Introduction
This chapter explores modern cosmological concepts using only Newtonian mechanics, avoiding the complexities of General Relativity It derives the Friedmann equations, which describe the universe's velocity and acceleration, and introduces the cosmological constant within a Newtonian context Additionally, it presents a straightforward derivation of equations of state and discusses key applications, including conservation laws, the age of the universe, and the various epochs dominated by inflation, radiation, and matter.
Equation of State
Matter
Writing the density of matter as ρ= M
3 πr 3 (1.6) it follows that ˙ ρ≡ dρ drr˙=−3ρr˙ r (1.7) so that by comparing to equation (1.5), it follows that the equation of state for matter is p= 0 (1.8)
The findings align with those derived from the ideal gas law at absolute zero temperature, emphasizing that this analysis does not incorporate kinetic energy, thereby focusing solely on conditions present at zero temperature.
Radiation
The equation of state for radiation is derived by examining radiation modes within a cavity, drawing an analogy to a violin string This concept is illustrated through the behavior of standing waves on a string that is fixed at both ends.
Velocity and Acceleration Equations
In the context of wave mechanics, the relationship between the length of a string (L), wavelength (λ), and frequency (f) is essential, where n represents a positive integer (n = 1, 2, 3 ) Radiation propagates at the speed of light (c), leading to the equation c = fλ = f(2Ln) By substituting f = (2Ln)/c into Planck's formula, we can derive significant insights into the behavior of electromagnetic radiation.
U = ¯hω=hf, whereh is Planck’s constant, gives
Using equation (1.2) the pressure becomes p≡ −dU dV = 1 3
Usingρ=U/V, the radiation equation of state is p= 1
It is customary to combine the equations of state into the form p= γ
The equation of state, represented as 3ρ (1.14), distinguishes between different components of the Universe, with γ set to 1 for radiation and γ set to 0 for matter Understanding these equations is essential for analyzing the radiation-dominated and matter-dominated epochs in the evolution of the Universe.
The Friedmann equation, which specifies the speed of recession, is obtained by writing the total energy E as the sum of kinetic plus potential energy terms (and using M = 4 3 πr 3 ρ )
The Hubble constant (H) is defined as the ratio of the rate of expansion (ṙ) to the distance (r) from the center of a dust mass (M) to a test particle In this context, the mass of the test particle is denoted as 'm', while 'G' represents Newton's gravitational constant.
Recall that the escape velocity is justv escape q 2GM r q 8πG
The equation ˙r² = v_escape² - k₀(1 - 2) illustrates the relationship between a particle's motion and its total energy, where k₀ is defined as -2E/m The value of k₀ can be negative, zero, or positive, indicating whether the total energy E is negative, zero, or positive For a particle near Earth, this translates to three scenarios: escaping (unbound), orbiting (critical case), or returning (bound) based on whether the speed ˙r exceeds, equals, or falls below the escape speed v_escape This concept can later be applied to describe the dynamics of an open, flat, or closed universe.
Defining k≡ − ms 2E 2 and writing the distance in terms of the scale factor R and a constant length sas r(t) ≡R(t)s, it follows that r r ˙ = R R ˙ and ¨ r r = R R ¨ , giving the Friedmann equation
The equation R 2 (1.18) defines the rate of cosmic recession, highlighting that in General Relativity, it is the fabric of space itself that undergoes expansion This principle applies not only to matter but also to radiation and even vacuum energy, represented by the cosmological constant Λ, which substitutes ordinary density The same relationship can be derived from Einstein's field equations Guth explains that the curvature parameter k can be adjusted to take values of -1, 0, or +1, corresponding to unbound, critical, or bound trajectories from a Newtonian perspective, while from a geometric standpoint, it indicates an open, flat, or closed universe.
In elementary mechanics the speed v of a ball dropped from a height r is evaluated from the conservation of energy equation as v = √
The acceleration of the ball can be expressed as a = g, where g represents the acceleration due to gravity This calculation follows a similar derivation to the one previously discussed, utilizing Newton's equation F = ma, where F denotes the force acting on the object and m is its mass.
Cosmological Constant
Einstein Static Universe
The cosmological constant, known for its repulsive properties, can be precisely calculated for a static universe In the Einstein static universe model, where the scale factor R remains constant (R = R0), both the first and second time derivatives of R are zero Focusing on the scenario where the second derivative of R is also zero, we find that the cosmological constant Λ is expressed as Λ = 4πG(ρ + 3p) = 4πG(1 + γ)ρ, highlighting the relationship between density (ρ), pressure (p), and the parameter γ.
In the absence of a cosmological constant (Λ = 0), the universe would either be empty (ρ = 0) or require negative pressure (p = -1/3 ρ), both of which Einstein found unacceptable Consequently, he concluded that a cosmological constant must exist, indicating that Λ is not equal to zero This leads to the implication that the energy density (ρ) is equivalent to the cosmological constant (ρ = Λ).
4πG(1 +γ) (1.32) and because ρ is positive this requires a positive Λ Substituting equa- tion (1.32) into equation (1.28) it follows that Λ = 3(1 +γ)
Now imposing ˙R = 0 and assuming a matter equation of state (γ = 0) implies Λ = R k 2
However the requirement that Λ be positive forcesk= +1 giving Λ = 1
The cosmological constant is specifically defined as the inverse of the square of the scale factor, which remains constant in this static model.
Conservation laws
The conservation of energy is implied by the equations of velocity and acceleration, similar to how Maxwell's equations indicate charge conservation The energy-momentum conservation equation can be derived by equating the covariant derivative of the energy-momentum tensor to zero This outcome is also obtained by differentiating equation (1.29) with respect to time, resulting in the expression ˙ ρ + 3(ρ + p)R˙.
The connection between thermodynamics and General Relativity, as highlighted in equation (1.5), has garnered recent attention We derived our velocity and acceleration equations using thermodynamic principles, which ultimately reappear in our findings Notably, these equations can also be directly derived from the Einstein field equations, indicating that the Einstein equations inherently support this thermodynamic relationship.
The above equation can also be written as d dt(ρR 3 ) +pdR 3 dt = 0 (2.2) and from equation (1.14), 3(ρ+p) = (3 +γ)ρ, it follows that d dt(ρR 3+γ ) = 0 (2.3)
R 3+γ (2.4) wherecis a constant This shows that the density falls as R 1 3 for matter and
Later we shall use these equations in a different form as follows From equation (2.1), ρ 0 + 3(ρ+p)1
R = 0 (2.5) where primes denote derivatives with respect to R, i.e x 0 ≡dx/dR Alter- natively d dR(ρR 3 ) + 3pR 2 = 0 (2.6) so that
R 3+γ d dR(ρR 3+γ ) = 0 (2.7) which is consistent with equation (2.4)
Age of the Universe
Recent observations from the Hubble Space Telescope indicate that the universe is younger than previously thought, particularly in relation to globular clusters This discrepancy may be addressed through the concept of the cosmological constant.
We illustrate this as follows.
3 (ρ+ρ vac )R 2 −k (2.8) the present day value of kis k= 8πG
3 (ρ 0+ρ 0vac )R 2 0 −H 0 2 R 0 2 (2.9) withH 2 ≡( R R ˙ ) 2 Present day values of quantities have been denoted with a subscript 0 Substituting equation (2.9) into equation (2.8) yields
Inflation
Integrating gives the expansion age
The presence of a non-zero cosmological constant, represented as ρ vac = ρ 0vac, results in an increased age of the universe compared to a scenario without this constant, particularly when R² < R² 0 This demonstrates that incorporating a cosmological constant leads to a greater estimated age of the universe.
In this section only a flat k = 0 universe will be discussed Results for an open or closed universe can easily be obtained and are discussed in the references [13].
The current phase of the universe is dominated by matter, which contributes significantly to its energy density In contrast, the early universe experienced a radiation-dominated phase, while the very early universe was characterized by vacuum dominance By setting k = 0, the dynamics of the universe can be described by a single term in the Friedmann equation, depending on the prevailing energy component In a matter-dominated universe (γ = 0) or a radiation-dominated universe (γ = 1), the equation simplifies to the form R^(3+γ), excluding vacuum energy Conversely, in a vacuum-dominated universe, the equation results in a constant value The solution to the Friedmann equation specifically for a radiation-dominated universe is thus derived from these considerations.
R∝t 1 2 , while for the matter dominated case it will beR∝t 2 3 One can see that these results give negative acceleration, corresponding to a decelerating expanding universe.
Inflation occurs when vacuum energy significantly outweighs ordinary density and curvature terms, leading to a scenario where R is proportional to the exponential of time (R ∝ exp(t)) This results in positive acceleration, characterizing an expanding universe known as the inflationary universe.
Quantum Cosmology
Derivation of the Schr¨ odinger equation
The Wheeler-DeWitt equation is derived by drawing an analogy with the one-dimensional Schrödinger equation, which is also presented here for completeness The Lagrangian \( L \) describes the motion of a single particle within a potential \( V \).
The relationship between kinetic energy (T) and potential energy (V) is expressed as L = T - V, where T is defined as T = 1/2 mẋ² The action, represented by S, is calculated using the integral S = ∫ L dt By varying the action to satisfy δS = 0, we derive the Euler-Lagrange equation, which describes the equations of motion as d/dt(∂L).
(Note P is the momentum but p is the pressure.) The Hamiltonian H is defined as
H(P, x)≡Px˙ −L( ˙x, x) (2.16) For many situations of physical interest, such as a single particle moving in a harmonic oscillator potential V = 1 2 kx 2 , the Hamiltonian becomes
The equation 2m + V = E (2.17) represents the total energy (E) in a quantum system Quantization is implemented through operator substitutions, specifically P → Pˆ = −i ∂x ∂ and E → Eˆ = i ∂t ∂, focusing solely on the one-dimensional case while omitting factors of ¯h The Schrödinger equation emerges from this framework by expressing the Hamiltonian as an operator.
Hˆ acting on a wave function Ψ as in
HˆΨ = ˆEΨ (2.18) and making the above operator replacements to obtain
∂tΨ (2.19) which is the usual form of the 1-dimensional Schr¨odinger equation written in configuration space.
Wheeler-DeWitt equation
The Wheeler-DeWitt equation is often discussed in the context of the minisuperspace approximation, typically focusing on closed (k = +1) and empty (ρ = 0) universes However, this article expands the discussion to include closed, open, flat, and non-empty universes, emphasizing the significance of incorporating matter and radiation, as their presence could alter the conclusions drawn This paper presents a derivation of the Wheeler-DeWitt equation within the minisuperspace approximation that accounts for matter, radiation, and various values of k.
3 (ρ+ρ vac )] (2.20) withκ≡ 4G 3π The momentum conjugate to R is
By substituting L and P into the Euler-Lagrange equation, the equation ˙P − ∂L/∂R = 0 is obtained, leading to the recovery of equation (1.29) The calculation of ∂L/∂R is streamlined using the conservation equation (2.5) alongside equation (1.14), which results in ρ₀ + ρ₀ vac − (3 + γ)ρ/R Additionally, the Hamiltonian is defined as H ≡ PR˙ − L.
The equation 3(ρ + ρ_vac) = 0 demonstrates that the Hamiltonian is identically zero when expressed in terms of ˙R, distinguishing it from the total energy as previously indicated in equation (1.29) This highlights the relationship between the Hamiltonian and the conjugate momentum in the context of the system's dynamics.
The Wheeler-DeWitt equation in the minisuperspace approximation, which incorporates matter or radiation through the density term (ρ), can be derived by substituting P with -i ∂R/∂ and setting HΨ to zero, resulting in the equation [3(ρ + ρ_vac)] = 0, confirming that it equals zero for arbitrary k.
Using equation (2.4) the Wheeler-DeWitt equation becomes
This just looks like the zero energy Schr¨odinger equation [21] with a potential given by
In the scenario of an empty universe devoid of matter or radiation (c = 0), the potential V(R) is illustrated for different curvature cases: k = +1 (closed), k = 0 (flat), and k = -1 (open) Notably, only the closed universe presents a potential barrier that allows for quantum tunneling, emphasizing that such universes can emerge through this process When radiation is introduced (γ = 1 and c ≠ 0), it results in a downward shift of the potential graph without altering the tunneling conclusions, as the term R^(1 - γ) remains constant Similarly, the inclusion of matter (γ = 0 and c ≠ 0) introduces a term that grows with R, but this only affects very small R values and does not change the overall conclusions In summary, quantum tunneling can only lead to the formation of closed universes, regardless of the presence of matter or radiation.
Problems
Answers
Solutions
Contravariant and Covariant Vectors
In a given coordinate system, an ordinary 2-dimensional vector is represented by its components (x, y) or (x₁, x₂) When this coordinate system is rotated by an angle θ while keeping the vector fixed, the components transform according to a specific relationship, denoted as [1] Ã x y.
In this article, we introduce the concept of contravariant vectors, denoted by superscripts (xi) rather than the traditional subscripts used in introductory physics It is important to note that these contravariant vectors are essentially the same ordinary vectors encountered in freshman physics courses.
Expanding the matrix equation we have x=xcosθ+ysinθ (3.2) y=−xsinθ+ycosθ from which it follows that
∂yy which can be written compactly as x i = ∂x i
∂x j x j (3.5) where we will always be using the Einstein summation convention for doubly repeated indices (i.e x i y i ≡ P i x i y i )
Rather than defining an ordinary contravariant vector as a simple arrow indicating direction, we define it as an object whose components transform according to a specific equation This approach is an advanced interpretation of how vector components behave during rotation, moving away from the traditional arrow representation Additionally, it's noteworthy that a differential version of this transformation could be derived using fundamental calculus principles, specifically utilizing the infinitesimal \(dx^i\).
(instead ofx i ) it follows immediately that dx i = ∂x i
∂x j dx j (3.6) which is identical to (3.5) and therefore we must say that dx i forms an ordinary or contravariant vector (or an infinitessimally tiny arrow).
In the realm of calculus and infinitesimals, the concept of ∂x ∂i can be viewed as the 'inverse' of dx i According to calculus, if we consider a function f represented as f(x, y), and express x and y as functions of x and y, this relationship is encapsulated in the equation (3.3).
Let’s ’remove’f and just write
The expression ∂/∂x j resembles the previous equation (3.5), suggesting that ∂/∂x i may represent the components of a 'non-ordinary' vector To simplify notation, we will denote ∂/∂x i as x i, recognizing that the index appears in the denominator Therefore, we can define x i as ∂ for clarity and convenience.
So now let’s define acontravariantvectorA à as anything whose components transform as (compare (3.5))
(3.12) and a covariant vector A à (often also called a one-form, or dual vector or covector)
In calculus we have two fundamental objectsdx i and the dual vector∂/∂x i
In differential geometry, forming the dual dual vector ∂/∂(∂/∂x i ) results in the original differential form dx i A manifold is defined as a set of points in a smooth space, where dx i represents a space and ∂/∂x i denotes the corresponding dual space The dual of the dual space returns to the original space, dx i, illustrating that contravariant and covariant vectors are duals of one another Additional examples of dual spaces include row and column matrices, such as (x y) and à x y.
! and the kets< a|and bras|a >used in quantum mechanics [3].
Before proceeding let’s emphasize again that our definitions of contravari- ant and covariant vectors in (3.13) and (3.13) are nothing more than fancy versions of (3.1).
Higher Rank Tensors
In the realm of tensors, scalar quantities, such as time (t) or temperature (T), are classified as tensors of rank zero due to their lack of indices In contrast, vectors, which possess a single index, are categorized as tensors of rank one Matrices, characterized by two indices (A_ij), represent tensors of rank two, while contravariant tensors of rank two take the form A_ν Additionally, tensors of rank three are denoted as A_νγ A mixed tensor, such as A_ν, exhibits both covariant and contravariant properties.
In order for an object to be called a tensor it must satisfy the tensor transformation rules, examples of which are (3.13) and (3.13) and
A matrix with two indices, A_ij, is not automatically classified as a second rank tensor unless it adheres to specific tensor transformation rules Nonetheless, all second rank tensors can be represented as matrices.
Higher rank tensors can be constructed from lower rank ones by forming what is called theouter productortensor product[14] as follows For instance
T γδ αβ ≡A α γ B δ β (3.18) The tensor product is often written simply as
(do Problem 3.1) (NNN Next time discuss wedge product - easy - just introduce antisymmetry).
Lower rank tensors can be derived from higher rank tensors through a process known as contraction, which involves setting a covariant index equal to a contravariant index This process utilizes the Einstein summation convention, where equal or repeated indices are summed over, facilitating the creation of lower-dimensional tensor representations.
Review of Cartesian Tensors
indices are summed over Thus contraction represents setting two indices equal and summing For example
Thus contraction over a pair of indices reduces the rank of a tensor by two [14].
The inner product [14] of two tensors is defined by forming the outer product and then contracting over a pair of indices as
Clearly the inner product of two vectors (rank one tensors) produces a scalar (rank zero tensor) as
A à B à =constant≡A.B (3.22) and it can be shown thatA.Bas defined here is a scalar(do Problem 3.2).
A scalar is a tensor of rank zero with the very special transformation law of invariance c=c (3.23)
It is easily shown, for example, thatA à B à is no good as a definition of inner product for vectors because it is not invariant under transformations and therefore is not a scalar.
Let us review the scalar product that we used in freshman physics We wrote vectors as A=A i eˆ i and defined the scalar product as
A B ≡ABcosθ (3.24) where A and B are the magnitudes of the vectors A and B and θ is the angle between them Thus
≡g ij A i B j (3.25) where the metric tensorg ij is defined as the dot product of the basis vectors.
A Cartesian basis is characterized by the condition g ij ≡ δ ij, which indicates that the basis vectors are unit length and orthogonal to one another This relationship is derived from the equation ˆ e i ˆe j = |eˆ i ||eˆ j |cosθ, leading to cosθ = δ ij.
=A x B x +A y B y + (3.26) where the sum (+ ) extends to however many dimensions are being consid- ered and
A A ≡A 2 =A i A i (3.27) which is just Pythagoras’ theorem, A A ≡A 2 = A i A i =A 2 x +A 2 y +
Notice that the usual results we learned about in freshman physics, equa- tions (3.26) and (3.27), resultentirely from requiring g ij =δ ij à 1 0
We could easily have defined a non-Cartesian space, for example, g ij à 1 1
! in which case Pythagoras’ theorem would change to
The metric tensor \( g_{ij} \equiv \hat{e}_i \cdot \hat{e}_j \), derived from the scalar product of unit vectors, plays a crucial role in defining the vector space under consideration This leads us back to the concepts of vectors and one-forms, which represent contravariant and covariant vectors, respectively.
Metric Tensor
Special Relativity
Whereas the 3-dimensional Cartesian space is completely characterized by g àν =δ àν or g àν
Obviously for unit matrices there is no distinction betweenδ à ν andδ àν The 4-dimensional spacetime of special relativity is specified by η àν
If a contravariant vector is specified by
A à = (A 0 , A i ) = (A 0 , A) (3.38) it follows that the covariant vector is A à =η àν A ν or
Exercise: Prove equation (3.39) using (3.38) and (3.37).
Thus, for example, the energy momentum four vector p à = (E, p) gives p 2 = E 2 − p 2 Of course p 2 is the invariant we identify as m 2 so that
Because of equation (3.38) we must have
Christoffel Symbols
Note that∂ 0 =∂ 0 = ∂t ∂ (with c≡1) We define
(Note that some authors [30] instead define2 2 ≡ 5 2 − ∂t ∂ 2 2).
Let us now briefly discuss the fourvelocity u à and proper time We shall write outcexplicitly here.
The invariant interval is expressed as \( ds^2 \equiv dx^\mu dx_\mu = c^2 dt^2 - dx^2 \), highlighting the relationship between time and space Proper time \( \tau \) is defined by the equation \( ds \equiv c d\tau = c dt \gamma \), aligning with the time dilation effect, as it represents the time measured in an observer's rest frame The four-velocity is defined as \( u^\mu \equiv \frac{dx^\mu}{d\tau} \equiv (\gamma c, \gamma v) \), leading to the definition of four-momentum as \( p^\mu \equiv mu^\mu = (E c, p) \), where \( m \) is the rest mass.
Exercise: Check that (mu à ) 2 =m 2 c 2 (This must be true so thatE 2 (p c) 2 + (mc 2 ) 2 ).
Some good references for this section are [7, 14, 8] In electrodynamics in flat spacetime we encounter
B=5 ×~ A (3.48) whereEandBare the electric and magnetic fields andφandAare the scalar and vectors potentials 5~ is the gradient operator defined (in 3 dimensions) as
Clearly then φ and A are functions of x, y, z, i.e φ = φ(x, y, z) and
In the context of fields, φ is defined as a scalar field, while A represents a vector field, with both E and B also categorized as vector fields due to their positional dependence For instance, the electric field generated by a point charge diminishes with increasing distance The equations (3.47) and (3.48) indicate that the derivatives of φ and A transform as vectors This raises the question of whether the derivatives of tensors in a general curved spacetime also transform in accordance with tensorial properties.
Consider a vector fieldA à (x ν ) as a function of contravariant coordinates. Let us introduce a shorthand for the derivative as
We want to know whether the derivativeA à,ν is a tensor That is doesA à,ν transform according toA à,ν = ∂x ∂x α à ∂x β
∂x ν A α,β ? To find out, let’s evaluate the derivative explicitly
∂x ν ∂x à A α (3.51) butA α is a function ofx ν not x ν , i.e A α =A α (x ν ) 6=A α (x ν ) Therefore we must insert ∂A ∂x ν α = ∂A ∂x γ α ∂x γ
We see therefore that the tensor transformation law for A à,ν is spoiled by the second term Thus A à,ν is not a tensor [8, 7, 14].
To see why this problem occurs we should look at the definition of the derivative [8],
A à (x+dx)−A à (x) dx ν (3.53) or more properly [7, 14] as lim dx γ → 0 A à (x γ +dx dx γ ν ) − A à (x γ )
The issue with equation (3.53) arises because the vectors A at (x + dx) and A at (x) are situated at different points, making their difference not a vector For two vectors to yield a valid difference, they must be located at the same point, as the transformation laws (3.12) and (3.13) are position-dependent In introductory physics, when representing vectors A and B as arrows, the difference A − B cannot be defined if they are at different locations To illustrate their difference, we instruct students to slide one vector to the location of the other, a process known as parallel transport, which is straightforward in flat space Therefore, to accurately compute A − B, both vectors must first be aligned at the same spacetime point.
To calculate the difference A(x + dx) - A(x), we must first understand parallel transport in curved space In flat space, a vector's components remain unchanged during transport, but this is not the case in curved space For example, if you stand on the Earth's surface in Paris holding a vertical arrow (vector A) and walk to Moscow while keeping the arrow pointed upward, an astronaut observing from space will see the arrow pointing in a different direction upon arrival in Moscow Despite your efforts to maintain its orientation, the astronaut concludes that the arrow represents a different vector, labeled C, due to its altered direction Thus, parallel transport in curved space results in a change of the vector.
To resolve the situation, the astronaut uses radio communication to guide you, aligning your arrow with a mark on her spacecraft window As you walk from Paris to Moscow, she draws parallel lines on her window, instructing you to keep your arrow aligned with these lines Upon reaching Moscow, she confirms that your arrow remains unchanged in relation to her markings This process illustrates that when a vector is parallel transported from an 'absolute' perspective, it retains its identity, even as it is relocated to a different point.
Let’s denoteδA à as the change produced in vectorA à (x α ) located atx α by an infinitessimal parallel transport by a distance dx α We expect δA à to be directly proportional todx α δA à ∝dx α (3.54)
We also expectδA à to be directly proportional toA à ; the bigger our arrow, the more noticeable its change will be Thus δA à ∝A ν dx α (3.55)
The only appropriate constant of proportionality must include covariant indices à and α, along with a contravariant index ν, represented as δA à ≡ Γ ν àα A ν dx α Here, Γ ν àα refers to Christoffel symbols, also known as coefficients of affine connection or connection coefficients As highlighted by Narlikar, while the metric tensor defines the distance between neighboring points, the connection coefficients establish the concept of parallelism among these points.
Parallel transport is defined in Equation (3.56), which describes how an infinitesimal change in vector A, denoted as δA, occurs during transport over a distance dx to create a new vector C ≡ A + δA For contravariant vector B, it is crucial that the scalar product A · B remains unchanged during parallel transport, leading to the equation δ(A · B) = 0 (3.57) This implies that the change in vector A can be expressed as δA ≡ −Γνα Aν dxα (3.58), where Γ represents the Christoffel symbols.
We shall also assume [8] symmetry under exchange of lower indices, Γ α àν = Γ α νà (3.59)
(We would have a truly crazy space if this wasn’t true [8] Think about it !)
Continuing with our consideration of A à (x α ) parallel transported an in- finitessimal distancedx α , the new vector C à will be
The equation C à = A à + δA à illustrates the relationship between the old vector A à (x α) and its new position A à (x α + dx α) The difference between these two vectors is represented by dA à = A à (x α + dx α) − [A à (x α) + δA à], which is defined as a vector This leads us to a new definition of the derivative, which is classified as a tensor.
Using (3.53) in (3.61) we havedA à = ∂A ∂x ν à dx ν −δA à = ∂A ∂x ν à dx ν −Γ ² àα A ² dx α and (3.62) becomes A à;ν ≡ dA dx ν à = ∂A ∂x ν à −Γ ² àν A ² (because dx dx α ν =δ α ν ) which we shall henceforth write as
The covariant derivative, denoted as A à;ν, is defined as ∂A/∂x ν à and is recognized as a second rank tensor, a property that will be confirmed in Problem 3.5.
(3.64) For tensors of higher rank the results are, for example, [14, 8]
Christoffel Symbols and Metric Tensor
We shall now derive an important formula which gives the Christoffel symbol in terms of the metric tensor and its derivatives [8, 14, 7] The formula is Γ α βγ = 1 2 g α² (g ²β,γ +g ²γ,β −g βγ,² ).
(3.69) Another result we wish to prove is that Γ ² ಠ= (ln√
Note thatg6=|g àν | Let us now prove these results.
The covariant differentiation process must not alter the length of a vector, which necessitates that the covariant derivative of the metric tensor remains identically zero, expressed as g àν;λ ≡ 0.
3.6 CHRISTOFFEL SYMBOLS AND METRIC TENSOR 37
Thus g àν,λ = Γ ² àλ g ²ν + Γ ² νλ g ಠ(3.74) and permuting theàνλindices cyclically gives g λà,ν = Γ ² λν g ²à + Γ ² àν g λ² (3.75) and g νλ,à = Γ ² νà g ²λ + Γ ² λà g ν² (3.76) Now add (3.75) and (3.76) and subtract (3.74) gives [8] g λà,ν +g νλ,à −g àν,λ = 2Γ ² àν g λ² (3.77) because of the symmetries of (3.59) and (3.34) Multiplying (3.77) by g λα and using (3.34) and (3.35) (to giveg λ² g λα =g ²λ g λα =δ α ² ) yields Γ α àν = 1
2g λα (g λà,ν +g νλ,à −g àν,λ ) (3.78) which gives (3.69) (do Problems 3.5 and 3.6).
Proof of equation (3.70) [14] (Appendix II) Using g α² g ²β,α = g α² g βα,²
(obtained using the symmetry of the metric tensor and swapping the names of indices) and contracting overαν, equation (3.69) becomes (first and last terms cancel) Γ α β,α = 1
Definingg as the determinant |g àν |and using (3.35) it follows that
=gg àν (3.80) a result which can be easily checked (do Problem 3.7) Thus (3.79) be- comes Γ α β,α = 1
∂x β (3.81) which is (3.70), where in (3.70) we write ln(−g) instead of lng because g is always negative.
Riemann Curvature Tensor
The Riemann curvature tensor is a fundamental component in general relativity, indicating the geometric properties of space A zero value of this tensor signifies flat space, while a non-zero value indicates curvature The tensor can be effectively derived by examining the order of double differentiation on tensors.
∂x α ∂x β (3.82) and also when we writeA à ;αβ we again mean second derivative Many authors instead writeA à ,αβ ≡A à ,α,β orA à ;αβ ≡A à ;α;β We shall use either notation.
In general, while it holds that A à ,αβ = A à ,βα, it is not true that A à ;αβ = A à ;βα To explore this further, we start with the second derivative of a scalar φ, which remains unchanged under parallel transport, leading to φ ;à = φ ,à According to equation (3.63), we find that φ ;à;ν = φ ,à;ν = φ ,à,ν − Γ ² àν φ ,² (3.83) Since Γ ² àν = Γ ² νà, it follows that φ ;àν = φ ;νà, indicating that the order of differentiation does not affect a scalar Next, we consider a vector and differentiate equation (3.64), noting that A à ;ν is a second rank tensor, thereby applying (3.67) accordingly.
=A à ,ν,λ + Γ à ν²,λ A ² + Γ à ν² A ² ,λ + Γ à λ² A ² ;ν −Γ ² νλ A à ;² (3.84) Now interchange the order of differentiation (just swap theν and λindices)
≡A ² R λ²ν à (3.86) with the famous Riemann curvature tensor defined as
Summary
Exercise: Check that equations (3.86) and (3.87) are consistent.
The Riemann tensor tells us everything essential about the curvature of a space For a Cartesian spcae the Riemann tensor is zero.
The Riemann tensor has the following useful symmetry properties [9]
All other symmetry properties of the Riemann tensor may be obtained from these For example
Finally we introduce the Ricci tensor [9] by contracting on a pair of indices
In empty space, it is established that R αβ = 0, highlighting a unique property of the Riemann tensor's contraction The Ricci tensor can be defined in various forms, such as R ² ²αβ, R ² α²β, or R ² αβ², resulting in equivalent outcomes, though a sign difference may occur Consequently, variations in sign can be found across different texts.
However all authors agree on the definition of the Riemann scalar (ob- tained by contracting R α β )
Finally the Einstein tensor is defined as
After discussing the stress-energy tensor in the next chapter, we shall put all of this tensor machinery to use in our discussion of general relativity following.
Problems
3.1IfA à and B ν are tensors, show that the tensor product (outer product) defined byT ν à ≡A à B ν is also a tensor.
3.2 Show that the inner product A.B ≡A à B à is invariant under transfor- mations, i.e show that it satisfies the tensor transformation law of a scalar (thus it is often called the scalar product).
3.3Show that the inner product defined byA.B ≡g àν A à B ν is also a scalar (invariant under transformations), where g àν is assumed to be a tensor.
3.5 Derive the transformation rule for Γ α βγ Is Γ α βγ a tensor ?
3.6 Show thatA à;ν is a second rank tensor.
Answers
3.10 Answers no answers; only solutions
Solutions
To prove that T ν à is a tensor we must show that it satisfies the tensor transformation lawT à ν = ∂x ∂x à α ∂x β
First let’s recall that if f = f(θ, α) and θ = θ(x, y) and α α(x, y) then ∂f ∂θ = ∂f ∂x ∂x ∂θ + ∂f ∂y ∂y ∂θ = ∂x ∂f i ∂x i
It is important to emphasize that our discussion in this chapter is based entirely on Special Relativity.
Euler-Lagrange and Hamilton’s Equations
Newton’s second law of motion is
F= d p dt (4.1) or in component form (for each componentF i )
F i = dp i dt (4.2) where p i =mq˙ i (with q i being the generalized position coordinate) so that dp i dt = ˙mq˙ i +mq¨ i If ˙m = 0 then F i = m¨q i = ma i For conservative forces
F=−5V whereV is the scalar potential Rewriting Newton’s law we have
Let us define the LagrangianL(q i ,q˙ i )≡T−V whereT is the kinetic energy.
In freshman physics, the kinetic energy \( T \) is defined as a function of velocity \( \dot{q}_i \), expressed as \( T = \frac{1}{2} m \dot{q}_i^2 \), while the potential energy \( V \) is a function of position \( q_i \), specifically for a harmonic oscillator given by \( V(q_i) = \frac{1}{2} k q_i^2 \) The Lagrangian \( L(q_i, \dot{q}_i) \) is then formulated as \( L = T(\dot{q}_i) - V(q_i) \) From this, we derive that \( \frac{\partial L}{\partial q_i} = -\frac{dV}{dq_i} \) and \( \frac{\partial L}{\partial \dot{q}_i} = m\dot{q}_i = p_i \), leading to the conclusion that these relationships are consistent with Newton's laws of motion.
) (4.4) with the canonical momentum [1] defined as p i ≡ ∂L
The second equation of (4.4) is known as the Euler-Lagrange equations of motion and serves as an alternative formulation of mechanics [1] It is usually written d dt(∂L
We have obtained the Euler-Lagrange equations using simple arguments A more rigorous derivation is based on the calculus of variations [1] as discussed in Section 7.3.
We now introduce the HamiltonianH defined as a function ofpandqas
For the simple case T = 1 2 mq˙ i 2 and V 6=V( ˙q i ) we havep i ∂L
T = 2m p 2 i and p i q˙ i = p m 2 i so that H(p i , q i ) = 2m p 2 i +V(q i ) =T +V which is the total energy Hamilton’s equations of motion immediately follow from (4.8) as
∂p i = ˙q i (4.9) because L6=L(p i ) and ∂H ∂q i =− ∂q ∂L i so that from (4.4)
Classical Field Theory
Classical Klein-Gordon Field
In order to illustrate the foregoing theory we shall use the example of the classical, massive Klein-Gordon field defined with the Lagrangian density
The covariant momentum density is more easily evaluated by re-writing
= 1 2 (δ à α ∂ à φ+∂ ν φδ ν α ) = 1 2 (∂ α φ+∂ α φ) = ∂ α φ Thus for the Klein-Gordon field we have Π α =∂ α φ (4.20) giving the canonical momentum Π = Π 0 =∂ 0 φ =∂ 0 φ= ˙φ, Π = ˙φ (4.21)
Evaluating ∂ ∂φ L =−m 2 φ, the Euler-Lagrange equations give the field equation as∂ à ∂ à φ+m 2 φ or
Principle of Least Action
which is the Klein-Gordon equation for a free, massive scalar field In mo- mentum space p 2 =−2 2 , thus
(Note that some authors [30] define 2 2 ≡ 5 2 − ∂t ∂ 2 2 different from (3.42), so that they write the Klein-Gordon equation as (2 2 −m 2 )φ= 0 or (p 2 +m 2 )φ 0.)
The energy momentum tensor is
Therefore the Hamiltonian density is H ≡ T 00 = ˙φ 2 − 1 2 (∂ α φ∂ α φ−m 2 φ 2 ) which becomes [31]
2[Π 2 + (5φ) 2 +m 2 φ 2 ] (4.25) where we have relied upon the results of Section 3.4.1.
4.3 Principle of Least Action derive EL eqns properly for q and φ(do later) Leave out for now.
Energy-Momentum Tensor for Perfect Fluid
The best references for this section are [9](Pg 124-125), [7], and [32](Pg.
D’Inverno's book offers an insightful discussion on the Navier-Stokes equation, highlighting its relevance to the material covered in this section For further reading, additional references include sources [8] (Pg 83), [15] (Pg 330), [33] (Pg 259), [34] (Pg 38), and [2].
These references show that the energy-momentum tensor for a perfect fluid is
(4.26) whereρis theenergy density and pis the pressure We shall now work this out for several specific cases [9] Fig 2.5 of Narlikar’s book [7] is particularly helpful.
Motionless dust represents a collection of particles at rest Thus u à (c, 0), so that T 00 = ρ The equation of state for dust is p = 0 so that
Motionless fluid consists of a collection of particles that move randomly, exerting pressure while remaining at rest This scenario is akin to a gas of particles at a non-zero temperature, contained within a stationary vessel In this context, the energy-momentum tensor retains the form T^00 = ρ, indicating energy density, while T^ii = p represents pressure, and T^ij = 0 signifies no momentum flux between different directions.
Motionless radiation is characterized by the equation of state p = 1 3 ρ.
Again the radiation is confined to a container not in motion so that u à (γc, 0) (The 1 3 just comes from randomizing the pressure in 3 dimensions [7].) Thus
In the general case, the energy-momentum tensor for a motionless fluid is described by equation (4.28) Specific instances, such as motionless dust and motionless radiation, can be derived by substituting p = 0 and p = 1/3 ρ, respectively, into this equation.
Continuity Equation
In classical electrodynamics, the four-current density is defined as \( j^\mu \equiv (c\rho, \mathbf{j}) \), and it follows the covariant conservation law \( \partial_\mu j^\mu = 0 \), leading to the continuity equation \( \frac{\partial \rho}{\partial t} + \nabla \cdot \mathbf{j} = 0 \) This relationship can also be derived from the Maxwell equations by applying the divergence to Ampère’s law Consequently, the four Maxwell equations can be represented equivalently by three Maxwell equations along with the continuity equation.
In Chapters 1 and 2, we discovered that the equations for velocity and acceleration are fundamentally linked to the conservation equation, demonstrating their equivalence This means that the combined velocity and acceleration equations can be effectively represented by the single velocity equation along with the conservation law.
In analogy with electrodynamics the conservation law for the energy- momentum tensor is
In the next chapter we shall show how equation (2.1) can be derived from this.
Interacting Scalar Field
This article discusses the interaction of a scalar field with a scalar potential V(φ) It highlights the transition from discrete coordinates ˙q i to continuous field variables φ(x), where φ serves as the generalized coordinate and the discrete index i is substituted by a continuous index x Consequently, the potential V(q i ) is naturally transformed into V(φ), with φ defined as φ(x).
Thus for an interacting scalar field we simply tack on−V(φ) to the free Klein-Gordon Lagrangian of equation (4.19) to give
≡ L O +L I (4.31) where L O ≡ L KG and L I ≡ −V(φ) Actually the Lagrangian of (4.31) refers to a minimally coupled scalar field as opposed to conformally coupled
In the context of the free particle case, it is crucial to note that the potential V(φ) lacks derivative terms like ∂ à φ As a result, the covariant momentum density and canonical momentum are consistent with equations (4.20) and (4.21), represented as Π α = ∂ α φ and π = ˙φ, respectively By applying the Euler-Lagrange equations, we can derive the necessary outcomes for the system.
The energy-momentum tensor is the same as for the free particle case, equation (4.24), except for the addition ofg àν V(φ) as in
2(∂ α φ∂ α φ−m 2 φ 2 )−V(φ)] (4.34) yielding the Hamiltonian density the same as for the free particle case, equa- tion (4.25), except for the addition ofV(φ) as in
The purely spatial components areT ii =∂ i φ∂ i φ−g ii [ 1 2 (∂ α φ∂ α φ−m 2 φ 2 )−
V(φ)] and with g ii =−1 (i.e assume Special Relativity NNN) we obtain
Note that even though T ii has repeated indices let us not assume P i is implied in this case That is T ii refers to T ii = T 11 = T 22 = T 33 and not
In the context of scalar fields, it is important to note that T ii = T 11 + T 22 + T 33, a convention supported by some authors like Serot and Walecka, which may lead to a disagreement of 1/3 with our results Assuming the scalar field effects are averaged to resemble a perfect, motionless fluid, we can draw a comparison with equation (4.28) to establish the necessary identification.
Cosmology with the Scalar Field
Alternative derivation
The equation of motion (4.46) for the scalar field can be derived more efficiently, as noted in [29] (Pg 73), but this method is valid only when both m = 0 and ∇φ = 0 are initially set An exercise is suggested to explore the complications that arise when m ≠ 0 and ∇φ ≠ 0.
Consider a Lagrangian forφwhichalreadyhas the scale factor built into it as
The R 3 factor, derived from the Robertson-Walker metric as √−g=R 3, will be elaborated in Chapter 7 and is also present outside the Friedmann Lagrangian in equation (2.20) The equation of motion, φ¨− 5 2 φ+ 3Hφ˙+m 2 φ+V 0 = 0 (4.50), differs from equation (4.45) Notably, when m= 0 and 5φ= 0, it aligns with equation (4.46).
2φ˙ 2 −V(φ)] (4.51) which results from setting m = 0 and 5φ = 0 in (4.49) The equation of motion is φ¨+ 3Hφ˙+V 0 = 0 (4.52)
We achieved this result much more rapidly than the lengthy process required to obtain (4.46), without even utilizing the energy-momentum tensor It's important to note that since 5φ= 0, the Lagrangian formalism we used is essentially equivalent to our traditional approach, where we focused on q i (t); in this case, we only have φ=φ(t).
(notφ=φ(x)), and so we only havei= 1, i.e q i ≡φ.
The Lagrangian is defined as \( L = R^3 (T - V) \), leading to the total energy density \( \rho = T + V = \frac{1}{2} \dot{\phi}^2 + V(\phi) \) By taking the time derivative \( \dot{\rho} = \dot{\phi} \ddot{\phi} + V' \dot{\phi} = -3H \dot{\phi}^2 \) and substituting it into the conservation equation \( \dot{\rho} = -3H(\rho + p) \), we derive the pressure as \( p = \frac{1}{2} \dot{\phi}^2 - V(\phi) \) This demonstrates that our expressions for energy density and pressure align with previous results, confirming that pressure can be expressed as \( p = \frac{R L}{3} \).
Limiting solutions
Assuming that k= Λ = 0 the Friedmann equation becomes
The coupled equations formed by this equation and equation (4.46) allow us to find the solutions for φ(t) and R(t) To solve these equations, we first eliminate one variable, then solve one equation, and subsequently substitute the solution back into the other equation to determine the remaining variable We will express equation (4.46) solely in terms of φ by eliminating the other variable.
R which appears in the form H = R R ˙ We eliminate R by substituting H from (4.53) into (4.46) to give φ¨+ q 12πG( ˙φ 2 + 2V) ˙φ+V’=0 φ¨ 2 + 2 ¨φV 0 −12πG( ˙φ 2 + 2V) ˙φ 2 +V 0 2 = 0 (4.54)
This article addresses the challenges of solving a non-linear differential equation for φ, which is generally complex We will explore solutions for specific limiting cases After determining φ(t) from equation (4.54), we will substitute it back into equation (4.53) to derive R(t).
Setting V = 0 we then haveρ = 1 2 φ˙ 2 =p Thus our equation of state is p=ρ (4.55) orγ = 3.
12πGφ˙ 2 = 0 (4.56) which has the solution (do problem 4.4) φ(t) =φ o + 1
4.7 COSMOLOGY WITH THE SCALAR FIELD 57
(Note that the solution is equation (9.18) of [29] is wrong.) Upon substitut- ing this solution back into the Friedmann equation (4.53) and solving the differential equation we obtain (do problem 4.5)
This result may be understood from another point of view Writing the Friedmann equations as
R m (4.60) then the solution is always
(corresponding to m= 0) then the solution is
(do problem 4.6) Note that for m < 2, one obtains power law inflation. For ordinary matter (m = 3), or radiation (m = 4) we have R ∝ t 2/3 and
R∝t 1/2 respectively Returning to the scalar field solution (4.57) the density isρ= 1 2 φ˙ 2 forV = 0 Thus φ(t) =˙ φ˙ o
(4.67) corresponding to m = 6 and thus R ∝t 1/3 in agreement with (4.58) Note also that this density ρ∝ R 1 6 also gives ρ∝ t 1 2.
Thus for a scalar field with V = 0, we have p =ρ (γ = 3) and ρ ∝ R 1 6. Contrast this with matter for which p= 0(γ = 0) and ρ∝ R 1 3 or radiation for which p= 1 3 ρ (γ = 1/3) and ρ∝ R 1 4
However equation (4.67) may not be interpreted as a decaying Cosmo- logical Constant becausep6=ρ (see later).
Here we take ˙φ= ¨φ= 0, so thatρ=V and p=−V giving p=−ρ (4.68) or γ = −3 which is a negative pressure equation of state Our equation of motion for the scalar field (4.54) becomes
V =V o (4.70) which is constant Substituting the solution into the Friedmann equation
3 V o (4.71) which acts as aCosmological Constantand which has the solution (do prob- lem 4.7)
3 V o (t − t o ) (4.72) which is an inflationary solution, valid for anyV.
We have found that if k = Λ = 0 and if ρ ∝ R 1 m then R∝t 2 for any value ofm All of this iscorrect To check this we might substitute into the
4.7 COSMOLOGY WITH THE SCALAR FIELD 59 and say R R ˙ ∝ 1 t giving R R 1 dR dt dt ∝ R dt t which yields lnR ∝ lnt and thus
R∝t 2/m The result R∝t is wrong because we have left out an important constant.
Actually if R R ˙ = c t then lnR=clnt= lnt c givingR ∝t c instead of R∝t. Let’s keep our constants then Write ρ = R d m 2 then R = ( md 2 ) 2/m t 2/m and ρ = d 2
( md 2 ) 2 t 2 = (2/m) t 2 2 Substituting into the Friedmann equation gives ( R R ˙ ) 2 = (2/m) t 2 2 or R R ˙ = (2/m) t with the above constant C = m 2 yielding
R∝t 2/m inagreement with the correct result above.
The lesson is be careful of constants when doing back-of-the-envelope calculations.
Exactly Solvable Model of Inflation
Due to the complexity of the non-linear equation (4.54), finding exactly solvable models is quite rare One notable exception is Barrow's model, which allows for exact solutions and results in power law inflation The benefit of utilizing an exactly solvable model lies in its capacity to enhance physical intuition A brief introduction to Barrow's model can be found in Islam's work.
Any scalar field model is specified by writing down the potential V(φ). Barrow’s potential is
V(φ)≡βe − λφ (4.74) where β and λare constants to be determined Barrow [35] claims that a particular solution to (4.54) is (which was presumably guessed at, rather then solving the differential equation) φ(t) =√
2Ais just some constant We check this claim by substituting (4.74) and (4.75) into (4.54) From this we find (do problem 4.9) that λ r2
Note that Barrow is wrong when he writes λA = √
2 Also he uses units with 8πG = 1, so that the second solution (4.78), he writes correctly as β =A(3A−1) Also Barrow doesn’t use the first solution (4.77) for reasons we shall see shortly.
Having solved for φ(t) we now substitute into (4.53) to solve for R(t).
(Recall φ(t) and R(t) are the solutions we seek to our coupled equations
(4.46) and (4.53).) Substituting V = t Λ 2 and ˙φ = √ 2A t (see solution to problem 4.9) we have
Clearly we see why we reject the first solution (4.77) withβ =−A It would give zero density Using the second solution (4.78) withβ =A(24πGA−1) yields ρ= 24πGA 2 t 2 (4.81)
Solving the Friedmann equation (4.79) gives
R∝t 8πGA (4.82) whereD is some constant Setting 8πG≡1 we have
R∝t A (4.83) in agreement with Barrow’s solution Power law inflation results for
Inverting the solution (4.83) we havet 2 =C 0 R 2/A whereC 0 is some constant. Substituting into (4.81) we have ρ∝ 1
R 2/A (4.85) which corresponds to aWeak decaying Cosmological Constant (See sections
4.7.4 and 4.7.5) For the inflationary resultA >1 we have A 2 ≡m