general relativity and einstein's field equations

General Relativity and Solutions to

Einstein’s Field Equations

Abhishek Kumar

Department of Physics and Astronomy, Bates College, Lewiston,

ME 04240

General Relativity andSolutions to Einstein’s Field Equations

A Senior Thesis

Presented to the Department of Physics and Astronomy

Bates College

in partial fulfillment of the requirements for the

Degree of Bachelor of Arts

by

Abhishek Kumar

Lewiston, Maine

January 13, 2009

Contents

Acknowledgments iii

Introduction iv

Chapter 1. Einstein’s Concept of Gravity 1

Chapter 2. Consequences of General Relativity 3

1. Gravitional Lensing 3

2. Black Holes 4

3. The Expanding Universe 7

Chapter 3. Mathematical Tools of General Relativity 10

1. Indices and Summation Convention 10

2. Defining Metrics and Tensors 10

3. Vectors in Curved Spacetime and Gradients 11

Chapter 4. The Einstein Curvature Tensor 14

1. Introduction to Curvature Tensors 14

2. The Schwarzschild Metric 15

3. The Robertson-Walker Metric 22

Bibliography 30

ii

Acknowledgments

First and foremost, I would like to thank my wonderful thesis advisor, Profes-

sor Tom Giblin. The high level of technicality required to understand a subject

like General Relativity requires quite an amount of patience. Professor Giblin has

been exceedingly understanding and supportive in helping me understand Einstein’s

beautiful theory.

I would also like to acknowledge my father, mother and brother, who were sup-

portive throughout my efforts. My family helped mitigate the frustration that some-

times arose with not obtaining the correct results or if calculations were not going

my way. And when results were coming out favorably, my family was also there to

congratulate me.

The Physics Department also deserves an overwhelming thank you for putting

together such a wonderful and understanding team of professors. My skills in Physics

have grown to a point where I can begin comprehending General Relativity only

because of the excellent teaching of the professors in the Physics Department.

iii

Introduction

Albert Einstein’s theory of General Relativity has revolutionized modern physics.

General Relativity is a theory that relates space and time to gravitation. Einstein’s

theory of General Relativity challenged Isaac Newton’s theory of gravity. In Newto-

nian gravity, space and time were absolute; they are unchangeable and completely

static. In Einsteinian gravity, however, space and time are intertwined into one single

fabric, that can be distorted and manipulated. This distortion is what creates the

phenomenon known as gravity.

With the introduction of Einstein’s theory of General Relativity, various new

doors began opening up in physics. We were able to better understand various

phenomenon occuring as a consequence of General Relativity. Some examples are

gravitational lensing, black holes and the expansion of our universe. The explanation

of these phenomenon can be found within Einstein’s Field Equations.

This paper aims to show how the Einstein’s gravity works differently from New-

tonian gravity, while providing substantial discussion on various consequences of his

theory. The paper then seeks to provide the reader with various mathematical tools

required to understand Einstein’s Field Equations. Lastly, the paper will show the

calculation of Einstein’s curvature tensor for metrics modeling phenomenon such as

non-rotating black holes and the expanding universe.

iv

CHAPTER 1

Einstein’s Concept of Gravity

The concept of gravity was truly first understood by Isaac Newton. The famous

incident of him sitting under a tree and an apple falling down in front of him led

him to unite the motion of terrestrial objects and celestial objects in a single theory

he called Gravity. He postulated that the force that keeps us on the ground, made

that apple fall to the ground and the force that keeps the planets in orbit around the

sun are both the same. In other words, any two masses essentially attract each other

with a force that is given by the formula,

(1.1) Fgravity =Gm1m2

r212

,

where m1 and m2 are the masses of the first and second mass, respectively, r212 is the

distance between the two masses, and G is Newton’s Gravitational Constant. This

theory remained the prominent theory of gravity for more than two-hundred years

until Albert Einstein proposed his theory of gravity.

Einstein’s rumminations on light while forming his theory of Special Relativity

led him to find a dire contradiction in Newton’s theory of gravity. Einstein discovered

that nothing can travel faster than the speed of light; the speed of light is essentially a

universal speed limit. However, Newton’s gravity given in Equation (1.1) has no time-

dependence. The gravitational force of one mass on another acts instantaneously. Say,

for instance, if the sun in our solar system were to, hypothetically, vanish, Newton’s

time-independent version of gravity says that the planets would immediately spin

out of orbit. This, however, cannot be true Einstein thought because if nothing can

travel faster than the speed of light, how can a gravitational affect travel faster than

1

1. EINSTEIN’S CONCEPT OF GRAVITY 2

the speed of light? To solve this issue, Einstein developed the General Theory of

Relativity, where gravity was not instantaneous, but its effects traveled at the speed

of light.

Einstein proposed a revolutionary idea in which the three dimensions of space

and time were intertwined in one single four-dimensional fabric that he would call

spacetime. This fabric of spacetime can be distorted by massive objects. Celestial

objects such as planets, stars, blackholes, galaxies and etc. curves this fabric of

spacetime, akin to how a bowling ball curves the rubber sheet it is placed on. Figure

1.1 visually shows how massive celetial objects may distort and curve the spacetime

around it. With this description of spacetime curvature, it is possible to formulate

Figure 1.1. An artist’s rendition of a massive object curving the fab-ric of spacetime [1].

Einstein’s version of gravity. Einstein’s gravity is not a force, but rather curvature

in spacetime created by the massive object defines the path that objects near it will

follow. Figure 1.1 shows a smaller object following a path defined by the curvature

in spacetime created by the more massive object.

Using this defintion of gravity, let us run the the similar scenario again, where

the sun vanishes. Once the sun has dissapeared, in Einstein’s theory of General

Relativity, the planets would not immediately spin out of orbit. The removal of the

1. EINSTEIN’S CONCEPT OF GRAVITY 3

sun would essentially create a ripple in space time almost as if a pebble were dropped

into a still water pond. This ripple will take the form of a wave that will travel at

the speed of light, and take eight minutes to reach the earth. Thus, the gravitational

effect of the sun dissapearing would not be instantaneous; it would be dependent on

time.

The more massive the object, the more it will curve spacetime and visa-versa. The

greater the curvature of spacetime, the stronger the gravitational field that object

will create. Thus, Einstein’s theory of gravity is unlike Newton’s where space and

time are static; Einstein’s General Theory of relativity allows space and time to both

be combined in one fabric and be manipulated by matter.

CHAPTER 2

Consequences of General Relativity

The contradictions between Newtonian Gravity and Einstein’s Gravity led to a

better understanding of fundamental physics. The interpertation that gravity is not a

force but rather the path that matter follows in curved space time, has explained phe-

nomena such as gravitational lensing, black holes and the expansion of our universe.

This chapter will discuss these predictions of General Relativity in some depth.

1. Gravitional Lensing

In his famous short article, “Lens-Like Action of a Star by the Deviation of Light

in teh Gravitational Field,” Albert Einstein introduces the concept of Gravitational

Lensing [2]. In this article, Einstein puts forth the idea that Gravitational Lensing is

the bending of light around massive objects. This gravitational lens sometimes may

produce multiple images of the distant object emmitting the light as it travels around

the intervening massive object.

Figure 2.1. A drawing of light emitted by the source, getting bentby the intervening mass on its way to the observer.

Matter traveling through curved spacetime will follow a path defined by that cur-

vature. In Figure 2.1, the intervening mass curves the spacetime around it. As light

nears this intervening mass, the photons’ path gets altered by this curved spacetime.

4

2. BLACK HOLES 5

This altered path is what we see as a “bending” of light. Moreover, this gravitational

lens may also cause the observer to see multiple images of the same object. Figure

2.1 depicts a scenario that occurs sometimes: observing dual images; the light takes

multiple paths to get to the observer as it passes the gravitational lens. If there are

multiple paths of light for the same source, the observer may see multiple images of

that source. The famous 1979 twin-quasar experiment served as evidence of gravi-

tational lensing, when two images of a quasar were observed as its light passed by a

lensing galaxy.

2. Black Holes

Black holes are regions of very large amounts of mass, usually resulting from a

gravitional collapse of a star. The mass of a black hole is so great that it creates a

very deep, almost cone-like curvature in the fabric of spacetime. Figures 2.2 and

Figure 2.2. An object with mass M curving spacetime [3].

2.3 depict a scenario very similar to a gravitational collapse of an object such as

a star. Figure 2.2 depicts an object with mass M curving the fabric of spacetime.

During a gravitational collapse, the mass of the object stays the same, but its size

decreases as its density increases. In the case of a black hole, the entire mass of the

original object (before the gravitational collapse) gets focused into a singularity [4].

As seen in Figure 2.3, when the same mass is concentrated and more dense, it creates

a deeper curvature in spacetime. In fact, spacetime around a black hole is so curved

2. BLACK HOLES 6

Figure 2.3. An object with the same mass M, but much smaller di-ameter, curving spacetime [3].

that light not only bends as it enters it (as seen in gravitational lensing), it cannot

even escape the gravitational field of the black hole. The “blackness” of black holes

results precisely from this phenomenon.

Newtonian gravity states thats for very massive and dense objects, the escape

velocity required to break free from their gravitational fields would be equal to the

speed of light [5]. Going with this assumption, light should be able to escape the

gravity of a black hole, as its speed is, well, equal to the speed of light. This is

not the case, however, since light cannot escape a black hole. Einstein’s theory of

General Relativty provides a more indepth reasoning for why light cannot escape the

gravitational grips of a black hole, for which we must study the embedding diagrams

of a black hole.

Horizon

Far Away

Figure 2.4. Spatial Embedding Diagram [6].

2. BLACK HOLES 7

Figure 2.4 shows a spatial embedding diagram of a black hole. In the diagram,

the top part is the structure of spacetime far away from the black hole, whereas

lower (narrower) part is at a location very close to the event horizon of a black hole.

Roughly speaking, the event horizon is the boundry of a black hole, which, if crossed,

is the point of no return; if an observer passes the event horizon, they have no way

of returning: a concept that will be discussed shortly. Note the relative flatness of

spacetime far away from the black hole as opposed to the extreme curvature seen

very close to the horizon of the black hole. Thus, as you get closer to a black hole,

spacetime begins to curve deeply. To understand why light (or any object for that

matter) cannot escape a black hole, however, we need to look at a different ebedding

diagram. In Figure 2.5, the vertical lines are the time-like lines and the horizontal lines

-6-4

-20

24

68

10

Y

-6-4

-20

24

6

X

-6

-4

-2

0

2

4

6

T

Figure 2.5. Space-time Embedding Diagram [6].

are the space-like lines. Figure 2.6 shows a bird’s eye view of a spacetime embedding

diagram. At either ends of the diagram, labeled as the flanges, the observer is far

away from the black hole. Spacetime is very flat at the flanges. However, as the

observer moves away from the flanges, toward Y = X = T = 0, spacetime begins to

2. BLACK HOLES 8

Flange

Flange0

+20

Y

0X

0T

Figure 2.6. Bird’s Eye view of Spacetime Embedding Diagram [6].

curve heavily. If the observer is approaching the singularity of the black hole, they

will need to essentially climb out of the black hole’s spacetime curvature by moving

towards the flanges of the embedding diagram. However, this poses the question:

why are observors, even those traveling at the speed of light, unable to move back

out to the flanges of the embedding diagram?

The answer to this question lies in the two separate timelike and spacelike lines

in the embedding diagram. While we are far away from the black hole, in the flanges

shown in Figure 2.6, the timelike paths do not lead to the center of the black hole.

As such, the obsever is free to travel anywhere along those timelike lines (note,

however, as a rule of General Relativity, the observer is only free to travel forward

in time, not backwards). However, notice that as you enter the curved space (shown

by the curving spacelike lines near the horizon of the black hole in Figure 2.6) the

timelike lines also start becoming distorted. Along with the spacelike paths, the

timelike paths also start tilting towards the center of the black hole’s mass as the

observer approaches the event horizon. As the distance between the center of mass

3. THE EXPANDING UNIVERSE 9

and the observer decreases, the distortion in the timelike paths becomes more and

more pronounced [4]. Figure 2.6 illustrates that exceedingly many timelike paths

lead to the center of mass as the observer moves away from the flanges and towards

the black hole. Once the observer has reached the event horizon of the black hole,

all timelike paths lead to the center of the black hole’s mass. Since the observer

cannot move backwards on these timelike paths, they must move forward along the

paths, but all possible paths lead to the center of mass. As such, once the observer

(whether it be light or any other object) has reached the event horizon they are

forever trapped in the black hole because they have no timelike paths available to

them that lead outside the black hole.

Thus, the concept of escape velocity in Newtonian gravity only gets us half-way

in understanding why an object cannot escape the gravitational field of a black hole.

The inherent contradiction of an object traveling at the speed of light not being to

escape a black hole is only answered through General Relativity with the help of

timelike paths, spacelike paths and their curvatures.

3. The Expanding Universe

For years scientists have wondered about the ultimate fate of the universe. With

the help of General Relativity, they came one step closer to answering that question.

Upon observing other galaxies in the universe, physicists discovered that their light

is redshifted [5]. That is, the wavelenghts of the light coming from those galaxies

are bigger than they are in nearby galaxies, indicating that they are moving away

from us. Now, this could merely mean that the galaxies were moviing away from us,

we were the center of the universe and the universe was not expanding. However,

Hubble’s Law, given by,

(2.1) V = H0d,


gives light to the idea that all galaxies are moving away from each other at the same

speed. In Hubble’s law, H0 is Hubble’s constant, and V is the velocity of the receding

galaxy, which is related to the shift in wavelength ∆λλ

by the Doppler relation,

(2.2)V

c=

∆λ

λ≡ z,

where z is the cosmological redshift. In Equation (2.1), it is the observation that H0

is constant that vindicates that our universe is expanding.

Hubble’s constant can be given by the relation,

(2.3) H0 ≡ H(t0) ≡a(t0)

a(t0),

where a(t) is the scale factor that represents the relative expansion of the universe

and it is also a function of time. As such, specifically, a(t0) is the scale factor at the

time the light signal is recieved from another galaxy. Observation has shown that

regardless of the distance between the two galaxies, a(t0)a(t0)

is a constant of 72±7 (km/s)Mpc

.

This essentially means that the change in the scale factor during a given time divided

by the scale factor at the time of reception is always constant, indicating that the

universe is expanding uniformally. Moreover, a relation that combines Equations

(2.2) and (2.3), gives further insight into the expansion of the universe:

(2.4) z ≡ ∆λ

λ=

a(te)

a(t0).


Simplifying,

z =∆λ

λ(2.5)

z =λ0 − λe

λe

(2.6)

z =λ0

λe

− 1(2.7)

1 + z =λ0

λe

.(2.8)

This simplification allows us to reform Equation (2.4) as,

(2.9) 1 + z ≡ λ0

λe

=a(te)

a(t0).

The observation that light coming from other galaxies is redshifted implies that wave-

lengths are becoming longer; λ0

λewill be greater than one. This then suggests that

a(te)a(t0)

> 1, which allows us to infer that the relative expansion of the universe at

the time of reception, a(t0), is larger than the relative expansion of the unvierse at

the time of emission, a(te). In less mathematical terms, the the distance between

the galaxies grows larger over time, which vindicates the theory that our universe is

expanding.

Going a bit deeper, we can understand the different possibilities of spacetime

geometries for our universe. While the metric,

(2.10) ds2 = −dt2 + a2(t)[

dr2 + r2(dθ2 + sin2 θdφ2)]

,

given by physicists Friedmann, Robertson and Walker is for a flat universe, a more

general version of this metric is given by:

(2.11) ds2 = −dt2 + a2(t)

[

dr2

1 − kr2+ r2(dθ2 + sin2 θdφ2)

]

.


The general FRW metric gives room for different spacetime gemoetries for our uni-

verse: closed, open or flat. Figure 2.7 shows visually what we mean by closed, open

or flat, respectively. The spacetime curvature of the universe is positive, negative or

Figure 2.7. Different possibilities of the spacetime curvature of our universe [7].

flat (or equal to zero) if the universe is closed, open or flat, respectively. However,

regardless of these three different possible curvatures of the universe, the spacetime

geometry still remains homogenous and isotropic such that the curvature is the same

at one point in the universe as it is in another. In the general FRW metric, closed,

open and flat universes can be represented by k = +1,−1, 0 [5]. The physicists Fried-

mann, Robertson and Walker described the evolution of our universe with respect to

the relative expansion of the universe in time, given by a(t).

CHAPTER 3

Mathematical Tools of General Relativity

For us to be able to calculate the curvatures of structures such as black holes

and the expanding universe, we need to build a foundation the mathematics re-

quired behind calculating their curvature. This chapter’s aim is to introduce various

mathematical concepts required to work with Einstein’s Field Equations, particularly

Einstein’s curvature tensor.

1. Indices and Summation Convention

Indicies are a more convenient, short-hand way of writing the components of a

vector. Say we have the following vector,

(3.1) r = r0e0 + r1e1 + r2e2,

which has three components. These components can be written in a simpler form

using indices,

(3.2) r = rµeµ.

To show that Equation (3.2) is equivalent to Equation (3.1), we introduce the sum-

mation convention, which states that repeated upper and lower indicies are to be

summed over in any given expression. Using this defintion, we obtain the following

representation of Equation (3.2),

(3.3) r =2∑

µ=0

rµeµ,

13

2. DEFINING METRICS AND TENSORS 14

which expands back into Equation (3.1).

2. Defining Metrics and Tensors

All of the work done with Einstein’s Field Equation starts with analyzation of

metrics. A metric generally provides information required to calculate distance be-

tween two points, which is given by ds2, defined as:

(3.4) ds2 = gµνdxµdxν .

In the Equation (3.4), the two indices µ and ν represent the components of the metric.

Writing metrics in matrix form make the concept of two different indicies clearer. So,

for example, if we are talking about flat spacetime, these indicies will be dependent

upon the components x, y, z and t (time). We can write the metric in matrix form

by making use of the summation convention to obtain the matrix:

(3.5) gµν =

gtt gtx gty gtz

gxt gxx gxy gxz

gyt gyx gyy gyz

gzt gzx gzy gzz

.

We are now in a position to provide an example of a spacetime metric: flat

spacetime is represented by the Minkowski Metric. The Minkowski Metric has four

components x, y, z and time t:

(3.6) ds2 = −(cdt)2 + dx2 + dy2 + dz2.

3. VECTORS IN CURVED SPACETIME AND GRADIENTS 15

Looking at the setup in Equation (3.5), we can write this metric in matrix form as:

(3.7) gµν =

−c 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

.

Because this metric only has the components, gtt, gxx, gyy and gzz, it is a diagonal

metric.

A tensor contains information about various vectors on a geometric space. It

creates linear maps from pairs of vectors to real numbers. More specifically, the rank

of the tensor, specified by the number of indicies, maps that number of vectors into

real numbers. So, a tensor Jαβ is a rank two tensor, and will map two vectors into real

numbers. Some tensors may also be written in matrix form, very much like metrics.

So, if a tensor Jαβ has the vector components t, r, θ and φ, it could be written in

matrix form using the summation convention, as shown:

(3.8) Jαβ =

Jtt Jtr Jtθ Jtφ

Jrt Jrr Jrθ Jrφ

Jθt Jθr Jθθ Jθφ

Jφt Jφr Jφθ Jφφ

.

We now have basic defintions of metrics and tensors. We will now elaborate

on some of the mathematical techniques required to manipulate these metrics and

tensors in order to calculate Einstein’s curvature tensor.

3. Vectors in Curved Spacetime and Gradients

An understanding of vectors in curved spacetime and gradients is a crucial in

obtain the Einstein curvature tensor. Consider a directional derivative of a function,

where f(xα) represents the function and its curve is represented by xα(σ). Using this


notation, the directional derivative is thus given by:

(3.9)df

dσ=

∂xα

∂σ

∂f

∂xα.

The vector tangent to the curve xα(σ) is given by:

(3.10) tα =∂xα

∂σ.

Visually, the tangential vector could be shown roughly by Figure 3.1: Given the

Curved Geometry

Tangential Vector

Figure 3.1. Rough representation of a vector tangent to a curved surface.

previous two definitions of the directional derivative and the tangent to the cuvrve,

allows us to re-write the definition of the directional derivative as:

(3.11)d

dσ= tα

∂

∂xα.

More generally, we write for any vector a the corresponding directional derivative as:

(3.12) a = aα ∂

∂xα.


As such, every directional derivative has a corresponding tangential vector to the

curve and visa-versa. The setup that we have worked out above will allow us to

introduce both the gradient much easier in curved spacetime.

Looking back the function in curved spacetime that we introduced earlier, f(xα),

the derivatives of that function formulate the gradient of that function. More specif-

ically, the linear map from the tangential vector t to real numbers given by the

deriviatives of f(xα) is known as the gradient of a function, often denoted as ∇f .

Finding the gradient of a function involves taking the partial derivative of the given

function with respect to each of the variables. For example, if we’re given a function,

f(xα) = ct2 + 3x2 − y2 + z2, it’s gradient must then be given by

(3.13) ∇f(t, x, y, z) =

(

∂f

∂t,∂f

∂x,∂f

∂y,∂f

∂z

)

.

Using the expression above, the gradient would then be worked out as

(3.14) ∇f = (2ct, 6x,−2y, 2z).

With this definition of a gradient, we are ready to apply the mathematics we have

introduced so far to calculating a key component of the Einstein curvature tensor.

The affine connection, represented by the Christoffel Symbols, is calculated with

the understanding of metrics, vectors in curved spacetime, partial derivatives and

gradients. The Chrisfoffel Symbol is written as Γ, and is calculated with the following

formula:

(3.15) gαδΓδβα =

1

2

(

∂gαβ

∂xγ+

∂gαγ

∂xβ− ∂gβγ

∂xα

)

.

Here, each index, α, β and γ will range from 0 to 3, encompassing each component

of the metric that we are working with. As such, this expression for the Christoffel

Symbols essentially asks us to take the gradient of the entire metric, taking partial


derivatives of each coefficient of the metric with respect to the components of the

metric.

We have laid the ground work for understanding the mathematics that will be

applied in the next chapter, which discusses how to calculate the spacetime curvature

of different geometries. At the moment, the connections between these different

concepts may not be completely clear, but the reader will soon notice how metrics,

tensors and gradients work together in computing Einstein’s curvature tensor.

CHAPTER 4

The Einstein Curvature Tensor

1. Introduction to Curvature Tensors

Our goal is to understand how two entities, (a) non-rotating spherically symetric

blackholes and (b) an expanding universe, create curvature in spacetime. On the

bigger scale of things, it is the Einstein curvature tensor that describes this curvature

in spacetime; the tensor is given by

(4.1) Gµν = Rµν −1

2gµνR,

where Gµν is the Einstein curvature tensor; the part of the equation that contains

a description of the curvature in spacetime [5]. However, to gain a further under-

standing of this tensor, we will build it from scratch and describe each tensor and its

function.

There exist three components that go form the Einstein curvature tensor: Rµν ,

gµν and R. Out of the three, the more straight-forward one to explain is gµν . This the

metric of the structure we considering; the Einstein curvature tensor, Gµν , describes

the curvature in spacetime created by the object represented by the metric, gµν . For

example, say we would like to see the curvature in spacetime created by a spherically

symetric black hole without angular moment (an example we will look at more in

depth in the next section), we would use the Schwarzschild metric, which describes

the gravitational field around such an object. So, for this example, the gµν component

19

1. INTRODUCTION TO CURVATURE TENSORS 20

of the Einstein curvature tensor would be written as

(4.2) gµν = −(

1 − 2M

r

)

dt2 +

(

1

1 − 2Mr

)

dr2 + r2dθ2 + r2sin2θdφ,

or in matrix form,

(4.3) gµν =

−(

1 − 2Mr

)

0 0 0

0(

11− 2M

r

)

0 0

0 0 r2 0

0 0 0 r2sin2θ

.

The Ricci curvature tensor, Rµν , roughly speaking, describes the volume of the

object in consideration relative to the number of dimensions of a manifold. The Ricci

curvature tensor is a contracted version of the Riemann curvature tensor. Mathe-

matically, the contraction is defined as:

(4.4) Rµν = Rαµαν ,

where Rαµαν is the Riemman curvature tensor, which is a measurement of the intrinsic

curvature of a geometry; a measurement that is not transformed away by changing

coordinates [8]. The Riemann curvature tensor is given by

(4.5) Rαµαν = Γα

µν,α − Γαµα,ν + Γα

βαΓβµν − Γα

βνΓβµα.

Using the definition of the Riemann curvature tensor, we may apply the summation

convention to contract the Riemann tensor to form the Ricci tensor. A contraction

involves summing over the the repeated upper and lower indices:

(4.6) Rµν = Rαµαν = R0

µ0ν + R1µ1ν + R2

µ2ν + R3µ3ν .

2. THE SCHWARZSCHILD METRIC 21

The last component of the Einsten curvature tensor is the Ricci scalar (or scalar

curvature) given by R. The scalar curvature is, well, as it says, a scalar. It represents

the curvature of the geometry we are considering by a single real number. This scalar

curvature is given by the following mathematical definition:

(4.7) R = gµνRµν .

Using the summation convention, sum over the indicies µ and ν:

(4.8) R = g00R00 + g01R01 + g02R02 + ... + g13R13 + g32R32.

This summation will result in a scalar, which will be a real number.

Recapitulating: we take the metric of the geometry we are considering, find its

Christoffel symbols, compute the Riemann curvature tensor, and through that ten-

sor, we find both the Ricci curvature tensor and the Ricci scalar. Taking all these

components and putting them together yields the Einstein curvature tensor!

(4.9) Gµν = Rµν −1

2gµνR.

We now have the tools to take the metric of a geometric structure and see how it

curves spacetime. In the next sections, we will provide a few examples of how to

actually carry out the calculations to find the Einstein curvature tensor.

2. The Schwarzschild Metric

The first example we will consider is the Schwarzschild metric. The Schwarzschild

metric representats the gravitational field around a symetrically spherical object with

not angular moment (has no rotation). An example would be a non-rotating black

hole. The Einstein curvature tensor shall be used to determine the spacetime cur-

vature of objects that may be represented by the Schwarzschild metric. Begin by


writing out the metric in its basic form:

(4.10) ds2 = −(

1 − 2M

r

)

dt2 +

(

1

1 − 2Mr

)

dr2 + r2dθ2 + r2sin2θdφ.

However, for convenience, it would be prudent to write this metric in matrix form,

(4.11) ds2 =

−(

1 − 2Mr

)

0 0 0

0(

11− 2M

r

)

0 0

0 0 r2 0

0 0 0 r2sin2θ

.

Reproducing the Einstein curvature tensor from the previous section below:

(4.12) Gµν = Rµν −1

2gµνR.

The process of computing the Einsten tensor will involve finding the unknown com-

ponents of this equation. We already know that gµν is the metric we are working

with, the Schwarzschild metric. So, the first logical step would be to compute the

Ricci curvature tensor Rµν . Computing the Ricci tensor, however, is not trivial; it

involves finding the Christoffel Symbols Γ for our metric. In fact, there are a total of

64 Christoffel Symbols because Γ because λ, µ and ν each represent four components

of the matrix: t, r, θ and φ. So, the full set of Christoffel Symbols must contain a

combination of all the possibilities of these coordinates. And, because each λ, µ and

ν represent four coordinates, the total number of possiblities is 4 × 4 × 4 = 43 = 64.

We shall now begin finding the sixty-four Christoffel Symbols, begining with Γ000.

For convenience, let us reproduce the expression or finding the Christoffel Symbols:


1

2

(

∂gαβ

∂xγ+

∂gαγ

∂xβ− ∂gβγ

∂xα

)

.


So, if we are trying to find the Christoffel Symbol Γ000, we must plug in 0 for all the

coordinates, except for δ, as such:

(4.14) g0δΓδ00 =

1

2

(

∂g00

∂x0+

∂g00

∂x0− ∂g00

∂x0

)

,

where we must solve for Γ000. To do so, the right side of the equation must be solved

first. The RHS of the equation has three partial differentials, each with a gµν where

the µ and ν represent the coordinates of the cell in the metric we are working with.

So, for instance, if we have g22, we would be referring to the 2-2 cell in the metric,

namely the metric component r2. For example, ∂g00

∂x0 is essentially∂( 2M

r−1)

∂t. Now, we

try to find Γ000:

g0δΓδ00 =

1

2

(

∂g00

∂x0+

∂g00

∂x0− ∂g00

∂x0

)

(4.15)

=1

2

(

∂(

2Mr

− 1)

∂t+

∂(

2Mr

− 1)

∂t−

∂(

2Mr

− 1)

∂t

)

(4.16)

=1

2(0 + 0 − 0)(4.17)

= 0.(4.18)

So, now are left with:

(4.19) g0δΓδ00 = 0.

And, to find Γ000, we sum over all values of δ, as so:

(4.20) g00Γ000 + g01Γ

100 + g02Γ

200 + g03Γ

300 = 0.


Refering to Equation (4.11), we can see that g01, g02 and g03 all equal zero except for

g00. So, we are left with the expression:

(4.21) g00Γ000 = 0.

And solving for Γ000, gives us:

(4.22) Γ000 = 0.

Thus, we have our first Christoffel Symbol, Γ000. All this work must be repeated

to find each Christoffel Symbol in the set for this metric. However, one interesting

property of Christoffel Symbols cuts down some of our work, which is:

(4.23) Γδβγ = Γδ

γβ.

We will now similarly move on to finding Γ100, for which we will set α = 1 and

both β and γ to zero.

g1δΓδ00 =

1

2

(

∂g10

∂x0+

∂g10

∂x0− ∂g00

∂x1

)

(4.24)

=1

2

(

0 + 0 −∂(

2Mr

− 1)

∂r

)

(4.25)

=1

2(−(−2M/r2))(4.26)

=M

r2.(4.27)

(4.28)


Then, we sum over all values of δ:

g10Γ000 + g11Γ

100 + g12Γ

200 + g13Γ

300 =

m

r2(4.29)

g11Γ100 =

M

r2(4.30)

(

1

1 − 2Mr

)

Γ100 =

M

r2(4.31)

Γ100 =

M

r2

(

1 − 2M

r

)

.(4.32)

(4.33)

Now that we know the method for finding the Christoffel Symbols, we will show

the working for the values of Γ that have non-zero values. So, for Γ010:

g0δΓδ10 =

1

2

(

∂g01

∂x0+

∂g00

∂x1− ∂g10

∂x0

)

(4.34)

=1

2

(

0 +∂(

2Mr

− 1)

∂r− 0

)

(4.35)

=1

2(−2M/r2)(4.36)

= −M

r2.(4.37)

(4.38)

Then, we sum over all values of δ:

g00Γ010 + g01Γ

110 + g02Γ

220 + g03Γ

330 = −M

r2(4.39)

g10Γ010 = −M

r2(4.40)

−(

1 − 2M

r

)

Γ010 = −M

r2(4.41)

Γ010 = − M

(2M − r)r.(4.42)

(4.43)


Now that the working for three Christoffel Symbols has been shown, we will simply

list all the values of Γ for the Schwarzschild Metric (Note the property mentioned in

Equation (4.23):

Γ010 = Γ0

01 = − M

(2M − r)r(4.44)

Γ100 =

M(r − 2M)

r3(4.45)

Γ111 =

M

(2M − r)r(4.46)

Γ122 = 2M − r(4.47)

Γ133 = (2M − r) sin2 θ(4.48)

Γ221 = Γ2

12 =1

r(4.49)

Γ233 = − cos θ sin2 θ(4.50)

Γ331 = Γ3

13 =1

r(4.51)

Γ332 = Γ3

23 = cot θ(4.52)

With the Christoffel Symbols found, the next step in solving Einstein’s Equation

would be to find the Ricci Coefficient Rµν . Let us reproducing the Ricci curvature

tensor for convenience:

(4.53) Rµν = Rαµαν = Γα

µν,α − Γαµα,ν + Γα

βαΓβµν − Γα

βνΓβµα.

Where, Rµν is the Ricci Tensor and Rαµαν is the Reimann Tensor. Let us demonstrate

how to obtain one of the non-zero Riemann Coefficients. Say we wanted to find R0330;

we would write Equation (4.53) as follows:

(4.54) R0330 = Γ0

30,3 − Γ033,0 + Γ0

β0Γβ30 − Γ0

β0Γβ33.


With the equation for the Riemann Coefficient in the discrete terms, we can refer to

the values of Γ to see if there are any we can cancel out right away. The first term

Γ030,3 maybe be canceled because Γ0

30 = 0. Similarly, Γ033,0 may also be taken out, as

Γ033 = 0. So, now we’re left with the following expression:

(4.55) R30 = R0330 = Γ0

β0Γβ30 − Γ0

β0Γβ33.

The next step is to expand Γ0β0Γ

β30 −Γ0

β0Γβ33. Let us begin with the first term Γ0

β0Γβ30:

Γ0β0Γ

β30 = Γ0

03Γ030 + Γ0

13Γ130 + Γ0

23Γ230 + Γ0

33Γ330(4.56)

Γ0β0Γ

β30 = (0)(0) + (0)(0) + (0)(0) + (0)(0)(4.57)

Γ0β0Γ

β30 = 0.(4.58)

Moving on to the second term Γ0β0Γ

β33:

Γ0β0Γ

β33 = Γ0

00Γ033 + Γ0

10Γ133 + Γ0

20Γ233 + Γ0

30Γ333(4.59)

Γ0β0Γ

β30 = 0 + Γ0

10Γ133 + 0 + 0(4.60)

Γ0β0Γ

β30 = −

(

M

2Mr − r2

)

[

(2M − r) sin2 θ]

(4.61)

Γ0β0Γ

β30 = −M sin2 θ

r.(4.62)

Plugging in the components that we have found of the Ricci Coefficient:

R0330 = 0 −

(

−M sin2 θ

r

)

(4.63)

⇔ R0330 =

M sin2 θ

r.(4.64)

With one example demonstrate, let us proceed to list all the Riemann Coefficients.

However, we must note before hand that only half the amount of Riemann Coefficients


will be listed because of the following property:

(4.65) Rαµαν = −Rα

µνα.

The list of non-zero Riemann Coefficient is as follows:

R0110 =

2M

(2M − r)r2(4.66)

R0220 =

M

r(4.67)

R0330 =

M sin2 θ

r(4.68)

R1010 =

2M(2M − r)

r4(4.69)

R1221 =

M

r(4.70)

R1331 =

M sin2 θ

r(4.71)

R2020 =

M(−2M + r)

r4(4.72)

R2121 =

M

(2M − r)r2(4.73)

R2332 = −2M sin2 θ

r(4.74)

R3030 =

M(−2M + r)

r4(4.75)

R3131 =

M

(2M − r)r2(4.76)

R3232 =

2M

r.(4.77)

We now have all the coefficients of the Riemann tensor. Using these coefficients,

we are able to derive the Ricci tensor, which would put us very far ahead in finding

a solution to the Einstein tensor. The Ricci tensor is a contraction of the Riemann

tensor:

Rµν = Rαµαν .


Using the summation convention, we obtain:

(4.78) Rµν = Rαµαν = R0

µ0ν + R1µ1ν + R2

µ2ν + R3µ3ν .

Let us attempt to find the first coefficient of the Ricci tensor, R00:

Rµν = Rαµαν(4.79)

R00 = R0000 + R1

010 + R2020 + R3

030(4.80)

R00 = 0 +2M(2M − r)

r4+

M(−2M + r)

r4+

M(−2M + r)

r4(4.81)

R00 =2M(2M − r)

r4+

2M(−2M + r)

r4(4.82)

R00 = 0.(4.83)

Now that we have one of the Ricci coefficients, let us try another one. We shall find

R01:

Rµν = Rαµαν(4.84)

R01 = R0001 + R1

011 + R2021 + R3

031(4.85)

R01 = 0 + 0 + 0 + 0(4.86)

R01 = 0.(4.87)

The Schwarzschild metric, so far, has two zero Ricci coefficients. In fact, all of the

Ricci coefficients for this metric are zero. As such, the Ricci tensor is zero as well.

Putting this into mathematical terms:

(4.88) Rµν = 0.


We are almost done! The next step in solving the Einstein tensor would be to

calculate the Ricci curvature scalar, R. This scalar is given by the following formula:

(4.89) R = gµνRµν .

Visually, it would be more effective to write the above-found Ricci tensor in Equation

(4.88) in matrix form:

(4.90) Rµν =

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

.

From the matrix representation and the definition of the Ricci curvature scalar, it

is evident that the curvature scalar will be zero. However, for the sake of providing

an example of how the calcualtion is done, we will demonstrate the working behind

finding the curvature scalar. To continue demonstrating the working, we need to

write the inverse metric; the inverse of the Schwarzschild metric is gµν , given by:

(4.91) gµν =

r2M−r

0 0 0

0 1 − 2Mr

0 0

0 0 1r2 0

0 0 0 csc2(θ)r2

.


Using the summation convention with the Ricci curvature scalar, the matrix repre-

sentations makes it easier to visually see the value of each coefficient in the expansion:

R = gµνRµν

R = g00R00 + g01R01 + g02R02 + g03R03 +

g11R11 + g21R21 + g31R31 + g22R22 +

g23R23 + g33R33 + g10R10 + g20R20 +

g30R30 + g12R12 + g13R13 + g32R32

R = 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0

R = 0

Because all values of the Ricci coefficients were zero, each term in the above sum

became zero. As such, the value of the Ricci curvature scalar is zero. We now finally

have all the components necessary to solve the Einstein tensor. We have the Ricci

tensor Rµν , the metric we are working with gµν (the Schwarzschild metric in this

case), and finally, the Ricci curvature scalar R. Having obtained all these values, we

will write once again the Einstein curvature tensor, and proceed to solving it:

Gµν = Rµν −1

2gµνR(4.92)

Gµν = 0 − 1

2gµν(0)(4.93)

Gµν = 0.(4.94)

Our long journey has finally come to an end! We have found the value of the Einstein

tensor, which in this case, is zero.

3. THE ROBERTSON-WALKER METRIC 32

3. The Robertson-Walker Metric

Now that the ground-work for solving the Einsten Curvature Tensor has been laid

out, we shall now pursue solving it for the FRW Metric:

(4.95) ds2 = −dt2 + a2(t)

[

dr2

1 − kr2+ r2(dθ2 + sin2 θdφ2)

]

.

Re-writing the metric in matrix form:

(4.96) ds2 =

−1 0 0 0

0 a2(t)1−kr2 0 0

0 0 r2a2(t) 0

0 0 0 r2a2(t)sin2θ

.

The Einstein curvature tensor once again is given by the following expression:

(4.97) Gµν = Rµν −1

2gµνR.

The first step in solving this tensor would be to find the Christoffel Symbols Γ; the

formula for which is once again given by:


1

2

(

∂gαβ

∂xγ+

∂gαγ

∂xβ− ∂gβγ

∂xα

)

.

The first Christoffel Symbol we will find is Γ011:

g0δΓδ11 =

1

2

(

∂g01

∂x1+

∂g01

∂x1− ∂g11

∂x0

)

(4.99)

=1

2

(

− 2aa

1 − kr2

)

(4.100)

= − aa

1 − kr2.(4.101)


The next step would be to sum over all values of δ:

g00Γ011 + g01Γ

111 + g02Γ

211 + g03Γ

311 = − aa

1 − kr2(4.102)

g00Γ011 = − aa

1 − kr2(4.103)

(−1)Γ011 = − aa

1 − kr2(4.104)

Γ011 =

aa

1 − kr2.(4.105)

One of the non-zero Christoffel Symbols has been found. We will now demonstrate

the working for two more non-zero values of Γ. Moving on to Γ022:

g0δΓδ22 =

1

2

(

∂g02

∂x2+

∂g02

∂x2− ∂g22

∂x0

)

(4.106)

=1

2

(

− ∂

∂tr2a2(t)

)

(4.107)

=1

2(−2r2aa)(4.108)

= −r2aa.(4.109)

Once again, equating this value and summing over all values of δ:

g00Γ022 + g01Γ

122 + g02Γ

222 + g03Γ

322 = −r2aa(4.110)

g00Γ022 = −r2aa(4.111)

(−1)Γ022 = −r2aa(4.112)

Γ022 = r2aa.(4.113)


We now have two values of Γ, and will proceed to show the working for one last

value, Γ033:

g0δΓδ33 =

1

2

(

∂g03

∂x3+

∂g03

∂x3− ∂g33

∂x0

)

(4.114)

=1

2

(

− ∂

∂tr2a2(t) sin2 θ

)

(4.115)

=1

2(−2r2aa sin2 θ)(4.116)

= −r2aa sin2 θ.(4.117)

Summing over all values of δ:

g00Γ033 + g01Γ

133 + g02Γ

233 + g03Γ

333 = −r2aa sin2 θ(4.118)

g00Γ033 = −r2aa sin2 θ(4.119)

(−1)Γ033 = −r2aa sin2 θ(4.120)

Γ033 = r2aa sin2 θ.(4.121)

Hence, the third Christoffel Symbol is Γ033 = r2aa sin2 θ.

The working behind the first three Christoffel Symbols has now been demon-

strated. With the working in mind, we will proceed to list all the non-zero Christoffel


Symbols for the FRW metric:

Γ011 =

aa

1 − kr2(4.122)

Γ022 = r2aa(4.123)

Γ033 = r2aa sin2 θ(4.124)

Γ110 = Γ1

01 =a

a(4.125)

Γ111 =

kr

1 − kr2(4.126)

Γ122 = r(kr2 − 1)(4.127)

Γ133 = r(kr2 − 1) sin2 θ(4.128)

Γ220 = Γ2

02 =a

a(4.129)

Γ221 = Γ2

12 =1

r(4.130)

Γ233 = − cos θ sin θ(4.131)

Γ330 = Γ3

03 =a

a(4.132)

Γ331 = Γ3

13 =1

r(4.133)

Γ332 = Γ3

23 = cot θ.(4.134)

We now have all the Christoffel Symbols for the FRW metric, and we can proceed to

calculating the other components of the Einstein curvature tensor.

The next step would be obtain the coefficients of the Riemann curvature tensor.

Let us work one example of a non-zero Riemann coefficient, R1010:

R1010 = Γ1

00,1 + Γ101,0 + Γ1

β1Γβ00 − Γ1

β0Γβ01(4.135)

R1010 = 0 +

a

a+ 0 + 0(4.136)

R1010 =

a

a.(4.137)


With one example shown, a list of all the non-zero Riemann Coefficients is given

below, but we must bear in mind that these are not all the coefficients. The identity

Rαµαν = −Rα

µνα implies that there will in fact be double the amount of Riemann

coefficients displayed below.

R0110 =

aa

kr2 − 1(4.138)

R0220 = −r2aa(4.139)

R0330 = −r2aa sin2 θ(4.140)

R1010 = − a

a(4.141)

R1221 = −r2(k + a2)(4.142)

R1331 = −r2 sin2 θ(k + a2)(4.143)

R2020 = − a

a(4.144)

R2121 =

k + a2

1 − kr2(4.145)

R2332 = −r2 sin2 θ(k + a2)(4.146)

R3030 = − a

a(4.147)

R3131 =

k + a2

1 − kr2(4.148)

R3232 = r2(k + a2)(4.149)

With a list of all the non-zero Riemann coefficients, much of the hard work has

been done. The next step in obtaining Einstein’s curvature is to compute the Ricci

curvature tensor.

Unlike the Schwarzschild metric, the FRW metric does not have an all-zero Ricci

curvature tensor; it is a defined diagonal tensor. Let us calcuate the values of the


four components of the diagonal Ricci tensor, R00, R11, R22 and R33:

R00 = R0000 + R1

010 + R2020 + R3

030(4.150)

= − a

a− a

a− a

a(4.151)

= −3a

a(4.152)

R11 = R0101 + R1

111 + R2121 + R3

131(4.153)

=aa

−(kr2 − 1)+

k + a2

1 − kr2+

k + a2

1 − kr2(4.154)

=2k + 2a2 + aa

1 − kr2(4.155)

R22 = R0202 + R1

212 + R2222 + R3

232(4.156)

= r2aa + r2(k + a2) + r2(k + a2)(4.157)

= r2(2(k + a2) + aa)(4.158)

R33 = R0303 + R1

313 + R2323 + R3

333(4.159)

= r2aa sin2(θ) + r2 sin2(θ)(k + a2) + r2 sin2(θ)(k + a2)(4.160)

= r2 sin2(θ)(2(k + a2) + aa).(4.161)

The Ricci curvature tensor is thus:

(4.162) Rµν =

−3aa

0 0 0

0 2k+2a2+aa1−kr2 0 0

0 0 r2 [2(k + a2) + aa] 0

0 0 0 r2 sin2(θ) [2(k + a2) + aa]

.

We are almost done finding all the components of the Einstein cuvature tensor. The

last one that remains is the curvature scalar, R.


Recall the definition of the curvature scalar, given by:

(4.163) R = gµνRµν .

We must sum over both µ and ν, but observing that the Ricci tensor is diagonal,

allows us to eliminate all the zero components of this summation. This leaves us with

the following expression:

(4.164) R = gµνRµν = g00R00 + g11R11 + g22R22 + g33R33.

The inverse metric gµν is:

(4.165) gµν =

−1 0 0 0

0 1−kr2

a(t)20 0

0 0 1r2a(t)2

0

0 0 0 csc2(θ)r2a(t)2

.

As such, the scalar curvature is computed as follows:

R = g00R00 + g11R11 + g22R22 + g33R33(4.166)

=6(k + a2 + aa)

a2.(4.167)


We now have all the components necessary to solve the Einstein curvature tensor!

Let us begin with 12gµν :

1

2gµν =

1

2

−1 0 0 0

0 a2(t)1−kr2 0 0

0 0 r2a2(t) 0

0 0 0 r2a2(t)sin2θ

(4.168)

=

−1/2 0 0 0

0 a2(t)2(1−kr2)

0 0

0 0 12r2a2(t) 0

0 0 0 12r2a2(t)sin2θ

.(4.169)

Multiplying Equation (4.169) by R, we obtain:

(4.170)

−3(k+a2+aa)a2 0 0 0

0 3(k+a2+aa)1−kr2 0 0

0 0 3r2(k + a2 + aa) 0

0 0 0 3r2 sin2(θ)(k + a2 + aa)

.

Lastly, we subract Tensor (4.170) from Rµν , resulting in the Einstein Curvature Ten-

sor:

(4.171)

Gµν =

3(k+a2)a2 a 0 0 0

0 k+a2+2aakr2

−10 0

0 0 −r2(k + a2 + 2aa) 0

0 0 0 −r2 sin2(θ)(k + a2 + 2aa)

.


To simplify this tensor, we can use the following orthonormal basis:

(et)α = [1, 0, 0, 0]

(er)α = [0,

√1 − kr2, 0, 0]

(eθ)α = [0, 0,

1

r, 0]

(eφ)α = [0, 0, 0,

1

r sin θ]

By using the above orthonormal basis, we get the following expressions for the Ein-

stein curvature tensor:

(4.172) Gtt =3(k + a2)

a2

(4.173) Grr = Gθθ = Gφφ = −(k + a2 + 2aa)

a2.

The above given expressions thus make up the Einstein curvature tensor for the FRW

metric!

Given the Einstein curvature tensor, we can use it to derive the Friedmann Equa-

tions, from which the FRW metric was created. The full Einstein Field Equation is

given by:

(4.174) Gµν = 8πGTµν ,

where Tµν is the stress-energy tensor of a manifold and G is Newton’s gravitational

constant. To derive the freedman equations, we begin with the stress-energy tensor


for a simple manifold, a perfect fluid:

(4.175) Tµν =

ρ 0 0 0

0 p 0 0

0 0 p 0

0 0 0 p

[5].

Solving the right-hand side of Equation (4.174), yields following results in geometrized

units (where G = 1):

(4.176) Tµν =

8πρ 0 0 0

0 8πp 0 0

0 0 8πp 0

0 0 0 8πp

.

With the right had side of Einstein’s Field Equation Solved, we equate it to Einstein’s

Curvature Tensor, yielding the following equations:

(4.177) Gtt =3(k + a2)

a2= 8πρ

(4.178) Grr = Gθθ = Gφφ = −(k + a2 + 2aa)

a2= 8πp.

The Friedmann Equations are:

(4.179) a2 − 8πρ

3a2 = −k [5].

(4.180)a

a= −4π

3(3p + ρ) [9].


Let us start with algebraically manipulating Equation (4.177):

3

a2(k + a2) = 8πρ(4.181)

3(k + a2) = 8πρa2(4.182)

3k + 3a2 = 8πρa2(4.183)

3a2 − 8πρa2 = −3k(4.184)

3a2 − 8πρa2

3= −k(4.185)

a2 − 8πρ

3a2 = −k.(4.186)

Hence, shown, because our result is equal to the first Friedmann Equation given in

Equation (4.179). Now, let us algabraically manipulate the second equation to see if

it yields the second Friedmann Equation:

−(k + a2 + 2aa)

a2= 8πp(4.187)

−2a

a− a2

a2− k

a2= 8πp,(4.188)

and using Equation (4.178) to simplify,

−2a

a− 8πρ

3= 8πp(4.189)

−6a − 8πρa

3a= 8πp(4.190)

−2a

a= 8πp +

8πρ

3(4.191)

a

a= −4π(p + ρ/3)(4.192)

a

a= −4π

3(3p + ρ)(4.193)


Manipulation of the second Einstein equation has yielded Equation (4.180), satisfying

our initial condition that both Einstein field equations should be able to lead to

Friedmann’s equations.

Bibliography

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[3] Soshichi Uchii. Embedding diagram, 2001. http://www.bun.kyoto-u.ac.jp/-

suchii/embed.diag.html.

[4] P. K. Townsend. Black holes, 1997. Lecture Notes.

[5] James B. Hartle. Gravity: An Introduction to Einstein’s General Relativity. Addison Wesley,

CA, 2003.

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general relativity and einstein's field equations

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