Homework 1 Solutions

ASTR/PHYS 4080: Introduction to Cosmology Spring 2019

1. Cosmic strings are hypothetical ∼1-dimensional topological defects in space- time that may have formed in the early universe. In the novel Tides of Light by Gregory Benford (also a physicist), an advanced alien civilization uses a cosmic string to bore a hole through the center of a planet to mine heavy metals in its core. The main character is tossed down the shaft through the planet and calculates how long the journey will take him, which is given as this equation in the 1989 paperback edition: " # π v 3 time = − tan−1 , (1) 2 q 4π 4πGρ R 3 Gρ

where v is his initial velocity, R is the radius of the planet, G is the gravitational constant, and ρ is the planet’s average density. Use this equation (and common sense reasoning) to estimate the length of the journey for an astronaut falling through an Earth-like planet.

Solution

Common sense immediately tells us there’s a problem, as the units of 1/Gρ is 2 √v not seconds, but s . The expression 4π must be unitless (since it’s inside R 3 Gρ −1 −1 q 4π the tan () function), and since v/R has units of t , the term 3 Gρ must also have inverse time units. The inverse square of that is outside the brack- eted expression, giving units of t2. Since this equation is found in a paperback science ﬁction novel, the most likely mistake is that the editor accidentally dropped a 1/2 power around the expression outside of the brackets.

To check this, we can attempt to rederive the equation. The equation of mo- GM 3 tion is mg = ma, where g = r2 and M = ρV = 4πr ρ/3. Putting that 4π together and recognizing a =r ¨, we getr ¨ − 3 Gρr = 0, which is the equa- q 4π tion of a harmonic oscillator of frequency ω = 3 Gρ. The solution is of the form r = R cos(ωt − φ). From here, with some manipulation and taking the derivative of r, the equation can be derived, but it should be clear enough that indeed, it’s simply that the square root is missing.

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If the character takes the equation at face value and lazily plugs and chugs and assumes his answer will be in seconds, given an Earth-like average density of ∼ 5500 kg m−3, he would estimate his trip to take over 200 hours (assuming v is small—it would have to be dangerously large to matter). Neglecting v, the π time elapsed is 2ω , which from the harmonic oscillator equation is obviously one-fourth of the total cycle, or the time it takes to reach the center of the planet, so to calculate how long it takes to get to the other side, we need to π double the number given by the equation, so ∆t = ω .

Considering the high temperature at the centers of planets, his miscalculation would lead to a prediction of an ordeal that would not only be longer, but also quite worse, than James Franco’s in that movie about the idiot who goes hiking alone without telling anyone. Based on this calculation, the character may have chosen to commit a more painless suicide rather than slowly bake to death, when the correct form of the equation would have told him to just wait 4π −8 −1 3 −2 out the ∆t ∼ π/ω ∼ 42 min trip (where 1/ω = ( 3 · 6.67 × 10 g cm s · 5.51 g cm−3)−1/2 ∼ 800 s).

The equation assumes the planet’s density is constant, which is generally not true and therefore the equation will not provide an exactly correct answer. However, the density of the Earth only varies by a factor of a few: (3 − 13) × 103 kg m−3, and the variation with density goes as ρ1/2, so corrections should be on the order of (131/2 − 31/2)/6 or ∼30%, which pales in comparison to the error invoked by a missing 1/2 power.

2. Suppose you are in an inﬁnitely large, inﬁnitely old universe in which the 9 −3 average number density of stars is n∗ = 10 Mpc and the average stellar 8 radius is equal to the Suns radius of R∗ = R = 7 × 10 m. (a) How far, on average, could you see in any direction before your line of sight struck a star? (Assume standard Euclidean geometry holds true in this universe.) Put your answer in units of Mpc.

Solution

To decide how far, on average, one can see in a universe ﬁlled with spher- ical objects of radius R, it is simplest to think of a long cylinder along the

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line of sight. If an object is closer than R to the line of sight, then the line of sight intersects the object’s surface.

For a distance l, the cylindrical volume which would contain such objects is πR2ln, where n is the density of objects. Therefore, the volume that will, on average, contain one object is deﬁned by πR2ln = 1, and the average distance to which one can see is given by: 1 l = . (2) πR2n You have probably seen a similar equation for the mean-free-path between collisions of, for example, atoms in a gas. Another way to derive equation 2 is to calculate the probability of each mean-free-path when the probability of collision in each dl along the line of sight is dlπR2. Calculating the average value of l using the probability distribution yields equation 2. Given a stellar density of n = 109 Mpc−3 and converting the stellar radius of R = 7 × 108 m to Mpc the average distance is:

1 17 l = 8 = 6.2 × 10 Mpc π( 7×10 m )2(109Mpc−3) 3.086×1022 m Mpc This distance is much larger than the distance light has traveled since the Big Bang, ∼ 6000 Mpc, which is a large part of the reason why the night sky is dark.

−3 (b) If the stars are clumped into galaxies with a number density ng = 1 Mpc and average radius Rg = 2000 pc, how far, on average, could you see in any direction before your line of sight hit a galaxy? Put your answer in units of Mpc. Solution If galaxies have an average density of 1 Mpc−3 and an average radius of R = 2000 pc= 2.0 × 10−3 Mpc, then the average distance to a galaxy along the line of sight is:

1 4 l = π(2.0×10−3Mpc)2(1Mpc−3) = 8.0 × 10 Mpc Therefore, we might expect to see a galaxy everywhere on the sky. How- ever, this distance is large enough that the expansion and geometry of the universe must be taken into account.

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3. A hypothesis once used to explain the Hubble relation is the tired light hypothesis. The tired light hypothesis states that the universe is not expanding, but that photons simply lose energy as they move through space (by some unexplained means), with the energy loss per unit distance being given by the law dE = −KE, (3) dr where K is a constant.

(a) Show that this hypothesis gives a distance-redshift relation that is linear in the limit z 1.

Solution

If photons lose energy according to equation 3, then

dE E = −Kdr.

Integrating the left side of the equation from E0 to E(r), and the right side from 0 to r yields

−Kr E(r) = E0e The energy and frequency of a photon are related by Planck’s constant, E = hν (or E = hf in the textbook’s nomenclature), so the relation between energy and wavelength is given by:

hc E = λ Therefore, in the “tired light hypothesis” the wavelength of light increases Kr with distance as λ(r) = λ0e . This equation implies a relation between the redshift, z, and the distance traveled: λ − λ λ(r) − λ z = obs em = 0 = eKr − 1 (4) λem λ0

In the limit Kr << 1 (hence, z << 1), eKr approximates as 1 + Kr + ... and equation 4 reduces to z = Kr to ﬁrst order.

(b) What must the value of K be in order to yield a Hubble constant of −1 −1 H0 = 70 km s Mpc ? Solution

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H0 Since z = c r (see Ryden eq. 2.8), the “tired light hypothesis” requires K = H0 = 1 Mpc−1, where I have used the value for c from page 17 c 4380 H0 of Ryden. More explicitly, K = 2.3 × 10−4 Mpc−1.

4. In a Steady State Universe, H0 is predicted to be constant with time and matter must be created at a constant rate to maintain a constant density. In this model, matter must be created in the solar system at the same constant rate. The solar system is virialized (gravitationally bound) and does not follow cosmological expansion. Instead, whatever is created in the solar system stays in the solar system and contributes to the mass of the sun. For a Hubble −1 −1 constant of H0 = 70 km s Mpc , assume that all matter created in a sphere with radius 1000 AU was used to create the sun. How many years does it take to generate enough material in this steady state universe to form the sun?

Solution

From the Ryden textbook (equation 2.23),

dM M˙ = = 3ρ H V (5) SS dt 0 0

−27 kg where ρ0 = 3×10 cm3 (the critical density). We want a volume corresponding to R = 1000 AU = 1.496 × 1014 m radius to account for the mass of the sun, 30 M = 1.98 × 10 kg. So the time it takes to form the sun is the mass divided dM by dt M 4π t = = M (3ρ H R3)−1 (6) dM 0 0 3 dt Using values for the solar mass, critical density, Hubble constant, and approx- imate radius used to create the sun, t = 2.2 × 1014 Gyr. This is a very long time compared to the age of the sun, indicating that structure formation is very diﬃcult to explain with a steady state universe.