A Bianchi Type IV Viscous Model of The Early Universe

A Bianchi Type IV Viscous Model of The Early Universe

Public Lecture August 28, 2012 - 10:00 AM - PSE258

Ikjyot Singh Kohli M.Sc. Candidate

Department of Physics and Astronomy York University

Quantum Gravity FLRW Metric - Present-Day Model

Early Universe

L = Nph�R+KabK

ab �K2 � 2⇤�

ds2 = �dt2 + a2(t)

dr2

1� kr2+ r2

�d✓2 + sin2 ✓d�2

��

The Early Universe

The Early Universe

We are interested in developing a cosmological model that describes the conditions of the early universe.

The Early Universe

We are interested in developing a cosmological model that describes the conditions of the early universe.

Before we continue, we note that a cosmological model necessarily implies the use of General Relativity

General Relativity: The “Brief” Version!


Basically, General Relativity is Einstein’s Theory of Gravitation. It is a geometrization of gravity, “GEOMETRODYNAMICS”.



The basic idea is that matter induces curvature upon spacetime, and spacetime “tells” that matter how to move.



The basic idea is that matter induces curvature upon spacetime, and spacetime “tells” that matter how to move.

In general, we would like to obey the following requirements:

(a)Any field equations describing the theory should be independent of coordinate systems of the laws of nature, and, hence, should be TENSOR equations


(b)Similar to other field equations of physics, the field equations should be PDEs of at most second-order for the functions to be determined, which are LINEAR in the highest derivatives.



(c)In the appropriate limit, they should reduce the Newton’s law of gravitation




(d)The energy-momentum tensor Tab is the S.R. analogue of mass density, and hence, should be the source of the gravitational field.




(d)The energy-momentum tensor Tab is the S.R. analogue of mass density, and hence, should be the source of the gravitational field.

(e)Indeed, if the spacetime is flat, Tab should vanish.

Implications:

Implications:a) => Need a tensor that has only derivatives of the metric tensor up to second order.

Implications:a) => Need a tensor that has only derivatives of the metric tensor up to second order.

b) => This tensor, The Einstein tensor Gmn must then be built out of the curvature tensor, metric tensor, and ℇ-tensor.

Implications Continued...

Implications Continued...Conservation of energy requires that:


Tmn;n = 0 ) Gmn

;n = 0


Tmn;n = 0 ) Gmn

;n = 0

One can show rigorously that there is only ONE 2nd-rank tensor which is linear

(requirement (b)) in the components of the curvature tensor and which satisfies this

relationship:


Tmn;n = 0 ) Gmn

;n = 0



relationship:

Gmn = Rmn � 1

2gmnR+ ⇤gmn = Tmn


Tmn;n = 0 ) Gmn

;n = 0



relationship:

Gmn = Rmn � 1

2gmnR+ ⇤gmn = Tmn

⋀ - Cosmological Constant (We will set this to zero)

The Einstein Field Equations


We thus arrive at The Einstein Field equations:



Rmn � 1

2gmnR = Tmn



Rmn � 1

2gmnR = Tmn

Derived by Einstein after 10 years of research in 1915.



Rmn � 1

2gmnR = Tmn

Derived by Einstein after 10 years of research in 1915.

These constitute a system of ten, nonlinear, coupled, hyperbolic PDEs to determine the ten metric functions. In general, no unique solution

exists.


Historical Note: In 1915, Hilbert also derived the field equations in a more

rigorous manner, but, Einstein beat him to the punch!

Just for kicks...

Just for kicks...SG =

1

2

Z

MR

p�gd

4x


1

2

Z

MR

p�gd

4x

SG =1

2

Z

M

�Ruvg

uvp�g

�d

4x


1

2

Z

MR

p�gd

4x

SG =1

2

Z

M

�Ruvg

uvp�g

�d

4x

�SG =1

2

Z

M

�g

uvp�g�Ruv +Ruv�⇥g

uvp�g

⇤�d

4x


1

2

Z

MR

p�gd

4x

SG =1

2

Z

M

�Ruvg

uvp�g

�d

4x

�SG =1

2

Z

M

�g


uvp�g

⇤�d

4x

After Much Computation...


1

2

Z

MR

p�gd

4x

SG =1

2

Z

M

�Ruvg

uvp�g

�d

4x

�SG =1

2

Z

M

�g


uvp�g

⇤�d

4x


�SG =1

2

Z

M

p�g

✓Ruv �

1

2Rguv

◆�g

uvd

4x


1

2

Z

MR

p�gd

4x

SG =1

2

Z

M

�Ruvg

uvp�g

�d

4x

�SG =1

2

Z

M

�g


uvp�g

⇤�d

4x


�SG =1

2

Z

M

p�g

✓Ruv �

1

2Rguv

◆�g

uvd

4x

�SG = 0 ) Ruv �1

2Rguv = 0


1

2

Z

MR

p�gd

4x

SG =1

2

Z

M

�Ruvg

uvp�g

�d

4x

�SG =1

2

Z

M

�g


uvp�g

⇤�d

4x


�SG =1

2

Z

M

p�g

✓Ruv �

1

2Rguv

◆�g

uvd

4x

�SG = 0 ) Ruv �1

2Rguv = 0 Vacuum Field

Equations!

Just for kicks...

Just for kicks...If you want to include matter, you must consider a variational principle

of the form:


of the form:� (SG + Sm) = 0



One can show that:



One can show that:�Sm = 0 = �1

2

Z

MTuv

p�g�g

uvd

4x




2

Z

MTuv

p�g�g

uvd

4x

Combining this with our previous derivation, we obtain:




2

Z

MTuv

p�g�g

uvd

4x

Combining this with our previous derivation, we obtain:

Ruv �1

2Rguv = Tuv

A Quick Note on Curvature


We have so far heavily relied on the notion of a curvature tensor, but have not really explained why we need it, except for the bit about matter inducing curvature, and so forth...

It has to do with the underlying geometry of General Relativity.

In classical physics, the underlying geometry is Euclidean, which is a flat-space geometry.

That is, when we wish to make measurements in a lab, we place a coordinate system over our experiment, and make computations with respect to elements of the Euclidean vector space.



x


x

y


x

y

z


x

y

z

P1


x

y

z

P1P2


x

y

z

P1P2

In Euclidean space, distance between any two points:


x

y

z

P1P2

In Euclidean space, distance between any two points:ds

2 = (dx1)2 + (dx2)2 + (dx3)2 + ...


Can Generalize this to Minkowsi Space of Special Relativity:



ds

2 = ⌘abdxadx

b = �dt

2 + dx

2 + dy

2 + dz

2



ds

2 = ⌘abdxadx

b = �dt

2 + dx

2 + dy

2 + dz

2

⌘ab = ⌘ab = diag(�1, 1, 1, 1)



ds

2 = ⌘abdxadx

b = �dt

2 + dx

2 + dy

2 + dz

2

⌘ab = ⌘ab = diag(�1, 1, 1, 1)

Einstein: In a more general theory, the metric tensor is actually a function of the coordinates, and NOT constant:



ds

2 = ⌘abdxadx

b = �dt

2 + dx

2 + dy

2 + dz

2

⌘ab = ⌘ab = diag(�1, 1, 1, 1)


ds

2 = gabdxadx

b



ds

2 = ⌘abdxadx

b = �dt

2 + dx

2 + dy

2 + dz

2

⌘ab = ⌘ab = diag(�1, 1, 1, 1)


ds

2 = gabdxadx

b

g

ab = f(x0, x

1, x

2, x

3)

General Relativity is then a theory in which the metric tensor is not constant, which implies that a pseudo-Riemannian manifold must be used as the underlying geometry.

General Relativity is then a theory in which the metric tensor is not constant, which implies that a pseudo-Riemannian manifold must be used as the underlying geometry.Rn

General Relativity is then a theory in which the metric tensor is not constant, which implies that a pseudo-Riemannian manifold must be used as the underlying geometry.Rn :Succession of n real numbers:

General Relativity is then a theory in which the metric tensor is not constant, which implies that a pseudo-Riemannian manifold must be used as the underlying geometry.Rn :Succession of n real numbers: (x1

, . . . , x

n)


, . . . , x

n)

We define the mapping:


, . . . , x

n)

We define the mapping: f : M ! N


, . . . , x

n)


M N


, . . . , x

n)


M NThis is a 1:1 mapping:


, . . . , x

n)


M NThis is a 1:1 mapping: x 6= y ) f(x) 6= f(y)


, . . . , x

n)



Which means that f has a well-defined inverse:


, . . . , x

n)



Which means that f has a well-defined inverse:

f�1 : N ! M

A manifold M is a space satisfying:


9 a family of open neighbourhoods Ui together with continuous 1-to-1 mappings

fi : Ui ! Rnwith a continuous inverse for a number n




The family of open neighbourhoods cover the whole of M :

[

i

Ui = M





[

i

Ui = M

There exist mappings � : U ! Rn.





[

i

Ui = M


Q 2 M ) �(Q) = (x1, . . . , x

n) 2 Rn





[

i

Ui = M


Q 2 M ) �(Q) = (x1, . . . , x

n) 2 Rn

This mapping is called a coordinate system, U is the coordinate region in M





[

i

Ui = M


Q 2 M ) �(Q) = (x1, . . . , x

n) 2 Rn

This mapping is called a coordinate system, U is the coordinate region in M

Can go forward and define coordinate transformations, tangent space, basis vectors, and so

on, but, won’t have enough time!

Summary of Prerequisites


The main point is that all of this geometric structure is the fundamental, underlying structure for General Relativity.

It naturally defines a metric tensor on a pseudo-Riemannian manifold:

pseudo-Riemannian:


The main point is that all of this geometric structure is the fundamental, underlying structure for General Relativity.

It naturally defines a metric tensor on a pseudo-Riemannian manifold:

pseudo-Riemannian:pg = ±1

⇠a;b + ⇠b;a = 0

Summary of PrerequisitesThe metric tensor leads to the definition of

fundamental quantities:

⇠a;b + ⇠b;a = 0



Christoffel Symbols�amn =

1

2gab (gbm,n + gbn,m � gmn,b)

⇠a;b + ⇠b;a = 0




1


Geodesic Equationx

a + �abcx

bx

c = 0

⇠a;b + ⇠b;a = 0




1


Geodesic Equationx

a + �abcx

bx

c = 0

Covariant Derivative

⇠a;b + ⇠b;a = 0




1


Geodesic Equationx

a + �abcx

bx

c = 0

Covariant DerivativeTa;m = Ta,m � �n

amTn

⇠a;b + ⇠b;a = 0




1


Geodesic Equationx

a + �abcx

bx

c = 0


amTnCurvature Tensors

⇠a;b + ⇠b;a = 0




1


Geodesic Equationx

a + �abcx

bx

c = 0


amTnCurvature Tensorsam;s;q � am;q;s = abR

bmsq

⇠a;b + ⇠b;a = 0




1


Geodesic Equationx

a + �abcx

bx

c = 0



bmsq

Riemann

⇠a;b + ⇠b;a = 0




1


Geodesic Equationx

a + �abcx

bx

c = 0



bmsq

Riemann

Rbmsq = �b

mq,s � �bms,q + �b

ns�nmq � �b

nq�nms

⇠a;b + ⇠b;a = 0




1


Geodesic Equationx

a + �abcx

bx

c = 0



bmsq

Riemann

Rbmsq = �b


ns�nmq � �b

nq�nms

Ricci

⇠a;b + ⇠b;a = 0




1


Geodesic Equationx

a + �abcx

bx

c = 0



bmsq

Riemann

Rbmsq = �b


ns�nmq � �b

nq�nms

RicciRmq = Rs

msq = �Rsmqs ; Rm

m = R⇠a;b + ⇠b;a = 0




1


Geodesic Equationx

a + �abcx

bx

c = 0



bmsq

Riemann

Rbmsq = �b


ns�nmq � �b

nq�nms

RicciRmq = Rs

msq = �Rsmqs ; Rm

m = R⇠a;b + ⇠b;a = 0

Killing’s Equation

Developing a Cosmological Model


The present-day universe is modeled with great accuracy by The Friedmann-LeMaitre-

Robertson-Walker (FLRW) metric:




ds2 = �dt2 + a2(t)

dr2

1� kr2+ r2

�d✓2 + sin2 ✓d�2

��




ds2 = �dt2 + a2(t)

dr2

1� kr2+ r2

�d✓2 + sin2 ✓d�2

��

Cosmological scale factor - Describes expansion/collapse behaviour of the

universe.




ds2 = �dt2 + a2(t)

dr2

1� kr2+ r2

�d✓2 + sin2 ✓d�2

��

Cosmological scale factor - Describes expansion/collapse behaviour of the

universe.

These models are spatially homogeneous

and isotropic

Cosmological Principle: The large-scale universe can be foliated into a family of space-like hypersurfaces, in which, physical quantities remain constant on each hypersurface, and are only functions of time.

Cosmological Principle: The large-scale universe can be foliated into a family of space-like hypersurfaces, in which, physical quantities remain constant on each hypersurface, and are only functions of time. ta

Cosmological Principle: The large-scale universe can be foliated into a family of space-like hypersurfaces, in which, physical quantities remain constant on each hypersurface, and are only functions of time. ta

⌃1

⌃2

⌃3

⌃n

R4

R⇥ ⌃t

Early-Universe Cosmology

Early-Universe CosmologyEven though the universe is observed to be 99.999 (1-10-5)% isotropic today, minor anisotropies indicate that after the Big Bang, the universe was in a chaotic state, and, in general, was anisotropic.


Somehow, these anisotropies smoothed out to the present-day universe.


Somehow, these anisotropies smoothed out to the present-day universe.

Early-universe cosmology, assumes that the universe, although, spatially homogeneous, was anisotropic.

Foundations of Spatial Homogeneity


Basically, a space is homogeneous if it admits a group of motions. That is, a space is topologically homogeneous if you can carry one point to any

other point via an isometry.



other point via an isometry.Isom(M) ⌘ {� : M ! M |� isometry}




We can therefore say that a space is homogeneous if for each pair of points




We can therefore say that a space is homogeneous if for each pair of points

(p, q) 2 M, 9� 2 Isom(M) such that �(p) = q


We will proceed by choosing a basis:


We will proceed by choosing a basis: ei at p 2 M



L⇠j [ei, ek] = 0



L⇠j [ei, ek] = 0

This basis then spans a Lie Algebra



L⇠j [ei, ek] = 0

This basis then spans a Lie AlgebraTo construct a homogeneous space, we define

a left-invariant frame:



L⇠j [ei, ek] = 0

This basis then spans a Lie AlgebraTo construct a homogeneous space, we define

a left-invariant frame:

[ei, ej ] = Ckijek

[ei, ej ] = Ckijek

[ei, ej ] = Ckijek

Structure Constants of The Group

[ei, ej ] = Ckijek


Now, define:

[ei, ej ] = Ckijek


Now, define: !k

[ei, ej ] = Ckijek


Now, define: !k Dual basis to:

[ei, ej ] = Ckijek


Now, define: !k Dual basis to:ek

[ei, ej ] = Ckijek



ds2 = gij!i ⌦ !j

[ei, ej ] = Ckijek



ds2 = gij!i ⌦ !j

This is the metric tensor of a homogeneous space.

The Bianchi Classifications


There are 9 such classifications of homogeneous spacetimes (with an isometry group of dimension 3), these are The Bianchi Classifications, named after Luigi Bianchi (1898).


There are 9 such classifications of homogeneous spacetimes (with an isometry group of dimension 3), these are The Bianchi Classifications, named after Luigi Bianchi (1898).

Essentially, one solves for the Killing vectors/differential one-forms, which then give the form of the metric tensor.

Table 8.2. Killing vectors and reciprocal group generators by Bianchi type

Expressions are given in canonical coordinates: for full explanation, see text.

I II IV V VI (including III ) VIIξA ∂x ∂x ∂x − y∂y − (y + z)∂z ∂x − y∂y − z∂z ∂x + (z −Ay)∂y ∂x + (z −Ay)∂y

+ (y −Az)∂z − (y + Az)∂z∂y ∂y ∂y ∂y ∂y ∂y∂z ∂z + y∂x ∂z ∂z ∂z ∂z

ηA ∂x ∂x ∂x ∂x ∂x ∂x∂y ∂y + z∂x e−x(∂y − x∂z) e−x∂y e−Ax(coshx ∂y + sinhx ∂z) e−Ax(cosx ∂y − sinx ∂z)∂z ∂z e−x∂z e−x∂z e−Ax(sinhx ∂y + coshx ∂z) e−Ax(sinx ∂y + cosx ∂z)

ωA dx dx− z dy dx dx dx dxdy dy exdy exdy eAx(coshxdy − sinhxdz) eAx(cosxdy − sinxdz)dz dz ex(dz + xdy) exdz eAx(− sinhxdy + coshxdz) eAx(sinxdy + cosxdz)

VIII IXξA sech y cosh z ∂x + sinh z ∂y − tanh y cosh z ∂z sec y cos z ∂x + sin z ∂y − tan y cos z ∂z

sech y sinh z ∂x + cosh z ∂y − tanh y sinh z ∂z − sec y sin z ∂x + cos z ∂y + tan y sin z ∂z∂z ∂z

ηA ∂x ∂x− sinx tanh y ∂x + cosx ∂y − sinx sech y ∂z sinx tan y ∂x + cosx ∂y − sinx sec y ∂zcosx tanh y ∂x + sinx ∂y + cosx sech y ∂z − cosx tan y ∂x + sinx ∂y + cosx sec y ∂z

ωA dx− sinh y dz dx + sin y dzcosxdy − sinx cosh y dz cosxdy − sinx cos y dzsinxdy + cosx cosh y dz sinxdy + cosx cos y dz

Table 8.2. Killing vectors and reciprocal group generators by Bianchi type

Expressions are given in canonical coordinates: for full explanation, see text.

I II IV V VI (including III ) VIIξA ∂x ∂x ∂x − y∂y − (y + z)∂z ∂x − y∂y − z∂z ∂x + (z −Ay)∂y ∂x + (z −Ay)∂y

+ (y −Az)∂z − (y + Az)∂z∂y ∂y ∂y ∂y ∂y ∂y∂z ∂z + y∂x ∂z ∂z ∂z ∂z

ηA ∂x ∂x ∂x ∂x ∂x ∂x∂y ∂y + z∂x e−x(∂y − x∂z) e−x∂y e−Ax(coshx ∂y + sinhx ∂z) e−Ax(cosx ∂y − sinx ∂z)∂z ∂z e−x∂z e−x∂z e−Ax(sinhx ∂y + coshx ∂z) e−Ax(sinx ∂y + cosx ∂z)

ωA dx dx− z dy dx dx dx dxdy dy exdy exdy eAx(coshxdy − sinhxdz) eAx(cosxdy − sinxdz)dz dz ex(dz + xdy) exdz eAx(− sinhxdy + coshxdz) eAx(sinxdy + cosxdz)

VIII IXξA sech y cosh z ∂x + sinh z ∂y − tanh y cosh z ∂z sec y cos z ∂x + sin z ∂y − tan y cos z ∂z

sech y sinh z ∂x + cosh z ∂y − tanh y sinh z ∂z − sec y sin z ∂x + cos z ∂y + tan y sin z ∂z∂z ∂z

ηA ∂x ∂x− sinx tanh y ∂x + cosx ∂y − sinx sech y ∂z sinx tan y ∂x + cosx ∂y − sinx sec y ∂zcosx tanh y ∂x + sinx ∂y + cosx sech y ∂z − cosx tan y ∂x + sinx ∂y + cosx sec y ∂z

ωA dx− sinh y dz dx + sin y dzcosxdy − sinx cosh y dz cosxdy − sinx cos y dzsinxdy + cosx cosh y dz sinxdy + cosx cos y dz

From Stephani’s Text - Exact Solutions of Einstein’s Field Equations, 2003

Bianchi Type IV

Bianchi Type IVWe are interested in a Bianchi Type IV (BIV) model, so we take the “naive” approach of forming the metric tensor from the differential forms from the previous page:


ds

2 = �dt

2 + �ij(t)dxidx

j


ds

2 = �dt

2 + �ij(t)dxidx

j

Time-dependent scaling function


dx

1= dx, dx

2= expxdy, dx

3= expx (dz + xdy)

ds

2 = �dt

2 + �ij(t)dxidx

j



dx

1= dx, dx

2= expxdy, dx

3= expx (dz + xdy)

ds

2= �dt

2+ a

2(t)dx

2+ dy

2�b

2(t) exp(2x) + c

2(t) exp(2x)x

2�+ dz

2c

2(t) exp(2x)+

2c

2(t) exp(2x)xdzdy

ds

2 = �dt

2 + �ij(t)dxidx

j



dx

1= dx, dx

2= expxdy, dx

3= expx (dz + xdy)

ds

2= �dt

2+ a

2(t)dx

2+ dy

2�b

2(t) exp(2x) + c

2(t) exp(2x)x

2�+ dz

2c

2(t) exp(2x)+

2c

2(t) exp(2x)xdzdy

For:

ds

2 = �dt

2 + �ij(t)dxidx

j



dx

1= dx, dx

2= expxdy, dx

3= expx (dz + xdy)

ds

2= �dt

2+ a

2(t)dx

2+ dy

2�b

2(t) exp(2x) + c

2(t) exp(2x)x

2�+ dz

2c

2(t) exp(2x)+

2c

2(t) exp(2x)xdzdy

For:

�ab(t) = diag�a2(t), b2(t), c2(t)

�

ds

2 = �dt

2 + �ij(t)dxidx

j


Computing The Einstein Field Equations in this coordinate basis gives very complicated results!

These expressions are too complicated for our liking... But, G.R. is a COVARIANT

theory, it shouldn’t matter what coordinate system we use, we will therefore take the approach of

ORTHONORMAL FRAMES.



ORTHONORMAL FRAMES.

That is, we choose a basis such that all four basis vectors:



ORTHONORMAL FRAMES.

That is, we choose a basis such that all four basis vectors: {eu}



ORTHONORMAL FRAMES.


are mutually orthonormal. Thus,



ORTHONORMAL FRAMES.


are mutually orthonormal. Thus,g(eu, ev) = ⌘uv = �uv = diag(�1, 1, 1, 1)



ORTHONORMAL FRAMES.



The structure constants from before, are now functions, and satisfy:



ORTHONORMAL FRAMES.




[eu, ev] = cauvea



ORTHONORMAL FRAMES.




[eu, ev] = cauvea cauv = �avu � �a

uv



ORTHONORMAL FRAMES.





uv �auv =1

2

�gabc

bvu + guvc

bav � gvbc

bau

�



ORTHONORMAL FRAMES.





uv �auv =1

2

�gabc

bvu + guvc

bav � gvbc

bau

�

These are all now independent of coordinate functions!

Before we continue, it is necessary to discuss the matter source of interest, particularly, we need to develop the theory behind a viscous fluid tensor.


Start with a pressure-less fluid:


Start with a pressure-less fluid:T00 = µu0u0 ) Tab = µuaub



Assuming there is a pressure (which only acts on the spatial components):




Tab = µuaub + ↵phab (↵ = ±1)





Where:





hab ⌘ gab � uaub

ucucWhere:






ucucWhere: is the projection

tensor.







tensor.Define: ↵ = �ucuc








T ab = (µ+ p)uaub � ucucgabp








T ab = (µ+ p)uaub � ucucgabp

This is the energy-momentum tensor for a perfect fluid.

Let the viscous contributions be denoted as:

Let the viscous contributions be denoted as: Vab


Tab = wuaub � ucucgabp+ Vab



From classical fluid mechanics, the Euler equation is:



From classical fluid mechanics, the Euler equation is:(⇢ui),t = �⇧ik,k




Where:




Where:⇧ik = p�ik + ⇢uiuk





is the momentum flux tensor.





is the momentum flux tensor.We will add on a term that represents the viscous

momentum flux:






momentum flux:⇧ik = p�ik + ⇢uiuk � ⌃0

ik = �⌃ik + ⇢uiuk








Where:








Where:⌃ik = �p�ik + ⌃0

ik









ikStress tensor









ikStress tensor

Viscous stress tensor

One can show that the most general viscous tensor can be formed as:


⌃0ik = ⌘

✓ui,k + uk,i �

2

3�ikul,l

◆+ ⇠�ikul,l


⌃0ik = ⌘

✓ui,k + uk,i �

2

3�ikul,l

◆+ ⇠�ikul,l

Shear viscosity


⌃0ik = ⌘

✓ui,k + uk,i �

2

3�ikul,l

◆+ ⇠�ikul,l

Shear viscosity Bulk viscosity


⌃0ik = ⌘

✓ui,k + uk,i �

2

3�ikul,l

◆+ ⇠�ikul,l


Note that:


⌃0ik = ⌘

✓ui,k + uk,i �

2

3�ikul,l

◆+ ⇠�ikul,l


Note that:

�ikul,l ⌘ ✓

✓ui,k + uk,i �

2

3�ikul,l

◆⌘ �ab

Expansion Rate Tensor

Shear Rate Tensor

We then have:

We then have:Vab = �2⌘�ab � ⇠✓hab


And Finally:


And Finally:Tab = (µ+ p)uaub � ucu

cgabp� 2⌘�ab � ⇠✓hab




Using the energy conservation law:




Using the energy conservation law:T ab;b = 0 = T b

a;b





a;b

We derive an equation of motion for the fluid:





a;b


ua�(µ+ p)uau

b + �bap� ⇡ba � ⇠✓

��ba + uau

b��

;b= 0





a;b


ua�(µ+ p)uau

b + �bap� ⇡ba � ⇠✓

��ba + uau

b��

;b= 0

Where we have defined:





a;b


ua�(µ+ p)uau

b + �bap� ⇡ba � ⇠✓

��ba + uau

b��

;b= 0

Where we have defined: ⇡ab = 2⌘�ab

ua�(µ+ p)uau

b + �bap� ⇡ba � ⇠✓

��ba + uau

b��

;b= 0

ua�(µ+ p)uau

b + �bap� ⇡ba � ⇠✓

��ba + uau

b��

;b= 0

We evaluate this expression term-by-term, and obtain:

ua�(µ+ p)uau

b + �bap� ⇡ba � ⇠✓

��ba + uau

b��

;b= 0


µ+ (µ+ p)✓ � 4⌘�2 � ⇠✓2 = 0

ua�(µ+ p)uau

b + �bap� ⇡ba � ⇠✓

��ba + uau

b��

;b= 0


µ+ (µ+ p)✓ � 4⌘�2 � ⇠✓2 = 0✓�2 =

1

2�ab�ab ) �ab�ab = 2�2

◆

ua�(µ+ p)uau

b + �bap� ⇡ba � ⇠✓

��ba + uau

b��

;b= 0


µ+ (µ+ p)✓ � 4⌘�2 � ⇠✓2 = 0✓�2 =

1

2�ab�ab ) �ab�ab = 2�2

◆

We assume that cosmological fluids obey the barotropic equation of state:

ua�(µ+ p)uau

b + �bap� ⇡ba � ⇠✓

��ba + uau

b��

;b= 0


µ+ (µ+ p)✓ � 4⌘�2 � ⇠✓2 = 0✓�2 =

1

2�ab�ab ) �ab�ab = 2�2

◆


p = wµ,w 2 R

ua�(µ+ p)uau

b + �bap� ⇡ba � ⇠✓

��ba + uau

b��

;b= 0


µ+ (µ+ p)✓ � 4⌘�2 � ⇠✓2 = 0✓�2 =

1

2�ab�ab ) �ab�ab = 2�2

◆


p = wµ,w 2 RFor relativistic particles in the early universe, w = 1/3, (See: Landau and Lifshitz, “Fluid Mechanics” or Ellis,

Maartens, MacCallum “Relativistic Cosmology”)

ua�(µ+ p)uau

b + �bap� ⇡ba � ⇠✓

��ba + uau

b��

;b= 0


µ+ (µ+ p)✓ � 4⌘�2 � ⇠✓2 = 0✓�2 =

1

2�ab�ab ) �ab�ab = 2�2

◆


p = wµ,w 2 RFor relativistic particles in the early universe, w = 1/3, (See: Landau and Lifshitz, “Fluid Mechanics” or Ellis,

Maartens, MacCallum “Relativistic Cosmology”)

µ+

✓4

3µ

◆✓ � 4⌘�2 � ⇠✓2 = 0

We will also assume energy conditions for our fluid, which will allow us to build the arguments for the

existence/non-existence of a past singularity for this cosmological model (at the end).



The Weak Energy Condition (WEC) is:



The Weak Energy Condition (WEC) is:T abuaub � 0




This basically says that any observer will always observe a positive energy density.





The Strong Energy Condition (SEC) is:





The Strong Energy Condition (SEC) is:✓T ab � 1

2Tgab

◆uaub � 0






2Tgab

◆uaub � 0

By The Einstein Field Equations, we have:






2Tgab

◆uaub � 0

By The Einstein Field Equations, we have:Rabuaub � 0






2Tgab

◆uaub � 0

By The Einstein Field Equations, we have:Rabuaub � 0

More on these things later.

Back To The Einstein Field Equations!


The properties of a fluid flow in a spacetime can be described by the decomposition:


The properties of a fluid flow in a spacetime can be described by the decomposition:ua;b = �aaub + �ab + !ab +

1

3hab✓



1

3hab✓

Congruence of fluid lines moving with 4-velocity:



1

3hab✓

Congruence of fluid lines moving with 4-velocity:ua

ua;b = �aaub + �ab + !ab +1

3hab✓


3hab✓

✓ ⌘ ua;a


3hab✓

✓ ⌘ ua;a

Expansion Tensor: Measures divergence of the fluid congruence.


3hab✓

✓ ⌘ ua;a


aa ⌘ ua;bub


3hab✓

✓ ⌘ ua;a


aa ⌘ ua;bub Fluid Acceleration: Measures how

much the fluid congruence can be taken to be non-geodesic.


3hab✓

✓ ⌘ ua;a




�ab =

1

2(um;n + un;m)� 1

3uc;chmn

�hma hn

b


3hab✓

✓ ⌘ ua;a




�ab =

1

2(um;n + un;m)� 1

3uc;chmn

�hma hn

b

Shear Tensor: Measures how much the fluid congruence deforms.


3hab✓

✓ ⌘ ua;a




�ab =

1

2(um;n + un;m)� 1

3uc;chmn

�hma hn

b


!ab =1

2(um;n � un;m)hm

a hnb


3hab✓

✓ ⌘ ua;a




�ab =

1

2(um;n + un;m)� 1

3uc;chmn

�hma hn

b


!ab =1

2(um;n � un;m)hm

a hnb

Vorticity Tensor: Measures how much the fluid congruence rotates.

Connecting this to our orthonormal frame approach, we assume that

the fluid is non-tilted:


the fluid is non-tilted:!uv = uu;vu

v = 0



v = 0

✓uv = uu;v =1

3✓huv + �uv



v = 0

✓uv = uu;v =1

3✓huv + �uv

ckij = ✏ijlnlk + al

��li�

kj � �lj�

ki

�



v = 0

✓uv = uu;v =1

3✓huv + �uv


��li�

kj � �lj�

ki

�

The structure constants from before (now functions) are purely spatial, and are constant on each spatial

slice.



v = 0

✓uv = uu;v =1

3✓huv + �uv


��li�

kj � �lj�

ki

�


slice.

So:



v = 0

✓uv = uu;v =1

3✓huv + �uv


��li�

kj � �lj�

ki

�


slice.

So:nlk, ai are purely functions of time.

Applying The Jacobi identity as applied to the set of vectors:


(U = @t, ea, eb)


(U = @t, ea, eb)

We get:


(U = @t, ea, eb)

We get:[U, [ea, eb]] + [ea, [eb,U]] + [eb, [U, ea]] = 0


(U = @t, ea, eb)


@t(ckab) + cktdc

dab + ckadc

dbt + ckbdc

dta = 0


(U = @t, ea, eb)


@t(ckab) + cktdc

dab + ckadc

dbt + ckbdc

dta = 0

One can show that upon applying The Jacobi identity to the three

spatial vectors, we obtain:


(U = @t, ea, eb)


@t(ckab) + cktdc

dab + ckadc

dbt + ckbdc

dta = 0


spatial vectors, we obtain:nijai = 0


(U = @t, ea, eb)


@t(ckab) + cktdc

dab + ckadc

dbt + ckbdc

dta = 0


spatial vectors, we obtain:nijai = 0

Condition for Lie Algebra property to hold.

It is important to note that we always take to be a symmetric matrix and as such we can diagonalize it using a orientation of our choice for the spatial frame. It has been

conventional to assume:



nij



nij

nij = diag(n1, n2, n3), , ai = (0, 0, a)



nij

nij = diag(n1, n2, n3), , ai = (0, 0, a)

Further, using the definitions:



nij

nij = diag(n1, n2, n3), , ai = (0, 0, a)

Further, using the definitions:catb = �✓ab + ✏abc⌦

c



nij

nij = diag(n1, n2, n3), , ai = (0, 0, a)


c

⌦a =1

2✏abcdubec · ed



nij

nij = diag(n1, n2, n3), , ai = (0, 0, a)


c

⌦a =1

2✏abcdubec · ed

In combination with the previous definitions in the Jacobi identity, we get:



nij

nij = diag(n1, n2, n3), , ai = (0, 0, a)


c

⌦a =1

2✏abcdubec · ed


ai +1

3✓ai + �ija

j + ✏ijkaj⌦k = 0



nij

nij = diag(n1, n2, n3), , ai = (0, 0, a)


c

⌦a =1

2✏abcdubec · ed


ai +1

3✓ai + �ija


˙nab +1

3✓nab + 2nk

(a✏b)kl⌦l � 2nk(a�b)

k = 0

One classifies the different Bianchi types by the signs of the eigenvalues:


n11, n22, n33 = n1, n2, n3


n11, n22, n33 = n1, n2, n3 and


n11, n22, n33 = n1, n2, n3 and a


n11, n22, n33 = n1, n2, n3 and a15.4 The orthonormal frame approach to the Bianchi models 411

Class Type a n1 n2 n3

A I 0 0 0 0II 0 + 0 0

VI0 0 + − 0VII0 0 + + 0VIII 0 + + −IX 0 + + +

B V + 0 0 0IV + + 0 0

VIh + + − 0VIIh + + + 0

Table 15.2: The Bianchi types in terms of the algebraic properties of the structurecoefficients.

For the structure coefficients eq. (15.62) to correspond to a Lie algebra, thevector ai must according to eq. (15.65) be in the kernel4 of the matrix nij . Forthe class A model, ai = 0 and this equation is identically satisfied. For the classB models, ai must be an eigenvector of the matrix nij with zero eigenvalue. Inany case, since nij is a symmetric matrix, we can diagonalise it using a specificorientation of the spatial frame. Thus, without any loss of generality we canassume that

nij = diag(n1, n2, n3), ai = (0, 0, a) (15.68)

by a suitable choice of frame. The Jacobi identity then implies n3a = 0.The eigenvalues of a matrix are invariant properties of a matrix under con-

jugation with respect to rotations. The Bianchi models can now be charac-terised by the relative signs of the eigenvalues n1, n2, n3 and a. In Table 15.2the classification of the Bianchi types in terms of these eigenvalues is listed.For the types VIh and VIIh the group parameter is defined by the equation

hn1n2 = a2. (15.69)

In this table III=VI−1.Note that for some of the Bianchi types, two or three eigenvalues are equal

to zero. Hence, for these we have unused degrees of freedom to choose theorientation of the spatial frame. For example, the type I case has vanishingstructure coefficients. Thus we have an unused SO(3) rotation for the spatialframe. Since the shear is symmetric, we can choose to diagonalise σab instead.So for a Bianchi type I universe model we can without any loss of generality choosethe shear to be diagonal.

Einstein’s Field Equations for Bianchi type universes

We can use the results from the previous chapter to find the field equationsfor the Bianchi type universe models. The Ricci tensor can be found fromcontracting the Riemann tensor eq. (7.45). Using the four-dimensional Riccitensor we can show that the tt-equation yields Raychaudhuri’s equation, eq.(14.31), and the spatial ab-equations yield the shear propagation equations, eq.(14.40), and the generalised Friedmann equation, eq (14.34). The off-diagonal

4Consider a matrix M and a vector v. The vector v is in the kernel of M if and only if Mv = 0.




A I 0 0 0 0II 0 + 0 0

VI0 0 + − 0VII0 0 + + 0VIII 0 + + −IX 0 + + +

B V + 0 0 0IV + + 0 0

VIh + + − 0VIIh + + + 0



nij = diag(n1, n2, n3), ai = (0, 0, a) (15.68)



hn1n2 = a2. (15.69)






We are interested in Bianchi IV:




A I 0 0 0 0II 0 + 0 0

VI0 0 + − 0VII0 0 + + 0VIII 0 + + −IX 0 + + +

B V + 0 0 0IV + + 0 0

VIh + + − 0VIIh + + + 0



nij = diag(n1, n2, n3), ai = (0, 0, a) (15.68)



hn1n2 = a2. (15.69)






We are interested in Bianchi IV:

ai = a�i3 > 0, and n1 > 0, n2 = n3 = 0

So far, we have reduced The Einstein Field equations into a set of first-order dynamical

equations:


equations:µ+

✓4

3µ

◆✓ � 4⌘�2 � ⇠✓2 = 0


equations:µ+

✓4

3µ

◆✓ � 4⌘�2 � ⇠✓2 = 0

ai +1

3✓ai + �ija



equations:µ+

✓4

3µ

◆✓ � 4⌘�2 � ⇠✓2 = 0

ai +1

3✓ai + �ija


˙nab +1

3✓nab + 2nk


k = 0


equations:µ+

✓4

3µ

◆✓ � 4⌘�2 � ⇠✓2 = 0

ai +1

3✓ai + �ija


˙nab +1

3✓nab + 2nk


k = 0

We are not done! The system above has unspecified functions for the expansion scalar, and the shear tensor. To close the system, we need dynamical

equations for these as well.

The Raychaudhuri Equation for Viscous Flow


We are interested in analyzing the equation:


We are interested in analyzing the equation:d✓

dt=

�ua;a

�;bub = ua

;a;bub



dt=

�ua;a

�;bub = ua

;a;bub

By the definition of The Riemann curvature tensor:



dt=

�ua;a

�;bub = ua

;a;bub

By the definition of The Riemann curvature tensor:ua;d;b � ua

;b;d = �Racdbu

c



dt=

�ua;a

�;bub = ua

;a;bub


;b;d = �Racdbu

c

Can show that:



dt=

�ua;a

�;bub = ua

;a;bub


;b;d = �Racdbu

c

Can show that:d✓

dt= �Rbcu

buc + aa;a + 2!2 � 2�2 � 1

3✓2



dt=

�ua;a

�;bub = ua

;a;bub


;b;d = �Racdbu

c

Can show that:d✓

dt= �Rbcu

buc + aa;a + 2!2 � 2�2 � 1

3✓2

Since:



dt=

�ua;a

�;bub = ua

;a;bub


;b;d = �Racdbu

c

Can show that:d✓

dt= �Rbcu

buc + aa;a + 2!2 � 2�2 � 1

3✓2

Since:

Rab =

✓Tab �

1

2Tgab

◆

We have that:

We have that:d✓

dt= �1

2 (µ+ 3p) +

3

2⇠✓ � 2�2 � 1

3✓2

We have that:d✓

dt= �1

2 (µ+ 3p) +

3

2⇠✓ � 2�2 � 1

3✓2

(After some work!) (See Kohli and Haslam:arXiv:1207.6132v2 [gr-qc])

http://arxiv.org/abs/1207.6132v2

We have that:d✓

dt= �1

2 (µ+ 3p) +

3

2⇠✓ � 2�2 � 1

3✓2


Using the ultrarelativistic equation of state:


We have that:d✓

dt= �1

2 (µ+ 3p) +

3

2⇠✓ � 2�2 � 1

3✓2


Using the ultrarelativistic equation of state: p =1

3µ


We have that:d✓

dt= �1

2 (µ+ 3p) +

3

2⇠✓ � 2�2 � 1

3✓2



3µ

d✓

dt= �µ+

3

2⇠✓ � 2�2 � 1

3✓2


We have that:d✓

dt= �1

2 (µ+ 3p) +

3

2⇠✓ � 2�2 � 1

3✓2



3µ

d✓

dt= �µ+

3

2⇠✓ � 2�2 � 1

3✓2

This is The Raychaudhuri equation with our modification for viscous flow.


The Generalized Friedmann Equation


ua;b = ✓ab = Kab


ua;b = ✓ab = KabDefine:


ua;b = ✓ab = KabDefine: Extrinsic Curvature tensor: Measure “warping” of spatial slices.



2.4 The extrinsic curvature 33

δna

δna ∝−Kabna

Figure 2.2 The extrinsic curvature of a hypersurface in an enveloping spacetime measures how much normalvectors to the hypersurface differ at neighboring points. It therefore measures the rate at which the hypersurfacewarps as it is carried forward along a normal vector.

The 3-dimensional covariant derivative can be expressed in terms of 3-dimensionalconnection coefficients, which, in a coordinate basis, are given by

!abc = 1

2γ ad (∂cγdb + ∂bγdc − ∂dγbc). (2.44)

The 3-dimensional Riemann tensor associated with γi j is defined by requiring that8

2D[a Db]wc = Rdcbawd Rd

cband = 0 (2.45)

for any spatial vector wd . In a coordinate basis, the components of the Riemann tensor canbe computed from

R dabc = ∂b!

dac − ∂a!

dbc + !e

ac!deb − !e

bc!dea. (2.46)

Contracting the Riemann tensor yields the 3-dimensional Ricci tensor Rab = Rcacb and the

3-dimensional Ricci scalar R = Raa .

Einstein’s equations (1.32) relate contractions of the 4-dimensional Riemann tensor(4) Ra

bcd to the stress–energy tensor. Since we want to rewrite these equations in terms of3-dimensional objects, we decompose (4) Ra

bcd into spatial tensors. Not surprisingly, thisdecomposition involves its 3-dimensional cousin Ra

bcd , but obviously this cannot containall the information needed. Rd

abc is a purely spatial object and can be computed from spatialderivatives of the spatial metric alone, while (4) Rd

abc is a spacetime creature which alsocontains time derivatives of the 4-dimensional metric. Stated differently, the 3-dimensionalcurvature Ra

bcd only contains information about the curvature intrinsic to a slice $, but itgives no information about what shape this slice takes in the spacetime M in which it isembedded. This information is contained in a tensor called the extrinsic curvature.

2.4 The extrinsic curvature

The extrinsic curvature Kab can be found by projecting gradients of the normal vectorinto the slice $ (see Figure 2.2). We will also see that the extrinsic curvature is related to

8 See equation (1.20) for the 4-dimensional analog of this expression.




δna

δna ∝−Kabna



!abc = 1




cband = 0 (2.45)


R dabc = ∂b!

dac − ∂a!

dbc + !e

ac!deb − !e

bc!dea. (2.46)













Recall from differential geometry, the embedding relation:




δna

δna ∝−Kabna



!abc = 1




cband = 0 (2.45)


R dabc = ∂b!

dac − ∂a!

dbc + !e

ac!deb − !e

bc!dea. (2.46)













Recall from differential geometry, the embedding relation:(4)R =(3) R+K2 �K↵�K↵� � 2(4)R↵�u

↵u�




δna

δna ∝−Kabna



!abc = 1




cband = 0 (2.45)


R dabc = ∂b!

dac − ∂a!

dbc + !e

ac!deb − !e

bc!dea. (2.46)














↵u�

) T↵�u↵� =

1

2

⇣(3)R�K↵�K↵� +K2

⌘




δna

δna ∝−Kabna



!abc = 1




cband = 0 (2.45)


R dabc = ∂b!

dac − ∂a!

dbc + !e

ac!deb − !e

bc!dea. (2.46)














↵u�

) T↵�u↵� =

1

2

⇣(3)R�K↵�K↵� +K2

⌘

(From Baumgarte et.al. - “Numerical Relativity”)

Upon applying the decomposition equation, we get (after some work!):


1

3✓2 =

1

2�ab�

ab � 1

2

(3)

R+ µ


1

3✓2 =

1

2�ab�

ab � 1

2

(3)

R+ µ

This is The Generalized Friedmann Equation.

The Shear Propagation Equations


Describe the evolution of the anisotropy in a cosmological model as a function of time.



Derivation is time-consuming!



Derivation is time-consuming!

The interested reader is encouraged to consult the texts by: Ellis and Wainwright, Hervik and Gron, or Plebanski.

The shear propagation equations are given as:


˙�ab + ✓�ab � �da✏bcd⌦

c � �db ✏acd⌦

c +(3) Rab �1

3�(3)ab R = 2⌘�ab




c +(3) Rab �1

3�(3)ab R = 2⌘�ab

The beauty of the orthonormal frame approach is that the Ricci tensor is now expressed in canonical

coordinate-free form:




c +(3) Rab �1

3�(3)ab R = 2⌘�ab


coordinate-free form:(3)Rab = �✏cda nbcad � ✏cdb nacad + 2nadn

db � nnab � �ab

✓2a2 + ncdn

cd � 1

2n2

◆




c +(3) Rab �1

3�(3)ab R = 2⌘�ab




✓2a2 + ncdn

cd � 1

2n2

◆

Ricci scalar:




c +(3) Rab �1

3�(3)ab R = 2⌘�ab




✓2a2 + ncdn

cd � 1

2n2

◆

Ricci scalar:(3)R =(3) Ra

a = �✓6a2 + ncdn

cd � 1

2n2

◆




c +(3) Rab �1

3�(3)ab R = 2⌘�ab




✓2a2 + ncdn

cd � 1

2n2

◆


a = �✓6a2 + ncdn

cd � 1

2n2

◆

But, at the expense of some constraint equations:




c +(3) Rab �1

3�(3)ab R = 2⌘�ab




✓2a2 + ncdn

cd � 1

2n2

◆


a = �✓6a2 + ncdn

cd � 1

2n2

◆

But, at the expense of some constraint equations:3a�33 +

�n11 � n22

��21 = 0

3a�31 + n22�32 = 0

3a�32 � n11�31 = 0

Applying this set of equations to The Bianchi Type IV model:


n11 = N, and n22 = n33 = 0, where N > 0


n11 = N, and n22 = n33 = 0, where N > 0

The constraint equations give:


n11 = N, and n22 = n33 = 0, where N > 0


3a�33 +N�21 = 0

3a�31 = 0

3a�32 � �31 = 0


n11 = N, and n22 = n33 = 0, where N > 0


3a�33 +N�21 = 0

3a�31 = 0

3a�32 � �31 = 0These restrict the form of the shear tensor (symmetric

and traceless):


n11 = N, and n22 = n33 = 0, where N > 0


3a�33 +N�21 = 0

3a�31 = 0


and traceless):For the diagonal components (a = b):


n11 = N, and n22 = n33 = 0, where N > 0


3a�33 +N�21 = 0

3a�31 = 0



�ab =⇣�+ +

p3��,�+ �

p3��,�2�+

⌘


n11 = N, and n22 = n33 = 0, where N > 0


3a�33 +N�21 = 0

3a�31 = 0



�ab =⇣�+ +

p3��,�+ �

p3��,�2�+

⌘

For the off-diagonal components:


n11 = N, and n22 = n33 = 0, where N > 0


3a�33 +N�21 = 0

3a�31 = 0



�ab =⇣�+ +

p3��,�+ �

p3��,�2�+

⌘

For the off-diagonal components:

�21 = �12 =�3a�33

N=

6a�+

N

With this shear tensor, the shear propagation equations become:


�� = � N2

2p3+

24p3a2�2

+

N2� ��✓ + 2��⌘


�� = � N2

2p3+

24p3a2�2

+

N2� ��✓ + 2��⌘

�+ = �N2

6� �+✓

5+ 2⌘�+

We have therefore reduced The Einstein Field equations for The Bianchi Type-IV model to a system of first-

order ordinary differential equations!



✓ =1

2

�12a2 + 2µ+N2 � 2✓2 + 3✓⇠

�



✓ =1

2

�12a2 + 2µ+N2 � 2✓2 + 3✓⇠

�

µ = �12a2⌘ � 4µ⌘ � ⌘N2 � 4µ✓

3+

4⌘✓2

3+ ✓2⇠



✓ =1

2

�12a2 + 2µ+N2 � 2✓2 + 3✓⇠

�

µ = �12a2⌘ � 4µ⌘ � ⌘N2 � 4µ✓

3+

4⌘✓2

3+ ✓2⇠

�� = � N2

2p3+

24p3a2�2

+

N2� ��✓ + 2��⌘



✓ =1

2

�12a2 + 2µ+N2 � 2✓2 + 3✓⇠

�

µ = �12a2⌘ � 4µ⌘ � ⌘N2 � 4µ✓

3+

4⌘✓2

3+ ✓2⇠

�� = � N2

2p3+

24p3a2�2

+

N2� ��✓ + 2��⌘

�+ = �N2

6� �+✓

5+ 2⌘�+



✓ =1

2

�12a2 + 2µ+N2 � 2✓2 + 3✓⇠

�

µ = �12a2⌘ � 4µ⌘ � ⌘N2 � 4µ✓

3+

4⌘✓2

3+ ✓2⇠

�� = � N2

2p3+

24p3a2�2

+

N2� ��✓ + 2��⌘

�+ = �N2

6� �+✓

5+ 2⌘�+

a = �a

✓✓

3� 2�+

◆



✓ =1

2

�12a2 + 2µ+N2 � 2✓2 + 3✓⇠

�

µ = �12a2⌘ � 4µ⌘ � ⌘N2 � 4µ✓

3+

4⌘✓2

3+ ✓2⇠

�� = � N2

2p3+

24p3a2�2

+

N2� ��✓ + 2��⌘

�+ = �N2

6� �+✓

5+ 2⌘�+

a = �a

✓✓

3� 2�+

◆

N = N

✓

3+ 2

⇣�+ +

p3��

⌘�

This six-dimensional system has no exact solution, so numerical methods must be employed. However, we can

simplify the system of equations, by re-writing it in Expansion-Normalized Form:



Define the quantities:




✓ ⌘ 3H




✓ ⌘ 3H where, H is the Hubble scalar:




✓ ⌘ 3H where, H is the Hubble scalar: H =s

s





sq = � ss

s2





sq = � ss

s2A deceleration parameter





sq = � ss


Clearly:





sq = � ss


Clearly:H = �(1 + q)H2





sq = � ss



s = s0e⌧





sq = � ss



s = s0e⌧ A dimensionless time variable





sq = � ss




We can show that:





sq = � ss




We can show that:dt

d⌧=

1

H) dH

d⌧⌘ H 0 = �(1 + q)H

Define the additional quantities:


Normalized density parameter



⌃2 =�2

3H2



⌃2 =�2

3H2Normalized shear parameter



⌃2 =�2


K = �(3)R

6H2



⌃2 =�2


K = �(3)R

6H2Normalized scalar curvature



⌃2 =�2


K = �(3)R


By comparing the two expansion-rate equations, we have:



⌃2 =�2


K = �(3)R



�(1 + q)H2 = �1

3µ+ ⇠H � 2

3�2 �H2



⌃2 =�2


K = �(3)R



�(1 + q)H2 = �1

3µ+ ⇠H � 2

3�2 �H2

) q = ⌦� ⇠

H� 2⌃2



⌃2 =�2


K = �(3)R



�(1 + q)H2 = �1

3µ+ ⇠H � 2

3�2 �H2

) q = ⌦� ⇠

H� 2⌃2

In addition:



⌃2 =�2


K = �(3)R



�(1 + q)H2 = �1

3µ+ ⇠H � 2

3�2 �H2

) q = ⌦� ⇠

H� 2⌃2

In addition:⇠

H= 3⇠0,

⌘

H= 3⌘0, where ⇠0, ⌘0 > 0



⌃2 =�2


K = �(3)R



�(1 + q)H2 = �1

3µ+ ⇠H � 2

3�2 �H2

) q = ⌦� ⇠

H� 2⌃2

In addition:⇠

H= 3⇠0,

⌘

H= 3⌘0, where ⇠0, ⌘0 > 0, n =

N

H,A =

a

H,⌃± =

�±H



⌃2 =�2


K = �(3)R



�(1 + q)H2 = �1

3µ+ ⇠H � 2

3�2 �H2

) q = ⌦� ⇠

H� 2⌃2

In addition:⇠

H= 3⇠0,

⌘

H= 3⌘0, where ⇠0, ⌘0 > 0, n =

N

H,A =

a

H,⌃± =

�±H

⌦ =µ

3H2� 0

By making these substitutions, the dynamical system takes the 5-D form:


⌃0� = � n2

2p3+ 24

p3A2⌃2

+

n2� 3⌃� � 6⌃�⌘0 + (1 + q)⌃�


⌃0� = � n2

2p3+ 24

p3A2⌃2

+

n2� 3⌃� � 6⌃�⌘0 + (1 + q)⌃�

⌃0+ = �1

6n2 � 3

5⌃+ + 6⌘0⌃+ + (1 + q)⌃+


⌃0� = � n2

2p3+ 24

p3A2⌃2

+

n2� 3⌃� � 6⌃�⌘0 + (1 + q)⌃�

⌃0+ = �1

6n2 � 3

5⌃+ + 6⌘0⌃+ + (1 + q)⌃+

A0 = �A+ 2⌃+A+ (1 + q)A


⌃0� = � n2

2p3+ 24

p3A2⌃2

+

n2� 3⌃� � 6⌃�⌘0 + (1 + q)⌃�

⌃0+ = �1

6n2 � 3

5⌃+ + 6⌘0⌃+ + (1 + q)⌃+

A0 = �A+ 2⌃+A+ (1 + q)A

n0 = n+ 2n⇣⌃+ +

p3⌃�

⌘+ (1 + q)n


⌃0� = � n2

2p3+ 24

p3A2⌃2

+

n2� 3⌃� � 6⌃�⌘0 + (1 + q)⌃�

⌃0+ = �1

6n2 � 3

5⌃+ + 6⌘0⌃+ + (1 + q)⌃+

A0 = �A+ 2⌃+A+ (1 + q)A

n0 = n+ 2n⇣⌃+ +

p3⌃�

⌘+ (1 + q)n

⌦0 = ⌘0��12A2 � 12⌦� n2

�� 4⌦+ 12⌘0 + 9⇠0 + 2⌦(1 + q)


⌃0� = � n2

2p3+ 24

p3A2⌃2

+

n2� 3⌃� � 6⌃�⌘0 + (1 + q)⌃�

⌃0+ = �1

6n2 � 3

5⌃+ + 6⌘0⌃+ + (1 + q)⌃+

A0 = �A+ 2⌃+A+ (1 + q)A

n0 = n+ 2n⇣⌃+ +

p3⌃�

⌘+ (1 + q)n

⌦0 = ⌘0��12A2 � 12⌦� n2

�� 4⌦+ 12⌘0 + 9⇠0 + 2⌦(1 + q)

Where:


⌃0� = � n2

2p3+ 24

p3A2⌃2

+

n2� 3⌃� � 6⌃�⌘0 + (1 + q)⌃�

⌃0+ = �1

6n2 � 3

5⌃+ + 6⌘0⌃+ + (1 + q)⌃+

A0 = �A+ 2⌃+A+ (1 + q)A

n0 = n+ 2n⇣⌃+ +

p3⌃�

⌘+ (1 + q)n

⌦0 = ⌘0��12A2 � 12⌦� n2

�� 4⌦+ 12⌘0 + 9⇠0 + 2⌦(1 + q)

Where:q = ⌦� 3⇠0 � 2⌃2,


⌃0� = � n2

2p3+ 24

p3A2⌃2

+

n2� 3⌃� � 6⌃�⌘0 + (1 + q)⌃�

⌃0+ = �1

6n2 � 3

5⌃+ + 6⌘0⌃+ + (1 + q)⌃+

A0 = �A+ 2⌃+A+ (1 + q)A

n0 = n+ 2n⇣⌃+ +

p3⌃�

⌘+ (1 + q)n

⌦0 = ⌘0��12A2 � 12⌦� n2

�� 4⌦+ 12⌘0 + 9⇠0 + 2⌦(1 + q)

Where:q = ⌦� 3⇠0 � 2⌃2, ⌃2 = ⌃2

+ + ⌃2�,


⌃0� = � n2

2p3+ 24

p3A2⌃2

+

n2� 3⌃� � 6⌃�⌘0 + (1 + q)⌃�

⌃0+ = �1

6n2 � 3

5⌃+ + 6⌘0⌃+ + (1 + q)⌃+

A0 = �A+ 2⌃+A+ (1 + q)A

n0 = n+ 2n⇣⌃+ +

p3⌃�

⌘+ (1 + q)n

⌦0 = ⌘0��12A2 � 12⌦� n2

�� 4⌦+ 12⌘0 + 9⇠0 + 2⌦(1 + q)

Where:q = ⌦� 3⇠0 � 2⌃2, ⌃2 = ⌃2

+ + ⌃2�, ⌦ � 0


⌃0� = � n2

2p3+ 24

p3A2⌃2

+

n2� 3⌃� � 6⌃�⌘0 + (1 + q)⌃�

⌃0+ = �1

6n2 � 3

5⌃+ + 6⌘0⌃+ + (1 + q)⌃+

A0 = �A+ 2⌃+A+ (1 + q)A

n0 = n+ 2n⇣⌃+ +

p3⌃�

⌘+ (1 + q)n

⌦0 = ⌘0��12A2 � 12⌦� n2

�� 4⌦+ 12⌘0 + 9⇠0 + 2⌦(1 + q)

Where:q = ⌦� 3⇠0 � 2⌃2, ⌃2 = ⌃2

+ + ⌃2�, ⌦ � 0

The expansion variable is now decoupled from the system of equations, with the “cost” of assuming expansion.

Solutions of The Dynamical System


In our numerical experiments, we expect the following reasonable results:

The models should become asymptotic to a FLRW model with vanishing anisotropy

The energy density should decrease as the universe model expands

Which values of the viscosity parameters lead to these results?


In our numerical experiments, we expect the following reasonable results:

The models should become asymptotic to a FLRW model with vanishing anisotropy

The energy density should decrease as the universe model expands

Which values of the viscosity parameters lead to these results?

⇠0, ⌘0

A Note on The Singularity Theorem

Recall The Raychaudhuri Equation:


✓ = �1

2 (µ+ 3p) +

3

2⇠✓ � 2�2 � 1

3✓2


✓ = �1

2 (µ+ 3p) +

3

2⇠✓ � 2�2 � 1

3✓2

And The SEC:


✓ = �1

2 (µ+ 3p) +

3

2⇠✓ � 2�2 � 1

3✓2

And The SEC:Rabuaub � 0


✓ = �1

2 (µ+ 3p) +

3

2⇠✓ � 2�2 � 1

3✓2


) �1

2 (µ+ 3p) +

3

2⇠✓ 0


✓ = �1

2 (µ+ 3p) +

3

2⇠✓ � 2�2 � 1

3✓2


) �1

2 (µ+ 3p) +

3

2⇠✓ 0

Assuming The Equation of State:


✓ = �1

2 (µ+ 3p) +

3

2⇠✓ � 2�2 � 1

3✓2


) �1

2 (µ+ 3p) +

3

2⇠✓ 0


⇠ =C

✓, C 2 R+


✓ = �1

2 (µ+ 3p) +

3

2⇠✓ � 2�2 � 1

3✓2


) �1

2 (µ+ 3p) +

3

2⇠✓ 0


⇠ =C

✓, C 2 R+

) ✓ +1

2 (µ+ 3p)� 3

2C +

1

3✓2 = �2�2 0


✓ = �1

2 (µ+ 3p) +

3

2⇠✓ � 2�2 � 1

3✓2


) �1

2 (µ+ 3p) +

3

2⇠✓ 0


⇠ =C

✓, C 2 R+

) ✓ +1

2 (µ+ 3p)� 3

2C +

1

3✓2 = �2�2 0

) ✓ +1

3✓2 0

) ✓ �1

3✓2

Assuming a Strict Inequality:


) 1

✓(t) 1

✓i+

t

3


) 1

✓(t) 1

✓i+

t

3

) 8 t 2 R, 9 tsing such that:1

✓(tsing)= 0


) 1

✓(t) 1

✓i+

t

3

) 8 t 2 R, 9 tsing such that:1

✓(tsing)= 0

) 1

✓(tsing)= 0 ) ✓(tsing) = 1

Assuming a strict equality, one can employ the IVT to prove the existence of a singularity point:


F (t) ⌘ 1

✓(t)=

1

✓i+

t

3


F (t) ⌘ 1

✓(t)=

1

✓i+

t

3

Lemma:


F (t) ⌘ 1

✓(t)=

1

✓i+

t

3

Lemma:Let F (t) ⌘ 1

✓(t) =1✓i+

t3 be continuous on [a,b]. If F (a) < 0 < F (b)_F (b) <

0 < F (a) ) 9 d, such that a < d < b for which F (d) = 0.


F (t) ⌘ 1

✓(t)=

1

✓i+

t

3


✓(t) =1✓i+



“Proof:”


F (t) ⌘ 1

✓(t)=

1

✓i+

t

3


✓(t) =1✓i+



“Proof:”Suppose that F (a) < 0 < F (b). Since F (a) < 0, by the continuity of F (t),

9 h 2 R such that F < 0 on [a, h).


F (t) ⌘ 1

✓(t)=

1

✓i+

t

3


✓(t) =1✓i+





Define:


F (t) ⌘ 1

✓(t)=

1

✓i+

t

3


✓(t) =1✓i+





Define: d = sup{h : F (t) < 0 on [a, h)}


F (t) ⌘ 1

✓(t)=

1

✓i+

t

3


✓(t) =1✓i+





Define: d = sup{h : F (t) < 0 on [a, h)}Clearly:


F (t) ⌘ 1

✓(t)=

1

✓i+

t

3


✓(t) =1✓i+





Define: d = sup{h : F (t) < 0 on [a, h)}Clearly: d b


F (t) ⌘ 1

✓(t)=

1

✓i+

t

3


✓(t) =1✓i+






One cannot have F (d) > 0, since this would mean that F > 0 on someinterval to the left of d.


F (t) ⌘ 1

✓(t)=

1

✓i+

t

3


✓(t) =1✓i+






One cannot have F (d) > 0, since this would mean that F > 0 on someinterval to the left of d.CONTRADICTION!

We also cannot have F (d) < 0, since this would imply that there would be

an interval [a, t), with t > d which would in turn imply that F (t) < 0



CONTRADICTION!



CONTRADICTION!Therefore, it must be that F (d) = 0.




Therefore, this model of the universe emerged from an initial singularity point.





This has been a demonstration of the famous Penrose-Hawking Singularity Theorem:





This has been a demonstration of the famous Penrose-Hawking Singularity Theorem:

More or less, we can say that if matter as described by the energy-momentum

tensor satisfies the strong energy condition, and there exists a

↵ 2 R+such that ✓(t) > ↵,

everywhere in the past of a specific spatial slice, then, there exists a singularity

point where all geodesics end.

THE END!

A Bianchi Type IV Viscous Model of The Early Universe

Science

Transcript of A Bianchi Type IV Viscous Model of The Early Universe