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 \[ r_H = 3  c t_0 \]

Geometry of the Universe: Blog 31: Worksheet 10.1, Question 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 \[ ds_{2d}^2 = dr^2 + r^2 dθ^2\] 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.

With these new definitions of R and dr, our relationship is:  \[ ds_{2d}^2 = 0 + R^2 dθ^2\] or \[ ds_{2d} = R dθ\] Taking the integral of each side gives us an expression for circumference:  \[ \int_0^C ds = \int_0^{2 \pi} R dθ\] \[ C = 2 \pi R \] This looks pretty familiar, now we can move on to less familiar geometries. \[ \frac{C}{R} = \frac{2 \pi R}{R} = 2 \pi \]

(b) For a closed geometry, we calculated the analogous two-dimensional part of the metric in Problem (1). This can be written as: \[ds_{2d}^2 = dξ^2 + sin(ξ)^2 dθ^2\] 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.)

Radius:
With these new definitions of R and dr, the metric goes from: \[ds_{2d}^2 = dξ^2 + sin(ξ)^2 (dθ^2 + sin(θ)^2 d \phi^2) \] to \[ds_{2d}^2 = dξ^2 \]  or \[ ds_{2d} = dξ\] Taking the indefinite integral of each side gives us an expression for radius: \[ \int ds_{2d} = \int dξ\] \[ s =  ξ = R \]
Circumference:
Once again, our relationship simplifies nicely. \[ds_{2d}^2 = dξ^2 + sin(ξ)^2 dθ^2\]  \[ds_{2d}^2 = sin(ξ)^2 dθ^2 \] \[ds_{2d} = sin(ξ) dθ \] Taking the integral with the outlined bounds to find circumference:   \[ \int_0^C ds = \int_0^{2 \pi} sin(ξ)  dθ\] \[ C = 2 \pi sin(ξ)  \] This expression is less familiar to us, but does resemble \( C = 2 \pi R\) with our definition of R in this case.
Ratio: \[ \frac{C}{R} = \frac{2 \pi sin(ξ)}{ξ}  \] Because this circumference is dependent on a sine term, the circumference will never be greater than \( 2 \pi \) or less than \( -2 \pi \) so this ratio will always be equal to or less than the ratio found in Euclidian Space in part a.

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

Radius:
In an open geometry, the procedure for radius is the same. \[ds_{2d}^2 = dξ^2 + sin(ξ)^2 (dθ^2 + sin(θ)^2 d \phi^2) \] \[ ds_{2d} = dξ\]  \[ \int ds_{2d} = \int dξ\] \[ s =  ξ = R \] Circumference: In the metric for open geometry is R = sinh(x) so the metric simplifies to  \[ds_{2d}^2 = sinh(ξ)^2 dθ^2 \]  \[ \int_0^C ds = \int_0^{2 \pi} sinh(ξ)  dθ\] \[ C = 2 \pi sinh(ξ)  \] Ratio:  \[ \frac{C}{R} = \frac{2 \pi sinh(ξ)}{ξ}  \] This ratio will always be larger or equal to the ratio in the flat, and closed universe.

(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?

The two ratios we found are :  \[ \frac{C}{R} = \frac{2 \pi sin(ξ)}{ξ}  \] \[ \frac{C}{R} = \frac{2 \pi sinh(ξ)}{ξ}  \] As we approach 0, we can use the small angle approximation for both of these functions, (e.g. sin(ξ) is ξ) so as the limit goes to zero, both ratios go to zero, approaching the flat case (this is easy to visualize in the graph above). In class, Ashley compared this geometry to us being small human beings on a large Earth. On the whole, we know the Earth is a three-dimensional sphere, but at small scales, as the limit approaches zero, the Earth looks flat.

Representations of the three types on universes - Source

Blog 30: Galactic Rotation

Next week we will begin our second lab of the semester. We will use the millimeter-wave telescope, a radio telescope, in order to observe giant molecular clouds in the Milky Way. This is a lot of new information at once, so we are going to answer some questions before we actually head to lab.

Why? - We are observing GMCs in order to find their redshifts. From redshift, we can find the velocity of a GMC along our line of sight. With that velocity, we can find the circular velocity of a GMC rotating in the Milky Way using \[ V_{cir} = V_r(max) + V_{\odot} sin(l) \] where Vr is the radial velocity of the GMC moving away from us and the second term is the velocity of the sun in the direction of the cloud.

In previous blog posts  we explored why the rotation curve of the galaxy is important in astronomy. The basic idea is that we observe objects to be rotating too quickly, too far out in the galaxy just based on visible matter. This is our main indicator that there is another type of dark matter in the universe. Additionally, we can learn more about the Milky Way's visible matter from this experiment. For example, we can estimate the number of stars in the galaxy. 

How? - The telescope will detect radiation at the wavelength near 2.6 mm, or a frequency of 115.271 GHz. This wavelength corresponds with a strong emission line of Carbon Monoxide. We will obtain this spectrum from the telescope's spectrometer, one of many of the telescope's parts among an antenna for collecting and focusing the signal, a liquid helium cooler, and a computer for recording and interpreting data.

What is going to be the typical integration time per point? - What integration time would be needed to detect the peak of 12CO with SNR = 10 if I use a filter bank with 256 channels that are 0.5 MHz wide. TA for CO is about 2-3 K and Tsys = 500K?

In worksheet 9.2 we found \[ \tau = \frac{ SNR^2 T_{sys}^2 }{T_A^2 \Delta \nu } \] \[ \tau = \frac{ (10)^2 (500)^2 }{(2.5)^2 (0.5 \times 10^6) } \approx 8 \: seconds \]

Over what range of longitude do you plan to observe? - We are planning on observing from \(10^{\circ} \: \text{ to } 70^{\circ} \). 

How many positions do you plan to observe? - We plan on observing 4 GMCs, so we will most likely observe 4 positions.

At what LST are you going to start observing? At what EST? - I have been assigned to Monday's lab from 11:00 - 3:00 EST. Let's say that means we start observing at 11:30 EST. The LST on that date and time at our longitude (71.11 degrees west) would be 15:55.

Additionally, we can solve question number two on the worksheet to learn more about the experiment.  

Resolution of a single dish radio telescope: The spatial resolution of a telescope is θ = 1.2 λ/D , where D is the diameter of the dish, and λ is the observing wavelength. In radio astronomy, θ is also known as the half-power beam width (or full-width half-max of the beam). 

(a) Find an equation for the beam width, in arcminutes, of a single-dish radio in terms of frequency ν in GHz, and diameter D in meters. Use a calculator to determine this to 1 decimal place. Write an equation of the form \[ \theta = degrees \left( \frac{ \nu}{GHz} \right)^{-1} \left( \frac{D}{m} \right)^{-1} \] We can solve this by redefining theta: \[ \theta = \frac{c}{\nu D} = \frac{1.2 (3 \times 10^8) }{(1 \times 10^9)(1)} = 0.36 \: rad \left( \frac{180^{\circ}}{ \pi rad } \right) = 20.6^{\circ} \] \[ \theta = 20.6^{\circ} \left( \frac{ \nu}{GHz} \right)^{-1} \left( \frac{D}{m} \right)^{-1} \] 

(b) What is the beamwidth, in arcminutes, for the CfA 1.2 m telescope at the 12CO frequency?

Using the formula we just found, we know \[ \theta = 20.6^{\circ} \left( \frac{ \nu}{GHz} \right)^{-1} \left( \frac{D}{m} \right)^{-1} \] \[ \theta = 20.6^{\circ} \left( \frac{ 115.271 \: GHz}{GHz} \right)^{-1} \left( \frac{1.2 \: m}{m} \right)^{-1} = 0.149^{\circ} = 8.95 \: arcminutes \] 

(c) What linear dimension in pc does this correspond to at the Galactic center (8.5 kpc)?

We can call on our trusty parallax formula for this problem: \[ d = \frac{1}{ \theta} \] Where 8.95 arcminutes is 537 arcseconds, so \[ d = \frac{1}{ 537} = 1.9 \times 10^{-3} \: pc \] Now we should be all set to start lab.

Correction: The parallax is not a terribly useful number in this case. Rather the question is asking for resolvable size, which we could find with simple trigonometry. \[sin( \theta) = \frac{l}{8.5 \: kpc} \] With the small angle approximation, this is \[ l \approx 0.149 \cdot \frac{\pi}{180} \cdot 8,500 \approx 22 \: pc\]

More Friedmann Equations: Blog 29, Worksheet 9.1, Problem 2

GR modification to Newtonian Friedmann Equation:

In Question 1, you have derived the Friedmann Equation in a matter-only universe in the Newtonian 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. Use the first Friedmann equation: \[ \left( \frac{\dot{a}}{a} \right)^2 = \frac{8 \pi}{3} G \rho + \frac{ \kappa c^2 }{a^2} + \frac{ \Lambda }{3} \] and the second Friedmann equation: \[ \frac{\ddot{a}}{a}  = - \frac{4 \pi G}{3c^2} ( \rho c^2 + 3P) + \frac{ \Lambda }{3} \] To find the third equation in a flat universe \[ \dot{ \rho} c^2 = -3 \frac{\dot{a}}{a}( \rho c^2 + P) \]

With the universe being flat and multiplying the first equation by acceleration squared gives us: \[ \ddot{a}^2 = \frac{8 \pi}{3} G \rho a^2+ \frac{ \Lambda }{3} a^2 \] Taking the time derivative gives us \[ 2 \dot{a} \ddot{a} = \frac{8 \pi}{3} G \dot{\rho} a^2 + \frac{16 \pi}{3} G \rho \dot{a} a   + \frac{2  \Lambda }{3} \dot{a} a \] Rearranging this allows us to set it equal to the second equation: \[  \frac{ \ddot{a} }{a} = \frac{4 \pi a}{3  \dot{a} } G \dot{\rho}  + \frac{ 8 \pi}{3} G \rho + \frac{ \Lambda }{3} =  - \frac{4 \pi G}{3c^2} ( \rho c^2 + 3P) + \frac{ \Lambda }{3} \] Which simplifies with a few cancellations to: \[ \dot{ \rho} c^2 = -3 \frac{\dot{a}}{a}( \rho c^2 + P) \]

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 ρ as a function of the scale factor of the universe a. You can leave this equation in terms of ρ, ρ0, a and a0, where ρ0 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 ρ 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. Solve 2 Astronomy 17 - Galactic and Extragalactic Astronomy Fall 2015 this differentiation equation to show that \(a(t) \propto t^{\frac{3}{2}}\) . This is the characteristic expansion history of the universe if it is dominated by matter.

Without pressure, the third equation can be written as \[ \frac{\dot{\rho}}{\rho} = -3 \frac{\dot{a}}{a}\] Taking the derivative of both sides with respect to time starting at \( \rho_0 \: \text{and} \: a_0 \) produces:  \[ \frac{\rho}{\rho_0} = \left(\frac{a}{a_0} \right)^{-3}\] Plugging this relationship into the first equation gives: \[ \left( \frac{\dot{a}}{a} \right)^2 = \frac{8 \pi}{3} G \left( \frac{a_0^3 \rho_0}{a^3} \right) \] Taking the derivative of each side gives \[ a^{\frac{1}{2}} = c \: dt \] Where c is a constant (because we just want a relationship between a and t). Taking the integral of each side gives \[ \frac{2}{3} a^{\frac{3}{2} } = c t \] So \[ a(t) \propto t^{\frac{3}{2}} \]

C) Radiation dominated universe. Let us repeat the above exercise for a universe filled with radiation only. For radiation, \(P = \frac{1}{ 3} ρc^2 \)and Λ = 0. Again, use the third Friedmann equation to see how the density of the radiation changes as a function of scale factor. The result also has a simple interpretation. Imagine the radiation being a collection of photons. Similar to the matter case, the number density of the photon is diluted, inversely proportional to the volume. Now the difference is that, in contrast to the dust particle, each photon can be thought of as wave. As you learned last week, the wavelength of the photon is also stretched as the universe expands, proportional to the scale factor of the universe. According to quantum mechanics, the energy of each photon is inversely proportional to its wavelength: E = hν. Unlike the dust case where each particle has a fixed energy. So in an expanding universe, the energy of each photon is decreasing inversely proportional to the scale factor. Check that this understanding is consistent with the result you got. Again using the relation between ρ and a and the first Friedmann equation to show that \(a(t) \propto t^{\frac{1}{2}}\) for the radiation only universe.

With the new pressure, the third equation can be written as \[ \frac{\dot{\rho}}{\rho} = -4 \frac{\dot{a}}{a}\] Taking the derivative of both sides with respect to time starting at \( \rho_0 \: \text{and} \: a_0 \) produces:  \[ \frac{\rho}{\rho_0} = \left(\frac{a}{a_0} \right)^{-4}\] Plugging this relationship into the first equation gives: \[ \left( \frac{\dot{a}}{a} \right)^2 = \frac{8 \pi}{3} G \left( \frac{a_0^4 \rho_0}{a^4} \right) \] Taking the derivative of each side gives \[ a = c \: dt \] Where c is a constant (because we just want a relationship between a and t). Taking the integral of each side gives \[ \frac{1}{2} a^{2 } = c t \] So \[ a(t) \propto t^{\frac{1}{2}} \]

D) Cosmological constant/dark energy dominated universe. 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. As a digression, notice that we said “cosmological-constant-like” term. This is because the effect of the cosmological constant may be mimicked by a special content of the universe which has a negative pressure \(P = - ρc^2 \). Check that the effect of this content on the right-hand-side of third Friedmann equation is exactly like that of the cosmological constant. To be general we call this content the Dark Energy. How does the energy density of the dark energy change in time? Show that the scale factor of the cosmological-constant-dominated universe expands exponentially in time. What is the Hubble parameter of this universe?

From the first equation we have: \[ \left( \frac{\dot{a}}{a} \right)^2 =  \frac{ \Lambda }{3} \] Taking the derivative gives: \[ \frac{da}{a} = \sqrt{ \frac{ \Lambda }{3} } dt \] Ignoring constants, the integral of this relationship turns into \[ ln(a) \propto t \] \[ a \propto e^t \] And because \[ H=  \frac{\dot{a}}{a} \] The Hubble constant in this universe is  \[ H = \sqrt{ \frac{ \Lambda }{3} } \]

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?

Radiation will dominate because it falls off at a slower rate ( \(a(t) \propto t^{\frac{1}{2}}\) ) than matter ( \(a(t) \propto t^{\frac{2}{3}}\) ).

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?

Dark energy will be dominant as the universe expands and matter is scattered. Energy, and therefore normal matter, cannot be created or destroyed, by the law of conservation of energy. We have also shown that dark energy will increase at a faster rate ( \(a \propto e^t\) ) than radiation ( \(a(t) \propto t^{\frac{2}{3}}\) ). Eventually, our universe will be a very sparse, vast expanse of dark energy with few pockets of matter spread throughout.

Cosmology: Blog 28, Worksheet 9.1, Problem 1

A Matter-only Model of the Universe in Newtonian Approach 

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 ρ(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).

Because this acceleration is due to gravity, we know \[ ma  = F_g \] \[ m\dot{v} = - \frac{GMm}{R^2} \] \[ \dot{v} = - \frac{GM}{R^2} \]

(b) To derive an energy equation, it is a common trick to multiply both sides of your acceleration equation by v. You should arrive at the following equation: \[ \frac{1}{2} \dot{R}^2 - \frac{GM}{R} = C \]

Using what we found in a: \[ \frac{dR}{dt} \dot{v} = - \frac{GM}{R^2} \frac{dR}{dt} \] Which simplifies to \[v \: dv = - \frac{GM}{R^2} dR \] The integral of this is \[ \frac{1}{2} v^2 = \frac{GM}{R} + C \] Where C is a constant. Substituting \( v = \dot{R} \) we have: \[ \frac{1}{2} \dot{R}^2 - \frac{GM}{R} = C \]

(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 \( \left( \frac{ \dot{R}}{R} \right)^2 \), where \(  \dot{R} = \frac{dR}{dt}  \). \[ M = \rho V = \frac{4}{3} \pi R^3 \rho \] Plugging this into our expression from part B gives us:  \[ \frac{1}{2} \dot{R}^2 - \frac{G \left( \frac{4}{3} \pi R^3 \rho \right)}{R} = C \] Rearranging this makes it: \[ \left( \frac{\dot{R}}{R} \right)^2 = \frac{2C}{R^2} + \frac{8 \pi G}{3} \rho (t) \]

(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). \[  \frac{\dot{R}}{R} =  \frac{\dot{a}(t) r}{a(t) r} =  \frac{\dot{a}}{a} \] So our expression is  \[ \left( \frac{\dot{a}}{a} \right)^2 = \frac{2C}{R^2} + \frac{8 \pi G}{3} \rho (t) \]

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

(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 \( \frac{2C}{r} = \kappa c^2 \) 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. \[ \left( \frac{\dot{a}}{a} \right)^2 = \frac{8 \pi}{3} G \rho - \frac{ \kappa c^2 }{a^2} \]

(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 a: \( \frac{\ddot{a}}{a}  = - \frac{4 \pi G}{3c^2} \rho \), 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.
\[ \dot{v} = - \frac{GM}{R^2} = -\frac{4}{3} \pi G \rho (t) a(t) r \] We know \( \dot{v} = \ddot{R} \) so \(  \dot{v} = \ddot{a}r \) and \[\ddot{a}r = -\frac{4}{3} \pi G \rho (t) a(t) r \]  \[ \frac{\ddot{a}}{a}  = - \frac{4 \pi G}{3c^2} ( \rho c^2 + 3P) + \frac{ \Lambda }{3} \]