±è°ü¼®

20240818 20:38:36, Á¶È¸¼ö : 132 
 Download #1 : BC_2b.jpg (112.1 KB), Download : 0
2.3 Dynamics
So far, we have used the symmetries of the universe to fix its geometry and studied the propagation of particles in the expending spacetime. The scale factor 𝑎(𝑡) has remained an unspecified function of time. The evolution of the scale factor follows from the Einstein equation
(2.84) 𝐺_{𝜇𝜈} ( + 𝛬𝑔_{𝜇𝜈}) = 8¥ð𝐺/𝑐^{4} 𝑇_{𝜇𝜈}, [𝛬: cosmological constant, 𝑔_{𝜇𝜈}:metric tensor]
where 𝐺 = 6.67 ¡¿ 10^{11} m^{3} kg^{1} s^{2} is Newton's constant. This relates the Einstein tensor 𝐺_{𝜇𝜈} (a measure of the "spacetime curvature" of the universe) to the energymomentum tensor 𝑇_{𝜇𝜈} (a measure of "matter content" of the universe). We will first discuss the possible forms of cosmological energymomentum tensors, then compute the Einstein tensor for FRW background, and finally put them together to solve for the evolution of the scale factor 𝑎(𝑡) as a function of the matter content.
2.3.1 Perfect Fluids
We have seen how the spacetime geometry of the universe is constrained by homogeneity and isotropy. Now, we will determine which types of matter are consistent with these symmetries. We will find that the coarsegrained energymomentum tensor is required to be that of a perfect fluid.
(2.85) 𝑇_{𝜇𝜈} = (𝜌 + 𝑃/𝑐^{2})𝑈_{𝜇}𝑈_{𝜈} + 𝑃𝑔_{𝜇𝜈},
where 𝜌𝑐^{2} and 𝑃 are the energy density and pressure in the rest frame of the fluid, and 𝑈^{𝜇} is its 4velocity relative to a comoving observer. {Wikipedia Perfect fluid: Stressenergy tensor of a perfect fluid in the rest frame diag(𝜌𝑐^{2}, 𝑝, 𝑝, 𝑝)]
Number density _{}^{}
Let us a simpler object: the number current 4vector 𝑁^{𝜇}. 𝑁^{0} measures the number density particles, where a "particle" in cosmology may be an entire galaxy. 𝑁^{𝑖} is the flux of particle in the direction 𝑥^{𝑖}. We call 𝑁^{0} a three scalar because its value doen't change under a purely spatial coordinate transformation. But 𝑁^{𝑖} transforms like a 3vector under such a transformation. We want to determine which form of the number current is consistent with the homogeneity and isotropy seen by a comoving observer.
Isotropy requires that the mean value of any 3vector, such as 𝑁^{𝑖}, must vanish, and homogeneity require that the mean value of 3scalar such as 𝑁^{0}, is only a function of time. Hence
(2.86) 𝑁^{0} = 𝑐𝑛(𝑡), 𝑁^{𝑖} = 0,
where 𝑛(𝑡) is the number of galaxies per proper volume. A general observer (i.e. an observer in motion relative to the men rest frame of the particles), would measure the following number current 4vector
(2.87) 𝑁^{𝜇} = 𝑛𝑈^{𝜇},
where 𝑈^{𝜇} ¡Õ 𝑑𝑥^{𝜇}/𝑑𝜏 is relative 4velocity between particles and the observer. We recover (2.86) for a comoving observer, with 𝑈^{𝜇} = (𝑐, 0, 0, 0). Moreover, for 𝑈^{𝜇} = 𝛾(𝑐, 𝑣^{𝑖}, the expression (2.87) gives the correctly boosted result, where 𝛾 is the Lorentz factor for the fact that one of the dimensions is Lorentz contracted.
How does the number density evolve with time? If the number of particles is conserved, then the rate of of change of number density must equal the divergence of the flux of particles. In Minkowski space, this imply the following continuity equation ¡Ó_{0}𝑁^{0} = ¡Ó_{𝑖}𝑁^{𝑖}, or in relativistic notation,
(2.88) ¡Ó_{𝜇}𝑁^{𝜇} = 0.
Equation (2.89) is generalized to curved spacetimes by replacing the partial derivative with a covariant derivative (see Appendix A)
(2.89) ¡Ô_{𝜇}𝑁^{𝜇} = 0.
This covariant form of the continuity equation is valid independent of the choice of coordinates.
Covariant derivative The covariant derivative is an important object in differential geometry and it is fundamental importance in general relativity. But we will not get into these detail here and simply tell us that how the covariant derivatives acts on 4vectors with upstairs and downstairs indices:
(2.9091) ¡Ô_{𝜇}𝐴^{𝜈} = ¡Ó_{𝜇}𝐴^{𝜈} + 𝑁^{𝜇}, ¡Ô_{𝜇}𝛣_{𝜈} = ¡Ó_{𝜇}𝛣_{𝜈}  𝛤^{𝜆}_{𝜇𝜈}𝛣_{𝜆}.
We already encounter (2.91) in our discussion of the geodesic equation in (2.43), [Wikipedia Covariant derivative] The action on a general tensor is straightforward generalization of these two results. For example, the covariant derivative of a rank2 tensor with mixed indices is
(2.92) ¡Ô_{𝜇}𝛵^{𝜎}_{𝜈} = ¡Ó_{𝜇}𝛵^{𝜎}_{𝜈} + 𝛤^{𝜎}_{𝜇𝜆}𝛵^{𝜆}_{𝜈}  𝛤^{𝜆}_{𝜇𝜈}𝛵^{𝜎}_{𝜈}.
Using (2.91) in (2.90, we get
(2.93) ¡Ó_{𝜇}𝑁^{𝜇} = 𝛤^{𝜇}_{𝜇𝜆}𝑁_{𝜆}.
Let us evaluate this in the rest frame of fluid. Substituting the components of the number current in (2.86), we obtain
(2.94) 𝑑𝑛/𝑐 𝑑𝑡 = 𝛤^{𝜇}_{𝜇0} 𝑛 = 3/𝑐 𝑎̇/𝑎 𝑛, [𝑐 𝑑𝑛/𝑐 𝑑𝑡 = 𝛤^{𝜇}_{𝜇0} 𝑐 𝑛]
where we have used that 𝛤^{0}_{00} = 0 and 𝛤^{𝑖}_{𝑖0} = 3/𝑐 𝑎̇/𝑎. Hence, we find that
(2.95) 𝑛̇/𝑛 = 3𝑎̇/𝑎 ¢¡ 𝑛(𝑡) ¡ð 𝑎^{3} [¡ò 𝑛̇/𝑛 𝑑𝑡 = ¡ò 1/𝑛 𝑑𝑛 = log(𝑛) + const; ¡ò 3𝑎̇/𝑎 𝑑𝑡 = ¡ò 3/𝑎 𝑑𝑎 = log(1/𝑎^{3}) + const]
As expected, the number density decreases in proportion to the increase of the proper volumne, 𝑉 ¡ð𝑎^{3}.
Energymomentum tensor _{}^{}
We will now use a similar logic to determine which form of the energymomentum tensor 𝛵_{𝜇𝜈} is consistent with the requirement of homogeneity and isotropy. First, we decompose 𝛵_{𝜇𝜈} into a 3scalar 𝛵_{00}, two 3vectors 𝛵_{𝑖0} and 𝛵_{0𝑗} and a 3tensor 𝛵_{𝑖𝑗}. The physical meaning of these component is
(2.96) 𝛵_{𝜇𝜈} = (𝛵_{00}, 𝛵_{0𝑗}, 𝛵_{𝑖0}, 𝛵_{𝑖𝑗}) = (energy density, momentum density, energy flex, stress tensor).
In a homogeneous universe, the energy density must be independent of position, but can depend on time, 𝛵_{00} = 𝜌(𝑡)𝑐^{2}. Moreover, isotropy again requires the mean values of the 3vectors in the comoving frame, 𝛵_{𝑖0} = 𝛵_{0𝑗} = 0. Finally isotropy around a point 𝐱 = 0 constrains the mean value of 3tensor, such as 𝛵_{𝑖𝑗}, to be proportional to 𝛿_{𝑖𝑗}. Since the metric 𝑔_{𝑖𝑗} equals 𝑎^{2}𝛿_{𝑖𝑗} at 𝐱 = 0, we have
(2.97) 𝛵_{𝑖𝑗}(𝐱 = 0) ¡ð 𝛿_{𝑖𝑗} ¡ð 𝑔_{𝑖𝑗}(𝐱 = 0).
Homogeneity requires the proportionality coefficient to be only a function of time. Moreover, Since this is a proportionality between two tensors, 𝛵_{𝑖𝑗} and 𝑔_{𝑖𝑗}, it remains unaffected by transformation of the spatial coordinates, including those transformations that preserve the form of 𝑔_{𝑖𝑗} while taking this origin into any other point. Hence homogeneity and isotropy require the component s of energymomentum tensor everywhere to take the form
(2.98) 𝛵_{00} ¡Õ 𝜌(𝑡)𝑐^{2}, 𝛵_{𝑖0} ¡Õ 𝑐𝜋_{𝑖} = 0, 𝛵_{𝑖𝑗} = 𝑃(𝑡)𝑔_{𝑖𝑗}(𝑡, 𝐱).
Raising one of indices, we find ^{[for example?]}
(2.99) 𝛵^{𝜇}𝜈 = 𝑔_{𝜇𝜈}𝛵_{𝜇𝜈} = diag(𝜌𝑐, 𝑃, 𝑃, 𝑃).
which we recognize as the energymomentum tensor of a perfect fluid in the frame of a comoving observer. More generally, the energymomentum tensor can be written in the following, explicitly covariant, form [RE (2.84)]
(2.100) 𝑇_{𝜇𝜈} = (𝜌 + 𝑃/𝑐^{2})𝑈_{𝜇}𝑈_{𝜈} + 𝑃𝑔_{𝜇𝜈}
where 𝑈^{𝜇} is the relative 4velocity between the fluid and the observer. We can recover (2.98) for a comoving observer, 𝑈^{𝜇} = (𝑐, 0, 0, 0). ^{[Why not 𝑈𝜇?]}
Continuity equation _{}^{}
In Minkowski space energy and momentum are conserved. The energy density therefore satisfied the continuity equation ῤ = ¡Ó_{𝑖}¥ð^{𝑖}, i.e. the rate of change of the density equals the divergence of the energy flux ¥ð. Similarly the evolution of momentum satisfies the Euler equation 𝑑¥ð^{𝑖}/𝑑𝑡 = ¡Ó_{𝑖}𝛲. These conservation laws can be combined into a 4component conservation equation for the energymomentum tensor [Wikipedia Continuity equation: and Euler equations (fluid dynamics):]
(2.101) ¡Ó_{𝜇}𝑇^{𝜇}_{𝜈} = 0.
In GR this id promoted to the covariant conservation equation
(2.102) ¡Ô_{𝜇}𝑇^{𝜇}_{𝜈} = 0.
Using (2.92)
(2.103) ¡Ô_{𝜇}𝑇^{𝜇}_{𝜈} = ¡Ó_{𝜇}𝛵^{𝜇}_{𝜈} + 𝛤^{𝜇}_{𝜇𝜆}𝛵^{𝜆}_{𝜈}  𝛤^{𝜆}_{𝜇𝜈}𝛵^{𝜇}_{𝜆} = 0.
This corresponds to four separate equation (𝜈: 0~3). The evolution of the energy density is determined by the 𝜈 = 0 equation
(2.104) ¡Ó_{𝜇}𝛵^{𝜇}_{0} + 𝛤^{𝜇}_{𝜇𝜆}𝛵^{𝜆}_{0}  𝛤^{𝜆}_{𝜇0}𝛵^{𝜇}_{𝜆} = 0.
Since 𝛵^{𝑖}_{0} vanishes by isopropy, this reduces to [¡Ó_{𝜇}𝛵^{0}_{0} + 𝛤^{𝜇}_{𝜇𝜆}𝛵^{0}_{0}  𝛤^{𝜆}_{𝜇0}𝛵^{𝜇}_{𝜆} = 0.]
(2.105) 1/𝑐 𝑑(𝜌𝑐^{2})/𝑑𝑡 + 𝛤^{𝜇}_{𝜇0}(𝜌𝑐^{2})  𝛤^{𝜆}_{𝜇0}𝛵^{𝜇}_{𝜆} = 0,
where 𝛤^{𝜆}_{𝜇0} vanishes unless 𝜆 and 𝜇 are spatial indices 𝑖 and they are same, i.e. 𝜆 = 𝜇 = 𝑖. So we have 𝛤^{𝑖}_{𝑖0} = 3𝑐^{1} 𝑎̇/𝑎. [RE (2.45) and 𝛵^{𝜇}_{𝜆} = 𝛵^{𝑖}_{𝑖} = 𝑃.] The continuity equation (2.105) then becomes
(2.106) 𝜌̇ + 3 𝑎̇/𝑎 (𝜌 + 𝑃/𝑐^{2}) = 0. [𝑐 𝑑𝜌/𝑑𝑡 + 3𝑐 𝑎̇/𝑎 (𝜌 + 𝑃/𝑐^{2}) = 0.]
This important equation describes "energy conservation" in the cosmological context. Notice that the usual notion of energy conservation in flat space relies on a symmetry under the translations. This symmetry is broken in an expanding space, so the familiar energy conservation does not have to hold and is replaced by (2.106).
Exercise 2.4 Show that (2.106) can be obtain from the thermodynamic relation 𝑑𝑈 = 𝑃𝑑𝑉, where 𝑈 = (𝜌𝑐^{2})𝑉 and 𝑉 ¡ð𝑎^{3}.
[Solution] Using 𝑈 = (𝜌𝑐^{2})𝑉, we have 𝑑𝑈 = 𝑐^{2}𝑉(𝑑𝜌 + 𝜌 𝑑𝑉/𝑉) and with 𝑑𝑈 = 𝑃𝑑𝑉 becomes the following and then we used 𝑉 ¡ð𝑎^{3} in the final equality.
(a) 𝑑𝜌 =  𝑑𝑉/𝑉(𝜌 + 𝑃/𝑐^{2}) = 3 𝑑𝑎/𝑎 (𝜌 + 𝑃/𝑐^{2}),
(b) (𝜌 + 𝑃/𝑐^{2}) log 𝑉 + const = (𝜌 + 𝑃/𝑐^{2}) log 𝑎^{3} + const)
By taking a derivative to time 𝑡 we get the continuity equation,
(c) 𝑑𝜌/𝑑𝑡 = 3 𝑎̇/𝑎 (𝜌 + 𝑃/𝑐^{2}). ¢¡ 𝜌̇ + 3 𝑎̇/𝑎 (𝜌 + 𝑃/𝑐^{2}) = 0. ▮
Most cosmological fluids can be parameterized in terms of a constant equation of state, 𝜔 = 𝛲/(𝜌𝑐^{2}). In that case, we get
(2.107) 𝜌̇/𝜌 = 3(1 + 𝜔) 𝑎̇/𝑎 ¢¡ 𝜌 ¡ð 𝑎^{3(1 + 𝜔)} [¡ò 𝜌̇/𝜌 𝑑𝑡 = log(𝜌) + const; ¡ò [3(1 + 𝜔) 𝑎̇/𝑎] 𝑑𝑡 = log[𝑎^{3(1 + 𝜔)}] + const]
which shows how the dilution of the energy density depends on the equation of state.
2.3.2 Matter and Radiation
The term matter will be used to refer to a fluid whose pressure is much smaller than its energy density, ∣𝛲∣¡ì 𝜌𝑐^{2}. As we will see in Chapter 3, this is the case for a gas of nonrelativistic particles, for which the energy density is dominated by their rest mass. Setting 𝜔 = 0 in (2.107) gives
(2.108) 𝜌 ¡ð 𝑎^{3}.
This dilution of the energy density simply reflects that the fact the volume of a region of space increases as 𝑉 ¡ð 𝑎^{3}, while the energy within the region stays constant.
• Baryons Cosmologists refers to ordinary matter (nuclei and electrons) as baryons.Electrons are leptons, but nuclei are so much heavier than electrons that most of the mass is in this baryons.
• Dark Matter Most of the matter in the universe is in the form of dark matter. Figure 2.9 shows the orbital speeds of visible stars or gas in the galaxy M33 versus their radial distance from the center of the galaxy. We see that the rotation speeds stay constant for behind the extent of the visible disc. This suggests that there is an invisible halo of dark matter that holds the galaxy together. On larger scale, dark matter has been detected by gravitational lensing. As the CMB photons travel through the universe they get deflected by the total matter in the universe. The effect can be measured statistically and provides one of the best pieces of evidence for dark matter on a cosmic scale.
The precise nature of the dark matter is unknown. We usually think of it as a new particle species, but we don't know what it really is. There are many popular dark matter candidates such as WIMPs, axions, MACHOs, and primordial black holes. [Wikipedia Dark matter:] Wile finding a microscopic detail don't affect the largescale evolution of the universe. On large scales, all dark matter models are described by the same pressureless fluid. This cold dark matter (CDM) is an integral part of the standard model of cosmology.
The term radiation will be used to denote for which the pressure is on third of the energy density, 𝛲 = 1/3 𝜌𝑐^{2}. This is the case for a gas of relativistic particles, for which the energy density to dominated by the kinetic energy. Setting 𝜔 = 1/3 in (2.107) gives
(2.109) 𝜌 ¡ð 𝑎^{4}.
The dilution now includes not just the expansion, 𝑉 ¡ð 𝑎^{3}, but also the redshifting of the energy of the particles, 𝐸 ¡ð 𝑎^{1}.
• Light particles At early times, all particles of the Standard Model acted as relativistic, because the temperature of the universe was larger than the masses of the particles. As the temperature of the universe dropped, the massless of many particles became relevant and they started behave like matter.
• Photons Being massless, photons are always relativistic. Together with neutrinos, they ere the dominant energy density during Big Bang nucleosynthesis. Today, we observe these photons in the form of the CMB.
• Neutrinos For most of history of the universe neutrinos behaves like radiation. Only recently have their small masses become relevant, making them act like matter.
• Gravitons The early universe may also have produced a background of gravitons. The density of these gravitons is predicted to be very small, so they have a negligible effect on the expansion of the universe. Nevertheless, experimental efforts are underway to detect them (see Section 8.4.2).
2.3.3 Dark Energy
We have recently learned that matter and radiation aren't enough to describe the evolution of the universe. Instead the universe today to determined by a mysterious form of dark energy with negative pressure,𝛲 = 𝜌𝑐^{2} and hence constant energy density.
(2.110) 𝜌 ¡ð 𝑎^{0}
Since the energy density doesn't dilute, energy has to created on the universe expands. As described above, that doesn't violate the conservation of energy, as long as equation (2.106) is satisfied.
• Vacuum energy A natural candidate for a constant energy density is the energy associated to empty space itself.As the universe expands, more space is being created and this energy therefore increase in proportion to the volume. In quantum field theory, this socalled "vacuum energy" in actually predicted, leading to an energymomentum tensor of the form
(2.111) 𝛵^{vac}_{𝜇𝜈} = 𝜌_{vac}𝑐^{2}𝑔_{𝜇𝜈}.
AS expected from Lorentz symmetry, the energymomentum tensor associated to the vacuum to proportional to the spacetime metric. Comparison with (2.100) shows that this indeed gives 𝛲_{vac} = 𝜌_{vac}𝑐^{2}. Unfortunately, as it will be described below, quantum field theory also predicts the size of the vacuum energy 𝜌_{vac} to be much larger than the value inferred from cosmological observations.
• Cosmological constant The observation of dark energy has also revived the old concept of a "cosmological constant," originally introduced by Einstein to make the universe static. Note that the lefthand side of Einstein equation (2.94) isn't uniquely defined. We can add the term 𝛬𝑔_{𝜇𝜈} for some constant 𝛬, without changing the conservation of the energymomentum tensor, ¡Ô^{𝜇}𝛵_{𝜇𝜈} = 0. To see this, you need to convince yourself that ¡Ô^{𝜇}𝑔_{𝜇𝜈} = 0. In other words, we could have written the Einstein equation as
(2.112) 𝐺_{𝜇𝜈} + 𝛬𝑔_{𝜇𝜈} = 8¥ð𝐺/𝑐^{4} 𝑇_{𝜇𝜈}.
However it has been standard practice to move this extra terms to the righthand side and treat it as a contribution to the energymomentum tensor
(2.113) 𝛵^{(𝛬)}_{𝜇𝜈} = 𝛬𝑐^{4}/8¥ð𝐺 𝑔_{𝜇𝜈} ¡Õ 𝜌_{𝛬}𝑐^{2}𝑔_{𝜇𝜈},
which is of the same form as (2.111).
• Dark energy The therm "dark energy" is often used to describe a more general fluid whose equation of state is not exactly that of a cosmological constant 𝜔 ≈ 1, or may even be varying in time. It is the failure of quantum field theory to explain the size of the observed vacuum energy that has led theorists to consider those more exotic possibilities. Whatever it is, dark energy plays a key role in the standard cosmological modelthe 𝛬CDM model (see Section 2.4).
The cosmological constant problem^{*}
When the cosmological constant was discovered, it came as a relief to observers, but was a shock to theorists. While it reconciled the age of the universe with the age of the oldest starts , its observed value is much smaller than all particle physics scales, making it hard to understand from a more fundamental perspective. This cosmological constant problem is the biggest crisis in modern theoretical physics.
In Section 2.4. we will see that the observed value of the vacuum energy is
(2.114) 𝜌_{𝛬}𝑐^{2} ≈ 6 ¡¿ 10^{10} J m^{3}
To compare this to our expectation from particle physics, we had better to write it in units of highenergy physics, electron volts (eV). Using 1 J = 6.2 ¡¿ 10^{18} eV and ©¤𝑐 ≈ 2.0 ¡¿ 10^{7} eV m, we get (©¤𝑐)^{3}𝜌_{𝛬}𝑐^{2} = (10^{3} eV)^{4}. Written in natural units, with ©¤ = 𝑐 ¡Õ 1, we have then
(2.115) 𝜌_{𝛬} = (10^{3} eV)^{4}.
We see that the scale appearing in the vacuum energy, 𝑀^{𝛬} ¡ 10^{3} eV, is much smaller than the typical scales relevant to particle physics.
Quantum field theory (QFT) predicts the existence of vacuum energy. However, the expected value of this vacuum energy is vastly too large. To start, recall that that in QFT every particle arises as an expectation of a fundamental field. Moreover, each field can be represented by an infinite number of Fourier modes with frequencies 𝜔_{𝜅}. These Fourier modes satisfy the equation of a harmonic oscillator and therefore experience the same zeropoint fluctuations as a quantum harmonic oscillator (see Section 8.2.1). In other words, the vacuum energy receive contributions of the form 1/2 ℏ𝜔_{𝜅} for each mode. The sum over all Fourier modes actually diverges. This divergence arises because we extrapolated our QFT beyond its regime of validity. Let us therefore assumed that our theory is valid only up to a cutoff scale 𝛭_{*} and only contributions to the vacuum energy from modes with frequencies below the cutoff. The vacuum energy will now be finite, but depends on the arbitrary cutoff scale. The dependence on the cutoff can be removed by adding a counterterm in the form of a bare cosmological constant 𝛬_{0}. After this process of renormalization, the remaining vacuum energy is
(2.116) 𝜌_{QFT} ¡ 𝛴_{𝑖} 𝑚^{4}_{𝑖},
where the sum is over all particles with mass below the cutoff. In the absence of gravity this constant vacuum energy wouldn't affect any dynamics and is therefore usually discarded. The equivalence principle, however,implies that all forms of energy gravitate and hence affect the curvature of the spacetime. The vacuum energy is no exception.
The first to worry about the gravitational effects of zeropoint energies was Wolfgang Pauli in the 1920s. Considering the contribution from the electron in (2.1116), he estimated that the induced curvature radius of the universe would be 10^{6} km, so that "the world would not even reach to the moon". Including rest of the Standard Model only makes the problem worse:
(2.117) 𝜌_{QFT} ≳ (1 TeV)^{4} = 10^{60} 𝜌_{𝛬}, [TeV = 10^{12} eV]
where we have added contributions up to the TeV scale. The enormous discrepancy between this estimate and the observed value in (2.114) is the cosmological constant problem.
But we can add an arbitrary constant to the QFT estimate to match the observed value of this vacuum energy
(2.118) 𝜌_{𝛬} = 𝜌_{QFT} + 𝜌_{0}.
To explain the small size of the observed vacuum energy would require two large numbers,𝜌_{QFT} and 𝜌_{0} have 60 digits in common, but differ at the 61st digit, so that the sum of the two is smaller by 60 orders of magnitude. moreover, each time the universe goes through a phase transition, the value of 𝜌_{QFT} changes, while the value of 𝜌_{0} presumably doesn't. For example, we expect the electroweakland QCD phase transition to induce jumps in the order of ∆_{𝜌QFT} ¡ (200 GeV)^{4} and (0.3 GeV)^{4} respectively.Any tuning must explain the small value of the cosmological constant today and not at the beginning.
One way to explain the apparent finetuning to the vacuum energy is to appeal to the anthropic principle. Suppose that our universe is part of a much larger multiverse and that the fundamental physical parameters exhibit different values indifferent regions of space. In particular, imagine that the value of the vacuum energy varies across the multiverse. If the vacuum energy were much larger, dark energy would start to dominate before galaxies would have formed and complex structures could not have evolved. In other words, we shouldn't be shocked by the small value of the vacuum energy, since if it was any bigger w would not be around to ask the question. "We live where we can live" in a vast landscape of possibilities and the observed value of the cosmological constant is simply the accident of "environmental selection."
2.3.4 Spacetime Curvature
Having introduced the relevant types of matter and energy in the universe, we now want to see how they source the dynamic s of the spacetime. To do this, we have to compute the Einstein tensor on the lefthand side of the Einstein equation (2.84)
(2.119) 𝐺_{𝜇𝜈} = 𝑅_{𝜇𝜈}  1/2 𝑅𝑔_{𝜇𝜈},
where 𝑅_{𝜇𝜈} is the Ricci tensor and 𝑅 = 𝑅^{𝜇}_{𝜇} = 𝑔^{𝜇𝜈}𝑅_{𝜇𝜈} is its trace, the Ricci scalar. In Appendix A, we give the following definition of the Ricci tensor
(2.120) 𝑅_{𝜇𝜈} ¡Õ ¡Ó_{𝜆}𝛤^{𝜆}_{𝜇𝜈}  ¡Ó_{𝜈}𝛤^{𝜆}_{𝜇𝜆} + 𝛤^{𝜆}_{𝜆𝜌}𝛤^{𝜌}_{𝜇𝜈}  𝛤^{𝜌}_{𝜇𝜆}𝛤^{𝜆}_{𝜈𝜌}.
Now we are going to compute (2.120) and (2.119) for the FRW spacetime. 𝑅_{𝑖0} = 𝑅_{0𝑖} = 0, because it is a 3vector and therefore must vanish due to the isotropy of RobertsonWalker metric. The non vanishing components of Ricci tensor are
(2.121) 𝑅_{00} = 3/𝑐^{2} (𝑑𝑎̇/𝑑𝑡)/𝑎,
(2.122) 𝑅_{𝑖𝑗} = 1/𝑐^{2} [(𝑑𝑎̇/𝑑𝑡)/𝑎 + 2 (𝑎̇/𝑎)^{2} + 2 𝑘𝑐^{2}/𝑎^{2}𝑅_{0}^{2}] 𝑔_{𝑖𝑗}.
Example Setting 𝜇 = 𝜈 = 0 in (2.120), we have
(2.123) 𝑅_{00} ¡Õ ¡Ó_{𝜆}𝛤^{𝜆}_{00}  ¡Ó_{0}𝛤^{𝜆}_{0𝜆} + 𝛤^{𝜆}_{𝜆𝜌}𝛤^{𝜌}_{00}  𝛤^{𝜌}_{0𝜆}𝛤^{𝜆}_{0𝜌}.
Since Christoffel symbols with two time indices00 vanish, this reduces to
(2.124) 𝑅_{00} ¡Õ ¡Ó_{0}𝛤^{𝑖}_{0𝑖}  𝛤^{𝑖}_{0𝑗}𝛤^{𝑗}_{0𝑖}.
Using 𝛤^{𝑖}_{0𝑗} = 𝑐^{1}(𝑎̇/𝑎)𝛿^{𝑖}_{𝑗}, {RE (2.45)] we find
(2.125) 𝑅_{00} = 1/𝑐^{2} 𝑑/𝑑𝑡 (3 𝑎̇/𝑎)  3/𝑐^{2} (𝑎̇/𝑎)^{2} = 3/𝑐^{2} (𝑑𝑎̇/𝑑𝑡)/𝑎. ▮
Example Computing 𝑅_{𝑖𝑗} can be greatly simplified using the following trick: we will compute 𝑅_{𝑖𝑗} at 𝐱 = 0 where the spatial metric is 𝛾_{𝑖𝑗} = 𝛿_{𝑖𝑗} and then argue that the result for general 𝐱 must have the form 𝑅_{𝑖𝑗} ¡ð 𝑔_{𝑖𝑗}. [RE (2.49) In FRW metric 𝑔_{𝑖𝑗} = 𝑎^{2}𝛾_{𝑖𝑗}]
In Cartesian coordinates the spatial metric is [RE (2.914)]
(2.126) 𝛾_{𝑖𝑗} = 𝛿_{𝑖𝑗} + 𝑘𝑥_{𝑖}𝑥_{𝑗}/[𝑅_{0}^{2}  𝑘(𝑥_{𝑙}𝑥^{𝑙})]. [Why? (𝐱 ⋅ 𝑑𝐱)^{2} = 𝑥_{𝑖}𝑥_{𝑗}; 𝐱^{2} = 𝑥_{𝑙}𝑥^{𝑙}]
The key point is to think ahead and anticipate that we will set 𝐱 = 0 at the end. In this condition the Christoffel symbols contain derivative metric and the Ricci tensor has another derivative, so there will be terms in the final answer with two derivatives acting on the metric from the second term. We only need to keep terms to quadratic order in 𝑥: ^{[Why?]}
(2.127) 𝛾_{𝑖𝑗} = 𝛿_{𝑖𝑗} + 𝑘𝑥_{𝑖}𝑥_{𝑗}/𝑅_{0}^{2} + 𝑂(𝑥^{4}).
Plugging this into the definition of Christoffel symbol gives [RE (2.45) for Christoffel symbol in FRW metric]
(2.128) 𝛤^{𝑖}_{𝑗𝑘} = 𝑘/𝑅_{0}^{2} 𝑥^{𝑖}𝛿_{𝑗𝑘} + 𝑂(𝑥^{3}).
This vanishes at 𝐱 = 0, but the derivative does not. From the definition of Ricci tensor, we then have
(2.129) 𝑅_{𝑖𝑗}(𝐱 = 0) ¡Õ ^{(𝐴)}{¡Ó_{𝜆}𝛤^{𝜆}_{𝑖𝑗}  ¡Ó_{𝑗}𝛤^{𝜆}_{𝑖𝜆}} + ^{(𝐵)}{𝛤^{𝜆}_{𝜆𝜌}𝛤^{𝜌}_{𝑖𝑗}  𝛤^{𝜌}_{𝑖𝜆}𝛤^{𝜆}_{𝑗𝜌}}.
Let us start with (𝐵). Dropping terms that are zero at 𝐱 = 0, we find
(2.130) (𝐵) = 𝛤^{𝑙}_{𝑙0}𝛤^{0}_{𝑖𝑗}  𝛤^{0}_{𝑖𝑙}𝛤^{𝑙}_{𝑗0}  𝛤^{𝑙}_{𝑖0}𝛤^{0}_{𝑗𝑙} = 3/𝑐^{2} 𝑎̇/𝑎 𝑎𝑎̇𝛿_{𝑖𝑗}  1/𝑐^{2} 𝑎𝛿_{𝑖𝑙} 𝑎̇/𝑎 𝑎𝛿^{𝑖}_{𝑗}  1/ 𝑐^{2} 𝑎̇/𝑎 𝛿^{𝑙}_{𝑖}𝑎𝑎̇𝛿_{𝑗𝑙} = 𝑎̇^{2}/𝑐^{2} 𝛿_{𝑖𝑗}.
The two terms labeled (𝐴) can be evaluated by using (2.128),
(2.131) (𝐴) = ¡Ó_{0}𝛤^{0}_{𝑖𝑗} + ¡Ó_{𝑡}𝛤^{𝑙}_{𝑖𝑗}  ¡Ó_{𝑗}𝛤^{𝑙}_{𝑖𝑙} = 𝑐^{2} ¡Ó_{𝑡}(𝑎𝑎̇)𝛿_{𝑖𝑗} + 𝑘/𝑅_{0}^{2} 𝛿^{𝑙}_{𝑖}𝛿_{𝑖𝑗}  𝑘/𝑅_{0}^{2} 𝛿^{𝑙}_{𝑗}𝛿_{𝑖𝑙} = 𝑐^{2} (𝑎𝑑𝑎̇/𝑑𝑡 + 𝑎̇^{2} + 2 𝑘𝑐^{2}/𝑅_{0}^{2})𝛿_{𝑖𝑗}.
Hence we get
(2.132) 𝑅_{𝑖𝑗}(𝐱 = 0) = (𝐴) + (𝐵) = 1/𝑐^{2}[𝑎𝑎̇ + 2𝑎̇^{2} + 2 𝑘𝑐^{2}/𝑅_{0}^{2}]𝛿_{𝑖𝑗} = 1/𝑐^{2}[𝑎𝑑𝑎̇/𝑑𝑡 + 2𝑎̇^{2} + 2 𝑘𝑐^{2}/𝑅_{0}^{2}]𝛿_{𝑖𝑗} = 1/𝑐^{2}[(𝑑𝑎̇/𝑑𝑡)/𝑎 + 2(𝑎̇/𝑎)^{2} + 2 𝑘𝑐^{2}/𝑅_{0}^{2}]𝑔_{𝑖𝑗}(𝐱 = 0).
Since this is a relation between tensors, it holds for general 𝐱, so we get the result quoted in (2.122). ▮
Now we can calculate the Ricci scalar [𝑔^{𝜇𝜈} = diag(1, 1/𝑎^{2} 𝛾^{𝑖𝑗})
(2.133) 𝑅 = 𝑔^{𝜇𝜈}𝑅_{𝜇𝜈} = 𝑅_{00}  𝑔^{𝑖𝑗}𝑅_{𝑖𝑗} = 3/𝑐^{2} (𝑑𝑎̇/𝑑𝑡)/𝑎 + 𝑔^{𝑖𝑗}𝑔_{𝑖𝑗} 1/𝑐^{2}[(𝑑𝑎̇/𝑑𝑡)/𝑎 + 2(𝑎̇/𝑎)^{2} + 2 𝑘𝑐^{2}/𝑅_{0}^{2}] = 𝛿^{𝑖}_{𝑖}[(𝑑𝑎̇/𝑑𝑡)/𝑎 + 2(𝑎̇/𝑎)^{2} + 2 𝑘𝑐^{2}/𝑅_{0}^{2}] = 6/𝑐^{2} [(𝑑𝑎̇/𝑑𝑡)/𝑎 + (𝑎̇/𝑎)^{2} + 𝑘𝑐^{2}/𝑅_{0}^{2}],
and the nonzero components of the Einstein tensor, 𝐺^{𝜇}_{𝜈} ¡Õ 𝑔^{𝜇𝜆}𝐺_{𝜆𝜈}, are
(2.134) 𝐺^{0}_{0} = 3/𝑐^{2} [(ȧ/𝑎)^{2} + 𝑘𝑐^{2}/𝑎^{2}𝑅_{0}^{2}],
Derivation From (2.119) 𝐺^{0}_{0} = 𝑔^{00}𝐺_{00} = 𝐺_{00}. 𝐺_{00}= 𝑅_{00} + 1/2 𝑅𝑔_{00} = 3/𝑐^{2} (𝑑𝑎̇/𝑑𝑡)/𝑎 + 1/2 6/𝑐^{2} [(𝑑𝑎̇/𝑑𝑡)/𝑎 + (𝑎̇/𝑎)^{2} + 𝑘𝑐^{2}/𝑅_{0}^{2}](1) = 3/𝑐^{2} [(𝑎̇/𝑎)^{2} + 𝑘𝑐^{2}/𝑎^{2}𝑅_{0}^{2}]. ▮
(2.135) 𝐺^{𝑖}_{𝑗} = 1/𝑐^{2} [2(𝑑𝑎̇/𝑑𝑡)/𝑎 + (𝑎̇/𝑎)^{2} + 𝑘𝑐^{2}/𝑎^{2}𝑅_{0}^{2}]𝛿^{𝑖}_{𝑗}
Derivation 𝑔^{𝑖𝜆}𝐺_{𝜆𝑗} = 𝐺^{𝑖}_{𝑗}. From (2.122) and (2.133) 𝐺_{𝑖𝑗} = 𝑅_{𝑖𝑗} 1/2 𝑅𝑔_{𝑖𝑗} = 1/𝑐^{2} [(𝑑𝑎̇/𝑑𝑡)/𝑎 + 2 (𝑎̇/𝑎)^{2} + 2 𝑘𝑐^{2}/𝑎^{2}𝑅_{0}^{2}] 𝑔_{𝑖𝑗}  1/2 6/𝑐^{2} [(𝑑𝑎̇/𝑑𝑡)/𝑎 + (𝑎̇/𝑎)^{2} + 𝑘𝑐^{2}/𝑅_{0}^{2}]𝑔_{𝑖𝑗} = 1/𝑐^{2} [2(𝑑𝑎̇/𝑑𝑡)/𝑎 + (𝑎̇/𝑎)^{2} + 𝑘𝑐^{2}/𝑎^{2}𝑅_{0}^{2}] 𝑔_{𝑖𝑗}, ¢¡ 𝐺^{𝑖}_{𝑗} = 1/𝑐^{2} [2(𝑑𝑎̇/𝑑𝑡)/𝑎 + (𝑎̇/𝑎)^{2} + 𝑘𝑐^{2}/𝑎^{2}𝑅_{0}^{2}]𝛿^{𝑖}_{𝑗}. ▮


