and old universes depleted of energy and matter

and antimatter

does that matter to doctor who?

who knows....

## joi, 31 martie 2016

### WHEN THE SOMATORY OF ALL COEFFICIENTS OF CONTRACTION TENDS TO INFINITY YOU HAVE A PONTUAL UNIVERSE -the expansion of the universe started to accelerate recently (meaning a few billion ...BUT IS ONLY A PONTUAL MEMENTO MORI OF ALL UNIVERSES THE SOMATORY OF ALL EXPANSION IS BIGGER THAN ALL THEPONTUAL CONTRACTIONS IN THE FORM OF BLACK HOLES AND OTHER ASS HOLES DA GAMMA RAY ..... Homogeneity requires the proportionality coefficient to be only a function of time.

Expansion of the Universe
One of the main goals of cosmology is to figure out how the universe expands as a function of time.
1.1 Expansion and Conservation
To describe the evolution of the average universe, one needs only two kinds of equations:
1. The equation that relates the density and pressure of constituents of the universe (such as
baryons, cold dark matter, photons, neutrinos, dark energy) to the expansion of the universe,
and
2. The equation that describes the energy conservation of the constituents.
Consider a line connecting two arbitrary points in space (which is expanding), and call it L. As the
universe expands, L changes with time. As you will derive in homework using

General Relativity, the equation of motion for L is given by

L¨(t) = − 4πG 3 L(t) X i [ρi(t) + 3Pi(t)] ,

(1.1) where ρi(t) and Pi(t) are the energy and pressure of the ith component of the universe, respectively. Here, note that the absolute value of L does not affect the equation of motion for L. Therefore, one may define a dimensionless “scale factor,” a(t), such that L(t) ≡ a(t)x, where x is a timeindependent separation called a “comoving” separation, which is in units of length. In cosmology, 1 we often encounter the Hubble expansion rate, H(t), which is defined by H(t) ≡ a˙(t) a(t) . (1.2) The dimension of this quantity is 1/(time). The age of the universe can be calculated from the above definition of H, which gives H(t)dt = da/a. Now, if we know H as a function of a instead of t, we obtain t = Z da aH(a) . (1.3) Another interpretation of H is found by writing L˙(t) = H(t)L(t), which tells us that H(t) gives a relation between the distance, L, and the recession velocity, L˙ . For this reason, it is often convenient to write H(t) in the following peculiar units: H(t) = 100 h(t) km/s/Mpc, where h is a dimensionless quantity. The current observations suggest that the present-day value of h is h(ttoday) ≈ 0.7.∗ Dividing both sides of equation (1.1) by L and using L(t) = a(t)x, we find one of the key equations connecting the energy density and pressure to the expansion of the universe: a¨(t) a(t) = − 4πG 3 X i [ρi(t) + 3Pi(t)] (1.4) As expected, positive energy density and positive pressure slow down the expansion of the universe.† This equation cannot be solved unless we know how ρi and Pi depend on time. How ρi depends on time is given by the energy conservation equation, while how Pi depends on time is usually given by the equation of state relating Pi to ρi and other quantities. As you will derive in homework, the energy conservation equation is given by X i ρ˙i(t) + 3a˙(t) a(t) X i [ρi(t) + Pi(t)] = 0 (1.5) Equation (1.5) is general and does not assume presence or absence of possible interactions between different components. If we assume that each component is conserved separately, then we have ρ˙i(t) + 3a˙(t) a(t) [ρi(t) + Pi(t)] = 0, (1.6) ∗The most precise value of h(ttoday) to date from the direct measurement using low-z supernovae and Cepheid variable stars is h(ttoday) = 0.742 ± 0.036 (Riess, Macri, et al., ApJ, 699, 539 (2009)). † If we ignore the effect of pressure relative to that of the energy density (which is always a good approximation for non-relativistic matter), and write ρ(t) in terms of the total mass enclosed with a radius L, P i ρi(t) = 3M 4πL3 , then equation (1.1) becomes L¨ = − GM L2 , which is the familiar Newtonian inverse-square law. Although one must not apply the Newtonian mechanics to describe the evolution of space (because Newtonian mechanism assumes static space), this is a convenient way to understand equations (1.1) and (1.4)for each of the ith component. Note that the second term contains the pressure, and thus how the energy density evolves depends on the pressure.‡ Looking at equations (1.4) and (1.5), one might think that we cannot solve for a(t) unless we have the equation of state giving Pi(t) as a function of ρi(t) etc. While in general that would be true, for these equations a little mathematical trick lets us combine equations (1.4) and (1.5) without knowing the evolution of P(t)! First, rewrite equation (1.4) as a¨(t) a(t) = 8πG 3 X i ρi(t) − 4πGX i [ρi(t) + Pi(t)] . (1.7) Using equation (1.5) on the second term of the right hand side, we get a¨(t) a(t) = 8πG 3 X i ρi(t) + 4πG 3 a(t) a˙(t) X i ρ˙i(t) a˙(t)¨a(t) = 8πGa(t)˙a(t) 3 X i ρi(t) + 4πGa2 (t) 3 X i ρ˙i(t) 1 2 (˙a 2 ) · = 4πG(a 2 ) · 3 X i ρi(t) + 4πGa2 (t) 3 X i ρ˙i(t). (1.8) As this has the form of A˙ = BC˙ + BC˙ = (BC) · , it is easy to integrate and obtain: a˙ 2 (t) = 8πGa2 (t) 3 X i ρi(t) − κ, (1.9) where κ is an integration constant, which is in units of 1/(time)2 . (A negative sign is for a historical reason.) Dividing both sides by a 2 (t), we finally arrive at the so-called Friedmann equation: a˙ 2 (t) a 2(t) = 8πG 3 X i ρi(t) − κ a 2(t) . (1.10) ‡While it is a wrong explanation, it is useful to compare this equation to the first law of thermodynamics: T dS = dU + P dV, where T, S, U, and V are the temperature, entropy, internal energy, and volume, respectively. To a very good accuracy, the entropy is conserved in the universe, dS = 0. The internal energy is U ∝ ρa3 and the volume is V ∝ a 3 , and thus d(ρa 3 ) + P d(a 3 ) = 0, which gives ρ˙ + 3a˙ a (ρ + P) = 0. This is a wrong explanation because it assumes that the pressure is doing work as a increases. However, in the average universe, the pressure is the same everywhere, and thus there is no under-pressure region against which the pressure can do work. Equation (1.5) must be derived using GR, which you will do in homework, but the above thermodynamic argument is an amusing way to arrive at the same equation. Also, this gives us some confidence that it is not crazy to think that the evolution of ρ depends on P. 3 A beauty of this equation is that it is easy to solve, once a time dependence of ρi(t) is known, which is usually the case. General Relativity tells us that the integration constant, κ, is equal to ±c 2/R2 where R is the curvature radius of the universe (in units of length) and c the speed of light. When the geometry of the universe is flat (as suggested by observations), R → ∞ (giving κ → 0), and thus one can ignore this term. Since we have so much to learn, to save time we will not consider the curvature of the universe throughout (most of) this lecture: a˙ 2 (t) a 2(t) = 8πG 3 X i ρi(t) (1.11) 1.2 Solutions of Friedmann Equation In order to use solve equation (1.11) for a(t), one must know how ρi(t) depends on time. To find solutions for a(t), let us first assume that the universe is dominated by one energy component at a time, i.e., a˙ 2 (t) a 2(t) = 8πG 3 X i ρi(t) ≈ 8πG 3 ρi(t), (1.12) and further assume that ρi depends on a(t) via a power-law: ρi(t) ∝ 1 a ni (t) . (1.13) 4 Finding the solution is straightforward: a(t) ∝ t 2/ni . (1.14) This is usually an excellent approximation, except for the transition era where two energy components are equally important. There are 3 important cases: 1. Radiation-dominated (RD) era. A radiation component (photons, massless neutrinos, or any other massless particles) has a large pressure, PR = ρR/3,§ which gives ρR(t) ∝ 1/a4 (t), or nR = 4. We thus obtain aRD(t) ∝ t 1/2 . (1.15) The expansion of the universe decelerates. With this solution, we can relate the age of the universe to the Hubble expansion rate: H(t) = a˙(t) a(t) = 1 2t . (1.16) 2. Matter-dominated (MD) era. A matter component (baryons, cold dark matter, or any other non-relativistic particles) has a negligible pressure compared to its energy density, PM ρM, which gives ρM(t) ∝ 1/a3 (t), or nM = 3. We thus obtain aMD(t) ∝ t 2/3 . (1.17) §Again, a “wrong” derivation, but there is an intuitive way to get this result using the equation of state for non-relativistic ideal gas (this is obviously a wrong derivation because we are about to apply non-relativistic equation of state to relativistic gas!): P = nkBT = ρ kBT hEi , where n is the number density, T the temperature of gas, kB the Boltzmann constant, and hEi the mean energy per particle. For relativistic particles in thermal equilibrium, hEi ≈ 3kBT, which gives P ≈ ρ/3. Now, actually, it turns out that the error we are making by using non-relativistic equation of state for relativistic gas cancels out precisely the error we are making by using an approximate relation hEi ≈ 3kBT. This gives us the exact relation, P = ρ/3 for relativistic particles. More precisely, the equation of state for relativistic gas takes on the form P = (1 + )ρ kBT hEi with hEi = 3(1 + )kBT, giving P = ρ/3. Here, ' 0.05 and −0.10 for Fermions and Bosons, respectively. 5 The expansion of the universe decelerates. With this solution, we can relate the age of the universe to the Hubble expansion rate: H(t) = a˙(t) a(t) = 2 3t . (1.18) 3. Constant-energy-density-dominated (ΛD) era. A hypothetical energy component (let’s call it Λ) whose energy density is a constant over time, nΛ = 0. In this case we cannot use equation (1.14). Going back to equation (1.12) and setting ρΛ = constant, we get ˙a/a = constant, whose solution is aΛD(t) ∝ e Ht , (1.19) where an integration constant, H, is the same as the Hubble expansion rate (which is a constant for this model). The expansion of the universe accelerates, which must mean that, according to the acceleration equation (1.4), the pressure of this energy component is negative. The conservation equation (1.5) tells us that such a component indeed has an enormous negative pressure given by PΛ = −ρΛ. (1.20) While this looks quite strange, we now know that something like this may actually exist in our universe, as the current observations suggest that the present-day universe is indeed accelerating. 6 1.3 Equation of State of “Dark Energy” and Density Parameters The matter has PM ρM; the radiation has PR = ρR/3; and Λ has PΛ = −ρΛ. This motivates our writing the equation of state of the ith component in the following simple form: Pi = wiρi . (1.21) Here, wi is called the “equation of state parameter,” and can depend on time (although it is usually taken to be constant). Why this form? It is important to keep in mind that there is no fundamental reason why we should use this form. This form is often used either just for convenience, or simply for parametrizing something we do not know. At the very least, this form is exact for radiation, wR = 1/3, and for Λ, wΛ = −1. For matter, since wM 1, the exact value does not affect the results very much. The equation of state parameter is almost exclusively used for parametrizing “dark energy,” which is supposed to cause the observed acceleration of the universe. If we assume that w for dark energy, wDE, is constant, then the current observations suggest that (Komatsu, et al., ApJS, 192, 18 (2011)) wDE = −0.98 ± 0.05 (68% CL). (1.22) In other words, the energy density of dark energy is consistent with being a constant (wDE = wΛ = −1). Determining wDE with better accuracy may tell us something about the nature of dark energy, especially if wDE 6= 1 is found with high statistical significance, as it would tell us that dark energy is something dynamical (time-dependent). Ignoring a potential interaction between dark energy and other components in the universe (e.g., dark matter), the energy density of dark energy obeys (see equation (1.6)) ρ˙DE(t) + 3a˙(t) a(t) (1 + wDE) ρDE(t) = 0, (1.23) whose solution is ρDE(t) ∝ [a(t)]−3(1+wDE) . On the other hand, if we do not assume that wDE is a constant, then the energy density of dark energy obeys ρ˙DE(t) + 3a˙(t) a(t) [1 + wDE(t)] ρDE(t) = 0, (1.24) whose solution is ρDE(t) ∝ e −3 R d ln a[1+wDE(a)] . (1.25) Putting these results together, we obtain the Friedmann equation for our Universe containing radiation, matter, and dark energy (but not curvature) as a˙ 2 (t) a 2(t) = H2 (t) = 8πG 3 ρM(t0) a 3 (t0) a 3(t) + ρR(t0) a 4 (t0) a 4(t) + ρDE(t0)e −3 R a(t) a(t0) d ln a[1+wDE(a)] , (1.26) where t0 is some epoch, which is usually taken to be the present epoch.

Equation of State of “Dark Energy” and Density Parameters The matter has PM ρM; the radiation has PR = ρR/3; and Λ has PΛ = −ρΛ. This motivates our writing the equation of state of the ith component in the following simple form: Pi = wiρi . (1.21) Here, wi is called the “equation of state parameter,” and can depend on time (although it is usually taken to be constant). Why this form? It is important to keep in mind that there is no fundamental reason why we should use this form. This form is often used either just for convenience, or simply for parametrizing something we do not know. At the very least, this form is exact for radiation, wR = 1/3, and for Λ, wΛ = −1. For matter, since wM 1, the exact value does not affect the results very much. The equation of state parameter is almost exclusively used for parametrizing “dark energy,” which is supposed to cause the observed acceleration of the universe. If we assume that w for dark energy, wDE, is constant, then the current observations suggest that (Komatsu, et al., ApJS, 192, 18 (2011)) wDE = −0.98 ± 0.05 (68% CL). (1.22) In other words, the energy density of dark energy is consistent with being a constant (wDE = wΛ = −1). Determining wDE with better accuracy may tell us something about the nature of dark energy, especially if wDE 6= 1 is found with high statistical significance, as it would tell us that dark energy is something dynamical (time-dependent). Ignoring a potential interaction between dark energy and other components in the universe (e.g., dark matter), the energy density of dark energy obeys (see equation (1.6)) ρ˙DE(t) + 3a˙(t) a(t) (1 + wDE) ρDE(t) = 0, (1.23) whose solution is ρDE(t) ∝ [a(t)]−3(1+wDE) . On the other hand, if we do not assume that wDE is a constant, then the energy density of dark energy obeys ρ˙DE(t) + 3a˙(t) a(t) [1 + wDE(t)] ρDE(t) = 0, (1.24) whose solution is ρDE(t) ∝ e −3 R d ln a[1+wDE(a)] . (1.25) Putting these results together, we obtain the Friedmann equation for our Universe containing radiation, matter, and dark energy (but not curvature) as a˙ 2 (t) a 2(t) = H2 (t) = 8πG 3 ρM(t0) a 3 (t0) a 3(t) + ρR(t0) a 4 (t0) a 4(t) + ρDE(t0)e −3 R a(t) a(t0) d ln a[1+wDE(a)] , (1.26) where t0 is some epoch, which is usually taken to be the present epoch. 7 Now, taking t → t0, we find the present-day expansion rate H2 0 ≡ H2 (t0) = 8πG 3 [ρM(t0) + ρR(t0) + ρDE(t0)] ≡ 8πG 3 ρc(t0), (1.27) which has been determined to be H0 ≈ 70 km/s/Mpc. Here, ρc(t0) is the so-called “critical density” of the universe, which is equal to the total energy density of the universe when the universe is flat. The numerical value of the critical density is ρc(t0) ≡ 3H2 0 8πG = 2.775 × 1011 h 2 M Mpc−3 . (1.28) The critical density provides a natural unit for the energy density of the universe, and thus it is convenient to measure all the energy densities in units of ρc(t0). Defining the so-called density parameters, Ωi , as Ωi ≡ ρi(t0) ρc(t0) , (1.29) one can rewrite the Friedmann equation (1.26) in a compact form: H2 (t) H2 0 = ΩM a 3 (t0) a 3(t) + ΩR a 4 (t0) a 4(t) + ΩDEe −3 R a(t) a(t0) d ln a[1+wDE(a)] (1.30) Basically, most of the literature on cosmology (within the context of General Relativity) use this equation as the starting point.¶ Taking z = 0, one finds that all the density parameters must sum to unity: P i Ωi = 1. In summary, the Friedmann equation is a combination of two key equations: (1) the equation describing how the universe decelerates/accelerates depending on the energy density and pressure of the constituents, and (2) the equation describing the energy conservation of the constituents. Once the Friedmann equation is given with the proper right hand side containing the energy densities of the relevant constituents of the universe, we can find a(t) as a function of time easily. ¶An interesting possibility is that General Relativity may not be valid on cosmological scales. There are scenarios in which the form of the Friedmann equation is modified. One widely-explored example is the so-called DvaliGabadadze-Porrati (DGP) model (Dvali, Gabadadze & Porrati, Phys. Lett. B485, 208 (2000)). In this scenario, the Friedmann equation is modified to: H 2 (t) − H(t) rc = 8πG 3 X i ρi(t), where rc is some length scale below which General Relativity is restored. (For r rc, the potential is given by −GN m/r where GN is the ordinary Newtonian gravitational constant. For r rc, the potential is modified to −G5m/r2 and decays faster. G5 is the gravitational strength in the 5th dimension.) This model has attracted a huge attention of the cosmology community, as it was shown that this modified Friedmann equation gives an accelerating expansion without dark energy. Namely, even when the right hand side contains only matter, the solution for this equation can still exhibit an accelerating expansion. As this is a quadratic equation for H(t), we can solve it and find H(t) = 1 2 1 rc ± r 1 r 2 c + 32πG 3 ρM(t) . At late times when ρ(t) becomes negligible compared to the other term, one of the solutions is given by a(t) ∝ e t/rc , i.e., an exponential, accelerated expansion. 8 At present, the radiation is totally negligible compared to matter, ΩR/ΩM ' 1/3250, and the dark energy density is about 3 times as large as the matter density, ΩDE/ΩM ' 2.7 (with ΩM ' 0.27 and ΩDE ' 0.73). 1.4 Redshift As the universe expands, the wavelength of light, λ, is stretched linearly: λ(t) ∝ a(t), (1.31) which implies that photons lose energy as E(t) ∝ 1/a(t). This is something one can observe, by comparing, for example, the observed wavelength of a hydrogen line to the rest-frame wavelength that we know from the laboratory experiment. We often use the redshift, z, to quantify the stretching of the wavelength: 1 + z ≡ λ(t0) λ(temitted) . (1.32) The present-day corresponds to z = 0. Using equation (1.31), we can relate the observed redshift to the ratio of the scale factors: 1 + z = a(t0) a(temitted) . (1.33) Using this result in the Friedmann equation (1.30), we obtain the most-widely-used form of the Friedmann equation:

H2 (z) H2 0 = ΩM(1 + z) 3 + ΩR(1 + z) 4 + ΩDEe 3 R z 0 d ln(1+z)[1+wDE(z)From this result, it follows that the best way to determine the equation of state of dark energy is to measure H(z) over a wide range of z. If we can only measure the expansion rates at z 1, then Taylor expansion of equation (1.34) with ΩR ΩM and ΩDE ' 1 − ΩM gives H2 (z 1) H2 0 ≈ 1 + 3ΩMz + 3(1 + wDE)(1 − ΩM)z. (1.35) As we know from observations that |1 + wDE| is small (of order 10−1 or less), the third term is tiny compared to other terms, making it difficult to measure wDE. This is why we need to measure H(z) over a wide redshift range

General Relativity, the equation of motion for L is given by

L¨(t) = − 4πG 3 L(t) X i [ρi(t) + 3Pi(t)] ,

(1.1) where ρi(t) and Pi(t) are the energy and pressure of the ith component of the universe, respectively. Here, note that the absolute value of L does not affect the equation of motion for L. Therefore, one may define a dimensionless “scale factor,” a(t), such that L(t) ≡ a(t)x, where x is a timeindependent separation called a “comoving” separation, which is in units of length. In cosmology, 1 we often encounter the Hubble expansion rate, H(t), which is defined by H(t) ≡ a˙(t) a(t) . (1.2) The dimension of this quantity is 1/(time). The age of the universe can be calculated from the above definition of H, which gives H(t)dt = da/a. Now, if we know H as a function of a instead of t, we obtain t = Z da aH(a) . (1.3) Another interpretation of H is found by writing L˙(t) = H(t)L(t), which tells us that H(t) gives a relation between the distance, L, and the recession velocity, L˙ . For this reason, it is often convenient to write H(t) in the following peculiar units: H(t) = 100 h(t) km/s/Mpc, where h is a dimensionless quantity. The current observations suggest that the present-day value of h is h(ttoday) ≈ 0.7.∗ Dividing both sides of equation (1.1) by L and using L(t) = a(t)x, we find one of the key equations connecting the energy density and pressure to the expansion of the universe: a¨(t) a(t) = − 4πG 3 X i [ρi(t) + 3Pi(t)] (1.4) As expected, positive energy density and positive pressure slow down the expansion of the universe.† This equation cannot be solved unless we know how ρi and Pi depend on time. How ρi depends on time is given by the energy conservation equation, while how Pi depends on time is usually given by the equation of state relating Pi to ρi and other quantities. As you will derive in homework, the energy conservation equation is given by X i ρ˙i(t) + 3a˙(t) a(t) X i [ρi(t) + Pi(t)] = 0 (1.5) Equation (1.5) is general and does not assume presence or absence of possible interactions between different components. If we assume that each component is conserved separately, then we have ρ˙i(t) + 3a˙(t) a(t) [ρi(t) + Pi(t)] = 0, (1.6) ∗The most precise value of h(ttoday) to date from the direct measurement using low-z supernovae and Cepheid variable stars is h(ttoday) = 0.742 ± 0.036 (Riess, Macri, et al., ApJ, 699, 539 (2009)). † If we ignore the effect of pressure relative to that of the energy density (which is always a good approximation for non-relativistic matter), and write ρ(t) in terms of the total mass enclosed with a radius L, P i ρi(t) = 3M 4πL3 , then equation (1.1) becomes L¨ = − GM L2 , which is the familiar Newtonian inverse-square law. Although one must not apply the Newtonian mechanics to describe the evolution of space (because Newtonian mechanism assumes static space), this is a convenient way to understand equations (1.1) and (1.4)for each of the ith component. Note that the second term contains the pressure, and thus how the energy density evolves depends on the pressure.‡ Looking at equations (1.4) and (1.5), one might think that we cannot solve for a(t) unless we have the equation of state giving Pi(t) as a function of ρi(t) etc. While in general that would be true, for these equations a little mathematical trick lets us combine equations (1.4) and (1.5) without knowing the evolution of P(t)! First, rewrite equation (1.4) as a¨(t) a(t) = 8πG 3 X i ρi(t) − 4πGX i [ρi(t) + Pi(t)] . (1.7) Using equation (1.5) on the second term of the right hand side, we get a¨(t) a(t) = 8πG 3 X i ρi(t) + 4πG 3 a(t) a˙(t) X i ρ˙i(t) a˙(t)¨a(t) = 8πGa(t)˙a(t) 3 X i ρi(t) + 4πGa2 (t) 3 X i ρ˙i(t) 1 2 (˙a 2 ) · = 4πG(a 2 ) · 3 X i ρi(t) + 4πGa2 (t) 3 X i ρ˙i(t). (1.8) As this has the form of A˙ = BC˙ + BC˙ = (BC) · , it is easy to integrate and obtain: a˙ 2 (t) = 8πGa2 (t) 3 X i ρi(t) − κ, (1.9) where κ is an integration constant, which is in units of 1/(time)2 . (A negative sign is for a historical reason.) Dividing both sides by a 2 (t), we finally arrive at the so-called Friedmann equation: a˙ 2 (t) a 2(t) = 8πG 3 X i ρi(t) − κ a 2(t) . (1.10) ‡While it is a wrong explanation, it is useful to compare this equation to the first law of thermodynamics: T dS = dU + P dV, where T, S, U, and V are the temperature, entropy, internal energy, and volume, respectively. To a very good accuracy, the entropy is conserved in the universe, dS = 0. The internal energy is U ∝ ρa3 and the volume is V ∝ a 3 , and thus d(ρa 3 ) + P d(a 3 ) = 0, which gives ρ˙ + 3a˙ a (ρ + P) = 0. This is a wrong explanation because it assumes that the pressure is doing work as a increases. However, in the average universe, the pressure is the same everywhere, and thus there is no under-pressure region against which the pressure can do work. Equation (1.5) must be derived using GR, which you will do in homework, but the above thermodynamic argument is an amusing way to arrive at the same equation. Also, this gives us some confidence that it is not crazy to think that the evolution of ρ depends on P. 3 A beauty of this equation is that it is easy to solve, once a time dependence of ρi(t) is known, which is usually the case. General Relativity tells us that the integration constant, κ, is equal to ±c 2/R2 where R is the curvature radius of the universe (in units of length) and c the speed of light. When the geometry of the universe is flat (as suggested by observations), R → ∞ (giving κ → 0), and thus one can ignore this term. Since we have so much to learn, to save time we will not consider the curvature of the universe throughout (most of) this lecture: a˙ 2 (t) a 2(t) = 8πG 3 X i ρi(t) (1.11) 1.2 Solutions of Friedmann Equation In order to use solve equation (1.11) for a(t), one must know how ρi(t) depends on time. To find solutions for a(t), let us first assume that the universe is dominated by one energy component at a time, i.e., a˙ 2 (t) a 2(t) = 8πG 3 X i ρi(t) ≈ 8πG 3 ρi(t), (1.12) and further assume that ρi depends on a(t) via a power-law: ρi(t) ∝ 1 a ni (t) . (1.13) 4 Finding the solution is straightforward: a(t) ∝ t 2/ni . (1.14) This is usually an excellent approximation, except for the transition era where two energy components are equally important. There are 3 important cases: 1. Radiation-dominated (RD) era. A radiation component (photons, massless neutrinos, or any other massless particles) has a large pressure, PR = ρR/3,§ which gives ρR(t) ∝ 1/a4 (t), or nR = 4. We thus obtain aRD(t) ∝ t 1/2 . (1.15) The expansion of the universe decelerates. With this solution, we can relate the age of the universe to the Hubble expansion rate: H(t) = a˙(t) a(t) = 1 2t . (1.16) 2. Matter-dominated (MD) era. A matter component (baryons, cold dark matter, or any other non-relativistic particles) has a negligible pressure compared to its energy density, PM ρM, which gives ρM(t) ∝ 1/a3 (t), or nM = 3. We thus obtain aMD(t) ∝ t 2/3 . (1.17) §Again, a “wrong” derivation, but there is an intuitive way to get this result using the equation of state for non-relativistic ideal gas (this is obviously a wrong derivation because we are about to apply non-relativistic equation of state to relativistic gas!): P = nkBT = ρ kBT hEi , where n is the number density, T the temperature of gas, kB the Boltzmann constant, and hEi the mean energy per particle. For relativistic particles in thermal equilibrium, hEi ≈ 3kBT, which gives P ≈ ρ/3. Now, actually, it turns out that the error we are making by using non-relativistic equation of state for relativistic gas cancels out precisely the error we are making by using an approximate relation hEi ≈ 3kBT. This gives us the exact relation, P = ρ/3 for relativistic particles. More precisely, the equation of state for relativistic gas takes on the form P = (1 + )ρ kBT hEi with hEi = 3(1 + )kBT, giving P = ρ/3. Here, ' 0.05 and −0.10 for Fermions and Bosons, respectively. 5 The expansion of the universe decelerates. With this solution, we can relate the age of the universe to the Hubble expansion rate: H(t) = a˙(t) a(t) = 2 3t . (1.18) 3. Constant-energy-density-dominated (ΛD) era. A hypothetical energy component (let’s call it Λ) whose energy density is a constant over time, nΛ = 0. In this case we cannot use equation (1.14). Going back to equation (1.12) and setting ρΛ = constant, we get ˙a/a = constant, whose solution is aΛD(t) ∝ e Ht , (1.19) where an integration constant, H, is the same as the Hubble expansion rate (which is a constant for this model). The expansion of the universe accelerates, which must mean that, according to the acceleration equation (1.4), the pressure of this energy component is negative. The conservation equation (1.5) tells us that such a component indeed has an enormous negative pressure given by PΛ = −ρΛ. (1.20) While this looks quite strange, we now know that something like this may actually exist in our universe, as the current observations suggest that the present-day universe is indeed accelerating. 6 1.3 Equation of State of “Dark Energy” and Density Parameters The matter has PM ρM; the radiation has PR = ρR/3; and Λ has PΛ = −ρΛ. This motivates our writing the equation of state of the ith component in the following simple form: Pi = wiρi . (1.21) Here, wi is called the “equation of state parameter,” and can depend on time (although it is usually taken to be constant). Why this form? It is important to keep in mind that there is no fundamental reason why we should use this form. This form is often used either just for convenience, or simply for parametrizing something we do not know. At the very least, this form is exact for radiation, wR = 1/3, and for Λ, wΛ = −1. For matter, since wM 1, the exact value does not affect the results very much. The equation of state parameter is almost exclusively used for parametrizing “dark energy,” which is supposed to cause the observed acceleration of the universe. If we assume that w for dark energy, wDE, is constant, then the current observations suggest that (Komatsu, et al., ApJS, 192, 18 (2011)) wDE = −0.98 ± 0.05 (68% CL). (1.22) In other words, the energy density of dark energy is consistent with being a constant (wDE = wΛ = −1). Determining wDE with better accuracy may tell us something about the nature of dark energy, especially if wDE 6= 1 is found with high statistical significance, as it would tell us that dark energy is something dynamical (time-dependent). Ignoring a potential interaction between dark energy and other components in the universe (e.g., dark matter), the energy density of dark energy obeys (see equation (1.6)) ρ˙DE(t) + 3a˙(t) a(t) (1 + wDE) ρDE(t) = 0, (1.23) whose solution is ρDE(t) ∝ [a(t)]−3(1+wDE) . On the other hand, if we do not assume that wDE is a constant, then the energy density of dark energy obeys ρ˙DE(t) + 3a˙(t) a(t) [1 + wDE(t)] ρDE(t) = 0, (1.24) whose solution is ρDE(t) ∝ e −3 R d ln a[1+wDE(a)] . (1.25) Putting these results together, we obtain the Friedmann equation for our Universe containing radiation, matter, and dark energy (but not curvature) as a˙ 2 (t) a 2(t) = H2 (t) = 8πG 3 ρM(t0) a 3 (t0) a 3(t) + ρR(t0) a 4 (t0) a 4(t) + ρDE(t0)e −3 R a(t) a(t0) d ln a[1+wDE(a)] , (1.26) where t0 is some epoch, which is usually taken to be the present epoch.

Equation of State of “Dark Energy” and Density Parameters The matter has PM ρM; the radiation has PR = ρR/3; and Λ has PΛ = −ρΛ. This motivates our writing the equation of state of the ith component in the following simple form: Pi = wiρi . (1.21) Here, wi is called the “equation of state parameter,” and can depend on time (although it is usually taken to be constant). Why this form? It is important to keep in mind that there is no fundamental reason why we should use this form. This form is often used either just for convenience, or simply for parametrizing something we do not know. At the very least, this form is exact for radiation, wR = 1/3, and for Λ, wΛ = −1. For matter, since wM 1, the exact value does not affect the results very much. The equation of state parameter is almost exclusively used for parametrizing “dark energy,” which is supposed to cause the observed acceleration of the universe. If we assume that w for dark energy, wDE, is constant, then the current observations suggest that (Komatsu, et al., ApJS, 192, 18 (2011)) wDE = −0.98 ± 0.05 (68% CL). (1.22) In other words, the energy density of dark energy is consistent with being a constant (wDE = wΛ = −1). Determining wDE with better accuracy may tell us something about the nature of dark energy, especially if wDE 6= 1 is found with high statistical significance, as it would tell us that dark energy is something dynamical (time-dependent). Ignoring a potential interaction between dark energy and other components in the universe (e.g., dark matter), the energy density of dark energy obeys (see equation (1.6)) ρ˙DE(t) + 3a˙(t) a(t) (1 + wDE) ρDE(t) = 0, (1.23) whose solution is ρDE(t) ∝ [a(t)]−3(1+wDE) . On the other hand, if we do not assume that wDE is a constant, then the energy density of dark energy obeys ρ˙DE(t) + 3a˙(t) a(t) [1 + wDE(t)] ρDE(t) = 0, (1.24) whose solution is ρDE(t) ∝ e −3 R d ln a[1+wDE(a)] . (1.25) Putting these results together, we obtain the Friedmann equation for our Universe containing radiation, matter, and dark energy (but not curvature) as a˙ 2 (t) a 2(t) = H2 (t) = 8πG 3 ρM(t0) a 3 (t0) a 3(t) + ρR(t0) a 4 (t0) a 4(t) + ρDE(t0)e −3 R a(t) a(t0) d ln a[1+wDE(a)] , (1.26) where t0 is some epoch, which is usually taken to be the present epoch. 7 Now, taking t → t0, we find the present-day expansion rate H2 0 ≡ H2 (t0) = 8πG 3 [ρM(t0) + ρR(t0) + ρDE(t0)] ≡ 8πG 3 ρc(t0), (1.27) which has been determined to be H0 ≈ 70 km/s/Mpc. Here, ρc(t0) is the so-called “critical density” of the universe, which is equal to the total energy density of the universe when the universe is flat. The numerical value of the critical density is ρc(t0) ≡ 3H2 0 8πG = 2.775 × 1011 h 2 M Mpc−3 . (1.28) The critical density provides a natural unit for the energy density of the universe, and thus it is convenient to measure all the energy densities in units of ρc(t0). Defining the so-called density parameters, Ωi , as Ωi ≡ ρi(t0) ρc(t0) , (1.29) one can rewrite the Friedmann equation (1.26) in a compact form: H2 (t) H2 0 = ΩM a 3 (t0) a 3(t) + ΩR a 4 (t0) a 4(t) + ΩDEe −3 R a(t) a(t0) d ln a[1+wDE(a)] (1.30) Basically, most of the literature on cosmology (within the context of General Relativity) use this equation as the starting point.¶ Taking z = 0, one finds that all the density parameters must sum to unity: P i Ωi = 1. In summary, the Friedmann equation is a combination of two key equations: (1) the equation describing how the universe decelerates/accelerates depending on the energy density and pressure of the constituents, and (2) the equation describing the energy conservation of the constituents. Once the Friedmann equation is given with the proper right hand side containing the energy densities of the relevant constituents of the universe, we can find a(t) as a function of time easily. ¶An interesting possibility is that General Relativity may not be valid on cosmological scales. There are scenarios in which the form of the Friedmann equation is modified. One widely-explored example is the so-called DvaliGabadadze-Porrati (DGP) model (Dvali, Gabadadze & Porrati, Phys. Lett. B485, 208 (2000)). In this scenario, the Friedmann equation is modified to: H 2 (t) − H(t) rc = 8πG 3 X i ρi(t), where rc is some length scale below which General Relativity is restored. (For r rc, the potential is given by −GN m/r where GN is the ordinary Newtonian gravitational constant. For r rc, the potential is modified to −G5m/r2 and decays faster. G5 is the gravitational strength in the 5th dimension.) This model has attracted a huge attention of the cosmology community, as it was shown that this modified Friedmann equation gives an accelerating expansion without dark energy. Namely, even when the right hand side contains only matter, the solution for this equation can still exhibit an accelerating expansion. As this is a quadratic equation for H(t), we can solve it and find H(t) = 1 2 1 rc ± r 1 r 2 c + 32πG 3 ρM(t) . At late times when ρ(t) becomes negligible compared to the other term, one of the solutions is given by a(t) ∝ e t/rc , i.e., an exponential, accelerated expansion. 8 At present, the radiation is totally negligible compared to matter, ΩR/ΩM ' 1/3250, and the dark energy density is about 3 times as large as the matter density, ΩDE/ΩM ' 2.7 (with ΩM ' 0.27 and ΩDE ' 0.73). 1.4 Redshift As the universe expands, the wavelength of light, λ, is stretched linearly: λ(t) ∝ a(t), (1.31) which implies that photons lose energy as E(t) ∝ 1/a(t). This is something one can observe, by comparing, for example, the observed wavelength of a hydrogen line to the rest-frame wavelength that we know from the laboratory experiment. We often use the redshift, z, to quantify the stretching of the wavelength: 1 + z ≡ λ(t0) λ(temitted) . (1.32) The present-day corresponds to z = 0. Using equation (1.31), we can relate the observed redshift to the ratio of the scale factors: 1 + z = a(t0) a(temitted) . (1.33) Using this result in the Friedmann equation (1.30), we obtain the most-widely-used form of the Friedmann equation:

H2 (z) H2 0 = ΩM(1 + z) 3 + ΩR(1 + z) 4 + ΩDEe 3 R z 0 d ln(1+z)[1+wDE(z)From this result, it follows that the best way to determine the equation of state of dark energy is to measure H(z) over a wide range of z. If we can only measure the expansion rates at z 1, then Taylor expansion of equation (1.34) with ΩR ΩM and ΩDE ' 1 − ΩM gives H2 (z 1) H2 0 ≈ 1 + 3ΩMz + 3(1 + wDE)(1 − ΩM)z. (1.35) As we know from observations that |1 + wDE| is small (of order 10−1 or less), the third term is tiny compared to other terms, making it difficult to measure wDE. This is why we need to measure H(z) over a wide redshift range

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