:5 × 1079 . 4.