Astronomy Education and the Shape of the Universe

In science, half the battle is being able to communicate and share ideas and discoveries. It is an important skill to have with many fun methods of communication in the digital era. For my educational project, I decided to explain cosmological geometries, one of the only completely new subjects to me in astronomy 16 and 17 and, in my opinion, one of the most interesting. The video can be found at https://www.youtube.com/watch?v=Y1aODOgNXCU or watched below.

Youtube description: A basic cosmological introduction to the various theories of the shape of the universe. More explanations for the geometry of circles in different universes can be found here (http://ay16-dfrostig.blogspot.com/2015/11/geometry-of-universe-blog-31-worksheet.html) and more information on Friedmann Equations can be found here (http://ay16-dfrostig.blogspot.com/2015/11/more-friedmann-equations-blog-29.html).

Sources include http://csep10.phys.utk.edu/astr162/lect/cosmology/geometry.html, http://www.astro.ucla.edu/~wright/cosmo_03.htm, and Harvard's Astronomy 17 class worksheets.

Modeling the Entire Universe: Blog 37 - The Last Blog

From the very first week of Astronomy 16 to now, from estimating the earth's rotation to modeling the geometry of the universe, we have made it to the last homework blog post of the introductory astronomy sequence. And what better way to wrap up our exploration of all things big and small in the universe than to model the entire universe? 

This week in astronomy, we explored how small density changes in the early universe can determine the overall structure of our universe today. As we have seen in the blog posts, these ideas take a lot of math, and we have barely scratched the surface of all the math that can be done to guess at the entirety of the universe, most of which we cannot see. So what better way to tackle such a large problem than with powerful computers modeling the universe?

The Illustris Simulation models the dark matter, baryonic matter, and dark energy to show the evolution of the universe, or at least a part of it that fits into a 75 Mpc/h box. The simulation goes from a redshift of 127 (the early universe) to a redshift of zero (present day).

The simulation starts off with a view of the dark matter density of the universe, which is not evenly distributed. Not coincidentally, the dark matter demonstrates a similar pattern to to the dense clumps and connecting filaments associated with galaxy clusters. Using the "Spatial Query on Click" feature, we can explore a 400 kpc/h radius of the simulation and return information on up to 20 subhalos. 

Dark Matter Density of a 75 Mpc/h box

Analyzing a random selection of subhalos in an over dense region, we can see that low mass halos are more frequent than high mass halos. Additionally, on average, 9% of the mass of these halos is stellar mass, meaning that this mass is significantly composed of something other than visible matter.

A histogram plotting the frequency of halo masses in log(M) with bins of 2.5.

A histogram plotting the frequency of halo masses in log(M) with bins of width 0.5.

Beyond these snapshots in time, we can also explore our universe over time. This video speeds up the formation of the universe in both dark matter and gas temperature. Just from this one short video, we can learn a lot of about our early universe and its evolution.

The simulation starts off rather "dark" with the universe being dominated by dark matter until about z = 9 (about half a billion years after the big bang), when the very first hints of gas appear. This early period is called the "Dark Ages," when gas is neutral. The gas goes from blue to having some green around z=6 (less than a billion years after the big bang), symbolic of the "Epoch of Reionization," the official end of the Dark Ages when gas becomes increasingly ionized. This seems to be around the same time the first stars begin to form.

The early universe (z \( \approx\) 6). Dark matter is prominent and spread out while there is little prominence of hot gas.

Stars then begin to form rapidly, with growth even accelerating at times. The most rapid star growth then occurs around redshifts of 2.3 to about 0.6. Around this time we also being to see large bursts of energy in the gas temperature simulation.

The universe closer to today (z \( \approx\) 1). Dark matter has concentrated into dramatic over dense regions and star formation is occurring. 

Clear structures form throughout the simulation with smaller structures combining to form larger ones, rather than large objects breaking up. This hints at a trend of increasing gravity and decreasing gas pressure over time. With this increase in gravity, we seem the formation of over dense regions of dark matter and hot gas connected by filaments. As we explored in previous blog posts, this distribution is not uniform due to inconsistencies in pressure in the early universe that have been magnified with the passage of time. Theses dense regions collapse in on themselves due to the high levels of gravity found in dark matter.

The universe today (z=0) with many dense regions of dark matter and hot gasses.

These structures are not only found on a large scale, but are also surprisingly similar to a simulations at smaller scales. On both scales, dark matter and gas density tends to be clustered into regions connected by filaments. The dark matter is much more closely limited to these filaments. This follows along with our earlier observation that ionized gas forms as a consequence of dark matter. However, the gas is at a higher energy and is less massive, less confined by gravity, and therefore less limited to the filaments. On the smaller scales, all types of matter seem to be less limited to these filaments, most likely due to the smaller scales of mass and gravity.

Dark Matter Density                                  Gas Density        

Most of the medium or large galaxies are found in clusters and not in the field and within those galaxies, gas is densest towards the middle of the galaxies. This lines up exactly with what we learned at the beginning of the semester, when we explored the structure of the Milky Way. And within one of these over dense regions of the universe lies a galaxy cluster containing a very familiar spiral galaxy, with an average-sized star being orbited by an averaged-sized planet filled with astronomers looking up and out at our universe today.

Sources:
http://www.illustris-project.org/explorer/
http://earthsky.org/space/dark-matter-hairs-filaments-streams-gary-prezeau

Large Scale Structure of the Universe: Blog 36, Worksheet 12.1, Problems 1 and 2d.

Linear perturbation theory.

In this and the next exercise we study how small fluctuations in the initial condition of the universe evolve with time, using some basic fluid dynamics. In the early universe, the matter/radiation distribution of the universe is very homogeneous and isotropic. At any given time, let us denote the average density of the universe as \( \bar{\rho} (t) \). Nonetheless, there are some tiny fluctuations and not everywhere exactly the same. So let us define the density at comoving position r and time t as ρ(x, t) and the relative density contrast as \[δ(r, t) = \frac{ \rho(r, t) - \bar{\rho} (t)}{\bar{\rho} (t)} \] In this exercise we focus on the linear theory, namely, the density contrast in the problem remains small enough so we only need consider terms linear in δ. We assume that cold dark matter, which behaves like dust (that is, it is pressureless) dominates the content of the universe at the early epoch. The absence of pressure simplifies the fluid dynamics equations used to characterize the problem.

(a) In the linear theory, it turns out that the fluid equations simplify such that the density contrast δ satisfies the following second-order differential equation \[ \frac{d^2δ}{dt^2} + \frac{2 \dot{a}dδ}{adt} = 4 \pi G \bar{\rho}δ \] where a(t) is the scale factor of the universe. Notice that remarkably in the linear theory this equation does not contain spatial derivatives. Show that this means that the spatial shape of the density fluctuations is frozen in comoving coordinates, only their amplitude changes. Namely this means that we can factorize \[δ(x, t) =  D(t) \tilde{δ}(x) \] where ˜δ(x) is arbitrary and independent of time, and D(t) is a function of time and valid for all x. D(t) is not arbitrary and must satisfy a differential equation. Derive this differential equation.

Plugging in the second equation to the first equation gives: \[ \frac{d^2}{dt^2} D(t) \tilde{δ}(x) + \frac{2 \dot{a}d}{adt} D(t) \tilde{δ}(x) - 4 \pi G \bar{\rho} D(t) \tilde{δ}(x) = 0 \] Which is \[  \tilde{δ}(x)( \ddot{D}(t) + \frac{2 \dot{a}}{a} \dot{D}(t) - 4 \pi G \bar{\rho} D(t)) = 0 \] When \( \tilde{δ}(x) = 0\) this is \[   \ddot{D}(t) + \frac{2 \dot{a}}{a} \dot{D}(t) - 4 \pi G \bar{\rho} D(t) = 0 \]

(b) Now let us consider a matter dominated flat universe, so that \( \bar{\rho}(t) =  a^{-3} \rho_{c,0} \) where \( \rho_{c,0} \)is the critical density today, \( \frac{3H_0^2}{8 \pi G} \) as in Worksheet 11.1 (aside: such a universe sometimes is called the Einstein-de Sitter model). Recall that the behaviour of the scale factor of this universe can be written \( a(t) = (3H_0t/2)^{2/3} \) , which you learned in previous worksheets, and solve the differential equation for D(t). Hint: you can use the ansatz \( D(t) \propto t^q \) and plug it into the equation that you derived above; and you will end up with a quadratic equation for q. There are two solutions for q, and the general solution for D is a linear combination of two components: One gives you a growing function in t, denoting it as D+(t); another decreasing function in t, denoting it as D_(t).

Combining the given equations allows us to find a new expression for density: \[\bar{\rho}(t) =  \left( \frac{3H_0^2}{2} \right)^{-2} \frac{3H_0^2}{8 \pi G} \] We know from previous weeks that \[H_0 = \frac{ \dot{a}}{a} \] and \[ a \propto t^{\frac{2}{3}} \] which means \[ \dot{a} \propto \frac{2}{3} t^{- \frac{1}{3}}\] so \[ H_0 = \frac{ \dot{a}}{a} \propto \frac{2}{3t} \] Combining that with our answer from (a) gives:  \[ \ddot{D}(t) + \frac{4 }{3t} \dot{D}(t) - \frac{2}{3t^2} D(t) = 0 \] Using the anasatz \( D(t) \propto t^q \) simplifies this to \[q(q-1)t^{q-2} +  \frac{4 }{3t}qt^{q-1}  - \frac{2}{3t^2}t^q \] \[t^{q-2}(q^2 - q + \frac{4q}{3} - \frac{2}{3}) = 0 \] Using the quadratic equation, q can be -1 or 2/3 so the solutions to this problem are: \[D_-(t) = C_1 t^{-1} \] \[D_+(t) = C_2 t^{\frac{2}{3}}\]
(c) Explain why the D+ component is generically the dominant one in structure formation, and show that in the Einstein-de Sitter model, \(D_+(t) \propto a(t) \).

The \(D_+ \) component is generally the dominant one because it is proportional to \(t^{\frac{2}{3} }\), which, when t is large, grows much faster than the shriking \(D_- \propto t^{-1} \). Because both  \(D_+(t) \propto t^{\frac{2}{3}} \) and  \(a(t) \propto t^{\frac{2}{3}}\),  \(D_+(t) \propto a(t) \).

Spherical collapse
Gravitational instability makes initial small density contrasts grow in time. When the density perturbation grows large enough, the linear theory, such as the one presented in the above exercise, breaks down. Generically speaking, non-linear and non-perturbative evolution of the density contrast have to be dealt with in numerical calculations. We will look at some amazingly numerical results later in this worksheet. However, in some very special situations, analytical treatment is possible and provide some insights to some important natures of gravitational collapse. In this exercise we study such an example.

We consider a spherical patch of uniform over-density. Let us study the motion of a particle in terms of its distance r from the center of the sphere as a function of time t. Recall that in Worksheet 9, from Newtonian dynamics, we have derived that this function satisfies the following equation \[ \frac{1}{ 2} \left( \frac{dr}{dt} \right)^2 - \frac{GM }{r}  = C \] where C is a constant. We can study this in a closed case where r = A(1  - cos η) and t =B(η -  sin η), the open case where r = A(cosh η - 1) and t =B(sin η - η ), and the flat case where r = \(Aη^2 /2\) and t =  \(Bη^3 /6 \).

We found that in the closed case, \[ C = \frac{-A^2}{2B^2} \] in the open case \[ C = \frac{A^2}{B^2} \] and in the flat case \[ C = 0\] Plotting these three together with r as a function of t (with 0 < η < \( 2 \pi \) )  where in the closed case, the particle turns around and collapse; in the open case, the particle keeps expanding with some asymptotically positive velocity; and in the flat case, the particle reaches an infinite radius but with a velocity that approaches zero. We can see this in a closed universe:

An open universe:

And a flat universe:

Which is where we are theorized to live.

Cosmic microwave background: Blog 34, Worksheet 11.1, Problem 2

One of the successful predictions of the Big Bang model is the cosmic microwave background (CMB) existing today. In this exercise let us figure out the spectrum and temperature of the CMB today. In the Big Bang model, the universe started with a hot radiation-dominated soup in thermal equilibrium. In particular, the spectrum of the electromagnetic radiation (the particle content of the electromagnetic radiation is called photon) satisfies (1). At about the redshift z « 1100 when the universe had the temperature T = 3000K, almost all the electrons and protons in our universe are combined and the universe becomes electromagnetically neutral. So the electromagnetic waves (photons) no longer get absorbed or scattered by the rest of the contents of the universe. They started to propagate freely in the universe until reaching our detectors. Interestingly, even though the photons are no longer in equilibrium with its environment, as we will see, the spectrum still maintains an identical form to the Planck spectrum, albeit characterized by a different temperature.

Source

Part A: If the photons was emitted at redshift z with frequency ν, what is its frequency ν' today?

We know from two weeks ago that \[ z = \frac{v}{c} = \frac{ \lambda_0 - \lambda_e }{ \lambda_e } \]  Combining this with the fact that \[ \lambda = \frac{c}{\nu} \]  \[ z = \frac{ \frac{c}{\nu'}  -  \frac{c}{\nu}  }{  \frac{c}{\nu}  }= \frac{ \nu }{\nu'} \] \[ \nu' = \frac{\nu}{z + 1} \]

Part B:  If a photon at redshift z had the energy density uνdν, what is its energy density uν'dν' today? (Hint, consider two effects: 1) the number density of photons is diluted; 2) each photon is redshifted so its energy, E = hν, is also redshifted.)

We know energy density is a function of energy and volume \(u_ν'dν' = \frac{E}{V} \) so we just need to figure out how energy and volume have changed from the past to today.

Once again, we know \[ E = h \nu \] so \[ E' = \frac{h \nu}{z + 1} \] based on the conversion found above. We also know distances scale to volume as \( R^3 ~ V \) so we have the relationship: \[V_{today} \propto a_0^3 ( 1 + z )^3 V_{tomorrow} \] This means \[ n = n_0 a_0^{-3} ( 1 + z )^{-3} \] Because \[u_νdν = E n = h \nu n \] \[u_ν'dν' = \frac{h \nu}{z + 1}n_0 a_0^{-3} ( 1 + z )^{-3} \] So the energy density today would be \[u_ν'dν' = a_0^{-3} ( 1 + z )^{-4} u_νdν \]

Part C: Plug in the relation between ν and ν' into the Planck spectrum: \[u_ν dν  = \frac{ 8 \pi h \nu^3 }{c^3} \frac{1}{e^{\frac{h \nu}{kT}} - 1 } d \nu \] and also multiply it with the overall energy density dilution factor that you have just figured out above to get the energy density today. Write the final expression as the form uν'dν' . What is uν'? This is the spectrum we observe today. Show that it is exactly the same Planck spectrum, except that the temperature is now T' =  T(1 + z)^-1 .

Plugging in our scaling relation and multiplying by our dilution factor give the rather complicated expression: \[u_ν' dν' = \frac{ 8 \pi h ( \nu'(z+1))^3}{c^3} \frac{1}{e^{\frac{h( \nu'(z+1))}{kT}} - 1 } d \nu' ( 1 + z) \cdot a_0^{-3} ( 1 + z )^{-4} \] This simplifies to \[u_ν' dν' = \frac{ 8 \pi h  \nu'^3}{c^3} \frac{1}{e^{\frac{h( \nu'(z+1))}{kT}} - 1 } d \nu' \] This is the same result that we would have obtainted if we replaced T with \(\frac{T}{1 + z} \).

Part D: As you have just derived, according to Big Bang model, we should observe a black body radiation with temperature T' filled in the entire universe. This is the CMB. Using the information given at the beginning of this problem, what is this temperature T' today? (This was indeed observed first in 1964 by American radio astronomers Arno Penzias and Robert Wilson, who were awarded the 1978 Nobel Prize.)

This problem is now just a matter of plugging in things we know: \[ z = 1100 \] \[ T = 3000 \: K \] \[ T' = \frac{T}{1 + z} \] So our final temperature of the CMB today is \[ T' = \frac{3000}{1101} = 2.725 \: K \]  This is pretty close to what Penzias and Wilson found.

Baryon-to-photon ratio of our universe: Blog 35, Worksheet 11.1, Problem 3

Part A:  Despite the fact that the CMB has a very low temperature (that you have calculated above), the number of photons is enormous. Let us estimate what that number is. Each photon has energy hν. From equation (1), figure out the number density, nν, of the photon per frequency interval dν. Integrate over dν to get an expression for total number density of photon given temperature T. Now you need to keep all factors, and use the fact that \[ \int_0^{\infty} \frac{x^2}{e^x - 1} \approx 4.2 \] We already know \[ E = h \nu \] and \[ E n d \nu = \frac{ 8 \pi h \nu }{c^3} \frac{1}{e^{\frac{h \nu}{kT}} - 1 } d \nu \] The integral of this is equivalent to \[ \int E n d \nu = \int \frac{ 8 \pi h \nu }{c^3} \frac{1}{e^{\frac{h \nu}{kT}} - 1 } d \nu = E n \] Adding this with our definition of energy and rearranging the integral gives us: \[ n = \frac{ 8 \pi K^3 T^3 }{c^3 h^3} \int_0^{\infty} \frac{x^2}{e^x - 1}\] Where \[ x = \frac{ h\nu}{KT}\] and \[ d \nu = \frac{ KT dx}{h} \] With the approximation given above we get a cleaner answer for n \[ n_{\nu} = \frac{ 8 \pi K^3 T^3 }{c^3 h^3}(2.4) \]

Part B: Use the following values for the constants: kB =1.38 x 10^-16 erg/K, c = 3.00 x 10^10 cm/s, h =  6.62 x 10^-27 erg s, and use the temperature of CMB today that you have computed from 2d), to calculate the number density of photon today in our universe today (i.e. how many photons per cubic centimeter?)

In problem 2b, we found the temperature of CMB today to be about 2.725 Kelvin. Plugging this in with the constants specified above should give us: \[ n_{\nu} = \frac{ 8 \pi K^3 T^3 }{c^3 h^3}(2.4) = \frac{ 8 \pi (1.38 \times 10^{-16})^3 2.725^3 }{( 3 \times 10^{10})^3 (6.62 \times 10^{-27})^3}(2.4) \approx 170.62 \frac{photons}{cm^3}\]

Part C: Let us calculate the average baryon number density today. In general, baryons refer to protons or neutrons. The present-day density (matter + radiation + dark energy) of our Universe is 9.2 x 10^-30g/cm^3 . The baryon density is about 4% of it. The masses of proton and neutron are very similar (= 1.7 x 10^-24g). What is the number density of baryons?

This means that the density of our universe that is baryons is \[ 9.2 \times 10^{-30} \times 0.04 = 3.68 \times 10^{-31} \] Dividing this number by the mass of a typical baryon gives us the density: \[ n_b = \frac{\rho}{m} = \frac{ 3.68 \times 10^{-31}}{1.7 \times 10^{-24} } \approx 2.16 \times 10^{-7} \]

Part D:  Divide the above two numbers, you get the baryon-to-photon ratio. As you can see, our universe contains much more photons than baryons (proton and neutron).
\[ \frac{photons}{baryons} = \frac{2.16 \times 10^{-7} }{170.62} = 1.26 \times 10^{-9} \]

Blog 33: Free-form blog post

Last spring, during one very long night in the eighth floor of the science center, Barra, Sean, April and I remotely operated the Minerva telescope, loudly sang disney songs, and ate way too much junk food in order to obtain three light curves. Two of those light curves were of exoplanets transiting other stars (we wanted to be very sure of our grade in the class). The third exoplanet was not transiting a star at all, but a white dwarf. One of our TFs last semester, Andrew Vanderburg, had asked us to take a light curve of this curious system for a project he was working on. Fast forward about half a year, and that project is now been published in Nature, Sky and Telescope, and a few other astronomical publications. In light if its recent publicity, I figured I would actually read the Nature paper so I can gain a deeper understanding of what exactly we were observing that night atop the science center.

The object in question is a white dwarf, one of the most common fates for a star, in the Virgo region, about 570 light years away from Earth. White dwarf atmospheres are generally composed of lighter elements, with a carbon and oxygen core and a helium and hydrogen outer shell. However, a quarter to a half of all white dwarfs observed contain heavier elements in their atmospheric spectrum, a puzzling fact knowing heavy elements should sink to the core. This discrepancy caused astronomers to suspect that outside materials had been introduced to the white dwarf's atmosphere, either through surrounding dust or disrupted asteroids, but no evidence had been found.

This white dwarf in question, WD 1145 + 017, was being observed for a different reason. The object exhibited a telltale, periodic, dip in its light curve. This occurred about once every 4.5 hours and could obscure up to 40% of the white dwarf's light. This behavior is a telling sign of a transiting exoplanet, though no exoplanets had ever been discovered around a white dwarf before.

An artist's rendering of a disintegrating rock planet around a white dwarf.

However, the periodic dip in the light curve was not a typical one either. Unlike a consistently transiting exoplanet around a star, this light curve was more erratic. The results were puzzling, until researchers saw two asymmetric light curves that are indicative of a disintegrating planet, which had been observed on main sequence stars before. The asymmetry was evidence for a comet-like tail trailing the planet and the variable transit depths indicated a disintegrating system. Combining this with data showing evidence of heavy elements, like nickel and iron, indicated to researchers that the disintegrating planet was rocky.

Some of the main periodic light curves of WD 1145 + 017.

The planet is about twice the distance from the Earth to the Moon away from the white dwarf, and about the mass of Ceres. The origin of the planet is unclear; it was most likely disrupted from a previous orbit by the gravity of the white dwarf. There is still a lot to learn about this unique system. It is currently the best evidence we have for external white dwarf "pollution" and exoplanets transiting white dwarfs. This is definitely something I will keep an eye on in the future.

Sources:
https://www.cfa.harvard.edu/~avanderb/page1.html
http://www.nature.com/articles/nature15527.epdf?referrer_access_token=JQgS2CFJ63QHiqwEPtPBJNRgN0jAjWel9jnR3ZoTv0OabmFzXPDJ8_WF8JNyTjQSuCoJ78UVkpUoy_k_5w1_QoSdPQRTejjUKTX0cRCWjHcKxRGYqDGUtTqkJgFEKN4xBCrjmjgKIdaoaOP99aALkSQAZ5OmnFbTUeiSGH9Dk_EhDjR0a7Z65xPZ34YmCVHX7DtNfxjg2G-X9-H-TMLmdg%3D%3D&tracking_referrer=www.nature.com

Size of the Observable Universe: Blog 32, Worksheet 10.1, Problem 3

Observable Universe. It is important to realize that in our Big Bang universe, at any given time, the size of the observable universe is finite. The limit of this observable part of the universe is called the horizon (or particle horizon to be precise, since there are other definitions of horizon.) In this problem, we’ll compute the horizon size in a matter dominated universe in co-moving coordinates. To compute the size of the horizon, let us compute how far the light can travel since the Big Bang.

(a) First of all - Why do we use the light to figure out the horizon size?

Light is both the fastest known phenomena in the universe and our way of observing the universe. All we can see is determined by how quickly light can reach our eyes, so we are limited to a particle horizon determined by the speed of light.

(b) Light satisfies the statement that \(ds^2 =  0\). Using the FRW metric, write down the differential equation that describes the path light takes. We call this path the geodesic for a photon. Choosing convenient coordinates in which the light travels in the radial direction so that we can set dθ = dφ = 0, find the differential equation in terms of the coordinates t and r only.
\[ds^2 = -c^2dt^2 + a^2(t) [\frac{dr^2}{1-kr^2} + r^2 (d \theta + sin^2( \theta) d \theta ]  \] With the conditions for light and comoving radii this becomes \[0= -c^2dt^2 + a^2(t) [\frac{dr^2}{1-kr^2} + 0 ]  \] \[ \frac{c^2dt^2}{ a^2(t)} = \frac{dr^2}{1-kr^2} \]

c) Suppose we consider a flat universe. Let’s consider a matter dominated universe so that a(t) as a function of time is known (in the last worksheet). Find the radius of the horizon today (at t =  \(t_0 \)). (Hint: move all terms with variable r to the LHS and t to the RHS. Integrate both sides, namely r from 0 to \(r_{horizon}\) and t from 0 to \(t_0 \).) \[ \frac{c^2dt^2}{ a^2(t)} = \frac{dr^2}{1-kr^2} \] In a flat universe, this simplifies to \[ \frac{c}{ a(t)} dt = dr \] The integral is \[ \int_0^{t_0} \frac{c}{ a(t)} dt = \int_0^{r_H} dr \] We know from a previous blog post that \( a(t) \propto  t^{\frac{2}{3}} \) so the integral becomes \[ \int_0^{t_0} \beta  \frac{c}{  \left( \frac{t}{t_0}\right)^{\frac{2}{3}}} dt = \int_0^{r_H} dr \] where beta is a constant. So we have \[ r_H = 3 \beta c t_0 \]

Correction: we can solve this problem more accurately by replacing the relation \( a(t) \propto  t^{\frac{2}{3}} \) with the more accurate \( a(t) \propto \left( \frac{ t}{t_0} \right)^{\frac{2}{3}} \) in this way we can redefine beta in terms of \(a_0 \text{ and } t_0 \) where \(a_0 \) is 1 so our final answer cleans up to just be \[ r_H = 3  c t_0 \]