General Relativity and Solutions to
Einstein’s Field Equations
Abhishek Kumar
Department of Physics and Astronomy, Bates College, Lewiston,
ME 04240
General Relativity and
Solutions 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, Professor 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 supportive throughout my efforts. My family helped mitigate the frustration that sometimes 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 Newtonian 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 Newtonian 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 =
Gm1 m2
,
2
r12
2
where m1 and m2 are the masses of the first and second mass, respectively, r12
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 timedependence. 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 fabric 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 phenomena 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 bent
by the intervening mass on its way to the observer.
Matter traveling through curved spacetime will follow a path defined by that curvature. 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 gravitational 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 diameter, 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.
Far Away
Horizon
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
2
T0
-2
-4
-6
-4
-6
-6
-2
0
-4
2
-2
4
0
X
6
2
Y
8
4
6
10
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
0
Flange
T0
Flange
X
Y
0
+20
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 = H0 d,
3. THE EXPANDING UNIVERSE
10
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
(2.2)
∆λ
λ
by the Doppler relation,
V
∆λ
=
≡ z,
c
λ
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 ) ≡
ȧ(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,
ȧ(t0 )
a(t0 )
is a constant of 72 ± 7 (km/s)
.
M pc
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
a(te )
=
.
λ
a(t0 )
3. THE EXPANDING UNIVERSE
11
Simplifying,
∆λ
λ
λ0 − λe
z =
λe
λ0
z =
−1
λe
λ0
.
1+z =
λe
(2.5)
z =
(2.6)
(2.7)
(2.8)
This simplification allows us to reform Equation (2.4) as,
(2.9)
1+z ≡
λ0
a(te )
.
=
λe
a(t0 )
The observation that light coming from other galaxies is redshifted implies that wavelengths are becoming longer;
a(te )
a(t0 )
λ0
λe
will be greater than one. This then suggests that
> 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)
dr2
2
2
2
2
+ r (dθ + sin θdφ ) .
ds = −dt + a (t)
1 − kr2
2
2
2
3. THE EXPANDING UNIVERSE
12
The general FRW metric gives room for different spacetime gemoetries for our universe: 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 Friedmann, 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 required 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 = r 0 e0 + r 1 e1 + r 2 e2 ,
which has three components. These components can be written in a simpler form
using indices,
(3.2)
r = r µ eµ .
To show that Equation (3.2) is equivalent to Equation (3.1), we introduce the summation 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
X
µ=0
13
r µ eµ ,
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 between two points, which is given by ds2 , defined as:
ds2 = gµν dxµ dxν .
(3.4)
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
gtt gtx
gxt gxx
(3.5)
gµν =
g
yt gyx
gzt gzx
to obtain the matrix:
gty gtz
gxy gxz
.
gyy gyz
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 + dy 2 + dz 2 .
3. VECTORS IN CURVED SPACETIME AND GRADIENTS
Looking at the setup in Equation (3.5), we
−c
0
(3.7)
gµν =
0
0
15
can write this metric in matrix form as:
0 0 0
1 0 0
.
0 1 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:
Jt̂t̂ Jt̂r̂ Jt̂θ̂ Jt̂φ̂
J
J
J
J
r̂t̂ r̂r̂
r̂ θ̂
r̂φ̂
(3.8)
Jαβ =
.
J
θ̂t̂ Jθ̂r̂ Jθ̂θ̂ Jθ̂φ̂
Jφ̂t̂ Jφ̂r̂ Jφ̂θ̂ Jφ̂φ̂
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
3. VECTORS IN CURVED SPACETIME AND GRADIENTS
16
notation, the directional derivative is thus given by:
∂xα ∂f
df
=
.
dσ
∂σ ∂xα
(3.9)
The vector tangent to the curve xα (σ) is given by:
tα =
(3.10)
∂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
= tα α .
dσ
∂x
More generally, we write for any vector a the corresponding directional derivative as:
(3.12)
a = aα
∂
.
∂xα
3. VECTORS IN CURVED SPACETIME AND GRADIENTS
17
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 specifically, 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 − y 2 + z 2 , it’s gradient must then be given by
(3.13)
∇f (t, x, y, z) =
∂f ∂f ∂f ∂f
,
,
,
∂t ∂x ∂y ∂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)
gαδ Γ
δ
βα
1
=
2
∂gβγ
∂gαβ ∂gαγ
+
−
γ
β
∂x
∂x
∂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
3. VECTORS IN CURVED SPACETIME AND GRADIENTS
18
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)
1
Gµν = Rµν − gµν R,
2
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 understanding 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
!
1
2M
2
dr2 + r2 dθ2 + r2 sin2 θdφ,
dt +
(4.2)
gµν = − 1 −
2M
r
1− r
or in matrix form,
(4.3)
gµν
− 1 −
0
=
0
0
2M
r
0
1
1− 2M
r
0
0
0
0
0
0
r2
0
0
r2 sin2 θ
.
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. Mathematically, the contraction is defined as:
(4.4)
α
Rµν = Rµαν
,
α
where Rµαν
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)
α
Rµαν
= Γαµν,α − Γαµα,ν + Γαβα Γβµν − Γαβν Γβµα .
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)
α
0
1
2
3
Rµν = Rµαν
= Rµ0ν
+ Rµ1ν
+ Rµ2ν
+ Rµ3ν
.
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 = g µν Rµν .
Using the summation convention, sum over the indicies µ and ν:
(4.8)
R = g 00 R00 + g 01 R01 + g 02 R02 + ... + g 13 R13 + g 32 R32 .
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 tensor, 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)
1
Gµν = Rµν − gµν R.
2
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 curvature of objects that may be represented by the Schwarzschild metric. Begin by
2. THE SCHWARZSCHILD METRIC
22
writing out the metric in its basic form:
(4.10)
2M
ds2 = − 1 −
r
dt2 +
1
1 − 2M
r
!
dr2 + r2 dθ2 + r2 sin2 θdφ.
However, for convenience, it would be prudent to write this metric in matrix form,
2M
0
0
0
− 1 − r
1
0
0
0
2M
1− r
2
(4.11)
ds =
.
2
0
0
r
0
0
0
0 r2 sin2 θ
Reproducing the Einstein curvature tensor from the previous section below:
1
Gµν = Rµν − gµν R.
2
(4.12)
The process of computing the Einsten tensor will involve finding the unknown components 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 Γ0 00 .
For convenience, let us reproduce the expression or finding the Christoffel Symbols:
(4.13)
gαδ Γ
δ
βα
1
=
2
∂gβγ
∂gαβ ∂gαγ
+
−
γ
β
∂x
∂x
∂xα
.
2. THE SCHWARZSCHILD METRIC
23
So, if we are trying to find the Christoffel Symbol Γ0 00 , we must plug in 0 for all the
coordinates, except for δ, as such:
(4.14)
g0δ Γ
δ
00
1
=
2
∂g00 ∂g00 ∂g00
+
−
∂x0
∂x0
∂x0
,
where we must solve for Γ0 00 . 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,
−1)
∂ ( 2M
r
00
namely the metric component r2 . For example, ∂g
is
essentially
. Now, we
0
∂x
∂t
try to find Γ0 00 :
(4.15)
g0δ Γ
δ
00
1
=
2
(4.16)
1
=
2
(4.17)
=
(4.18)
= 0.
∂g00 ∂g00 ∂g00
+
−
∂x0
∂x0
∂x0
2M
−
1
−
1
∂ 2M
∂
∂
r
r
+
−
∂t
∂t
!
−1
∂t
2M
r
1
(0 + 0 − 0)
2
So, now are left with:
(4.19)
g0δ Γδ 00 = 0.
And, to find Γ0 00 , we sum over all values of δ, as so:
(4.20)
g00 Γ0 00 + g01 Γ1 00 + g02 Γ2 00 + g03 Γ3 00 = 0.
2. THE SCHWARZSCHILD METRIC
24
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:
g00 Γ0 00 = 0.
(4.21)
And solving for Γ0 00 , gives us:
Γ0 00 = 0.
(4.22)
Thus, we have our first Christoffel Symbol, Γ0 00 . 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:
Γδ βγ = Γδ γβ .
(4.23)
We will now similarly move on to finding Γ1 00 , for which we will set α = 1 and
both β and γ to zero.
(4.24)
(4.25)
(4.26)
(4.27)
(4.28)
g1δ Γ
δ
00
1
=
2
=
1
2
∂g10 ∂g10 ∂g00
+
−
∂x0
∂x0
∂x1
!
∂ 2M
−
1
r
0+0−
∂r
1
(−(−2M/r2 ))
2
M
= 2.
r
=
2. THE SCHWARZSCHILD METRIC
25
Then, we sum over all values of δ:
(4.29)
m
r2
M
= 2
r
g10 Γ0 00 + g11 Γ1 00 + g12 Γ2 00 + g13 Γ3 00 =
(4.30)
1
1 − 2M
r
(4.31)
g11 Γ1 00
!
Γ1 00 =
Γ
(4.32)
1
00
M
r2
M
= 2
r
2M
1−
r
.
(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 Γ0 10 :
(4.34)
g0δ Γ
δ
10
1
=
2
(4.35)
=
(4.36)
=
(4.37)
1
2
∂g01 ∂g00 ∂g10
+
−
∂x0
∂x1
∂x0
!
∂ 2M
−
1
r
0+
−0
∂r
1
(−2M/r2 )
2
M
= − 2.
r
(4.38)
Then, we sum over all values of δ:
(4.39)
(4.40)
(4.41)
(4.42)
(4.43)
M
r2
M
= − 2
r
M
= − 2
r
g00 Γ0 10 + g01 Γ1 10 + g02 Γ2 20 + g03 Γ3 30 = −
g10 Γ0 10
2M
Γ0 10
− 1−
r
Γ0 10 = −
M
.
(2M − r)r
2. THE SCHWARZSCHILD METRIC
26
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):
Γ0 10 = Γ0 01 = −
(4.44)
M
(2M − r)r
M (r − 2M )
r3
M
=
(2M − r)r
(4.45)
Γ1 00 =
(4.46)
Γ1 11
(4.47)
Γ1 22 = 2M − r
(4.48)
Γ1 33 = (2M − r) sin2 θ
Γ2 21 = Γ2 12 =
(4.49)
1
r
Γ2 33 = − cos θ sin2 θ
(4.50)
1
r
(4.51)
Γ3 31 = Γ3 13 =
(4.52)
Γ3 32 = Γ3 23 = cot θ
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)
α
Rµν = Rµαν
= Γαµν,α − Γαµα,ν + Γαβα Γβµν − Γαβν Γβµα .
α
Where, Rµν is the Ricci Tensor and Rµαν
is the Reimann Tensor. Let us demonstrate
0
how to obtain one of the non-zero Riemann Coefficients. Say we wanted to find R330
;
we would write Equation (4.53) as follows:
(4.54)
0
R330
= Γ030,3 − Γ033,0 + Γ0β0 Γβ30 − Γ0β0 Γβ33 .
2. THE SCHWARZSCHILD METRIC
27
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 Γ030 = 0. Similarly, Γ033,0 may also be taken out, as
Γ033 = 0. So, now we’re left with the following expression:
(4.55)
0
R30 = R330
= Γ0β0 Γβ30 − Γ0β0 Γβ33 .
The next step is to expand Γ0β0 Γβ30 − Γ0β0 Γβ33 . Let us begin with the first term Γ0β0 Γβ30 :
(4.56)
Γ0β0 Γβ30 = Γ003 Γ030 + Γ013 Γ130 + Γ023 Γ230 + Γ033 Γ330
(4.57)
Γ0β0 Γβ30 = (0)(0) + (0)(0) + (0)(0) + (0)(0)
(4.58)
Γ0β0 Γβ30 = 0.
Moving on to the second term Γ0β0 Γβ33 :
(4.59)
Γ0β0 Γβ33 = Γ000 Γ033 + Γ010 Γ133 + Γ020 Γ233 + Γ030 Γ333
(4.60)
Γ0β0 Γβ30 = 0 + Γ010 Γ133 + 0 + 0
M
β
0
Γβ0 Γ30 = −
(2M − r) sin2 θ
2
2M r − r
(4.61)
(4.62)
Γ0β0 Γβ30 = −
M sin2 θ
.
r
Plugging in the components that we have found of the Ricci Coefficient:
(4.63)
0
R330
M sin2 θ
= 0− −
r
(4.64)
0
R330
M sin2 θ
=
.
r
⇔
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
2. THE SCHWARZSCHILD METRIC
28
will be listed because of the following property:
α
α
.
Rµαν
= −Rµνα
(4.65)
The list of non-zero Riemann Coefficient is as follows:
2M
(2M − r)r2
M
r
M sin2 θ
r
2M (2M − r)
r4
M
r
M sin2 θ
r
M (−2M + r)
r4
M
(2M − r)r2
(4.66)
0
R110
=
(4.67)
0
R220
=
(4.68)
0
R330
=
(4.69)
1
R010
=
(4.70)
1
R221
=
(4.71)
1
R331
=
(4.72)
2
R020
=
(4.73)
2
R121
=
(4.74)
2
R332
= −
(4.75)
3
R030
(4.76)
3
R131
(4.77)
3
R232
2M sin2 θ
r
M (−2M + r)
=
r4
M
=
(2M − r)r2
2M
=
.
r
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:
α
Rµν = Rµαν
.
2. THE SCHWARZSCHILD METRIC
29
Using the summation convention, we obtain:
1
2
3
α
0
+ Rµ1ν
+ Rµ2ν
+ Rµ3ν
.
Rµν = Rµαν
= Rµ0ν
(4.78)
Let us attempt to find the first coefficient of the Ricci tensor, R00 :
(4.79)
α
Rµν = Rµαν
(4.80)
2
3
1
0
+ R020
+ R030
R00 = R000
+ R010
(4.81)
R00 = 0 +
(4.82)
R00
(4.83)
R00 = 0.
2M (2M − r) M (−2M + r) M (−2M + r)
+
+
r4
r4
r4
2M (2M − r) 2M (−2M + r)
=
+
r4
r4
Now that we have one of the Ricci coefficients, let us try another one. We shall find
R01 :
(4.84)
α
Rµν = Rµαν
(4.85)
3
2
0
1
+ R031
R01 = R001
+ R011
+ R021
(4.86)
R01 = 0 + 0 + 0 + 0
(4.87)
R01 = 0.
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.
2. THE SCHWARZSCHILD METRIC
30
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:
R = g µν Rµν .
(4.89)
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 g µν , given by:
r
0
0
0
2M −r
2M
0
0
0
1
−
r
(4.91)
g µν =
.
0
1
0
0
r2
2
0
0
0 cscr2(θ)
2. THE SCHWARZSCHILD METRIC
31
Using the summation convention with the Ricci curvature scalar, the matrix representations makes it easier to visually see the value of each coefficient in the expansion:
R = g µν Rµν
R = g 00 R00 + g 01 R01 + g 02 R02 + g 03 R03 +
g 11 R11 + g 21 R21 + g 31 R31 + g 22 R22 +
g 23 R23 + g 33 R33 + g 10 R10 + g 20 R20 +
g 30 R30 + g 12 R12 + g 13 R13 + g 32 R32
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:
(4.92)
(4.93)
(4.94)
1
Gµν = Rµν − gµν R
2
1
Gµν = 0 − gµν (0)
2
Gµν = 0.
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)
dr2
2
2
2
2
+ r (dθ + sin θdφ ) .
ds = −dt + a (t)
1 − kr2
2
2
2
Re-writing the metric in matrix form:
0
0
0
−1
a2 (t)
0
0
0
1−kr 2
2
(4.96)
ds =
.
0
0
r2 a2 (t)
0
2 2
2
0
0
0
r a (t)sin θ
The Einstein curvature tensor once again is given by the following expression:
1
Gµν = Rµν − gµν R.
2
(4.97)
The first step in solving this tensor would be to find the Christoffel Symbols Γ; the
formula for which is once again given by:
(4.98)
gαδ Γ
δ
βα
1
=
2
∂gβγ
∂gαβ ∂gαγ
+
−
γ
β
∂x
∂x
∂xα
The first Christoffel Symbol we will find is Γ011 :
(4.99)
(4.100)
(4.101)
g0δ Γδ11
1
=
2
∂g01 ∂g01 ∂g11
+
−
∂x1
∂x1
∂x0
1
2aȧ
=
−
2
1 − kr2
aȧ
.
= −
1 − kr2
.
3. THE ROBERTSON-WALKER METRIC
33
The next step would be to sum over all values of δ:
(4.102)
aȧ
1 − kr2
aȧ
= −
1 − kr2
aȧ
= −
1 − kr2
aȧ
=
.
1 − kr2
g00 Γ011 + g01 Γ111 + g02 Γ211 + g03 Γ311 = −
(4.103)
g00 Γ011
(4.104)
(−1)Γ011
(4.105)
Γ011
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 :
(4.106)
(4.107)
(4.108)
(4.109)
g0δ Γδ22
∂g02 ∂g02 ∂g22
+
−
∂x2
∂x2
∂x0
1
∂
=
− r2 a2 (t)
2
∂t
1
(−2r2 aȧ)
=
2
1
=
2
= −r2 aȧ.
Once again, equating this value and summing over all values of δ:
(4.110)
g00 Γ022 + g01 Γ122 + g02 Γ222 + g03 Γ322 = −r2 aȧ
(4.111)
g00 Γ022 = −r2 aȧ
(4.112)
(−1)Γ022 = −r2 aȧ
(4.113)
Γ022 = r2 aȧ.
3. THE ROBERTSON-WALKER METRIC
34
We now have two values of Γ, and will proceed to show the working for one last
value, Γ033 :
(4.114)
g0δ Γδ33
(4.115)
(4.116)
1
=
2
∂g03 ∂g03 ∂g33
+
−
∂x3
∂x3
∂x0
∂ 2 2
1
2
− r a (t) sin θ
=
2
∂t
1
=
(−2r2 aȧ sin2 θ)
2
= −r2 aȧ sin2 θ.
(4.117)
Summing over all values of δ:
(4.118)
g00 Γ033 + g01 Γ133 + g02 Γ233 + g03 Γ333 = −r2 aȧ sin2 θ
(4.119)
g00 Γ033 = −r2 aȧ sin2 θ
(4.120)
(−1)Γ033 = −r2 aȧ sin2 θ
(4.121)
Γ033 = r2 aȧ sin2 θ.
Hence, the third Christoffel Symbol is Γ033 = r2 aȧ sin2 θ.
The working behind the first three Christoffel Symbols has now been demonstrated. With the working in mind, we will proceed to list all the non-zero Christoffel
3. THE ROBERTSON-WALKER METRIC
35
Symbols for the FRW metric:
aȧ
1 − kr2
(4.122)
Γ011 =
(4.123)
Γ022 = r2 aȧ
(4.124)
Γ033 = r2 aȧ sin2 θ
ȧ
a
(4.125)
Γ110 = Γ101 =
(4.126)
Γ111 =
(4.127)
Γ122 = r(kr2 − 1)
(4.128)
Γ133 = r(kr2 − 1) sin2 θ
kr
1 − kr2
ȧ
a
1
=
r
(4.129)
Γ220 = Γ202 =
(4.130)
Γ221 = Γ212
Γ233 = − cos θ sin θ
(4.131)
ȧ
a
1
=
r
(4.132)
Γ330 = Γ303 =
(4.133)
Γ331 = Γ313
(4.134)
Γ332 = Γ323 = cot θ.
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.
1
Let us work one example of a non-zero Riemann coefficient, R010
:
(4.135)
1
R010
= Γ100,1 + Γ101,0 + Γ1β1 Γβ00 − Γ1β0 Γβ01
(4.136)
1
R010
= 0+
(4.137)
1
R010
=
ä
.
a
ä
+0+0
a
3. THE ROBERTSON-WALKER METRIC
36
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
α
α
Rµαν
= −Rµνα
implies that there will in fact be double the amount of Riemann
coefficients displayed below.
aä
−1
(4.138)
0
R110
=
(4.139)
0
R220
= −r2 aä
(4.140)
0
R330
= −r2 aä sin2 θ
(4.141)
1
R010
= −
(4.142)
1
R221
= −r2 (k + ȧ2 )
(4.143)
1
R331
= −r2 sin2 θ(k + ȧ2 )
(4.144)
2
R020
= −
(4.145)
2
R121
(4.146)
2
R332
= −r2 sin2 θ(k + ȧ2 )
(4.147)
3
R030
= −
(4.148)
3
R131
(4.149)
3
R232
= r2 (k + ȧ2 )
kr2
ä
a
ä
a
k + ȧ2
=
1 − kr2
ä
a
k + ȧ2
=
1 − kr2
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
3. THE ROBERTSON-WALKER METRIC
37
four components of the diagonal Ricci tensor, R00 , R11 , R22 and R33 :
(4.150)
(4.151)
(4.152)
(4.153)
2
3
0
1
+ R020
+ R030
R00 = R000
+ R010
ä ä ä
= − − −
a a a
3ä
= −
a
0
1
2
3
R11 = R101
+ R111
+ R121
+ R131
(4.154)
=
(4.155)
=
(4.156)
k + ȧ2
k + ȧ2
aä
+
+
−(kr2 − 1) 1 − kr2 1 − kr2
2k + 2ȧ2 + aä
1 − kr2
2
3
0
1
+ R232
R22 = R202
+ R212
+ R222
(4.157)
= r2 aä + r2 (k + ȧ2 ) + r2 (k + ȧ2 )
(4.158)
= r2 (2(k + ȧ2 ) + aä)
(4.159)
0
1
2
3
R33 = R303
+ R313
+ R323
+ R333
(4.160)
= r2 aä sin2 (θ) + r2 sin2 (θ)(k + ȧ2 ) + r2 sin2 (θ)(k + ȧ2 )
(4.161)
= r2 sin2 (θ)(2(k + ȧ2 ) + aä).
The Ricci curvature tensor is thus:
3ä
0
0
0
− a
2
2k+2ȧ +aä
0
0
0
2
1−kr
(4.162) Rµν =
.
0
2
2
0
r [2(k + ȧ ) + aä]
0
0
0
0
r2 sin2 (θ) [2(k + ȧ2 ) + aä]
We are almost done finding all the components of the Einstein cuvature tensor. The
last one that remains is the curvature scalar, R.
3. THE ROBERTSON-WALKER METRIC
38
Recall the definition of the curvature scalar, given by:
R = g µν Rµν .
(4.163)
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 = g µν Rµν = g 00 R00 + g 11 R11 + g 22 R22 + g 33 R33 .
The inverse metric g µν is:
(4.165)
g
µν
−1
0
=
0
0
0
0
1−kr 2
a(t)2
0
0
1
r 2 a(t)2
0
0
csc (θ)
r 2 a(t)2
As such, the scalar curvature is computed as follows:
(4.166)
(4.167)
0
0
.
0
2
R = g 00 R00 + g 11 R11 + g 22 R22 + g 33 R33
=
6(k + ȧ2 + aä)
.
a2
3. THE ROBERTSON-WALKER METRIC
39
We now have all the components necessary to solve the Einstein curvature tensor!
Let us begin with 12 gµν :
1
gµν
2
(4.168)
(4.169)
−1
1
0
=
2 0
0
−1/2
0
=
0
0
0
0
0
a2 (t)
1−kr 2
0
0
0
r2 a2 (t)
0
0
0
r2 a2 (t)sin2 θ
0
0
0
a2 (t)
2(1−kr 2 )
0
0
0
1 2 2
r a (t)
2
0
0
0
1 2 2
r a (t)sin2 θ
2
.
Multiplying Equation (4.169) by R, we obtain:
3(k+ȧ2 +aä)
0
0
0
a2
−
2
3(k+ȧ +aä)
0
0
0
1−kr 2
(4.170)
.
2
2
0
0
3r
(k
+
ȧ
+
aä)
0
2
2
2
0
0
0
3r sin (θ)(k + ȧ + aä)
Lastly, we subract Tensor (4.170) from Rµν , resulting in the Einstein Curvature Tensor:
(4.171)
Gµν
3(k+ȧ2 )
a2 a
=
0
0
k+ȧ2 +2aä
0
0
0
0
kr 2 −1
0
0
0
0
.
2
2
−r (k + ȧ + 2aä)
0
2
2
2
0
−r sin (θ)(k + ȧ + 2aä)
3. THE ROBERTSON-WALKER METRIC
40
To simplify this tensor, we can use the following orthonormal basis:
(et̂ )α = [1, 0, 0, 0]
(er̂ )α = [0,
√
1 − kr2 , 0, 0]
1
(eθ̂ )α = [0, 0, , 0]
r
(eφ̂ )α = [0, 0, 0,
1
]
r sin θ
By using the above orthonormal basis, we get the following expressions for the Einstein curvature tensor:
(4.172)
(4.173)
Gt̂t̂ =
3(k + ȧ2 )
a2
Gr̂r̂ = Gθ̂θ̂ = Gφ̂φ̂ = −
(k + ȧ2 + 2aä)
.
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 Equations, 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
3. THE ROBERTSON-WALKER METRIC
for a simple manifold, a perfect fluid:
(4.175)
Tµν
ρ
0
=
0
0
0 0
p 0
0 p
0 0
41
0
0
[5].
0
p
Solving the right-hand side of Equation (4.174), yields following results in geometrized
units (where G = 1):
(4.176)
Tµν
0
0
8πρ 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)
(4.178)
Gt̂t̂ =
3(k + ȧ2 )
= 8πρ
a2
Gr̂r̂ = Gθ̂θ̂ = Gφ̂φ̂ = −
(k + ȧ2 + 2aä)
= 8πp.
a2
The Friedmann Equations are:
8πρ 2
a = −k [5].
3
(4.179)
ȧ2 −
(4.180)
ä
4π
= − (3p + ρ) [9].
a
3
3. THE ROBERTSON-WALKER METRIC
42
Let us start with algebraically manipulating Equation (4.177):
3
(k + ȧ2 ) = 8πρ
2
a
(4.181)
(4.182)
3(k + ȧ2 ) = 8πρa2
(4.183)
3k + 3ȧ2 = 8πρa2
(4.184)
(4.185)
(4.186)
3ȧ2 − 8πρa2 = −3k
3ȧ2 − 8πρa2
= −k
3
8πρ 2
a = −k.
ȧ2 −
3
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:
(4.187)
(4.188)
(k + ȧ2 + 2aä)
= 8πp
a2
k
ä ȧ2
−2 − 2 − 2 = 8πp,
a a
a
−
and using Equation (4.178) to simplify,
(4.189)
(4.190)
(4.191)
(4.192)
(4.193)
ä 8πρ
−2 −
a
3
6ä − 8πρa
−
3a
ä
−2
a
ä
ȧ
ä
ȧ
= 8πp
= 8πp
= 8πp +
8πρ
3
= −4π(p + ρ/3)
= −
4π
(3p + ρ)
3
3. THE ROBERTSON-WALKER METRIC
43
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.
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to einstein’s 1936 science paper. Science, 275, 1997.
[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.
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44