Blog #35, Worksheet #11.1, Problem #3, Baryons and Photons

3. a) Despite the fact that the CMB has a very low temperature (that you have calculated above), the number of photon 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

We start with the given equation 1) of \[\frac{8 \pi}{c^3} \cdot \frac{v^2}{e^{\frac{hv}{kT}} - 1} dv = n_v dv\]To integrate dv, we will use a u substitution and call \[u = \frac{hv}{kT}\] Which gives us \[du = \frac{h}{kT}dv \qquad and \qquad v^2 = \frac{k^2 T^2 u^2}{h^2}\] Plugging these relationships into our original equation 1) we are left with \[\frac{8 \pi k^3 T^3}{c^3 h^3} \cdot \frac{u^2}{e^u - 1} du = n_v dv \]We then integrate the left side from 0 to \(\infty\) and using the relationship that \[\int^{\infty}_0 \frac{x^2}{e^x - 1} dx \thickapprox 2.4 \]We end up with \[n_v dv = \frac{19.2 \pi k^3 T^3}{c^3 h^3}\]

b) (b) Use the following values for the constants: kB = 1.38x10^-16 erg K^-1 , c = 3.00x10^10cm s^-1 , h = 6.62x10^27 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?)

For this part, we simply use our relationship from part a to solve for the number density of photons in our universe today. We solve \[\frac{19.2 \pi (1.38x10^{-16})^{3} (2.7)^{3}}{(3x10^10)^3 (6.62x10^{-27})^3} \thickapprox 400 \frac{photon}{cm^3}\]

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 9.2x10^30g cm^3 . The baryon density is about 4% of it. The masses of proton and neutron are very similar (1.7x 10^24g). What is the number density of baryons?

We are told that the total density of the universe is equal to \[\rho_{total} = 9.2x10^{-30} \frac{g}{cm^3}\]and we are told that baryons make up 4% of the total density meaning \[\rho_{baryon} = 3.68x10^{-31} \frac{g}{cm^3} \]We also know that the average baryon weighs \(1.7x10^{-24} g \) so the number density of baryons is \[n_{baryon} = \frac{\rho_{baryon}}{1.7x10^{-24} g } = 2.16x10^{-7} \frac{baryon}{cm^3}\]

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).

Finally, we compare the amount of baryons to photons and get \[\frac{400 \frac{photons}{cm^3}}{2.16x10^{-7} \frac{baryons}{cm^3}} \thickapprox 2x10^9 \frac {photon}{baryon}\] Showing that there are far far more photons in our universe than there are baryons, or there is a lot more light than matter!

Blog #34, Worksheet #11.1, Problem #4, Flatness Problem of The Big Bang Model

4. The flatness problem of the big bang model. Despite the success of the Big Bang model mentioned in the lecture, there are also problems with it. These problems mostly have to do with the initial conditions for the Big Bang model, and provide the motivations for scientists to seek deeper answers for the origin of the Big Bang. In this exercise we study one of those problems. 

a) Recall the Friedman equation we learned in a previous lecture \[H^2 = \frac{8 \pi G \rho}{3} - \frac{k c^2}{a^2}\]. Rewrite this in terms of the density parameter \(\Omega\). 

We began with the given equation \[H^2 = \frac{8 \pi G \rho}{3} - \frac{k c^2}{a^2}\]Solving this for \(\rho\) we get \[\rho = \frac{3 H^2}{8 \pi G} + \frac{3 k c^2}{8 a^2 \pi G}\]However we were told that when we assume that k = 0, we get our critical \(\rho\) which is \[\rho_c  = \frac{3H^2}{8 \pi G}\]We substitute in our critical rho to get \[\rho = \rho_c + \frac{k c^2}{a^2 H^2} \rho_c \] We then solve for the desired \(\Omega\) ratio of \(\frac{\rho}{\rho_c}\) and get \[\Omega = 1 + \frac{k c^2}{a^2 H^2}\]as expected.

b) Consider the epoch of our universe that is dominated by matter. We have already computed the time-dependence of a and H in such a universe (refer to Worksheet 9.1). Use this result and a) to show the time-dependence of Ω.

To show the time dependence we have to relate our \(\Omega\) ratio to the time dependence of a. Getting rid of the constants in the equation above, we know \[\Omega \propto (a \cdot H)^{-2} \qquad and \qquad H = \frac{\dot{a}}{a}\] meaning \[\Omega \propto {\dot{a}}^{-2}\] From a previous worksheet we found that \[a \propto t^{\frac{2}{3}} \qquad and \qquad \dot{a} \propto t^{-  \frac{1}{3}}\]Thus we have \[\Omega \propto t^{\frac{2}{3}} \]

c) Today, experiments have measured the density parameter of our universe is within 0.005 : Ω - 1 : 0.005, namely Ω is close to 1 today. Use your above result to determine the range of Ω - 1 at the time of CMB formation, using the fact that today the age of universe is about 13.7 billion years and the age of universe when the CMB formed was about 380, 000 years. From this exercise we see that in the early universe, the density parameter was very close to one. This is one of the issues of the Big Bang model – why is this special initial condition chosen? We provide one possible solution to this question in the next exercise.

Beginning with the equation we just solved for \[\Omega - 1 = \frac{kc^2}{a^2 H^2}\]and plugging in the time dependence we just found we are left with \[\Omega - 1 = C_0 t^{\frac{2}{3}}\]We then use the given information for \(\Omega -1\) and t to solve for \(C_0\). We solve that the time constant for light in this model is \[C_0 \thickapprox 8.73 x10^{-10} \]We then use this constant and the given time when the CMB was taken t = 380,000 to solve for the approximate density constant. \[\Omega - 1 = (8.73 x 10^{-10})(380,000)^{- \frac{2}{3}} = \pm 5 x 10^{-6} \]Because this answer is approximately zero, we can approximate that the density constant, \(\Omega\) at that time fits the estimation that \[\Omega \thickapprox 1 \]

Blog #33, Free-Form Post, The Blue Marble and Spaceship Earth

After a tragedy like the horrors experienced this past weekend in Paris, it is hard to step back and see humanity as one united, whole species. With a world filled with so much hate and animosity, how could we all possibly be united under some joining force of humanity?

This weekend I was fortunate enough to attend the annual Students for the Exploration and Development of Space (SEDS) conference at Boston University. At the conference, multiple speakers spoke of the future of space travel, living on mars, and a variety of amazing space technologies. But one speaker, Frank White, a storied author and Professor at the Extension School, spoke about his expertise, The Overview Effect. Frank has been working with astronauts for the past 30 years to document and detail the experience of traveling to space and how it changes the psyche.

Known colloquially as the Overview Effect, Professor White describes a cognitive shift in awareness in astronauts who have been beyond the boundaries of Earth and seen our planet from a far, simply a vessel carrying 7 billion people through the vast emptiness of space. This realization that humans are so small, so insignificant in the grand scheme of things, is said to bring on an immense truth and actually some sadness in astronauts. A sadness over the fact that not everyone on the planet has the opportunity to experience the truth and still quarrels over things like borders, religion, and ideology instead of coming together under the ties of humanity.

Luckily, also at this conference were several companies extending the offer to bring everyday humans into low orbit, to look back onto our planet and discover what it is like to escape from the grips of Earth's problems. I hope that one day every human, who chooses to, has the ability to go to space and learn what countless astronauts have felt when looking back on the small blue marble, sitting on the forever expanses of black space.

image:
http://eoimages.gsfc.nasa.gov/images/imagerecords/55000/55418/AS17-148-22727_lrg.jpg

Blog #32, Worksheet #10.1, Problem #3, Observable Universe

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? 
Well we classify the horizon as the edge of the observable universe and we can only observe as far as light has traveled. Thus if we want to calculate the size of the "observable" universe, we must use the fastest known observable element, which in our case is 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. 

We begin with the general FRW metric given earlier in the worksheet and set ds^2 = 0 giving us \[c^2 dt^2 = a^2 (t) [\frac{dr^2}{1-kr^2} + r^2 (d \theta^2 + sin^2 \theta d \phi^2)] \]We simplify given the information that dθ = dφ = 0 and get \[c^2 dt^2 = \frac{a^2 (t) dr^2}{1-kr^2}\]which gives us an equation \[c^2 = \frac{a^2}{1-kr^2} (\frac{dr}{dt})^2\]

(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 = t0). (Hint: move all terms with variable r to the LHS and t to the RHS. Integrate both sides, namely r from 0 to rhorizon and t from 0 to t0.)

We begin with the equation we just derived and put in the fact that for a flat universe, k = 0. Thus we are left with \[c^2 = a^2 (\frac{dr}{dt})^2 \rightarrow dr = \frac{c}{a(t)} dt \] In last week's worksheet we found that a(t) in modern times is proportional to \(t^{\frac{2}{3}}\). We then plug into our equation \[a(t) = a_0 (\frac{t}{t_0})^{2/3}\] \[dr = \frac{c t_0^{\frac{2}{3}}}{a_0 t^{\frac{2}{3}}}\]We then integrate both sides and get \[R = \frac{c}{a_0} t_0^{\frac{2}{3}} \int_{0}^{t_0} t^{- \frac{2}{3}} = \frac{c}{a_0} t_0^{\frac{2}{3}} \cdot 3 t_0^{\frac{1}{3}}\]Which gives us our final relationship of \[R = \frac{3c t_0}{a_0}\]

Blog #31, Worksheet #10.1, Problem #2, Circumference Ratios

2) . Ratio of circumference to radius. Let’s continue to study the difference between closed, flat and open geometries by computing the ratio between the circumference and radius of a circle. 

(a) To compute the radius and circumference of a circle, we look at the spatial part of the metric and concentrate on the two-dimensional part by setting dφ = 0 because a circle encloses a two-dimensional surface. For the flat case, this part is just \[{ds_{2D}}^2 = d r^2 + r^2 d \theta^2\] The circumference is found by fixing the radial coordinate (r = R and dr = 0) and both sides of the equation (note that θ is integrated from 0 to 2π). 
The radius is found by fixing the angular coordinate (θ, dθ = 0) and integrating both sides (note that dr is integrated from 0 to R). 
Compute the circumference and radius to reproduce the famous Euclidean ratio 2π. 

We first set r = R and dr = 0 to find the circumference. We get \[ds^2 = R^2 d \theta^2 \]We then take the root of both sides to get \[ds = R d \theta\]and when we integrate both sides we get \[\int_{0}^{s} ds = \int_{0}^{2 \pi} R d \theta\]Which gives us \[S_{circ} = 2 \pi R\]Then we solve for the radius by setting \(d \theta = 0 \) and get \[S_{rad} = \int_{0}^{R} dr = R \]Finally we take the ratio of our S factors to get the desired ratio \[\frac{S_{circ}}{S_{rad}} = \frac{2 \pi R}{R} = 2 \pi \]

(b) For a closed geometry, we calculated the analogous two-dimensional part of the metric in Problem (1). This can be written as: \[{ds_{2D}}^2 = d \xi^2 + sin \xi^2 d \theta^2\]  Repeat the same calculation above and derive the ratio for the closed geometry. Compare your results to the flat (Euclidean) case; which ratio is larger? (You can try some arbitrary values of ξ to get some examples.) 

We repeat a similar process compared to part a) although now our \(S_{circ}\) comes out to be \[S = \int_{0}^{2 \pi} sin X d \theta = 2 \pi sin X\]As in part a), our radius integral simply comes out to \[S_{rad} = X\]Then we place the same ratio as in part a) to get  \[\frac{S_{circ}}{S_{rad}} = \frac{2 \pi sin X}{X}\]We know that the function \(\frac{sin(X)}{X}\) can never be greater than one (see plot below). Thus the flat ratio will always be larger than the closed case, except at X = 0.

(c) Repeat the same analyses for the open geometry, and comparing to the flat case. 

The open case is identical to the to the closed case except where the closed case has a sin term, the open case has a sinh term. This is due to the \(\frac{dr^2}{1- kr^2}\) that reduces to sinusoidal terms for the closed case, yet remains in the open system. Thus our ratio for the the open case becomes \[\frac{S_{circ}}{S_{rad}} = \frac{2 \pi sinh X}{X}\]However the plot of \(\frac{sinh(X)}{X}\) is always rising and must always be larger than the flat case of \(2 \pi\) (see graph below).

(d) You may have noticed that, except for the flat case, this ratio is not a constant value. However, in both the open and closed case, there is a limit where the ratio approaches the flat case. Which limit is that? 

As the above graphs show, the only time that both cases approach the flat ratio of \(2 \pi\) is time 0. When we estimate the universe as taking different spacetime geometries, we still approximate all estimates as flat because the universe is so large and crosses such a expansive time, that any approximations we make are based on a very small, short time period and can be approximated as 0.

Blog #30, Lab 2 Preview

In the second lab we will be using the millimeter-wave telescope at the Harvard College Observatory to observe giant molecular clouds (GMCs) spread throughout the inner regions of the Milky Way. We will be pointing the telescope at a variety of GMCs to try and calculate orbital velocities in an attempt to us the data for a variety of calculations including finding redshifts of the clouds, number of stars interior to the sun, galactic age of the sun, and plot a rough orbit of the galaxy. 

- What is going to be the typical integration time per point?
Amazingly, the millimeter telescope only needs an integration time of 2-3 minutes per point and can see up to 50,000 light years away in that time. 

- Over what range of longitude do you plan to observe?
I plan to observe over galactic longitudes b ∼ 10° to ∼ 70°

- How many (l,b) positions do you plan to observe?
We plan to observe 4 different GMCs and their associated galactic (l,b) positions.

- Will all your target positions be above 30° when you plan to observe them?
We need all of our targets to be above 30° or we will get messy data from noise in the sky. Luckily we are observing around noon and all of our targets should be readily visible.

- At what LST are you going to start observing? At what EST? 

We will be observing around noon EST, which ends up being around 15:30 LST.

Blog #29, Worksheet #9.1, Question 2, GR modifications to Friedmann Equations

2. In Question 1, you have derived the Friedmann Equation in a matter-only universe in the New- tonian approach. That is, you now have an equation that describes the rate of change of the size of the universe, should the universe be made of matter (this includes stars, gas, and dark matter) and nothing else. Of course, the universe is not quite so simple. In this question we’ll introduce the full Friedmann equation which describes a universe that contains matter, radiation and/or dark energy. We will also see some correction terms to the Newtonian derivation.

(a) The full Friedmann equations follow from Einstein’s GR, which we will not go through in this course. Analogous to the equations that we derived in Question 1, the full Friedmann equations express the expansion/contraction rate of the scale factor of the universe in terms of the properties of the content in the universe, such as the density, pressure and cosmological constant. We will directly quote the equations below and study some important consequences.

Starting from these two equations, derive the third Friedmann equation, which governs the way average density in the universe changes with time.

As suggested in the assignment, much of the math has been shortened or skip due to tediousness but the processes will be described for each step. We begin by multiplying equation (1) on the sheet by a and taking the time derivative then plugging in equation 2 to cancel \(\ddot{a}\). We end up with \[\frac{4 \pi G a \dot{a}}{3c^2} (\rho c^2 + 3P) = \frac{8}{3} \pi a \dot{a} G \rho + \frac{4}{3} \pi a^2 G \dot{\rho}\]We then cancel and simplify both sides to get \[- \frac{3P \dot{a}}{c^2} = 3 \dot{a} \rho + a \dot{\rho}\]Finally we multiply both sides by \(c^2\) and divide by a to get our desired final answer of \[\dot{\rho} c^2 = -3 \frac{\dot{a}}{a} (\rho c^2 + P) \]
 

(b) Cold matter dominated universe. If the matter is cold, its pressure P = 0, and the cosmological constant Λ = 0. Use the third Friedmann equation to derive the evolution of the density of the matter \(\rho\) as a function of the scale factor of the universe a. You can leave this equation in terms of (\rho\) , (\rho\)0,a and a0, where (\rho\)0 and a0 are current values of the mass density and scale factor. 

The result you got has the following simple interpretation. The cold matter behaves like “cosmological dust” and it is pressureless (not to be confused with warm/hot dust in the interstellar medium!). As the universe expands, the mass of each dust particle is fixed, but the number density of the dust is diluted - inversely proportional to the volume.

Using the relation between (\rho\) and a that you just derived and the first Friedmann equation, derive the differential equation for the scale factor a for the matter dominated universe. 

We can derive from the third Friedmann equation that \[\frac{\dot{\rho}}{\rho} \propto \frac{\dot{a}}{a} \rightarrow \rho \propto a^{-3} \]We then solve the first Friedmann equation with \(\Lambda = 0\) and \(k = 0 \) giving us \[\dot{a}^2 = \frac{8 \pi}{3} G \rho a^2 \propto \frac{8 \pi}{3} G a^{-1} \]Which means \[\dot{a} \propto a^{- \frac{1}{2}} \propto \frac{da}{dt} \]We then separate and integrate both sides \[\int{a^{\frac{1}{2}} da} = \int{dt} \] \[a^{\frac{3}{2}} \propto t \rightarrow a \propto t^{\frac{2}{3}}\]

c) Let us repeat the above exercise for a universe filled with radiation only. For radiation, \(P = \frac{1}{3} \rho c^2 \) and Λ = 0. Again, use the third Friedmann equation to see how the density of the radiation changes as a function of scale factor.

As explained in the assignment, parts b, c, and d are very similar in their mathematical process. We follow the same steps as in part b to show \[\rho \propto a^{-4}\]Which, using the same reasoning would give us \[\dot{a} \propto a^{-1} \propto \frac{da}{dt} \] and when we integrate we would get \[a^2 \propto t \rightarrow a \propto t^{\frac{1}{2}}\]

d) Imagine a universe dominated by the cosmological-constant-like term. Namely in the Friedmann equation, we can set ρ = 0 and P = 0 and only keep Λ nonzero. 

To avoid repetition, we will skip the same steps seen in part b and c and give the answer as \[a(t) \propto e^t\]Meaning that in a dark energy dominated universe, the scale factor increases exponentially as time progresses. 

(e)  Suppose the energy density of a universe at its very early time is dominated by half matter and half radiation. (This is in fact the case for our universe 13.7 billion years ago and only 60 thousand years after the Big Bang.) As the universe keeps expanding, which content, radiation or matter, will become the dominant component? Why? 

Simply put, if we look at our answers for b) and c) we see that \[a_{matter} \propto t^{\frac{2}{3}} \qquad and \qquad a_{radiation} \propto t^{\frac{1}{2}}\]Thus we can say that the portion of the universe dominated by matter will expand much faster than that of radiation so over a long period of time, the matter will become dominant because it is constantly expanding faster. 

(f) Suppose the energy density of a universe is dominated by similar amount of matter and dark energy. (This is the case for our universe today. Today our universe is roughly 68% in dark energy and 32% in matter, including 28% dark matter and 5% usual matter, which is why it is acceleratedly expanding today.) As the universe keeps expanding, which content, matter or the dark energy, will become the dominant component? Why? What is the fate of our universe?
In a similar manner \[a_{matter} \propto t^{\frac{2}{3}} \qquad and \qquad a_{energy} \propto e^t\] \(e^t\) expands much much faster than \(t^{\frac{2}{3}}\) so our universe will soon be far dominated by dark matter. We have a very dark future to look forward to. 

Blog #28, Worksheet #9.1, Question 1, Matter-Only Universe

1) In this exercise, we will derive the first and second Friedmann equations of a homogeneous, isotropic and matter-only universe. We use the Newtonian approach. Consider a universe filled with matter which has a mass density \(\rho (t)\). Note that as the universe expands or contracts, the density of the matter changes with time, which is why it is a function of time t. Now consider a mass shell of radius R within this universe. The total mass of the matter enclosed by this shell is M. In the case we consider (homogeneous and isotropic universe), there is no shell crossing, so M is a constant.

(a) What is the acceleration of this shell? Express the acceleration as the time derivative of velocity, \(\dot{v}\) (pronounced v-dot) to avoid confusion with the scale factor a (which you learned about last week).

We start with the simple equation for force so that we can relate acceleration to constants in our system. We know that \[F = m \dot{v} = - \frac{GMm}{r^2}\] We then simply cancel m's and rearrange to solve \[\dot{v} = - \frac {GM}{r^2} \]

(b) To derive an energy equation, it is a common trick to multiply both sides of your acceleration equation by v. Turn your velocity, v, into dR/dt , cancel dt, and integrate both sides of your equation. This is an indefinite integral, so you will have a constant of integration; combine these integration constants call their sum C. 

Once we multiply both sides by v and cancel dt, we are left with \[\dot{v} dr = - \frac{GM}{r^2} dr \] and when we integrate both sides we get \[\frac{\dot{r}^2}{2} = \frac{GM}{r} + C\]Then we rearrange to get our desired final answer of \[\frac{\dot{r}^2}{2} - \frac{GM}{r} = C \]

(c) Express the total mass M using the mass density, and plug it into the above equation. Rearrange your equation to give an expression for \((\frac{\dot{R}}{R})^2\) where \(\dot{R}\) is equal to dR/dt.
We know that \[\rho = \frac{M}{\frac{4}{3} \pi r^3}\] which gives us \[M = \frac{4}{3} \pi r^3 \rho\]We then plug in to our equation from part b) and get \[\frac{\dot{r}}{2} - \frac{4}{3} \pi G \rho r^2 = C \] Finally we multiply through by 2 and divide by \(R^2\) to get the desired left side of the equation. Our final answer comes out to \[(\frac{\dot{r}}{r})^2 = \frac{8}{3} \pi G \rho + \frac{2C}{r^2}\]

(d) R is the physical radius of the sphere. It is often convenient to express R as R = a(t)r, where r is the comoving radius of the sphere. The comoving coordinate for a fixed shell remains constant in time. The time dependence of R is captured by the scale factor a(t). The comoving radius equals to the physical radius at the epoch when a(t) = 1. Rewrite your equation in terms of the comoving radius, R, and the scale factor, a(t).

(e) Rewrite the above expression so \((\frac{\dot{a}}{a})^2 appears alone on the left side of the equation.

This step only requires replacing the r's in part c) with a's to account for the comoving radius. 

\[(\frac{\dot{a}}{a})^2 = \frac{8}{3} \pi G \rho + \frac{2C}{(a(t)r)^2}\]

(f) Derive the first Friedmann Equation: From the previous worksheet, we know that \(H(t) = \frac{\dot{a}}{a} \). Plugging this relation into your above result and identifying the constant \(\frac{2C}{r^2} = -kc^2 \) where k is the “curvature” parameter, you will get the first Friedmann equation. The Friedmann equation tells us about how the shell of expands or contracts; in other words, it tells us about the Hubble expansion (or contracration) rate of the universe.

We know \[H(t) = \frac{\dot{a}}{a} \qquad and \qquad \frac{2C}{r^2} = -kc^2\]Plugging into our equation from part d) we get \[H(t) = \frac{8}{3} \pi G \rho - \frac{kc^2}{a^2}\]

(g) Derive the second Friedmann Equation: Now express the acceleration of the shell in terms of the density of the universe, and replace R with R = a(t)r. You should see that \(\frac{\ddot{a}}{a} = - \frac{4 \pi}{3} G \rho \) which is known as the second Friedmann equation.The more complete second Friedmann equation actually has another term involving the pressure following from Einstein’s general relativity (GR), which is not captured in the Newtonian derivation. If the matter is cold, its pressure is zero. Otherwise, if it is warm or hot, we will need to consider the effect of the pressure.

We begin with \[\dot{v} = \ddot{r} = \ddot{a}(t) r = - \frac{GM}{r^2}\]Then we substitute as we did earlier for density, replacing r with a(t)r and get \[\frac{-4 \pi G \rho a(t)r}{3} = \ddot{a}(t)r\]Finally we cancel r's and divide by a to get a final answer of \[-\frac{4}{3} \pi G \rho = \frac{\ddot{a}}{a}\]