Category: cosmology

Cosmology – Transit Method The Transit Method (Kepler Space Telescope) is used to find out the size. The dip in brightness of a star by a planet is usually very less unlike a binary system. F0 is flux of the star before the planet occults it. F1 is the flux after the entire planet is in front of star. The following image will be used for all the calculations. $$frac{F_0 – F_1}{F_0} = frac{pi r_p^{2}}{pi R^2_ast}$$ $$frac{Delta F}{F} cong frac{r^2_p}{R^2_ast}$$ $$left ( frac{Delta F}{F} right )_{earth} cong 0.001%$$ $$left ( frac{Delta F}{F} right )_{jupiter} cong 1%$$ This is not easy to achieve by ground based telescope. It is achieved by the Hubble telescope. Here, $t_T$ is time between position A and D and $t_F$ is time between position B and C. The geometry of a transit related to the inclination i of the system. Transit latitude and inclination are interchangeable. From the above images, we can write − $$frac{h}{a} = cos(i)$$ $$frac{h}{R_ast} = sin(delta)$$ $$cos(i) = frac{R_ast sin(delta)}{a}$$ $$y^2 = (R_ast + R_p)^2 – h^2$$ $$y = [(R_ast + R_p)^2 – h^2]^{frac{1}{2}}$$ $$sin(theta) = frac{y}{a}$$ $$theta = sin^{-1}left [ frac{(R_ast + R_p)^2 – a^2cos^2(i)}{a^2} right ]^{frac{1}{2}}$$ $$t_T = frac{P}{2pi} times 2theta$$ Here, $t_T$ is the fraction of a time-period for which transit happens and (2θ/2π) is fraction of angle for which the transit happens. $$sin(frac{t_Tpi}{P}) = frac{R_ast}{a}left [ left ( 1+ frac{R_p}{R_ast}right )^2 – left ( frac{a}{R_ast}cos(i)right )^2 right ]^{frac{1}{2}}$$ Usually, a >> R∗ >> Rp. So, we can write − $$sin(frac{t_Tpi}{P}) = frac{R_ast}{a}left [ 1- left ( frac{a}{R_ast}cos(i) right )^2right ]^{frac{1}{2}}$$ Here, P is the duration between two successive transits. The transit time is very less compared to the orbital time-period. Hence, $$t_T = frac{P}{pi}left [ left ( frac{R_ast}{a}right )^2 – cos^2(i)right ]^{frac{1}{2}}$$ Here, tT, P, R∗ are the observables, a and i should be found out. Now, $$sin(frac{t_Fpi}{P}) = frac{R_ast}{a}left [left (1 – frac{R_p}{R_ast} right )^2 – left ( frac{a}{R_ast}cos:i right )^2right ]^{frac{1}{2}}$$ where, $y^2 = (R_ast − R_p)^2 − h^2$. Let, $$frac{Delta F}{F} = D = left ( frac{R_p}{R_ast} right )^2$$ Now, we can express, $$frac{a}{R_ast} = frac{2P}{pi} D^{frac{1}{4}}(t^2_T – t^2_F)^{-frac{1}{2}}$$ For the main sequence stars, $$R_ast propto M^alpha_ast$$ $$frac{R_ast}{R_0} propto left ( frac{M_ast}{M_0}right )^alpha$$ This gives R∗. Hence, we get value of ‘a’ also. So, we get ‘Rp’, ’ap’ and even ‘i’. For all this, $$h leq R_ast + R_p$$ $$a: cos: i leq R_ast + R_p$$ For even 𝑖~89 degrees, the transit duration is very small. The planet must be very close to get a sufficient transit time. This gives a tight constraint on ‘i’. Once we get ‘i’, we can derive ‘mp′from the radial velocity measurement. This detection by the transit method is called as chance detection, i.e., probability of observing a transit. Transit probability (probability of observing) calculations are shown below. The transit probability is related to the solid angle traced out by the two extreme transit configurations, which is − $$Solid : angle :of :planet : = 2pi left ( frac{2R_ast}{a} right )$$ As well as the total solid angle at a semi-major axis a, or − $$Solid :angle :of :sphere : =: 4pi$$ The probability is the ratio of these two areas − $$= : frac{area :of: sky : covered :by: favourable : orientation}{area: of: sky :covered: by: all: possible: orientation: of: orbit}$$ $= frac{4pi a_pR_ast}{4pi a^2_p} = frac{R_ast}{a_p}$ $frac{area: of: hollow : cyclinder}{area: of: sphere}$ This probability is independent of the observer. Points to Remember The Transit Method (Kepler Space Telescope) is used to find out the size. Detection by Transit Method is a chance detection. The planet must be very close to get sufficient transit time. Transit probability is related to the solid angle of planet. This probability is independent of observer frame of reference. Learning working make money

Cosmology – Radial Velocity Method In the previous chapter, the Radial Velocity Method for the case in which the orbital plane and the plane of sky are perpendicular was discussed for circular orbits. Here, we deal with one more case, when the orbital plane and the plane of sky are not perpendicular for circular orbits. When the orbital plane is at an angle with respect to the plane of sky (not perpendicular), we have the following situation − In this case, when they were perpendicular, we had two points at which we could measure the true velocity. But here, that is not possible. At all points, we can measure only a component of the true velocity, v. $$v_r = v :sin(i)cos(theta)$$ where θ is the phase of the orbit which is a time-dependent quantity. The inclination angle i on the other hand, is time-independent. Hence, $$(v_r)_{max} = v: sin(i)$$ The observed radial velocity curve will be of the following form − When the orbital plane is perpendicular to the sky − $$m_p = left ( frac{P}{2pi G} right )^{frac{1}{3}}(M_ast)^{frac{2}{3}}v$$ where mp, P, G, M∗ are mass of the planet, orbital period, universal gravitational constant and mass of the star respectively. But in this case, we should modify it as follows − $$m_psin(i) = left ( frac{P}{2pi G} right )^{frac{1}{3}} (M_ast)^{frac{2}{3}}(v_r)_{max}$$ But, finding the value of i is a difficult task. We can impose certain constraints on the value of i using transit method. The passage of the planet between the star and the Earth is called a transit. We can obtain the light curve by observing a transit and a significant dip in observed flux of a light curve implies that i is close to 90 degrees. If such conditions are not satisfied, we cannot have any idea on the value of i. Then the value of mp that we find can serve as a lower limit for the mass of the planet, since it is actually mp sin(i) and sin(i) ≤ 1. To conclude, the Radial velocity method is more convenient than the transit method because radial velocity can be measured at any time but transit measurements can be made only during the transit which may not last for long. Points to Remember Finding inclination of the planet’s orbit is not achieved by Radial Velocity method. Radial Velocity Method is better than the Transit Method because radial velocity can be measured always unlike transits. Transits are short-lived and very easy to miss. Learning working make money

Anisotropy of CMB Radiation & Cobe In this chapter, we will discuss the anisotropy of CMB Radiation and COBE, i.e., Cosmic Background Explorer. Primary Anisotropies in the CMB To understand the observations from space and the primary anisotropies in the Cosmic Microwave Background Radiation, let us take the following equations and understand it as shown below. CMB Photon Number Density (nγ,0) $$n_{gamma,0} = frac{Total : energy : density}{Characteristic : energy :of :Photons}$$ $$n_{gamma,0} = frac{aT_0^4}{k_BT_0}$$ Where $k_B$ is Boltzmann Constant and $T_0$ is the present temperature of the universe. Using the present temperature $(T_0)$ as 2.7 K, we get the current CMB photon number density as 400 cm−3. The cosmic stellar photon number density is much smaller (∼= 10−3 cm−3) over large scales. Baryon to Photon ratio (η) If the stellar contributions from galaxies, which get mixed with CMB, are negligible, the baryon to proton ratio is − $$eta = frac{n_{b,0}}{n_{gamma,0}}$$ The present value is ∼5 × 10−10. Since both photon and baryon number densities are proportional to a−3, then η doesn’t evolve with time. Energy Density As opposed to the number density, the matter energy density is more dominated than photon energy density at present. The Energy density of baryonic matter = $rho_{b,0}c^2 = 0.04rho_cc^2 = 2 × 10^{−9} ergcm^{−3}$. While, the Energy density of radiation = $aT_0^4 = 4 times 10^{−13}ergcm{−3}$. Isotropy of CMB Radiation Penzias and Wilson found the CMB to be isotropic within the limits of observations. The limits are low angular resolution and sensitivity of instruments. They made observations from earth, due to this, observations cannot be made through all the spectrum as water vapor in the atmosphere absorbs many wavelengths ranging from 1mm to 1m. So, CMB can’t be asserted as a spectrum. The CMB is thought to be rotationally invariant (isotropic). Since there existed a time when matter and radiation were in equilibrium, then the formation of structures in the universe is unexplainable. Since the distribution of matter is not isotropic but is clumped together like a cosmic web with huge voids in between, CMB is thought to have an extragalactic origin. But, as the observations from the space began, anisotropies in the CMB were found, which lead to the reasoning that these anisotropies in matter lead to the formation of structures. Observation of CMB Radiation from Space The main satellites which were launched to observe the CMB were − Cosmic Microwave Background Explorer (COBE, 1989) Wilkinson Microwave Anisotropy Probe (WMAP, 2001) and Planck (2009). COBE (Cosmic Background Explorer) COBE mainly had two instruments. They were Far InfraRed Absolute Spectrometer (FIRAS) and Differential Microwave Radiometers (DMR Antennas). FIRAS measures intensity of the CMB as a function of wavelength along any specific direction. Whereas, DMR has 3 antennas to measure the difference in intensity of CMB from three different directions. The following pointers give us some more information on FIRAS and DMR. CMB observations from FIRAS show that the CMB radiation corresponds to black body spectrum at T = 2.72528±0.00065 K. The DMR measures three frequencies (31.5 GHz, 53 GHz, 90 GHz) in all directions in the sky. The “red batman symbol” in the DMR observations is noise from foreground emission (galactic diffused synchrotron emission). The intensity variations in the observations correspond to temperature variations. The presence of hot and cold spots proves that the CMB radiation is anisotropic. This anisotropy must be present at decoupling time as there are no distortions in CMB. So, matter should have some pockets with higher density than that of the others. COBE Results The CMB spectrum (intensity as a function of energy) is nearly a perfect black body corresponding to T = 2.7 K. The specific intensity of the CMB radiation is nearly the same for all directions. Confirmation that universe is isotropic at large scales (validates our assumption of cosmological principle). Analysis of the data showed that there are temperature anisotropies (“fluctuations”) in the CMB spectrum at the resolution of COBE (DMR). Resolution of COBE, WMAP, Planck The DMR instrument on-board COBE had a limiting (maximum) spatial resolution of ∼ 7 degrees. Wilkinson Microwave Anisotropy Probe (WMAP) had an average resolution of ∼ 0.7 degrees. Planck satellite has an angular resolution of ∼ 10 arc-minute. Points to Remember Cosmic stellar photon number density is much smaller than the CMB photon number density. We live in a matter dominated universe, since matter energy density is higher than the photon energy density. COBE, WMAP, Planck are efforts to measure and quantify anisotropies in the CMB. The formation of structure in the universe is a result of CMB anisotropies. Learning working make money

Cosmology – Age of Universe As discussed in the earlier chapters, the time evolution of the Hubble parameter is given by − $$H(z) = H_0E(z)^{frac{1}{2}}$$ Where z is the red shift and E(Z) is − $$E(z) equiv Omega_{m,0}(1+z)^3 + Omega(1+z)^4 +Omega_{k,0}(1+z)^2 + Omega^{wedge,0}$$ If the expansion of the universe is constant, then the true age of the universe is given as follows − $$t_H = frac{1}{H_0}$$ If it is the matter dominated universe, i.e., Einstein Desitter universe, then the true age of universe is given by − $$t_H = frac{2}{3H_0}$$ Scale and Redshift is defined by − $$a=frac{a_0}{1+z}$$ Age of the universe in terms of the cosmological parameter is derived as follows. The Hubble Parameter is given by − $$H = frac{frac{da}{dt}}{a}$$ Differentiating, we get − $$da = frac{-dz}{(1+z)^2}$$ Where a0 = 1 (present value of the scale factor) $$frac{mathrm{d} a}{mathrm{d} t} = frac{-1}{(1+z)^2}$$ $$frac{mathrm{d} a}{mathrm{d} t} = frac{mathrm{d} a}{mathrm{d} t}frac{mathrm{d} z}{mathrm{d} t}$$ $$H = frac{dot{a}}{a} = frac{mathrm{d} a}{mathrm{d} t}frac{mathrm{d} z}{mathrm{d} t} frac{1+z}{1}$$ $$frac{dot{a}}{a} = frac{-1}{1+z}frac{mathrm{d} z}{mathrm{d} t}frac{1}{1}$$ $$H(z) = H_0E(z)^{frac{1}{2}}$$ $$dt = frac{-dz}{H_0E(z)^{frac{1}{2}}(1+z)}$$ If we want to find the age of the universe at any given redshift ‘z’ then − $$t(z) = frac{1}{H_0}int_{infty}^{z_1} frac{-1}{E(z)^{frac{1}{2}}(1+z)}dz$$ Where k is the curvature density parameter and − $$E(z) equiv Omega_{m,0}(1+z)^3 + Omega_{rad,0}(1+z)^4 + Omega_{k,0}(1+z)^2 + Omega_{wedge,0}$$ To calculate the present age of the universe, take z1 = 0. $$t(z=0) = t_{age} = t_0 = frac{1}{H_0}int_{infty}^{z_1} frac{-1}{E(z)^{frac{1}{2}}(1+z)}dz$$ For the Einstein Desitter Model, i.e, $Omega_m = 1$, $Omega_{rad} = 0$, $Omega_k = 0$, $Omega_wedge = 0$, the equation for the age of the universe becomes − $$t_{age} = frac{1}{H_0}int_{0}^{infty} frac{1}{(1+z)^{frac{5}{2}}}dz$$ After solving the integral, we get − $$t_H = frac{2}{3H_0}$$ The night sky is like a Cosmic Time Machine. Whenever we observe a distant planet, star or galaxy, we are seeing it as it was hours, centuries or even millennia ago. This is because light travels at a finite speed (the speed of light) and given the large distances in the Universe, we do not see objects as they are now, but as they were when the light was emitted. The time elapsed between – when we detect the light here on Earth and when it was originally emitted by the source, is known as the Lookback Time (tL(z1)). So, the lookback time is given by − $$t_1(z_1) = t_0-t(z_1)$$ The lookback time for the Einstein Desitter Universe is − $$t_L(z) = frac{2}{3H_0}left [ 1- frac{1}{(1+z)^{frac{3}{2}}} right ]$$ Points to Remember Whenever we observe a distant planet, star or galaxy, we are seeing it as it was hours, centuries or even millennia ago. The time elapsed between – when we detect the light here on Earth and when it was originally emitted by the source, is known as the lookback time. Learning working make money

Cosmology – Angular Diameter Distance In this chapter, we will understand what the Angular Diameter Distance is and how it helps in Cosmology. For the present universe − $Omega_{m,0} : = : 0.3$ $Omega_{wedge,0} : = : 0.69$ $Omega_{rad,0} : = : 0.01$ $Omega_{k,0} : = : 0$ We’ve studied two types of distances till now − Proper distance (lp) − The distance that photons travel from the source to us, i.e., The Instantaneous distance. Comoving distance (lc) − Distance between objects in a space which doesn’t expand, i.e., distance in a comoving frame of reference. Distance as a Function of Redshift Consider a galaxy which radiates a photon at time t1 which is detected by the observer at t0. We can write the proper distance to the galaxy as − $$l_p = int_{t_1}^{t_0} cdt$$ Let the galaxy’s redshift be z, $$Rightarrow frac{mathrm{d} z}{mathrm{d} t} = -frac{1}{a^2}frac{mathrm{d} a}{mathrm{d} t}$$ $$Rightarrow frac{mathrm{d} z}{mathrm{d} t} = -frac{frac{mathrm{d} a}{mathrm{d} t}}{a}frac{1}{a}$$ $$therefore frac{mathrm{d} z}{mathrm{d} t} = -frac{H(z)}{a}$$ Now, comoving distance of the galaxy at any time t will be − $$l_c = frac{l_p}{a(t)}$$ $$l_c = int_{t_1}^{t_0} frac{cdt}{a(t)}$$ In terms of z, $$l_c = int_{t_0}^{t_1} frac{cdz}{H(z)}$$ There are two ways to find distances, which are as follows − Flux-Luminosity Relationship $$F = frac{L}{4pi d^2}$$ where d is the distance at the source. The Angular Diameter Distance of a Source If we know a source’s size, its angular width will tell us its distance from the observer. $$theta = frac{D}{l}$$ where l is the angular diameter distance of the source. θ is the angular size of the source. D is the size of the source. Consider a galaxy of size D and angular size dθ. We know that, $$dtheta = frac{D}{d_A}$$ $$therefore D^2 = a(t)^2(r^2 dtheta^2) quad because dr^2 = 0; : dphi ^2 approx 0$$ $$Rightarrow D = a(t)rdtheta$$ Changing r to rc, the comoving distance of the galaxy, we have − $$dtheta = frac{D}{r_ca(t)}$$ Here, if we choose t = t0, we end up measuring the present distance to the galaxy. But D is measured at the time of emission of the photon. Therefore, by using t = t0, we get a larger distance to the galaxy and hence an underestimation of its size. Therefore, we should use the time t1. $$therefore dtheta = frac{D}{r_ca(t_1)}$$ Comparing this with the previous result, we get − $$d_wedge = a(t_1)r_c$$ $$r_c = l_c = frac{d_wedge}{a(t_1)} = d_wedge(1+z_1) quad because 1+z_1 = frac{1}{a(t_1)}$$ Therefore, $$d_wedge = frac{c}{1+z_1} int_{0}^{z_1} frac{dz}{H(z)}$$ dA is the Angular Diameter Distance for the object. Points to Remember If we know a source’s size, its angular width will tell us its distance from the observer. Proper distance is the distance that photons travel from the source to us. Comoving distance is the distance between objects in a space which doesn’t expand. Learning working make money

Cosmology – Hubble & Density Parameter In this chapter, we will discuss regarding the Density and Hubble parameters. Hubble Parameter The Hubble parameter is defined as follows − $$H(t) equiv frac{da/dt}{a}$$ which measures how rapidly the scale factor changes. More generally, the evolution of the scale factor is determined by the Friedmann Equation. $$H^2(t) equiv left ( frac{dot{a}}{a} right )^2 = frac{8pi G}{3}rho – frac{kc^2}{a^2} + frac{wedge}{3}$$ where, ∧ is a cosmological constant. For a flat universe, k = 0, hence the Friedmann Equation becomes − $$left ( frac{dot{a}}{a} right )^2 = frac{8pi G}{3}rho + frac{wedge}{3}$$ For a matter dominated universe, the density varies as − $$frac{rho_m}{rho_{m,0}} = left ( frac{a_0}{a} right )^3 Rightarrow rho_m = rho_{m,0}a^{-3}$$ and, for a radiation dominated universe the density varies as − $$frac{rho_{rad}}{rho_{rad,0}} = left ( frac{a_0}{a} right )^4 Rightarrow rho_{rad} = rho_{rad,0}a^{-4}$$ Presently, we are living in a matter dominated universe. Hence, considering $rho ≡ rho_m$, we get − $$left ( frac{dot{a}}{a} right )^2 = frac{8pi G}{3}rho_{m,0}a^{-3} + frac{wedge}{3}$$ The cosmological constant and dark energy density are related as follows − $$rho_wedge = frac{wedge}{8 pi G} Rightarrow wedge = 8pi Grho_wedge$$ From this, we get − $$left ( frac{dot{a}}{a} right )^2 = frac{8pi G}{3}rho_{m,0}a^{-3} + frac{8 pi G}{3} rho_wedge$$ Also, the critical density and Hubble’s constant are related as follows − $$rho_{c,0} = frac{3H_0^2}{8 pi G} Rightarrow frac{8pi G}{3} = frac{H_0^2}{rho_{c,0}}$$ From this, we get − $$left ( frac{dot{a}}{a} right )^2 = frac{H_0^2}{rho_{c,0}}rho_{m,0}a^{-3} + frac{H_0^2}{rho_{c,0}}rho_wedge$$ $$left ( frac{dot{a}}{a} right )^2 = H_0^2Omega_{m,0}a^{-3} + H_0^2Omega_{wedge,0}$$ $$(dot{a})^2 = H_0^2Omega_{m,0}a^{-1} + H_0^2Omega_{wedge,0}a^2$$ $$left ( frac{dot{a}}{H_0} right )^2 = Omega_{m,0}frac{1}{a} + Omega_{wedge,0}a^2$$ $$left ( frac{dot{a}}{H_0} right )^2 = Omega_{m,0}(1+z) + Omega_{wedge,0}frac{1}{(1+z)^2}$$ $$left ( frac{dot{a}}{H_0} right)^2 (1+z)^2 = Omega_{m,0}(1+z)^3 + Omega_{wedge,0}$$ $$left ( frac{dot{a}}{H_0} right)^2 frac{1}{a^2} = Omega_{m,0}(1 + z)^3 + Omega_{wedge,0}$$ $$left ( frac{H(z)}{H_0} right )^2 = Omega_{m,0}(1+z)^3 + Omega_{wedge,0}$$ Here, $H(z)$ is the red shift dependent Hubble parameter. This can be modified to include the radiation density parameter $Omega_{rad}$ and the curvature density parameter $Omega_k$. The modified equation is − $$left ( frac{H(z)}{H_0} right )^2 = Omega_{m,0}(1+z)^3 + Omega_{rad,0}(1+z)^4+Omega_{k,0}(1+z)^2+Omega_{wedge,0}$$ $$Or, : left ( frac{H(z)}{H_0} right)^2 = E(z)$$ $$Or, : H(z) = H_0E(z)^{frac{1}{2}}$$ where, $$E(z) equiv Omega_{m,0}(1 + z)^3 + Omega_{rad,0}(1+z)^4 + Omega_{k,0}(1+z)^2+Omega_{wedge,0}$$ This shows that the Hubble parameter varies with time. For the Einstein-de Sitter Universe, $Omega_m = 1, Omega_wedge = 0, k = 0$. Putting these values in, we get − $$H(z) = H_0(1+z)^{frac{3}{2}}$$ which shows the time evolution of the Hubble parameter for the Einstein-de Sitter universe. Density Parameter The density parameter, $Omega$, is defined as the ratio of the actual (or observed) density ρ to the critical density $rho_c$. For any quantity $x$ the corresponding density parameter, $Omega_x$ can be expressed mathematically as − $$Omega_x = frac{rho_x}{rho_c}$$ For different quantities under consideration, we can define the following density parameters. S.No. Quantity Density Parameter 1 Baryons $Omega_b = frac{rho_b}{rho_c}$ 2 Matter(Baryonic + Dark) $Omega_m = frac{rho_m}{rho_c}$ 3 Dark Energy $Omega_wedge = frac{rho_wedge}{rho_c}$ 4 Radiation $Omega_{rad} = frac{rho_{rad}}{rho_c}$ Where the symbols have their usual meanings. Points to Remember The evolution of the scale factor is determined by the Friedmann Equation. H(z) is the red shift dependent Hubble parameter. The Hubble Parameter varies with time. The Density Parameter is defined as the ratio of the actual (or observed) density to the critical density. Learning working make money

Velocity Dispersion Measurements of Galaxies First direct evidence of dark matter came from Frids Ricky. He did some observations which revealed dark matter for the first time. His observations considered the overall motion within the galaxy cluster. Extended objects are galaxy clusters and they are considered bound structures. These galaxies are moving with respect to cluster center but do not fly off. We look at the overall motion of the galaxy. Assumption: Velocities are Represent of Underlying Potential Every galaxy will have its own proper motion within the cluster and Hubble Flow Component. The smaller galaxies are smaller, most of the light comes from M31 and MW, there are several dwarf galaxies. For our crude analysis, we can only use M31 and MW and evaluate dynamic mass of the local group. There is a relative velocity between us and M31. It is crude, but it is true. The story begins long back when M31 and MW were close to each other, because they were members of a cluster they were moving away from each other. After some time they reach the maximum separation, then come closer to each other. Let us say that the maximum separation that can ever reach is $r_{max}$. Now they have a separation called r. Let M be the combined mass of MW and M31. We do not know when $r_{max}$ is reached. $$frac{GM}{r_{max}} = :Potential : at :r_{max}$$ When these galaxies were coming close to each other at some instant r, then the energy of the system will be − $$frac{1}{2}sigma^2 = frac{GM}{r} = frac{GM}{r_{max}}$$ σ is relative velocity of both the galaxies. M is reduced mass only, but the test mass is 1. σ is velocity of any object at distance r from the centre of the cluster. We believe that this cluster is in dynamic equation because virial theorem holds. So, galaxies cannot come with different velocity. How much time would these galaxies take to reach the maximum distance? To understand this, let us consider the following equation. $$frac{1}{2}left ( frac{dr}{dt} right )^2 = frac{GM}{r} – frac{GM}{r_{max}}$$ $$t_{max} = int_{0}^{r_{max}} dt = int_{0}^{r_{max}} frac{dr}{sqrt{2GM}}left ( frac{1}{r} – frac{1}{r_{max}} right )^2$$ $$t_{max} = frac{pi r_{max}^{frac{3}{2}}}{2sqrt{2GM}}$$ Where, M = dynamical mass of local group. The Total time from the start till the end of collision is $2t_{max}$. Therefore, $$2t_{max} = t_0 + frac{D}{sigma}$$ And $t_0$ is the present age of universe. If actual $t_{max} < RHS$, then we have a lower limit for the time. $D/sigma$ is the time when they will collide again. Here, we have assumed σ to be constant. $$t_{max} = frac{t_0}{2} + frac{D}{2sigma}$$ $$r_{max} = t_{max} times sigma = 770K_{pc}$$ Here, σ = relative velocity between MW and M31. $$M_{dynamic} = 3 times 10^{12}M_0$$ $$M_{MW}^{lum} = 3 times 10^{10}M_0$$ $$M_{M31}^{lum} = 3 times 10^{10}M_0$$ But practically, dynamic mass is found out considering every galaxy within the cluster. The missing mass is the dark matter and Frids Ricky noticed that the galaxies in the coma cluster are moving too fast. He predicted the existence of neutron stars the year after neutron stars were discovered and used Palomar telescope to find the supernova. Points to Remember First direct evidence of dark matter came from Frids Ricky. Extended objects are galaxy clusters and they are considered bound structures. Dynamic mass is found out considering every galaxy within the cluster. Learning working make money

Cosmology – Radiation Dominated Universe In this chapter, we will discuss the Solutions to Friedmann Equations relating to the Radiation Dominated Universe. In the beginning, we compare the energy density of matter to that of radiation. This will enable us to see if our universe is matter dominated or radiation dominated. Energy Density of Radiation Radiation prevalent in the present universe can be attributed very little to the stellar sources, but it is mainly due to the remnant CMB (Cosmic Microwave Background). The energy density of the radiation, $epsilon_{gamma,0}$, can be expressed as follows − $$epsilon_{gamma,0} = aT_0^4$$ Here, a is the radiation constant which has the expression $(8pi^5k_B^4)/(15h^3c^2)$ equal to a = 7.5657 × 10−15erg: cm−3 K−4. The Temperature, T0, we consider here, corresponds to that of the black body corresponding to the CMB. Substituting the results, we have, $$epsilon_{gamma,0} = aT_0^4 = 4 times 10^{-13}erg: cm^{-3}$$ Energy Density of Matter In the following calculations, we have the assumption of working with a flat universe and K = 0. We consider the energy density of matter as $epsilon = rho c^2$. We consider the following − $$rho_{m,0}c^2 = 0.3rho_{c,0}c^2 = 0.3 times frac{3H_0^2}{8pi G} times c^2$$ $$rho_{m,0}c^2 simeq 2 times 10^{-8} erg :cm^{-3}$$ $$rho_{b,0}c^2 = 0.03rho_{c,0}c^2 = 0.03 times frac{3H_0^2}{8pi G} times c^2$$ $$rho_{b,0}c^2 simeq 2 times 10^{-9} erg: cm^{-3}$$ Thus, from the above calculation, we see that we live in a matter-dominated universe. This can be supported by the fact that the CMB is very cold. As we look back in time, we would have the CMB temperature getting hotter, and will be able to conclude that there might have been an epoch where the universe was dominated by radiation. Variation of Density and Scale Factor The fluid equation shows us that − $$dot{rho} + 3frac{dot{a}}{a}left ( rho + frac{P}{c^2} right ) = 0$$ If we consider a dusty universe, we would have P = 0. Setting aside the previous results, we consider the universe as dominated by radiation. $$dot{rho}_{rad} + 3 frac{dot{a}}{a}left ( rho_{rad} + frac{P}{c^2} right ) = 0$$ Using the pressure relation of $P_{rad} = rho c^{2/3}$ we have − $$dot{rho}_{rad} + 3 frac{dot{a}}{a}left ( rho_{rad} + frac{rho_{rad}}{3} right ) = 0$$ $$dot{rho}_{rad} + 4frac{dot{a}}{a}(rho_{rad}) = 0$$ On further simplification, we have, $$frac{1}{a^4}frac{mathrm{d}}{mathrm{d} t}(rho_{rad}a^4) = 0$$ $$rho_{rad}a^4 =: constant$$ $$rho_{rad} propto frac{1}{a^4}$$ The above result shows an inverse 4th power variation of a with $rho$. This can be physically interpreted as $a^{-3}$ coming in from the variation in volume as it increases. The remaining $a^{-1}$ can be treated as the energy lost by the photon due to the expansion of space in the universe (Cosmological redshift 1 + z = a-1). The following image shows the variation of matter and radiation density with time. For a flat, radiation dominated universe, we would have the Friedmann equation as follows − $$left ( frac{dot{a}}{a} right )^2 = frac{8pi Grho}{3}$$ $$left ( frac{dot{a}}{a} right)^2 = frac{8pi G}{3}frac{rho_0}{a^4}$$ On simplification and applying the solution to the differential equation, we have − $$(dot{a})^2 = frac{8pi Grho_0}{3a^2}$$ $$Rightarrow a(t) propto t^{frac{1}{2}}$$ Thus, we have − $$a(t) = a_0 left ( frac{t}{t_0} right )^{frac{1}{2}}$$ From the above equation, we see that the rate of increase of scale factor is smaller than that of the dusty universe. Points to Remember Radiation prevalent in the present universe can be attributed very little to the stellar sources. For a dusty universe, pressure is zero. CMB is very cold. Learning working make money

Cosmology – Matter Dominated Universe In this chapter, we will discuss the Solutions to Friedmann Equations relating to the Matter Dominated Universe. In cosmology, because we are seeing everything in a large scale, the solar systems, galaxies, everything happens to be like dust particles (that’s what we see it with our eyes), we can call it dusty universe or matter only universe. In the Fluid Equation, $$dot{rho} = -3left ( frac{dot{a}}{a} right )rho -3left ( frac{dot{a}}{a} right )left ( frac{P}{c^2} right )$$ We can see there is a pressure term. For a dusty universe, P = 0, because the energy density of the matter will be greater than radiation pressure, and matter is not moving with relativistic speed. So, the Fluid Equation will become, $$dot{rho} = -3left ( frac{dot{a}}{a} right )rho$$ $$Rightarrow dot{rho}a + 3dot{a}rho = 0$$ $$Rightarrow frac{1}{a^3}frac{mathrm{d}}{mathrm{d} t}(a^3 rho) = 0$$ $$Rightarrow rho a^3 =: constant$$ $$Rightarrow rho propto frac{1}{a^3}$$ There is no counter intuition in this equation because density should scale as $a^{-3}$ because Volume is increasing as $a^3$. From the last relation, we can say that, $$frac{rho (t)}{rho_0} = left [ frac{a_0}{a(t)} right ]^3$$ For the present universe, a, which is equal to a0 should be 1. So, $$rho(t) = frac{rho_0}{a^3}$$ In a matter dominated flat universe, k = 0. So, Friedmann equation will become, $$left ( frac{dot{a}}{a} right )^2 = frac{8 pi Grho}{3}$$ $$dot{a}^2 = frac{8pi G rho a^2}{3}$$ By solving this equation, we will get, $$a propto t^{2/3}$$ $$frac{a(t)}{a_0} = left ( frac{t}{t_0} right )^{2/3}$$ $$a(t) = left( frac{t}{t_0} right )^{2/3}$$ This means that the universe will keep on increasing with a diminishing rate. The following image show the expansion of a Dusty Universe. How ρ Changes with Time? Take a look at the following equation − $$frac{rho(t)}{rho_0} = left ( frac{t_0}{t} right )^2$$ We know that the scale factor changes with time as $t^{2/3}$. So, $$a(t) = left ( frac{t}{t_0} right )^{2/3}$$ Differentiating it, we will get, $$frac{(da)}{dt} = dot{a} = frac{2}{3} left ( frac{t^{-1/3}}{t_0} right )$$ We know that the Hubble Constant is, $$H(t) = frac{dot{a}}{a} = frac{2}{3t}$$ This is the equation for Einstein-de sitter Universe. If we want to calculate the present age of the universe then, $$t_0 = t_{age} = frac{2}{3H_0}$$ After putting the value of $H_0$ for the present universe, we will get the value of the age of the universe as 9 Gyrs. There are many Globular Cluster in our own milky way galaxy which have ages more than that. That was all about the dusty universe. Now, if you assume that the universe is dominated by radiation and not by matter, then the radiation energy density goes as $a^{-4}$ rather than $a^{-3}$. We will see more of it in the next chapter. Points to Remember In cosmology, everything happens to be like dust particles, hence, we call it dusty universe or matter only universe. If we assume that the universe is dominated by radiation and not by matter, then the radiation energy density goes as $a^{-4}$ rather than $a^{-3}$. Learning working make money

Friedmann Equation & World Models In this chapter, we will understand what the Friedmann Equation is and study in detail regarding the World Models for different curvature constants. Friedmann Equation This equation tells us about the expansion of space in homogeneous and isotropic models of the universe. $$left ( frac{dot{a}}{a} right )^2 = frac{8pi G}{3}rho + frac{2U}{mr_c^2a^2}$$ This was modified in context of General Relativity (GR) and Robertson-Walker Metric as follows. Using GR equations − $$frac{2U}{mr_c^2} = -kc^2$$ Where k is the curvature constant. Therefore, $$left ( frac{dot{a}}{a} right )^2 = frac{8 pi G}{3}rho – frac{kc^2}{a^2}$$ Also, $rho$ is replaced by energy density which includes matter, radiation and any other form of energy. But for representational purposes, it is written as $rho$. World Models for Different Curvature Constants Let us now look at the various possibilities depending on the curvature constant values. Case 1: k=1, or Closed Universe For an expanding universe, $da/dt > 0$. As expansion continues, the first term on the RHS of the above equation goes as $a^{-3}$, whereas the second term goes as $a^{-2}$. When the two terms become equal the universe halts expansion. Then − $$frac{8 pi G}{3}rho = frac{kc^2}{a^2}$$ Here, k=1, therefore, $$a = left [ frac{3c^2}{8 pi Grho} right ]^{frac{1}{2}}$$ Such a universe is finite and has finite volume. This is called a Closed Universe. Case 2: k=-1, or Open Universe If k < 0, the expansion would never halt. After some point, the first term on the RHS can be neglected in comparison with the second term. Here, k = -1. Therefore, $da/dt ∼ c$. In this case, the universe is coasting. Such a universe has infinite space and time. This is called an Open Universe. Case 3: k=0, or Flat Universe In this case, the universe is expanding at a diminishing rate. Here, k = 0. Therefore, $$left ( frac{dot{a}}{a} right )^2 = frac{8pi G}{3}rho$$ Such a universe has infinite space and time. This is called a Flat Universe. Points to Remember The Friedmann equation tells us about the expansion of space in homogeneous and isotropic models of the universe. Depending on different curvature constant values, we can have a Closed, Open or Flat Universe. Learning working make money