## Search the FAQ Archives

3 - A - B - C - D - E - F - G - H - I - J - K - L - M
N - O - P - Q - R - S - T - U - V - W - X - Y - Z # Relativity and FTL Travel--PART III (optional reading)

( Part1 - Part2 - Part3 - Part4 )
[ Usenet FAQs | Web FAQs | Documents | RFC Index | Airports ]
Archive-name: star-trek/relativity_FTL/part3
Posting-Frequency: bimonthly for r.a.s.tech, monthly for news.answers

`View all headers`
```=============================================================================
Relativity and FTL Travel

by Jason W. Hinson (hinson@physicsguy.com)
-----------------------------------------------------------------------------

Part III: A Bit About General Relativity

=============================================================================
Edition: 5.1
URL: http://www.physicsguy.com/ftl/
FTP (text version): ftp://ftp.cc.umanitoba.ca/startrek/relativity/

This is Part III of the "Relativity and FTL Travel" FAQ. It is an
"optional reading" part of the FAQ in that the FTL discussion in Part IV
does not assume that the reader has read the information discussed below. If
your only interest in this FAQ is the consideration of FTL travel with
relativity in mind, then you may only want to read Part I: Special
Relativity and Part IV: Faster Than Light Travel--Concepts and Their
"Problems".
In this part, we take a look at general relativity. The discussion is
rather lengthy, but I hope you will find it straight forward and easy to
follow. The subject of GR is still new to this FAQ, and your comments on the
usefulness, ease of reading, etc. for this part of the FAQ would be
appreciated.
the FAQ portion which should be distributed with this document.

Contents of Part III:
Chapter 5: Introduction to General Relativity
5.1 Reasoning for its Existence
5.2 The "New Inertial Frame"
5.3 The Global Break-Down of Special Relativity
5.4 Manifolds, Geodesics, Curvature, and Local Flatness
5.5 The Invariant Interval
5.7 The Metric Tensor and the Stress-Energy Tensor
5.8 Applying these Concepts to Gravity
5.8.1 The Basic Idea
5.8.2 Some Notes on the Physics and the Math
5.8.3 First Example: Back to SR
5.8.4 Second Example: Stars and Black Holes
5.9 Experimental Support for GR

Chapter 5: Introduction to General Relativity

Thus far, we have confined our talks to the realm of what is known as
Special Relativity (or SR). In this section I will introduce a few of the
main concepts in General Relativity (or GR). The difference between the two
is basically that GR deals with how relativity applies to gravitation. As it
turns out, our concept of how gravity works must be changed because of
relativity, and GR explains the new concept of gravity. It is called
"General" relativity because if you look at General Relativity in the case
where there is little or no gravity, you get Special Relativity (SR is a
special case of GR).
Now, GR is a heavily mathematical theory, and while I will try to
simply give the reader some understanding of the physical notions
underlining the theory, some mathematics will inevitably come into play. I
will, however, try to give simple, straight-forward explanations of where
the math comes from and how it helps explain the theory. I will start by
discussing why we might even think that gravity and relativity are related
in the first place. This will lead us to change our concepts of space and
time in the presence of gravity. To discuss this new concept of space-time,
we will need to introduce the idea of mathematical constructs known as
Tensors. The two tensors we will talk about in specific are called the
Metric Tensor and the Stress-Energy Tensor. Once we have discussed these
concepts, we will look at how it all comes together to produce the basic
ideas behind the theory of general relativity. We will also consider a
couple of examples to illustrate the use of the theory. Finally, we will
mention some of the experimental evidence which supports general relativity.

5.1 Reasoning for its Existence

To start off our discussion, I want to indicate why one would reason
that gravity and relativity are connected. While I could start with a
somewhat unrealistic thought experiment to explain the first point I want to
make, perhaps it will be better if I just tell you about actual experimental
evidence to support the point. We thus start by considering an experiment in
which a light beam is emitted from Earth and rises in the atmosphere to some
point where the light is detected. When one performs this experiment, one
finds that the energy of the light decreases as it rises.
So, what does this have to do with our view of relativity and gravity?
Well, let's reason through the situation: First, we note that the energy of
light is related to its frequency. (If you think of light as a wave with
crests and troughs, and if you could make note of the crests and troughs as
they passed you, then you could calculate the frequency of the wave as 1/dt,
where dt is the time between the point when one crest passes you and the
point when the next crest passes you.) So, if the energy of the light
decreases (and thus its frequency decreases), then dt (the time between
crests) must increase. Let's then consider a frame of reference sitting
stationary on the Earth. We will look at a space-time diagram in this frame
which shows the paths that two crests would take as the light travels away
from the Earth.
In Diagram 5-1 I have drawn indications of the paths the two crests
might take. The diagram shows distance above the Earth as distance in the
positive x direction, so as time goes on, the two crests rise (move in the
positive x direction) and eventually meet a detector. Now, we don't know
what the gravity of the Earth might do to the light. We thus want to
generalize our diagram by allowing for the possibility that the paths of the
crests might be influenced in some unknown way by gravity. So, I have drawn
a haphazard path for the two crests marked with question marks. The actual
paths don't matter for our argument, but what does matter is this: whatever
gravity does to the light, it must act the same way on both crests.
Therefore, the two haphazard paths are drawn the same way.

Diagram 5-1

t                        #  = detector's path
|                  #
|                  ?
|               ?  #
|      second ?    # dt-final
|   crest ?        ?
|      ?        ?  #
|   ?         ?    #
| ?       ?        #
?      ? first     #
dt-initial |   ? crest        #
| ?                #
------------?------------------#------> x (distance above surface)
|                  #
|                  #
dt-initial = dt-final

As we see in the diagram, because gravity acts the same way on both
crests, the time between them when they leave the surface (dt-initial) is
the same as the time between them when they are detected (dt-final). Thus,
our diagram does not predict that the energy of the light should change, but
experimental evidence shows it does. According to special relativity, this
frame of reference we have drawn is an inertial frame (that is, if we ignore
the Earth's motion, this frame of reference is stationary--it's just inside
a gravitational field). Thus our diagram (drawn for an inertial frame of
reference) should explain the geometry of the situation, but does not. That
indicates that SR must be changed in light of gravity. However, we have yet
to show that SR must be completely thrown out.
What if there were another way to define an inertial frame such that
its geometry would explain the above situation and other situations which
occur in the presence of a gravitational field? That is what we will
consider next.

5.2 The "New Inertial Frame"

Before starting this section, I want to mention something to the
reader: in the end, when gravity is concerned, we will not be able to find a
single inertial frame of reference which will correctly explain the geometry
of all situations. This will be the actual death-blow to special relativity.
In this section, it will start to look as if the situation is hopeful, and
that by defining a proper inertial frame, SR will be saved. However, in the
next section, we will see where this all falls apart, and I want the reader
to realize this from the beginning.
Now, in the previous section we showed that a space-time diagram drawn
for an inertial frame of reference doesn't explain the way things really are
for a frame of reference sitting stationary on the Earth's surface. If such
a frame cannot be called an inertial frame because of some effect of
gravity, then perhaps there is another way to define an inertial frame of
reference in the presence of gravity.
First, let's consider the properties of a frame which we know to be an
inertial frame without gravity. Consider a space ship sitting far from any
source of gravity. Here we will assume that the ship isn't
accelerating--it's just sitting there in the middle of space. Diagram 5-2
shows such a space ship at different times. Also shown is an observer and a
ball, both of which start out stationary in this frame of reference. Both
the observer and the ball are weightless along with the ship, and as time
goes on neither move with-respect-to the sides of the ship. This is
obviously what we would consider to be an ideal inertial frame of reference.

Diagram 5-2

--------------------- time -------------------->
1             2             3             4
+------+      +------+      +------+      +------+
|      |      |      |      |      |      |      |
|    O |      |    O |      |    O |      |    O |
|      |      |      |      |      |      |      |
| O    |      | O    |      | O    |      | O    |
|/|\   |      |/|\   |      |/|\   |      |/|\   |
| |    |      | |    |      | |    |      | |    |
|/ \   |      |/ \   |      |/ \   |      |/ \   |
|      |      |      |      |      |      |      |
|      |      |      |      |      |      |      |
|      |      |      |      |      |      |      |
|      |      |      |      |      |      |      |
+------+      +------+      +------+      +------+
Ship Floating in Space

Next, consider the same ship, but let it be sitting stationary on the
Earth. Diagram 5-3 shows such a ship at different times, and again there is
an observer and a ball shown as well. Obviously, the observer and the ball
in this case cannot remain stationary with respect to the ship--rather they
must fall in the Earth's gravity and accelerate towards the Earth's surface.
Note that because of the way gravity works, the observer and the ball and
anything else in the ship will accelerate downward at the same rate
regardless of their mass (as long as they are at relatively the same height
above the Earth's surface, and neglecting air resistance). This
distinguishes gravity from all other forces in nature. With the other three
forces (electromagnetism, the strong nuclear force, and the weak nuclear
force) the motion of an object in the presence of the force depends on the
composition of the object. For example, electromagnetism doesn't act on
neutral particles, but does act on charged ones. However, when we consider
gravity, the path taken by an object which is released with a given velocity
in a gravitational field does not depend on the composition of the
object--not even its mass. So, both the ball and the observer in Diagram 5-3
accelerate at the same, constant rate towards the bottom of the ship. In
step 3 on that diagram, the observer hits the bottom of the ship, and in
step 4 the ball reaches the bottom as well. Obviously this situation isn't
like the inertial frame of reference we described above, and the observer in
these two situations could easily tell the difference between the two cases.

Diagram 5-3

--------------------- time -------------------->
1             2             3             4
+------+      +------+      +------+      +------+
|      |   |  |      |      |      |      |      |
|    O | G |  |      |      |      |      |      |
|      | r |  |    O |      |      |      |      |
| O    | a |  |      |      |      |      |      |
|/|\   | v |  | O    |      |      |      |      |
| |    | i |  |/|\   |      |    O |      |      |
|/ \   | t |  | |    |      |      |      |      |
|      | y |  |/ \   |      | O    |      | O    |
|      |  \|/ |      |      |/|\   |      |/|\   |
|      |      |      |      | |    |      | |    |
|      |      |      |      |/ \   |      |/ \ O |
+------+      +------+      +------+      +------+
==========    ==========    ==========    ==========
Earth's       Earth's       Earth's       Earth's
Surface       Surface       Surface       Surface
Ship Sitting on the Earth's Surface

Further, consider the same ship again, this time letting it accelerate
at a constant rate in the middle of space. Diagram 5-4 shows such a ship at
different times (again with an observer and a ball). Note that in the
diagram, the observer and the ball start out at a constant speed (in steps
1, 2, and 3, both move one interval up during each step of time). However,
the acceleration of the ship causes it to move further between steps 2 and 3
than it did between steps 1 and 2, and so on. Therefore, at step 3 the
bottom of the ship meets with the observer, and the observer begins to be
pushed by the ship, accelerating along with the it from then on. This would
cause the observer to feel the force of the ship against him, "holding" him
against the floor. In the final step, the ball meets with the bottom of the
ship, and it too accelerates from then on because the ship is pushing
against it. This case thus looks very much like the case just above where
the ship was sitting on the Earth's surface--in both cases objects in the
ship will seem to accelerate at the same, constant rate towards the bottom
of the ship (regardless of their mass) and once there they will feel a force
against them as they sit on the floor of the ship. The observer in each of
these cases would find it hard to tell which of the two situations he was
really in.

Diagram 5-4

--------------------- time -------------------->
4
+------+
|      |
|      |
|      |
3          |      |
|      |
+------+      |      |
1             2          |      |      |      |
|      |      | O    |
accel                      |      |      |/|\   |
^          +------+      |      |      | |    |
|          |      |      |      |      |/ \ O |
+------+      |      |      |    O |      +------+
|      |      |    O |      |      |       \/  \/
|    O |      |      |      | O    |
|      |      | O    |      |/|\   |
| O    |      |/|\   |      | |    |
|/|\   |      | |    |      |/ \   |
| |    |      |/ \   |      +------+
|/ \   |      |      |       \/  \/
|      |      |      |
|      |      |      |
|      |      +------+
|      |       \/  \/
+------+
\/  \/
Ship Accelerating in Space

Given all three examples above, it seems obvious that a frame sitting
stationary on the Earth is much more like an accelerating frame than it is
like an inertial frame. Seeing that, it now seems perfectly reasonable for
us to find that an experiment performed on the surface of the Earth can't be
explained by a diagram drawn for an inertial frame.
But, can we now find a frame of reference in the presence of gravity
which DOES look like an inertial frame? Well, look back to Diagram 5-4
(where the ship is accelerating in space) and notice the state of the ball
and the observer during the first part of that illustration. Even though the
ship in that case is not an inertial frame because it is accelerating, the
observer and the ball don't begin to accelerate until the bottom of the ship
reaches them and begins to push them. Thus, until that point, the ball and
the observer are not accelerating. They are shown moving at a constant
velocity. Thus, until the bottom of the ship reaches them, the observer and
the ball are inertial observers. AH, but as we have pointed out, this
situation is supposed to be analogous to the one in Diagram 5-3 (where the
ship is sitting stationary on the Earth). If so, then we could argue that
the observer and the ball in the first part of Diagram 5-3 (which are in
free-fall in the Earth's gravitational field) are what we would now call our
inertial observers in the presence of gravity.
So, let's look at one last illustration in which the whole ship is in
free-fall as well as the observer and the ball. Diagram 5-5, shows such a
situation. Notice that the observer, the ball, and the ship all accelerate
at the same rate towards the earth. They each move the same distance during
each step shown. Now, look at just the ship and everything in it at each
step shown. The observer, the ball, and the sides of the ship are not moving
with respect to one another because they are all falling at the same rate.
At each step, the ball and the observer are at the same position inside the
ship. Therefore, until the ship in Diagram 5-5 reaches the surface of the
Earth, the observer wouldn't notice any difference between this situation
and the one in Diagram 5-2 (where the ship is floating in space).

Diagram 5-5

--------------------- time -------------------->
1
+------+         2
|      |      +------+
|    O | G |  |      |
|      | r |  |    O |         3
| O    | a |  |      |      +------+
|/|\   | v |  | O    |      |      |
| |    | i |  |/|\   |      |    O |
|/ \   | t |  | |    |      |      |
|      | y |  |/ \   |      | O    |         4
|      |  \|/ |      |      |/|\   |      +------+
|      |      |      |      | |    |      |      |
|      |      |      |      |/ \   |      |    O |
+------+      |      |      |      |      |      |
+------+      |      |      | O    |
|      |      |/|\   |
|      |      | |    |
+------+      |/ \   |
|      |
|      |
|      |
|      |
+------+
==========    ==========    ==========    ==========
Earth's       Earth's       Earth's       Earth's
Surface       Surface       Surface       Surface
Ship Falling in Earth's Gravitation

It certainly seems, then, that a frame which is freely falling in the
presence of gravity is actually an inertial frame of reference. As one final
test, let's go back to the experiment mentioned earlier in which light rises
in the presence of Earth's gravity. As it turns out (though I won't go into
the proof) if the light is detected while it is still relatively close to
the Earth, and we consider the experiment in a frame of reference which is
freely falling near the Earth's surface, then in that frame, the light does
not loose energy. Thus, in the freely falling frame of reference, Diagram
5-1 (which depicts an inertial frame of reference) can correctly depict the
geometry of the situation.
And so, things are looking deceptively hopeful. In every case we have
studied, it seems as if we can continue to use special relativity as-is,
even in the presence of gravity, if we simply define "inertial frame" to
mean a frame which is in free fall. Then the space-time diagrams we have
drawn throughout our discussions would work just fine in the presence of
gravity, as long as we understand that they are drawn in free falling
frames. However, as I warned earlier, there is a problem here which we
haven't solved.

5.3 The Global Break-Down of Special Relativity

Now that we have tried to argue that we can continue using special
relativity even when gravity is involved (by appropriately defining a new
inertial frame), we are now in a position to explain where the argument
breaks down.
Consider Diagram 5-6. There we see a ship which is much wider than the
ships we have shown thus far. It is in free fall towards the surface of the
Earth, and there are two observers shown, one at either side of the ship.
Now, according to our argument, both observers are said to be in inertial
frames of reference because they are both in free-fall. However, as they
each fall towards the center of the Earth, because they are at great
distances from one another, they accelerate in different directions as
shown. If one observer looks at the other, he will see that other observer
accelerating towards him. But if they are both supposed to be inertial
observers, then how can they also each be accelerating in the frame of the
other?

Diagram 5-6

+--------------------------------------------------------------+
|                                                              |
|    O                                                    O    |
|     \                                                  /     |
| O    \|                                              |/    O |
|/|\  ""`Acceleration                     Acceleration '""  /|\|
| |      toward center                    toward center      | |
|/ \ \     of Earth                         of Earth      / / \|
|     \|                                                |/     |
|    ""`                                                '""    |
+--------------------------------------------------------------+
_______==========_______
___==="""""                        """""===___
The Earth's Surface
Long Ship Falling in Earth's Gravitation

Also, consider Diagram 5-7 in which there is a ship which is much
taller than the ships we have been considering. Here, two observers are
again shown, one at the bottom of the ship and one at the top. Because the
one near the bottom is much closer to the surface of the Earth, he is
accelerating at a greater rate than the other observer. Again, these two
observers are both supposed to be inertial observers, yet each is
accelerating in the other observer's frame. Further, as the observer on the
top continues to accelerate downward, he will eventually be where the
observer at the bottom is now. Thus, as time passes, he will fall into a
stronger gravitational field, and he will be in a "different" inertial frame
than he his now.

Diagram 5-7

+------+
|    O |
|      |
| O    |
|/|\   |
|  | |    |
smaller\|/ |/ \   |
accel     |      | G |
|      | r |
|      | a |
|      | v |
|      | i |
|      | t |
|      | y |
|      |  \|/
|    O |
|      |
|  | O    |
|  |/|\   |
larger  |  | |    |
accel \|/ |/ \   |
|      |
+------+
==========
Earth's
Surface
Tall Ship Falling in Earth's Gravitation

What does all this say? Well, we have shown that for small distances
and over small amounts of time, a free falling frame has all the properties
we want in an inertial frame when gravity is present. However, in each of
the last two cases above, we have observers who are all free-falling and
thus (by our new definition of an inertial frame in the presence of gravity)
are all supposed to be in inertial frames. Yet, if we draw a space-time
diagram for one of the observers, and extend it so that the other observer
can be drawn on the diagram, that other observer will be accelerating on the
space-time diagram. Therefore, a space-time diagram which well describes an
inertial frame for all of space-time in special relativity can only well
describe an inertial frame of reference over a small distance in space and
time when a general gravitational field is involved.
This is analogous to the situation in which a flat map can well
describe a small, local piece of the curved surface of the Earth (such as a
city). However, globally, as you extend the map, it no longer describes the
curved surface of the Earth.
We therefore find that when gravity is involved, we can still define an
inertial frame of reference LOCALLY (meaning local in both space and time),
but globally, there is no way to define a single, rigid frame of reference
which describes an inertial frame of reference everywhere in space-time.
Therefore, globally we cannot use special relativity to describe space-time
in the presence of a general gravitational field. We must therefore re-think
relativity in the presence of gravity.
What we will find is that gravity is actually caused by a curvature of
space-time, and like the map trying in vain to describe the curved surface
of the Earth, special relativity cannot describe the curved space-time
caused by gravity. It is general relativity which describes curved
space-time, and for us to fully appreciate it, we will need to discuss some
basic ideas used to describe such a geometry.

5.4 Manifolds, Geodesics, Curvature, and Local Flatness

Before we discuss space-time in the presence of gravity, we need to
understand some basic geometric concepts which we will use. We will develop
these concepts by considering normal, spatial geometry which can be fully
grasped using common sense. Applying these concepts to space-time becomes
less intuitive (in part because we still aren't that used to thinking of
time as just another dimension); therefore, developing them using normal
spatial geometry will be beneficial.
First, we introduce the term "manifold". Basically, for our purposes,
you can think of a manifold as a fancy term for a space. The space around us
that you are used to thinking of can be called a three dimensional manifold.
The surface of a sheet of paper is a two dimensional manifold, as is the
surface of a cylinder or the surface of a sphere. Much of our focus on
manifolds will involve discussing their geometry. Understanding the geometry
of a manifold means understanding the relationships between various points
on the manifold and understanding various curves on the manifold as well as
knowing how to measure distances on the manifold. Thus, we want to define a
few specific notions which will help us understand and explain the geometry
of a manifold.
So, next we look at a particular type of path on a manifold called a
geodesic. A geodesic is essentially the path which takes the shortest
distance between two points on the manifold. On a piece of paper (a flat
manifold) the shortest distance between two points is found by following the
path of a straight line. However, for a sphere, the shortest distance
between two points would be traveled by following a curve known as a great
circle. If you imagine cutting a sphere directly in half and then putting it
back together, then the cut mark on the surface of the sphere would be a
great circle. If you move along the surface of a sphere between two points,
then the shortest path you could take would lie on a great circle. Thus, a
great circle on a sphere is basically equivalent to a line on a flat
manifold--they are both geodesics on their respective manifolds. Similarly,
on any other manifold there would be a path to follow between two points
such that you would travel the shortest distance. Such a path is a geodesic
on that manifold.
Next, we introduce the concept of the curvature of a manifold. There
are two different types of curvature: intrinsic and extrinsic. To
demonstrate the difference between the two, let's first consider a surface
which has only extrinsic curvature. Imagine taking a flat sheet of paper and
rolling it as if you were making a cylinder; however, don't let the two ends
touch to complete the cylinder. Now, while this two dimensional surface will
now look curved in our three dimensional perspective, the geometry of the
surface is still the same as the geometry of the flat sheet of paper from
which it was made. If you were a two dimensional creature confined to live
on this two dimensional surface, there would be no test you could perform to
prove you weren't on a flat sheet of paper rather than this cylinder-like
surface. Now if you did complete the cylinder, then a two dimensional
creature could tell that the global topology of the situation has changed
(for example, on a complete cylinder, he could follow a particular path
which would bring him around back to where he started). However, this
doesn't change the fact that throughout the cylinder, the internal geometry
is just like the geometry of a flat sheet of paper from which it was made.
So, for a two-dimensional cylinder, its curvature is only "visible"
when viewed from a higher dimensional space (our three-dimensional space).
We only say it is curved because a line on the 2-D cylinder can bend away
from a straight line in three dimensions. However, The cylinder has no
intrinsic curvature to its geometry, so its curvature is extrinsic.
Contrast this with the surface of a sphere. You cannot bend a flat
sheet of paper around a sphere without crumpling or cutting the paper. The
geometry on the surface of a sphere will then be different from the geometry
of a flat sheet of paper. To distinctly show this, let's consider a couple
of two dimensional creatures who are confined to the surface of a sphere.
Say that they stand facing the same direction at a given, small distance
apart from one another on the two dimensional surface, and then they begin
walking in the same direction parallel to one another. As they continue to
walk beside one another, each will continue in what seems to him to be a
straight line. If they do this--if each of them believes that he is
following a straight line from one step to the next--then each will follow
the path of a geodesic on the sphere. As we said earlier, this means that
they will each follow a great circle. But if they each follow a great circle
on the surface of a sphere, then they will each eventually notice that their
friend walking next to them is moving closer and closer, and eventually they
will meet. Now, they started out moving on parallel paths, and they each
believed that they were walking in a straight line, but their paths
eventually came together. This would not be the case if they performed this
experiment on a flat sheet of paper (or on a cylinder). Thus, creatures who
are confined to live on the two dimensional surface of a sphere could tell
that the geometry of their space was different from the geometry of a flat
piece of paper (even though they couldn't "see" the curvature because they
are trapped in only two dimensions). That intrinsic difference is due to the
intrinsic curvature of the sphere's surface.
This, then, is what we want to note about curvature: There are two
types of curvature, extrinsic and intrinsic. Extrinsic curvature is only
detectable from dimensions higher than the dimension of the manifold being
considered. Intrinsic curvature can be detected and understood even by
creatures who are confined to live within the dimensions of the manifold.
Thus, just because a manifold may looked "curved" in a higher dimension,
that doesn't mean that its intrinsic geometry is different from that of a
flat manifold (i.e. its geometry can still be flat--like the cylinder).
Thus, the test of whether a manifold is intrinsically curved does not have
anything to do with higher dimensions, but with experiments that could be
performed by beings confined on that manifold. (For example, if two parallel
lines do not remain parallel when extended on the manifold, then the
manifold possesses curvature). This is important to us in our discussion of
space-time in the presence of gravity. It means that the curvature of the
four dimensional manifold of space-time in which we live can be understood
without having to worry about or even speculate on the existence of any
other dimensions.
As a final note in this introduction to manifolds, I want to mention a
bit about local flatness. Note that even though a manifold can be curved, on
a small enough portion of that manifold, it will be fairly flat. For
example, we can represent a city on our curved Earth by using a flat map.
The map will be a very good representation of the city because it is a very
small piece of the curved manifold. Earlier I mentioned that over a small
enough piece of space-time in the presence of gravity, you can define a
frame of reference which is still very similar to an inertial reference
frame in special relativity. This gives an indication as to why the geometry
of space-time in special relativity is that of a flat manifold, while with
general relativity, space-time is said to be curved in the presence of
gravity. Still, the space-time of any observer being acted on only by
gravity is LOCALLY flat.
Later we will see how the concepts discussed here will help us in
explaining gravity and relativity. Next, however, we want to discuss another
property of manifolds which itself will tell us everything we want to know
about the geometry of a particular manifold. We will call this property the
invariant interval.

5.5 The Invariant Interval

Here we will basically be discussing distances on manifolds, and what
we can learn about a manifold based on how we calculate distances on that
manifold. We start by discussing the length of a random path on a manifold.
Consider a random path on a flat sheet of paper. We can use an x-y
coordinate system to specify any point on the paper and any point on the
path. With this coordinate system in place, how can we use it to measure the
length of that random path? One way is to break up the path into tiny parts,
each of which can be approximated with a straight line segment. Then, if we
know how to measure the length of a straight line, we can measure the length
of each line segment and add them up to find the approximate length of the
path. Now, since the random path doesn't have to be very straight, the line
segments we use might not be very good at approximating the path at some
point. However, if we break up the path into smaller pieces, then the
smaller line segments should do a better job of approximating the curve and
giving us the correct length for the path. The smaller we make the line
segments, the better our approximation of the path's length will be. The
ultimate result of this idea is to figure out what the calculated length
would be if we made the line segments infinitesimally small. That would give
us the actual length of the curve.
So, the next question is this: How do we calculate the length of a very
small (infinitesimal) line segment using our x-y coordinate system? Well,
each segment is made up of a component in the x direction (dx) and a
component in the y direction (dy) as shown in Diagram 5-8. These components
represent infinitesimal distances. The length of the infinitesimal line
segment (let's call the length ds) is then given by the following (using the
Pythagorean theorem):

(Eq 5:1)
ds^2 = dx^2 + dy^2

(Note that this is the length of a straight line--a geodesic on this
manifold--between an initial and a final position which are separated by a
distance dx in the x direction and dy in the y direction.)

Diagram 5-8

y
|       /.
|      / .
|   ds/  .
|    /   .dy
|   /    .
|  /......
|     dx
----+------------->x
|

This distance between two very-nearby points is what I call the
invariant interval. Why? Well, first I need to note that there are other
types of coordinate systems one could use to locate every point on a flat
surface, and that the equation for ds in terms of small changes in each
coordinate will depend on the coordinate system you use. However, though the
form of the equation will change, the actual distance between two points on
the manifold is a physical reality which won't change. The actual interval
is independent of the coordinate system you place on the manifold.
Now, Below, I will specifically use ds as defined here (in a flat, x-y
coordinate system) to make a comparison with an invariant interval defined
using a particular coordinate system on a curved manifold. However, all the
arguments I will make can also be made using any other coordinate system on
a flat manifold and any other coordinate system on a curved manifold. I
simply use two specific ones as solid examples.
So, to demonstrate how the equation for ds will tell us everything we
want to know about a manifold, we next need to consider a curved manifold.
We will use our old friend the sphere. Let's start by defining a coordinate
system on the sphere. Picture a sphere with a great circle drawn on it.
Let's call that great circle the equator. Next, consider a point on the
equator, and call that point our origin. We want to define two independent
coordinates which will allow us to locate any point on the sphere starting
from the origin (note: by "independent coordinates" I mean that you can
always change your position in one coordinate independent of any change in
the other). So, consider some other point on the sphere (call the point
"P"), and let's explain how to get to that point using two coordinates. We
start by moving either towards the "east" or "west" from our origin in the
general direction of "P" (you can define "east" and "west" however you
wish). We move along the equator until P is directly north or south of us,
and we call the distance we move "L" (L is positive if we move east). Next,
we need to move north or south on the sphere to reach P. The distance we
move north or south to reach P will be called "H" (H is positive if we move
north). That gives us our coordinate system. Every point on the sphere can
now be represented by an L-H coordinate pair. The "grid" on the surface of
the sphere which represents this coordinate system would be made of latitude
and longitude lines such as those on a globe.
Next, we need to figure out what infinitesimal distance (ds) would be
associated with moving a small distance in L (dL) and a small distance in H
(dH). For the sake of time, I'll just give the answer here. (Note, R is the
radius of the sphere we are considering):

(Eq 5:2)
ds^2 = dH^2 + [cos(H/R)]^2*dL^2

Remember what this represents. If you start at some point (L,H) on the
sphere, and you change your L coordinate by a small amount (dL) and your H
coordinate by a small amount (dH) then the shortest distance along the
sphere between your first position and your final position would be ds. Note
that this distance depends on your H position (because of the "cos(H/R)"
part of the equation). This is an interesting point because as soon as you
start moving from one position to the next, the equation for ds becomes
slightly different. We basically think of this difference as negligible as
long as dL is very small, but, in fact, the equation is only correct when dL
is truly "infinitesimal". Such concepts are generally covered in calculus,
and for our purposes, we will just claim that the equation is practically
true as long as dL is very small.
So now we come to an important statement to be made in this section:
THE FORM OF THE INVARIANT INTERVAL FULLY DEFINES THE INTRINSIC GEOMETRY OF A
MANIFOLD. For example, what if we tried to find another coordinate system on
the sphere using two independent coordinates (a and b) such that the
invariant interval on the sphere would be given by the following:

(Eq 5:3)
ds^2 = da^2 + db^2?

Well, because that invariant interval looks just like the formula for ds on
a flat sheet of paper (ds^2 = dx^2 + dy^2), then it should be impossible for
Equation 5:3 to be the invariant interval on the sphere (no matter how we
define "a" and "b"). If I drew a grid on a flat sheet of paper and labeled
the axes "a" and "b", then Equation 5:3 would appropriately describe the
relationship between every single point on that flat manifold given the "a"
and "b" coordinate system. Thus, if I define "a" and "b" to be independent
coordinates on a sphere, and I claimed that Equation 5:3 described the
invariant interval on the sphere given those coordinates, then I'd be saying
that Equation 5:3 describes the relationship between every single point on
the sphere given the "a" and "b" coordinate system. But that's saying that
by appropriately defining "a" and "b", I can make the relationship between
all points on the sphere be just like the relationship between every point
on a flat sheet of paper. We know that physically, this simply can't be
done, because there are intrinsic ways to tell the difference between the
geometry of a sphere and the geometry of a flat sheet of paper.
You might be looking back at Equation 5:2 and thinking, "but what if I
just define a new coordinate, L' such that dL'^2 = cos^2(H/R) dL^2? Then I
get ds^2 = dH^2 + dL'^2, which looks like the invariant interval for a flat
sheet of paper." Ah, but look at your definition for dL' and notice that it
involves your other coordinate, H. You see that H and L' are NOT independent
coordinates. To be valid in our discussion here, the coordinates you use on
a manifold must be independent.
So, Considering this example of a sphere and a flat sheet of paper,
let's make some general points: First, consider some manifold, M1. On M1, we
have some (valid) coordinate system, S1. Next we consider two very-nearby
points on M1 (call the points P and Q). If we know the distance between P
and Q along each of the coordinates (like dx and dy, for example), then we
can find some function for ds (the shortest distance on M1 between the
very-nearby points) using the coordinates in S1. Now, consider a second
manifold, M2. If a (valid) coordinate system, S2, can be defined on that
manifold such that ds has the same functional form in S2 as it did using the
S1 coordinate system on M1, then the geometry of the two manifolds must be
identical.
This indicates that the geometry of a manifold is completely determined
if one knows the form of the invariant interval using a particular
coordinate system on that manifold. In fact, starting with the form of the
invariant interval in some coordinate system on a manifold, we can determine
the curvature of the manifold, the path of a geodesic on the manifold, and
everything we need to know about the manifold's geometry.
Now, the mathematics used to describe these properties involves
geometric constructs known as tensors. In fact, the invariant interval on a
manifold is directly related to a tensor known as the metric tensor on the
manifold, and we will discuss this a bit later. First, I want to give a very
brief introduction to tensors in general.

In this section I will introduce just a few basic ideas which will give
the reader a feeling for what tensors are. This is simply meant to provide a
minimum amount of information to those who do not know about tensors.
Basically, a tensor is a geometrical entity which is identified by its
various components. To give a solid example, I note that a vector is a type
of tensor. In an x-y coordinate system, a vector has one component which
points in the x direction (its x component) and another component which
points in the y direction (its y component). If you consider a vector
defined in three dimensional space, then it will also have a z component as
well. Similarly a tensor in general is defined in a particular space which
has some number of dimensions. The number of dimensions of the space is also
called the number of dimensions of the tensor. Note that vectors have a
component for each individual (one) dimension, and they are called tensors
of rank 1. For other tensors, you have to use two of the dimensions in order
to specify one component of the tensor. In x-y space, such a tensor would
have an xx component, an xy component, a yx component, and a yy component.
In three-space, it would also have components for xz, zx, yz, zy, and zz.
Since you have to specify two of the dimensions for each component of such a
tensor, it is called a tensor of rank 2. Similarly, you can have third rank
tensors (which have components for xxx, xxy, ...), fourth rank tensors, and
so on.
So that you aren't confused, I want to explicitly note that the
dimensionality of a tensor (the number of dimensions of the space in which
the tensor is defined) is independent of the rank of the tensor (the amount
of those dimensions that have to be used to specify each component of the
tensor). In any dimensional space, we can have a tensor of rank 0 (just a
number by itself, because it is not associated in any way with any of the
dimensions), a tensor of rank 1 (like a vector--it has a component for every
one dimension you can specify), a tensor of rank 2 (it has a component for
every pair of dimensions you can specify), etc.
Now we look at a very important property of tensors. In fact, it is the
property which really defines whether a set of components make up a tensor.
This property involves the question of how the tensor's components change
when you change the coordinate system you are using for the space in which
the tensor is defined. So, let's consider an example in two dimensional
space where you go from some coordinate system (call the coordinates x and
y) to some other coordinate system (call these coordinates x' and y'). There
will be some sort of relationship between the two systems. For example, say
we start at some point in this space such that our coordinates are (x,y) and
(x',y') (depending on which coordinate system you are using). Now, say we
move an "infinitesimal distance" in x (using the first coordinate system).
Call that distance dx. When we do so, we may have changed our x' position
(using the second coordinate system) by some infinitesimal amount, dx'.
Also, we may have changed our y' position by some amount dy'. We can use
these concepts of infinitesimal changes to define some relationships between
the two systems. We can answer the question "how does x' change when x
changes at this point" by noting the ratio, dx'/dx. Similarly we can write
dx/dx' to denote how much x changes with changes in x' at some point, and
dy'/dx denotes how y' changes with changes in x.
Please understand that these are not simply ratios of definite numbers.
For example, dx'/dx is not necessarily the inverse of dx/dx' because dx in
one expression is NOT the same as dx in the other. The first expression uses
dx in the following context: "If I hold y constant and change x by an amount
dx, x' and y' might change by amounts dx' and dy'. Take the amount that x'
changes (dx') and divide it by the amount I changed x (dx)." The second
expression uses dx in the following context: "If I hold y' constant and
change x' by an amount dx', x and y might change by amounts dx and dy. Take
the amount that x changes (dx) and divide it by the amount I changed x'
(dx')." You can see that the dx in the former context does not have to be
the same amount as dx in the latter. So, when I write dx'/dx or dx/dx' or
dy/dx' etc, you must understand that the form of these ratios (what's on top
and what's on bottom) defines how they are produced, and they are not just
ratios of definite numbers. (Those who know something of calculus will
obviously recognize these terms as simple partial derivatives, but
anyway....)
Now, all together there are four of these ratios which denote how the
x' and y' coordinates change with changes in x and y:

dx'/dx, dx'/dy, dy'/dx, and dy'/dy.

Similarly, there are four more to denote how x and y change with changes in
x' and y':

dx/dx', dx/dy', dy/dx', and dy/dy'.

In general the values of these ratios can depend on where you are on a
manifold, so each ratio is generally a function of x and y (or x' and y', if
you like).
Now, we have these ratios which help us relate one coordinate system to
another. If we have a tensor defined in this space, then we must be able to
use those ratios to find out how the tensor's components themselves change
when we go from considering them in one coordinate system to considering
them in the other. Let's consider a tensor of rank 1 (a vector) in a two
dimensional space. Let the vector, call it V, have an x component (V_x) and
a y component (V_y). Then, the rules for finding the x' and y' components of
the vector at some point are the following:

(Eq 5:4)
V_x' = dx'/dx V_x + dx'/dy V_y
and
V_y' = dy'/dx V_x + dy'/dy V_y.

That is the way in which this type of first rank tensor must transform
from one coordinate system to another. Note that we can write both equations
in Equation 5:4 by using the following:

(Eq 5:5)
V_a = SUM(b = x,y) [da'/db V_b]

In that expression, "a" can be either x or y (so we actually have two
equations--those in Equation 5:4). Also, the right side of the equation is a
summation where the first term in the summation is found by letting b = x,
and the second term is found by letting b = y. Further, we could make this
expression more general by noting that it will be true for a space with
higher dimensions when we let "a" be any one of those dimensions and let the
sum with b extend over all the dimensions.
The fact that the physical components of a vector do actually transform
this way is what makes the vector a tensor. However, we should note that not
all types of vectors transform this way.
To show this is so, first we will consider a function which has a value
at every point in x-y space. Call the function f(x,y). Such a function is a
0 rank tensor, because at any point in the space, it has some single,
numerical value (it does not have components for x and y like a vector
does--you can't ask "what's its value in the x direction", or "what's its
value in the y direction", because it has only a single number at any
point). Note that if we change to another coordinate system, the value of f
at some physical point in the space will not change. Because it has no x or
y component, it is invariant when you change coordinate systems, as are all
0 rank tensors. This is the way all 0 rank tensors must transform when you
change coordinate systems--they must be invariant.
Now, back to the point that there are other types of vectors which do
not transform as discussed earlier. Let's take the function we were just
discussing, f(x,y), at some point and ask "how does it change with small
changes in x?" If the function changes by an amount df when we move to
another x location a distance dx away, then we can write the expression
df/dx to tell how f changes with x. We can do the same in y and have the
expression df/dy. Then we could define a vector (call it G) which has an x
component (G_x) equal to df/dx at every point in x and y, while it has a y
component (G_y) equal to df/dy at every point. Now, what if we do this same
procedure in the x'-y' coordinate system. First, we need to convert f into a
function f'. We do this such that if a point in our space has coordinates
(x,y) in one coordinate system while the same physical point has coordiantes
(x',y') in the other coordinate system, then we want f(x,y) = f'(x',y').
That way f' is the proper representation of f in the primed coordinate
system. Now we again find a vector, G, and we will end up with the x' and y'
components of the G vector such that G_x' = df'/dx' and G_y' = df'/dy'.
We now want to figure out how to transform G from one frame to another.
First, we will look at G_x' = df'/dx' which says that G_x' comes from
knowing how f' changes with respect to x' (i.e. df'/dx'). To transform this
component of G, we must know how to find df'/dx' using G_x and G_y. This
means we will be using information about how f changes with respect to x and
y (i.e., using df/dx and df/dy). We will also need to use information about
how x and y change with respect to x'. Without taking the time to fully
explain the calculus involved, perhaps the following equation will not be
too surprising:

(Eq 5:6)
df'   df'  dx     df'  dy
--  = -- * --  +  -- * --
dx'   dx   dx'    dy   dx'

Conceptually (though mathematicians would cringe a bit at this
explanation) one can imagine canceling out the dx in df'/dx * dx/dx' and
canceling out the dy in df'/dy * dy/dx' to see that in both parts of that
equation we are looking at information about df'/dx'. In the first case, we
are looking at how f' changes with respect to x' by way of how x changes
with respect to x', while in the second case we are looking at how f'
changes with respect to x' by way of how y changes with respect to x'.
Adding these two components together as we do in the above equation gives us
a full picture of how f' changes with respect to x' given information about
how f' changes with respect to x and y.
We further note that f' and f are actually the same physical function,
we just use the prime to indicate which coordinate system we are primarily
thinking of. Thus f and f' will both change in the same way with respect to
changes in x and y (i.e. df'/dx = df/dx and df'/dy = df/dy. We therefore
rewrite Equation 5:6 as

(Eq 5:7)
df'   df   dx     df   dy
--  = -- * --  +  -- * --
dx'   dx   dx'    dy   dx'

dx          dy
= G_x* --  +  G_y* --
dx'         dx'

Note that we have substituted G_x = df/dx and G_y = df/dy. The above
equation provides the transformation of G_x' given the components of G in
the (x,y) coordinate system. Similarly, we can also find the transformation
of G_y'. In the end, simply because of the way this vector is defined, it
transforms as follows:

(Eq 5:8)
G_x' = dx/dx' G_x + dy/dx' G_y
and
G_y' = dx/dy' G_x + dy/dy' G_y

As before, we can rewrite these two equations as follows:

(Eq 5:9)
G_a' = SUM(b = x, y) [db/da' G_b]

Note that we are using ratios like db/da' rather than da'/db (which we used
earlier). That means that this is a different type of vector (because it
transforms in a different way). The vector we discussed earlier (V) is
called a contravariant vector, and the fact that it transforms as shown in
Equation 5:5 is what defines it as that type of vector. The G vector is
called a covariant vector, and it is defined as such because it transforms
as shown in Equation 5:9. Usually, we express which type of vector we have
by the way we denote its components. For contravariant vectors, we denote
their components by putting their indexes (the x or the y) in superscripts:

x      y
V  and V   (or V^{x} and V^{y}),

While we denote the components of covariant vectors by putting their indices
in subscripts:

G  and G   (or G_x and G_y)
x      y

With this notation, the two different transformations begin to take on
an easy to remember form. See if you can figure out how the "upper" indices
and the "lower" indices match up on both sides of the two transformation
equations when they are written as follows:

(Eq 5:10)
a'               da'  b
V  = SUM(b = x,y) --  V
db

and

(Eq 5:11)
db
G  = SUM(b = x,y) --  G
a'               da'  b

Notice that the superscript (or subscript) on one side remains "upper" (or
"lower") in the ratio on the other side. Also, note that the summation is
always over the index which is repeated on the right side, once in an
"upper" position and once in a "lower" position. This basic "formula" helps
to produce equations for all transformation in tensor analyses (note this in
the next part of this section).
It is interesting to note that in the normal spatial coordinates we are
used to using (Cartesian coordinates), db/da' = da'/db, and there is no
distinction between covariant and contravariant vectors. However, in other
systems, the difference is there and must be considered.
Further, we note that with higher rank tensors, they are also defined
by the way they transform from one coordinate system to another. For
example, consider a second rank tensor, U. It could be that both of its
indices are associated with the contravariant type of transformation (note:
the following actually denotes four equations because a'b' can be set to
x'x', x'y', y'x', or y'y'):

(Eq 5:12)
a'b'    da'  db' xx    da'  db' xy    da'  db' yx    da'  db' yy
U      =  -- * -- U   +  -- * -- U   +  -- * -- U   +  -- * -- U
dx   dx        dx   dy        dy   dx        dy   dy

[ da'  db' ce ]
= SUM(c & e vary over all dimensions) [ -- * -- U   ]
[ dc   de     ]

Or they could both be associated with the covariant type of transformation:

(Eq 5:13)
[ dc   de      ]
U     = SUM(c,e) [ -- * --  U   ]
a'b'           [ da'  db'  ce ]

Or it could be a mix of the two:

(Eq 5:14)
a'               [ da'  de   c  ]
U      = SUM(c,e) [ -- * --  U   ]
b'             [ dc   db'   e ]

Finally, we will see in the next section that any contravariant tensor
also has a covariant form (and vice-versa), and we can transform from one
form to the other if we know the geometry of the manifold on which the
tensors are defined.
And that about ends our introduction to tensors. To sum up, they are
geometric entities which have components denoted by some number of indices.
Each index can be any of the dimensions in which the tensor is defined, and
the number of indices needed to specify a component of a tensor is called
the tensor's rank. We are familiar with 0 and 1 rank tensors (numbers--or
"scalars"--and vectors). Finally, the way one transforms a tensor from one
coordinate system to another depends on the type of tensor, and it (in fact)
defines what it actually is to be a tensor. Each index of a tensor will
transform in either a contravariant way or a covariant way.
These are the basic ideas behind tensors, and they allow us to define
some very powerful mathematics. If you are familiar with the usefulness of
vectors, then you have touched the surface of the usefulness of tensors in
general. In the following section, we will look at two particular tensors,
and we will see that they can be quite useful.

5.7 The Metric Tensor and the Stress-Energy Tensor

Now that we have had a glimpse at tensors, let's consider a couple that
will be important to us. The first is called the metric tensor. I mentioned
a couple of sections ago that this tensor is related to the invariant
interval for a certain coordinate system on a given manifold. So, let's go
back and look at a the two specific invariant intervals which we introduced.
First, in normal, x-y, Cartesian coordinates, we have Equation 5:1
duplicated here:

(Eq 5:15--Copy of Eq 5:1)
ds^2 = dx^2 + dy^2

Second, on the surface of a sphere, using the L-H coordinate system
which we defined, we have Equation 5:2 duplicated here:

(Eq 5:16--Copy of Eq 5:2)
ds^2 = dH^2 + [cos(H/R)]^2*dL^2

Now, let's make this more general by considering an arbitrary, two
dimensional manifold and an arbitrary coordinate system on that manifold.
Let's call the coordinates "a" and "b". Now, in general, the invariant
interval on this manifold is defined in terms of the square of that interval
ds^2. The equation for ds^2 involves the infinitesimal distances da and db
in second order combinations. By second order combinations, I mean, for
example, da^2 or da*db. Thus, in general, the invariant interval will have
the following form (note: the g components are generally formulas of "a" and
"b"):

(Eq 5:17)
ds^2 = g   *da^2 + g  *da*db + g  *db*da + g  *db^2
aa          ab          ba          bb

In that equation you see the four components of the metric tensor in
this two dimensional, a-b coordinate system. They are the "g's" in the
equation. For our x-y coordinate system, we have

(Eq 5:18)
g  = 1,     g  = 0,     g  = 0,     g  = 1
xx          xy          yx          yy

For our L-H coordinate system, we have

(Eq 5:19)
g  = 1,     g  = 0,     g  = 0,     g  = [cos(H/R)]^2
HH          HL          LH          LL

So, we can construct the invariant interval if we know the metric
tensor for a coordinate system on a manifold. Now, remember that we said
that the form of the invariant interval for a particular coordinate system
tells us everything there is to know about the manifold for which those
coordinates are valid. So, now we see that all we need to know is the form
of the metric tensor. Once we know g, we know the geometry of the manifold.
Using tensor analysis, we can take the metric tensor and find an equation
for geodesics on the manifold. We can use it to find out all about the
curvature of the manifold. We can even use it to find the dot product (we
will discuss this a bit later) of two vectors in a particular coordinate
system.
Another thing the metric allows us to do is something generally called
"raising" or "lowering" indices. Basically, if you consider a tensor with a
contravariant index (which transforms in a particular way as discussed
earlier), then there is another way to express the tensor as one which has a
covariant index (and vice versa). That is to say that the geometric entity
represented by the tensor with the contravariant index has another
representation which involves a covariant index. For example, consider the
tensor A^a, which has a contravariant index, a. There is a corresponding
covariant tensor, A_a, which can be found using the metric of the space (and
coordinate system) we are dealing with. Here is an example of how you find
it (finding A_x when you know A^x) for a coordinate system with some
arbitrary coordinates, x and y:

(Eq 5:20)
x          y
A  =  g   A   +  g   A
x     xx         xy

For a general space and coordinate system, you can write this rule as
follows (remember, "a" can be any one dimension in the space, so this
represents a number of equations):

(Eq 5:21)
b
A  =  SUM(b varies over all dimensions) g   A
a                                       ab

Similarly, if you know the covariant form of A (A_a) you can find the
contravariant form by using the following:

(Eq 5:22)
a                                       ab
A  =  SUM(b varies over all dimensions) g   A
b

But that equation involves the contravariant form of the metric g^ab. In the
invariant interval, the metric is expressed in its covariant form g_ab. It
is therefore important for the reader to remember as we discuss various
metrics below, that for all of them we have

(Eq 5:23)
ab    1
g   = ---   if a = b
g
ab
and
ab
g   =  0    if a doesn't = b

Thus, using the metric tensor, one can "raise" or "lower" any index of
a tensor. Remember, what one is really doing is finding a form of that
tensor which transforms in a different way.
With this example of how the metric can be used, we will end our
discussion of this tensor. To sum up, the metric tensor on a manifold is a
very important entity which not only tells us all about the manifold's
geometry, but which also provides a very powerful tool which allows us to
deal with that geometry mathematically.

The second tensor we want to mention is the stress-energy tensor. I
don't want to get too deep into a discussion of the stress-energy tensor,
but the reader should know a couple of key points. With the stress-energy
tensor, we see our first example of a tensor explicitly defined in four
dimensional space-time (though later we will look at the metric tensor
defined in 4-d space-time). The stress-energy tensor (T) is also a tensor of
rank 2 (like the metric tensor), which gives it 16 components in 4
dimensions. Sometimes we express such a tensor in the form of a matrix as
follows:

(Eq 5:24)
+-                          -+
|   tt     tx     ty     tz  |
|  T      T      T      T    |
|                            |
|   xt     xx     xy     xz  |
ab      |  T      T      T      T    |
T   =    |                            |
|   yt     yx     yy     yz  |
|  T      T      T      T    |
|                            |
|   zt     zx     zy     zz  |
|  T      T      T      T    |
+-                          -+

There you can see the 16 different components. Now, each of these components
tell us something about the distribution and "flow" of energy and momentum
in a region. More precisely, T contains information about all the stresses
and pressures and momenta in a region. For example, The "tt" component of
the stress-energy tensor would be the density of the energy in the region
(the amount of energy--including mass energy--per unit volume).
As to why the stress-energy tensor is important to us, that will be
discussed further in a bit. However, here we can note the following in order
to pull us back towards our discussion of relativity and gravity: In
Newtonian physics, gravity was caused by the density of mass in an area.
However, in SR we find that mass is just a form of energy, and so we might
think that the "tt" component of the stress-energy tensor would be the right
thing to look at when it comes to gravity. However, if we write a rule using
one component of a tensor, then because the value of that component will
depend on your coordinate system (or frame of reference in space-time) then
the rule will also be frame-dependent. In short gravity would not be an
invariant theory, and it would require a preferred frame if we based it only
on the "tt" component of T. However, if we use all the components of a
tensor to form our theory, then (as it turns out) the theory can be made
frame-independent. Einstein thus considered the possibility that the whole
stress-energy tensor would need to play a part as the source of gravity. Add
to this some insight on curved manifolds and you end up with general
relativity, as we will see.

5.8 Applying these Concepts to Gravity

Now that we have discussed manifolds and their properties along with
some of the basic concepts of tensors, let's see how all of this applies to
relativity and gravitation. First, I will go over the main ideas which lead
us from what we have discussed so far to a general relativistic theory.
After that, I want to mention a few notes on the physics and the mathematics
we will be using given the concepts we have gone over. Next, we will go back
and look again at special relativity while applying a bit of our new
knowledge. This will show that GR is indeed general, because when applied to
space-time without the presence of gravity it will explain a special
case--special relativity. Finally, we will look quickly at a specific
application of the GR concepts to a space-time in which there is a
gravitational field. This application will focus on a particular class of
stars and black holes.

5.8.1 The Basic Idea

Let's get started with the basic ideas which combine the concepts we
have discussed to produce GR. Here I will simply state the main ideas
without an explanation of their application. You will get some feel for
their application in our two examples to follow.
So, here are the main claims of GR which involve the concepts we have
discussed. First, the space-time in which we live is a four dimensional
manifold. On that manifold there is a metric tensor (or just "a metric")
which describes the geometry of space-time. The metric can be used to find
geodesics on the space-time manifold, and when an object (only being acted
on by gravity) goes from one point in space-time to another point in
space-time (note: these are not just two points in space, but two
points--i.e. events--in space-time), it moves between the points by
following a space-time geodesic. Therefore, all the information necessary
for us to determine how such objects move through space-time is held within
the form of the metric. How, then, do we determine the metric? Well, the
metric of space-time in a region is itself determined (in a not-too-trivial
way) from the stress-energy tensor (T) which is affecting the region. This
then is the new theory of gravity which relativity has produced. The
stresses and pressures and momenta in a nearby region produces a
stress-energy tensor which, in turn, changes the metric of the nearby
space-time (making its geometry "curved"). This forces objects in the region
to follow specific paths (geodesics) through the "curved" space-time, and we
attribute this motion to gravitational effects.
As a conceptual example, consider a football being thrown from the
surface of the earth. Because of the mass of the earth, the space-time the
football is traveling through is a curved manifold, and the football follows
a "straight line" geodesic in the four dimensional curved space-time. To us,
the football's path is curved through three-space, but if we could somehow
experience the time dimension as a spacial dimension (i.e. if we were four
dimensional beings) and if we followed the path of the football in the
four-space, we would seem to be following a straight line on our four
dimensional curved manifold. However, in reality, the fourth dimension of
time does not act like the other dimensions in our perception of the
space-time manifold. Thus we do not see the actual four dimensional path of
the football, we only see the path in three dimensions while the fourth
component of the path is revealed to us as a dynamic component of the ball's
motion through time. That's why we can't see that its path is a "straight
line" in curved space-time. The stright-line is revealed to us as curved
motion, and we attribute that motion to gravitational effects.

5.8.2 Some Notes on the Physics and the Math

Before we go on to our two examples, I wanted to mention a couple of
points about the mathematics which can be used to develop physics in a
particular space-time.
First, note that for any space-time there is a four dimensional metric
involved. This metric can be used to find the invariant interval between two
space-time points. That interval (recall) can generally be expressed as

(Eq 5:25)
ds^2 = SUM(a & b vary over space and time dimensions) g  *da*db
ab

Second, consider a vector in our four dimensional space. Such a vector
(usually called a four-vector) has four components, three relating to space
and one relating to time. Now, in general, the values for these components
will depend on the coordinate system/frame of reference in which you are
considering the vector. However, we can use the metric to act on two
four-vectors to produce an invariant number. In other words, if there are
two four-vectors in a space-time, then two different observers using two
different frames of reference will each find different x, y, z, and t
coordinates which represent those two vectors in their respective frames.
However, when they each act on those two vectors in a specific way using
their own coordinate systems and using their own representation of the
metric, they will each produce the same particular number. The action on the
two vectors is called the dot product of the vectors, and many of you may
have heard of and used it before (though perhaps you didn't realize you were
using the metric--if you have ever had to remember how to produce a dot
product in polar coordinates, then you have seen how the metric in that
coordinate system affects the way you produce the dot product).
So, consider two four vectors, U and V. Remember that these are simply
tensors with either contravariant or covariant components. Now, we can
produce the dot product of U with V as follows.

(Eq 5:26)
a  b
U (dot) V = SUM(a,b) g  *U *V
ab

This produces a frame invariant number (a scalar), and if U and V have
particular physical properties in space-time, then we can use the dot
product to produce frame invariant physical rules in a particular
space-time.
For our third note in this section, let's discuss the time between two
events. It will be useful for us to find a frame-independent way of
expressing that time. To explore this a bit, consider an observer who is not
being acted on by any forces other than gravity. Because of gravity, he will
simply follow a geodesic through space-time--being at certain points in
space at particular times. Now, consider two events which each occur at the
position of our observer, but which occur at two different times on our
observer's clock. For such events, the time on the observer's clock which
ticks off between the two events is called the "proper time" (T, though it
is usually denoted using the Greek letter "tau") between those two events.
The time this observer reads on his clock does not depend on what any other
observer sees or does, and T is therefore a frame-invariant way of
specifying a time between two such events. Of course, the time as measured
in other frames will be different from T, but every frame will agree that
for the one, unique observer who naturally follows space-time curvature to
be at the position of both events, T is the proper time which he measures on
his clock.
We should note that not all events can be connected by the natural
space-time path of an observer because no observer can travel faster than
light in that space-time. Any two events which can be connected by an
observer's natural space-time path are called "time-like separated", and T
can easily be defined for such events.
Now, consider the invariant interval for some observer's space-time
path between two particular points. Remember that in general the invariant
interval is a function of your position in space-time. Thus, as soon as you
start moving down a path, the invariant interval begins to change. We
discussed this fact briefly in Section 5.5 and decided that we would deal
with it by breaking up the path into small bits and consider the invariant
interval at each bit. Therefore, rather than discuss the entire interval
between the two events, it is better to consider just one point along our
observer's path and look the infinitesimal (ds) at that point. That
infinitesimal in four dimensional space-time is generally made up of an
infinitesimal change in space and an infinitesimal change in time. However,
remember that for the observer and the two events we are considering, both
of the events occur right at the observer's position. So, for him there is
no spatial distance (dx' = 0, dy' = 0, and dz' = 0) between any two points
on the path. Therefore, the invariant interval at any point on his path as
calculated using his coordinates must be made up of only changes in his time
coordinate (dt'). Thus, the value of the invariant interval at some point on
the observer's path is given totally by the infinitesimal change in the
proper time (dT = dt', the infinitesimal change in time on our observer's
watch). We can therefore write the following (taking the spatial components
out of Equation 5:25):

(Eq 5:27)
ds^2 = g    *dT^2
t't'

Notice that the component of the metric tensor in the above equation is
expressed in the coordinates of the observer we are considering (i.e. we are
specifically using t' and not t). This must be the case, because it is only
when we measure the infinitesimal invariant interval (ds) using his
coordinates that we can disregard any spatial component and write the
interval totally in terms of dT. However, since this observer is free
falling (only being acted on by gravity), then recall that his local
space-time is flat, regardless of the global geometry of the space-time he
is in. Thus, for small distances in space and time in his coordinate system
(i.e. for infinitesimals like dt') his space-time can be considered to be
that of special relativity (flat space-time). We will find out in the next
section what g_tt is for the flat space-time of SR, and when we plug this
into Equation 5:27 we will find that

(Eq 5:28)
dT^2 = -ds^2/c^2.

That equation is true for any space-time, because the space-time of the
observer is locally flat regardless of the global geometry of the space-time
we are considering.
So, how will this help us with the physics? Well, specifically, this
gives us a way to define the momentum of an object in any space-time.
Consider a free-falling object of mass m. In some coordinate system, the
object's position in one coordinate (say "a") can be changing. Note that "a"
could be x in an x-y-z coordinate system, r in polar coordinates (which we
will discuss later), etc. Now, as the object changes spatial coordinates in
this system, it will follow a natural geodesic path through space-time. As
the object's position in "a" changes by some infinitesimal amount (da) its
own "clock" will tick off some small time (dT--note that this is a proper
time because it is measured on the clock of the object itself). In that
case, the "a" component of the momentum for that object in this coordinate
system will be expressed as

(Eq 5:29)
a
p  = m*da/dT

Notice that if we consider the situation where "a" is the time coordinate
itself in our system, then we have a sort of "temporal momentum" who's
significance will be discussed in the next section. Thus, p^a actually has
four dimensions, and is, in fact, a four-vector. Combine this with our
discussion of four-vectors above, and we will find some useful physics, as
we will see in the following examples.

5.8.3 First Example: Back to SR

The most simple application of the ideas expressed in Section 5.8.2 is
one which we have already looked at (though without using the concepts
discussed in that section). It is the situation where there is no
gravitational field. That is exactly the situation we were considering when
we discussed special relativity. In special relativity, there is no
gravitational field. All the components of the stress-energy tensor are
identically zero.
Now, we will figure out the metric of space-time in such a case by
examining what we already know about special relativity. So, let's go back
to our space-time diagrams. (By the way, our diagrams only considered one of
the spatial dimensions, but we will incorporate the other two in this
section.) Consider two observers who start out moving parallel to one
another on the diagram. This would mean that they start out with the same
velocity in any inertial frame. Well, in special relativity (with no
gravitational field) the two observers will continue to remain on parallel
paths on the space-time diagram. This is the property of a flat manifold, so
in SR, space-time is "flat".
Before we go on, it will be helpful for us to redefine the time
variable in our space-time coordinates. Instead of "t", consider the
combination "c*t" (where c is the speed of light). For convenience, we will
simply define a new variable, w, where

(Eq 5:30)
w = c*t

Then we can use w in place of t in our coordinates. This is actually a
fairly natural substitution in a couple of ways: First, note that w has the
units of length, just like x, y, and z do. Second, using w on our space-time
diagrams makes them a little more general. Why? Well, remember how we
defined the units of length and time to be the light-second and the second?
We did this so that a light ray would make a line at a 45 degree angle on
our diagram. Well, with a w-x coordinate system, this will automatically be
the case, regardless of what units you use. To see this, note that the value
of t at a certain value of w is just the time it takes for light to travel
that length, w (because t = w/c). For example, the point x = 1 light-second
and t = 1 second corresponds to the point x = 1 light-second and w = 1
light-second. So, on both an x-t diagram and on an x-w diagram, a light beam
would make a 45 degree angle with the x axis by going through the point
(1,1). However, if we wanted to, we could now use a meter as our unit of
length. Then, when w = 1 meter, t would just be the time it takes for light
to travel 1 meter. So, the point x = 1 meter, w = 1 meter also lies on the
light path, and again, that light path would automatically make a 45 degree
angle with the x axis by going through the point (1,1). For consistency, we
will continue to use units of seconds and light-seconds, but we will now use
"w" in units of light-seconds to indicate time in our discussions and
diagram (remember, the length "w" just represents the time it takes light to
travel that length).
Now, let's look at a change in coordinates on the flat space-time of
SR. In space-time, a change in coordinates can represent a change in an
observer's frame of reference. So, when we discussed two observers who were
moving with respect to one another, we were looking at two different
coordinate systems (x-t and x'-t', or now, x-w and x'-w') which both
correctly described space-time in SR. This leads us to consider the
invariant interval, because we know it must be the same for each of these
two coordinate systems. So, let's take a closer look at these coordinate
systems on our diagrams and see if we can't define the invariant interval
(which, remember, is just another way of writing the metric).
We will specifically want to consider infinitesimal lengths like dx.
So, let's look at a small line segment which lies on a particular
geodesic--a geodesic we know a little about. That geodesic is the path which
light follows. Like anything else being acted on only by gravity, light must
follow a geodesic on the space-time manifold. So, for the particular case of
a light path, a small segment on that path would have an x component (dx)
and a t component (dt); however, we now want to begin thinking of w as the
unit which represents time, so we note that a small change in t (dt)
represents a change in w of dw = c*dt. Now, since the small distance light
travels (dx) divided by the time (dt) it took it to travel that distance is
defined as the speed of light, then we have the following:

(Eq 5:31)
dx
-- = c  (where c is the speed of light)
dt

which can be rewritten as

(Eq 5:32)
dx
-- = 1
dw

That means that dx = dw (for light). Now, since we always define the
invariant interval in terms of the infinitesimal lengths squared, we will
actually want to square both sides of that equation and then bring
everything to one side so as to get the following:

(Eq 5:33)
dx^2 - dw^2 = 0  (For light)

Now, because the speed of light is the same for all inertial observers, the
above equation must be true for all frames of reference. Thus, we might
consider the idea that the invariant interval for any small line segment
(not just for light) is given in SR by

(Eq 5:34)
ds^2 = dx^2 - dw^2,

and this turns out to be the case. The light path, then, is just the case
where ds^2 = 0.
independent of where you are in space-time. All that matters is the lengths
dx and dw, regardless of what actual x and w position you have. This means
that the distances (like dx) don't have to be infinitesimal, because the
equation remains true regardless of how far you extend dx and dw. Thus,
let's consider the case where one side of the line segment is at x = w = 0
(the origin). Then dx will be the x distance from the origin to the end of
the line segment (which in this case can be as far away as we like), and dw
will be the w distance to that point. In other words, for SR, dx and dw can
be replaced with x and w when we consider one side of the line segment to be
at the origin. Further, consider a point in space-time with coordinates
(x,w) in the o observer's coordinates and (x',w') in the o' observer's
coordinates. Since the value of the invariant interval is the same for any
frame of reference, the following must be true:

(Eq 5:35)
x^2 - w^2  =  x'^2 - w'^2

Let's see that this is the case on our space-time diagrams. Diagram 5-9
shows a space-time diagram with two coordinate systems indicated, one for an
observer o, and a second for an observer (o') moving with velocity 0.6 c
with respect to o. (Note that now we use w = ct for the time axes.) There is
also a point marked "*" on the diagram. The x'-w' coordinates for that point
are clearly shown to be x'= 1 light-second and w'= 2 light-seconds (i.e. t'=
2 second, remember?). The x-w coordinates are x = 2.75 light-seconds and w =
3.25 light-seconds, and I tried to show this as best I could with an ASCII
diagram.

Diagram 5-9

w                    w'
|                   /
|                  /
|                 /
w=3.25 |->              /         *
+               /      '  '
|              /   '     '
|         w'=2+'        '
|            /         '
|           /         '
+          /         '                    x'
|         /         '                   '
|        /         '                '
|       /         '             '
|      +         '         +'
|     /         '       '
+    /         '    '
|   /         + '
|  /        'x' = 1
| /     '
|/  '
--+-----------o---------+----------+---------+--->x
'  /|                          ^
'     / |                       x=2.75

We therefore find the following:

(Eq 5:36)
ds^2  =  x^2 - w^2  =  (2.75)^2 - (3.25)^2
=  -3 light-seconds^2
and

ds'^2 =  x'^2 - w'^2  =  (1)^2 - (2)^2
=  -3 light-seconds^2

we note that ds^2 = ds'^2, as it must be. In fact, it is the form of the
invariant interval and the fact that it must be invariant from one
coordinate system to another that causes the transformation from x-w to
x'-w' to look as it does. If the x' and w' axes didn't look the way they do
relative to the x and w axes in our diagrams, then the interval would not be
invariant. Note that if the "-" sign in the invariant interval were a "+"
sign, then the invariant interval would look just like the one for a normal,
space-only x-y coordinate system where ds^2 = dx^2 + dy^2. Then, the
coordinate transformation to x'-w' would be just like a rotation of
coordinates (see Diagram 5-10). The "-" sign in the SR interval causes one
of the axes to rotate in the opposite direction from the other when we do
our space-time coordinate transformation.
Second, note that the interval squared is, in fact, negative. This is
not too distressing, because we know that _physical_ lengths on our diagram
do not represent the space-time "lengths" which the invariant interval gives
us. If they did, then the invariant interval for special relativity would be
just like the x-y form of the invariant interval (since the physical lengths
on our diagrams are just normal lengths on the flat paper/screen we draw
them on). Now, the actual length of an infinitesimal interval on a manifold
is usually defined to be the square root of the absolute value of ds^2.
Thus, we can still make sense of lengths, even when the invariant interval
squared is negative.

Diagram 5-10

x'-y' is rotated from x-y, and the line segment
in the two diagrams are identical
y                                       y'
|                                      /
|                                     /
|               /                    /        /
|             / .                   /       / '
|        ds /   .                  /   ds /  '
|         /     . dy              /     /   '
|       /       .                /    /    'dy'
|     /..........               /   /     '
|          dx                  /    ' .  '
--+------------------ x         +    dx'  '
|                                 \
\
\
Note: the length of the line segment             \
doesn't change just because you rotated                x'
the coordinate system, so
dx^2 + dy^2 = dx'^2 + dy'^2

The reader may have noted that thus far in our look back at special
relativity we have still only included two of the four dimensions of
space-time. The other two (y and z) could actually replace x in any of our
discussions, and so they play the same roll in the invariant interval as x
does. Therefore, the total four dimensional invariant interval for special
relativity is given by

(Eq 5:37)
ds^2 = dx^2 + dy^2 + dz^2 - dw^2

Finally, let's talk about some physics in this space-time using the
concepts discussed in the previous section. First, consider the proper time
between two time-like separated events. Recall that we defined this time
such that:

(Eq 5:38)
ds^2 = g  (of SR)*dT^2
tt

We now know that g_ww = -1 for SR from the above, so g_tt = -c^2 for SR.
This is how we got Equation 5:28, which is duplicated here:

(Eq 5:39--Copy of Eq 5:28)
dT^2 = -ds^2/c^2.

in the previous section. However, since we are now working with w for our
time coordinate, we should define dW = c*dT, and rewrite Equation 5:39 as

(Eq 5:40)
dW^2 = -ds^2

Now, let's consider the observer which followed the t' axes in Diagram 5-9
such that his velocity was 0.6 c. Consider the O observer's frame of
reference, and note that if it takes O' a certain time (dw) to travel a
certain distance (dx) in the O observer's coordinates, then it must be the
case that dx/dt = 0.6 c. So dx/dw = 0.6, or

(Eq 5:41)
dx = 0.6*dw

This, then, is true all along the w' axes (the line that O' follows through
the O observer's coordinate system). So, the invariant interval (considering
only two dimensions once again) at any point along the w' axes must be given
by the following (using Equation 5:37 with only x and w coordinates and
substituting Equation 5:41):

(Eq 5:42)
ds^2 =         dx^2 - dw^2

= [0.6]^2*dw^2 - dw^2  = -[1 - 0.6^2]*dw^2

plugging this into Equation 5:40 we find that

(Eq 5:43)
dW^2 = [1 - 0.6^2] * dw^2

so,

(Eq 5:44)
1
dw = --------------- * dW = gamma*dW
SQRT[1 - 0.6^2]

Since dW just represents an infinitesimal time as measured on our "moving"
observer's clock, and dw an infinitesimal time measured on our clock,
Equation 5:44 is just the equation which shows time-dilation effects in SR,
and it was quickly derived using our new knowledge.
For another physics consideration, look at the momentum four-vector. We
defined this earlier (Equation 5:29) and it is duplicated here:

(Eq 5:45--Copy of Eq 5:29)
a
p  = m*da/dT

Again, we want to use dW = c*dT, and we thus find

(Eq 5:46)
a
p  = m*c*da/dW

For us, we consider the situation where "a" is the x dimension. Then, p^x'
for the "moving" observer himself is zero (because all along the w' axes we
have dx' = 0 by definition, i.e. he is not moving relative to himself).
However, for the O observer (for whom the "moving" observer moves a distance
dx in a time dw) we find the following from Equation 5:46 by substituting x
for a(Note that from Equation 5:44 we can write dW = dw/gamma, and we are
substituting that here. We also use dw = c*dt and v = dx/dt in this
equation.):

(Eq 5:47)
x
p  = m*c*dx/[dw/gamma]  =  gamma*m*c*dx/dw

= gamma*m*dx/dt = gamma*m*v.

This is exactly the definition of the momentum we saw in our discussions of
special relativity.
However, now we can also look at the time component of the momentum
four-vector and figure out what it represents. Again we use Equation 5:46,
but here we substitute w for x:

(Eq 5:48)
w
p  = m*c*dw/[dw/gamma] = gamma*m*c

But this is just the energy we had defined in SR (E = gamma*m*c^2) divided
by c:

(Eq 5:49)
w
p  = E/c.

And so, we now know all about the components of the momentum four-vector of
a particle: three are the spatial components of the momentum of the
particle, and the time component represents the energy of the particle
divided by c.
As a final bit of physics, consider the dot product (as defined in
Equation 5:26) of the momentum four-vector with itself:

(Eq 5:50)
w  w        x   x
p (dot) p = g  *p *p +  g  *p * p
ww          xx

= -[E/c]^2 + p^2

(Note that the total momentum of this observer is p^x, and so we write p^2
in the last line to mean the total momentum squared). Now, recall that the
dot product is invariant, so that if any observer measures the energy and
momentum of a particle and calculates the above equation in his frame of
reference, he must find the same number that any other observer would find
in any other frame of reference. This shouldn't come as too much of a
surprise if we look back for a moment. Back when we discussed energy and
momentum in special relativity, we found in Equation 1:7 that E^2 = m^2*c^4
+ p^2*c^2. Thus, we find that the dot product in Equation 5:50 is simply
equal to -m^2*c^2. Since m and c are invariant (remember, m is the rest
mass), we could have already known that the formula in Equation 5:50 would
be invariant.
We have therefore been able to find all the major physics equations we
saw in special relativity by simply apply some tensor analyses using the
metric of flat space-time.
So, to sum up, we have found the following: For SR, where there is no
gravitational field, space-time has the properties of a flat manifold. The
invariant interval of a flat space-time manifold is given by the following:

(Eq 5:51--Copy of Eq 5:37)
ds^2 = dx^2 + dy^2 + dz^2 - dw^2

That interval tells us all about the nature of space-time in SR. The fact
that the contribution of the time component (dw) is negative where as the
spatial components have positive contributions is what gives the coordinate
transformation between different frames of reference its unique form. Thus,
it is the negative sign which essentially causes time dilation and length
contraction effects, and it is the fact that the speed of light is invariant
which causes that sign to be negative.

5.8.4 Second Example: Stars and Black Holes

In this second example, we will briefly look at the description GR
gives us for the gravitational field of certain stars. We will also take a
look at one of the most widely publicized consequences of GR--black holes.
To make our discussion simpler, the types of stars we will be
considering will be spherically symmetric. What does that mean? Well,
consider an imaginary sphere with some radius. Place the center of that
sphere at the center of the star. If the star is spherically symmetric, then
the strength of the gravitational field everywhere on the surface of our
imaginary sphere will be exactly the same. For example, a star who's density
is spherically symmetric and which is not spinning would work.
Now, it will be helpful for us to discuss the space around the star in
terms of spherical coordinates; therefore, I should make sure the reader
knows what these coordinates are. Rather than using x, y, and z coordinates
for the three dimensional space around the star, we will use r, a, and b
coordinates, which I will define here. In Diagram 5-11 I have tried to draw
(in three dimensions) an z-y-z coordinate system, and I have marked a point
in space, *. There is a line segment drawn from the origin (o) to that
point, and the lengths of the x, y, and z components of the line segment are
the values for the x, y, and z coordinates of the point, *. These components
have been indicated on the diagram using "dotted" lines. Now, note that
there is one other dotted line which is not labeled. If you imagine a light
shining down on our line segment, then the unlabeled dotted line would be
the shadow that light produced on the x-y plane. It is called the projection
of the line segment on the x-y plane, but let's just call it "the x-y
component" for convenience.

Diagram 5-11

z
|
|
|      *
|     /'
|    / '
|_a /  'z-comp
| \/r  '
| /    '
|/     '
o------'----- y
/ '.    '  '
/__b/'.  ' ' x-comp
/       '.''
/'''''''''''
x   y-comp

Now we can define the r-a-b coordinates for the point, *. First, the
distance from the origin to the point (the length of the line segment) is
the "r" coordinate as indicated on the diagram. Next, the angle between the
z axes and the line segment is our "a" coordinate (though it is usually
denoted by the Greek letter "theta"). It too is indicated on the diagram.
Finally, there is the angle between x and the x-y component of the line
segment. That angle is our "b" coordinate (though it is usually denoted by
the Greek letter "phi"), and it is indicated on the diagram as well. Thus,
with r-a-b coordinates as defined here, we can specify any point in three
dimensional space.
metric of a flat 3-space using these coordinates. For an x-y-z system, the
metric is (of course) given by this invariant interval:

(Eq 5:52)
ds^2 = dx^2 + dy^2 + dz^2.

However, for our new coordinate system in the same flat 3-space, it is given
by the following:

(Eq 5:53)
ds^2 = dr^2 + r^2*da^2 + r^2*[sin(a)]^2*db^2.

For convenience, a new infinitesimal (call it du) is sometimes defined such
that:

(Eq 5:54)
du^2 = da^2 + [sin(a)]^2*db^2.

Then we can rewrite Equation 5:53 as

(Eq 5:55)
ds^2 = dr^2 + r^2*du^2.

We will therefore continue to use du throughout this discussion, but
remember it is just a convenient way to write the a and b components of the
invariant interval.
Next, let's look at some properties of the star we will be considering.
Basically, we will say it has a total mass of m(star) and a radius R. The
center of the star will be centered at the origin, o. Finally, we will only
be considering the gravitational field outside of the star itself. In
general, physicists are interested in the gravitational field inside the
star as well, but we will not worry about it that much.
We also want to define a new variable for mass using the Newtonian
gravitational constant G. In Newtonian gravitation, the force between two
objects of mass m1 and m2 which are a distance r apart is given by

(Eq 5:56)
F(Newtonian Gravity) = G * m1 * m2 / r^2

(where G = 6.672*10^-11 m^3/(s^2*kg) and we note that kg is the symbol for
kilogram). We will use G to define a new variable, M, such that

(Eq 5:57)
M = G*m(star)/c^2

Notice that M has the units of meters, and so M gives us a way of specifying
the mass of an object in units of meters (similar to the way w allows us to
specify time in units of meters). It is called the "geometrized" mass. So,
using M we can say that an object has a mass of 1 meter, and one can
decipher what mass we are talking about in terms of conventional units by
using Equation 5:57. As a note, a mass of M = 1 meter corresponds to
m(conventional) = 1.35E27 kg, the mass of the sun is M(sun) = 1477 meters
(1.989E30 kg), and the mass of the earth is M(earth) = 0.004435 meter
(5.973E24 kg).
Now, with this information in mind, the next step is to figure out what
the metric of the space-time around the star would be because of the
stress-energy tensor of the star. Generally, one uses the fact that we are
considering spherically symmetric stars in order to make some assumptions
about the form of the metric. One then uses this general form to calculate
the general form the stress-energy tensor would have. Finally, one uses what
we know physically about the star compared to the form of the stress-energy
tensor, and one can decipher what equations must have made up the metric in
the first place. In the end, one finds a metric for the space-time around
this type of star, and for our purposes, we will simply state that end
result. Thus, the metric is as follows (expressed in terms of the invariant
interval):

(Eq 5:58)
ds^2 = -(1 - 2*M/r)*dw^2 + [1/(1 - 2M/r)]*dr^2  + r^2 du^2

=        g    *dw^2 +          g  *dr^2    + g  *du^2
ww                    rr            uu

Note that we are using du as defined earlier, and we are using dw = c*dt as
our time component as discussed in the previous section. Also, we are using
M (as defined in Equation 5:57 ) to denote the mass of the star rather than
m(star). This metric is known as the Schwarzschild metric.
The next step, then, is to show that we can get useful physics by
considering this metric. We will again (as we did with the Special
Relativity discussion earlier) be looking at a particle of mass m, and here
we will be interested in its motion in the space-time around the star.
Because of the spherical symmetry of the space-time, the motion of such a
particle will remain within a plane, and we can orient our coordinate system
so that the plane is one where the angle "a" = 90 degrees (and sin(a) = 1).
Since the particle doesn't move out of that plane, there is never a change
in the angle "a" (da = 0). Thus, for this particle, we can consider the
metric as follows (putting sin(a) = 1 and da = 0 into Equation 5:58):

(Eq 5:59)
ds^2(particle's path)
= -(1 - 2*M/r)*dw^2 + [1/(1 - 2*M/r)]*dr^2 + r^2*db^2

=        g    *dw^2 +          g    *dr^2  + g  *db^2
ww                    rr            bb

In the interest of time (because we simply haven't been able to cover
everything we need to know about tensor analyses in this text), I will have
to simply state a couple of facts which we will use to produce the physics
we will look at. Namely, we notice that the form of the metric depends on
your particular position in r (because g_ww, g_rr, and g_bb are all
functions of r). However, none of the metric's components are functions of
w. Because of that, as it turns out, p_w (the covariant form of the time
component of the momentum four-vector) is constant throughout the motion of
the particle. The metric is also independent of the angle b. This, as it
turns out, implies that p_b is a constant. We can therefore define two
constants, E and L such that

(Eq 5:60)
p  = -E*m*c
w

and

(Eq 5:61)
p  = L*m*c
b

where m is the mass of the particle. These definitions will simplify the
equations we will produce below (and they are related to our usual concepts
of energy and angular momentum, so the fact that they are constant basically
say that energy and angular momentum are conserved as the particle moves).
Now, so far we have only defined the contravariant form of the
momentum, p^a. However, when we discussed the metric tensor we learned how
to use it to "raise" and "lower" indices. So, we can write the following
from Equation 5:22:

(Eq 5:62)
w     ww       wr       wb       wa
p  =  g  *p  + g  *p  + g  *p  + g  *p
w        r        b        a

Note that we are considering the case where the angle "a" is a constant so
that p^a = 0 in Equation 5:62. Also recall that in Equation 5:23 we noted
how to go from contravariant to covariant forms of the metric. For the
metrics we are discussing we thus have (note that the metric components come
from Equation 5:59).

(Eq 5:63)
ww    1         -1
g   = ---  =  ---------
g       1 - 2*M/r
ww

rr    1
g   = ---  =  1 - 2M/r
g
rr

bb    1       1
g   = ---  =  ---
g       r^2
bb

all other covariant metric components = 0

Thus, only the p_w part remains in Equation 5:62 giving us the following
(note that I substitute using Equation 5:60):

(Eq 5:64)
w       -1                    1
p  = ----------- * p   =  ----------- * E*m*c
(1 - 2*M/r)    w     (1 - 2*M/r)

Similarly we can find the equation for p^b:

(Eq 5:65)
b    bb            1              1
p =  g   *p    =   --- * p    =   --- * L*m*c
b       r^2    b       r^2

Now, recall that in the last section we found that p(dot)p was a
constant, -(m*c)^2. That remains true here, so we find the following:

(Eq 5:66)
w  w        r  r        b  b
p (dot) p = g  *p *p  + g  *p *p  + g  *p *p   = -(m*c)^2
ww          rr          bb

We can express each of the parts for that equation by substituting in the
metric components from Equation 5:59, using the above equations for p^w and
p^b, and writing p^r as m*c*dr/dW to get the following:

(Eq 5:67)
w  w                  [  (E*m*c)^2  ]
g  *p *p  = -(1 - 2*M/r) * [-------------]
ww                        [(1 - 2*M/r)^2]

-E^2*(m*c)^2
= ------------
(1 - 2*M/r)

r  r        1        [  dr] 2
g  *p *p  = ----------- * [m*--]    (NOTE: dr/dW = c*dr/dT)
rr         (1 - 2*M/r)   [  dW]

(dr/dT)^2*(m*c)^2
= -----------------
(1 - 2*M/r)

b  b         (L*m*c)^2
g  *p *p  = r^2 * ---------
bb                  r^4

L^2*(m*c)^2
= -----------
r^2

Substitute this into Equation 5:66 and the (m*c)^2 portions will cancel out
on both sides giving this:

(Eq 5:68)
-E^2        (dr/dT)^2     L^2
-1  = ----------- + ----------- + -----
(1 - 2*M/r)   (1 - 2*M/r)    r^2

From this, we can find the following equation which describes the orbits the
particle can take. It is the equation of motion of the particle:

(Eq 5:69)
(dr/dT)^2 = E^2 - (1 - 2*M/r)*(1 + L^2/r^2)

Now, it turns out that if one examine this equation for the case of a
circular orbit (where r is a constant and dr = 0) and for the case where the
mass is small or the orbit is large, we find things to be quite similar to
what Newtonian physics predicts. However, it is interesting to note that for
orbits for which r can change (elliptical orbits in Newtonian physics) GR
predicts something a bit different from Newtonian physics. Basically, in
Newtonian physics, the path of the particle in space is a true, closed
ellipse. However, with the above equation one finds that the "elliptical"
orbit in GR does not close in on itself. Instead, it's as if the ellipse
changes position as the particle's orbit goes on. We thus see a difference
in the predictions of the two theories, and we will mention this again in
the next section.
With this quick look at the physics one can derive using the metric for
such a star, we now want to go on and look at a very special case where this
metric comes into play. Consider for a moment what would happen if the
star's radius were to somehow become smaller than 2*M. Such a thing can
theoretically happen for certain stars at the end of their life cycle,
(though we won't get into how in our discussion).
So, consider the case where the radius of the star is smaller than 2*M.
We can then consider a point above the star for which r < 2*M. Now look back
at the metric of the star. If r < 2*M then g_tt becomes positive, while g_rr
becomes negative. That is to say that the time component of the invariant
interval will contribute to the interval in the same way that a space-like
coordinate did when r was greater than 2*M, and the radial component will
contribute in the same way as a time-like coordinate did when r was greater
than 2*M. Further, when r was greater than 2*M, we understood that all
particles followed a space-time path which took them "forward" in time.
Similarly, when g_rr becomes negative and d_tt becomes positive, (when r <
2*M) we find that all particles must continue along a space-time path for
which r continually decreases. In other words, the point r = 0 becomes part
of the "future" of every particle/observer for which r is less than 2*M.
Thus, such a particle will be doomed to fall in toward the center of the
star. One can then imagine that the star itself would be doomed to fall in
upon itself completely, becoming nothingness at r = 0.
This is known as a black hole (specifically, for the metric we are
considering, it is a spherically symmetric black hole), and the radius r=2*M
is called the Schwarzschild radius or the event horizon. Any observer with
an r coordinate less than 2*M must fall into the point r = 0. Note that at r
= 0 our metric becomes truly infinite, and as it turns out, that would be a
point where physical laws break down. Such a point is called a singularity.
We should also note that any signal (even a light signal) which the observer
tries to send outside of the event horizon must also fall into the
singularity (because all space-time geodesics for r < 2*M fall into the
singularity). Thus, there is no way to get any information from the
singularity to the "outside universe". There is no way for one to "see" the
singularity and its destruction of physical laws. In that sense, the
singularity's existence isn't a problem for our physical laws outside of the
event horizon.
As a last consideration about black holes, one might ask what would
happen to an observer who starts where his r coordinate is greater than 2*M
and then falls toward the event horizon. I won't go through the math, but
one finds that in our coordinates, the observer will take an infinite amount
of time to reach r = 2*M. However, if we ask about how much time the
observer himself reads on his watch as he falls (the proper time) we find
that in his coordinates, the time it takes for him to reach the event
horizon is finite. To try and understand how this can be, we will start by
considering the equation for p^w (the time component of the momentum
four-vector) as defined in Equation 5:46:

(Eq 5:70)
w       dw
p  = m*c*--
dW

However, if we look back at Equation 5:64, we can combine it with Equation
5:70 to find the following:

(Eq 5:71)
dw        E
-- = -----------
dW   (1 - 2*M/r)

Rewriting this, one finds that

(Eq 5:72)
(1 - 2*M/r)
dW = ----------- * dw.
E

So what does that tell us? Well, consider an observer at the coordinate
position r. If a small time ticks in our coordinate w = c*t, then the amount
of time which ticks on the observer's clock (dW = c*dT, where dT is the
proper time) depends on the r position of the observer. The smaller his r
position (as long as he is above the event horizon) the smaller dW will be
for a given dw. This is similar to time dilation in SR, but here it is
caused by the gravitational field and not by the relative motion of two
observers.
Applying this to our discussion of the observer falling towards the
event horizon, we find the following: In our coordinates (w) the clock of
the infalling observer (who is constantly falling to smaller and smaller r
values) takes longer and longer to tick its next tick. For example, let's
say that for the observer's clock, it ticks 10 ticks before it reaches the
event horizon. As we mentioned earlier, the coordinate time (w) will have to
become infinitely large before the observer will reach the horizon. However,
as the observer gets closer and closer to the event horizon, his clock takes
longer and longer to tick its next tick. Essentially, in our coordinate
system, the observer's clock will never be able to tick the 10th tick.
Meanwhile, for the observer, time goes on as usual. For him, therefore, the
10th tick will come, and he will enter the event horizon. However, once in
the horizon, he will not be able to send any signals out of the r = 2*M
event horizon (in our coordinates). Thus, no one with r greater than 2*M in
our coordinates will ever be able to see the infalling observer go into the
event horizon. This then explains how we can say that the infalling observer
never reaches the horizon according to our coordinate system.
As it was in SR, there are different explanations for how certain
outcomes come to be. The explanation depends on what coordinate system you
use to explain the occurrences (which means that it depends on your frame of
reference). The important point is that the end result of the explanations
agree with the each other as far as any physical laws can be applied. In the
twin paradox of SR, when the two twins come back together and stand next to
one another at the end of the trip, each explanation must agree as to which
twin is actually, physically older. For the question of whether an infalling
observer reaches the event horizon, regardless of which coordinate system we
use, we must agree that the observer is never seen to enter the horizon by
any observer outside of the event horizon. The fact that the infalling
observer "sees" himself enter the horizon has no physical consequences to
the outside world.

Thus, with spherically symmetric stars and black holes, we have found
the following: the metric of the surrounding space-time is given by the
following (using variables we have defined earlier):

(Eq 5:73--Copy of Eq 5:58)
ds^2 = -(1 - 2*M/r)*dw^2 + [1/(1 - 2M/r)]*dr^2  + r^2 du^2

=        g    *dw^2 +          g  *dr^2    + g  *du^2
ww                    rr            uu

Symmetries in this metric can be used along with the metric itself to find
the equations of motion for a particle which moves within this space-time.
Finally, the space-time has interesting consequences for the measurement of
space and time for observers at different points in the curved space-time
surrounding such stars and black holes.

That ends our look at some examples of the application of GR. The only
thing left in our discussion of this theory is to show some experimental
evidence for its existence, as we will do in the following section.

5.9 Experimental Support for GR

In this section we will take a look at a few experiments which agree
with the predictions of GR.
For the first experiment, we use the effect mentioned in the previous
section whereby orbits which were supposed to be elliptical according to
Newtonian physics didn't actually close in on themselves according to GR
predictions. This effect can be seen as a rotation (or precession) of the
"long axis" of the elliptical orbit, whereas under Newtonian theory, this
axes doesn't move. Now, for the orbits of most planets, this effect is too
small to measure. However, for Mercury (which is closest to the sun and
would thus be the most affected) the effect is measurable. In fact,
measurements taken during the 1800s showed that Mercury's orbit precessed.
Now, much of this could be attributed to effects from the gravity of the
other planets, however, after all those effects were taken into account,
there was still a small amount of precession which wasn't accounted for. The
predictions of GR accounted for the left-over difference. It was Einstein
who first pointed this out, and this was the first evidence in favor of GR.
For the second experiment we want to consider, note that light, just
like anything else being acted on only by gravity, must follow a geodesic in
space-time. One can use the metric introduced in the previous section to
figure out how light would travel when passing near an approximately
spherically symmetric star. What one finds is that the light would be bent
by the presence of the star's gravitational field. Now, one might try to
make an argument using special relativity by which light with an energy E
would be said to have a "relativistic mass" defined by "m" = E/c^2. One
could then figure out how much the light with this "mass" would bend in the
presence of a Newtonian-type gravitational field. This, one might hope,
could allow the explanation of how light could be bent without considering
GR. However, one finds that the amount of bending predicted by this
SR-Newtonian method is exactly half as much as the bending predicted by GR.
Thus, if we could actually measure the bending of the light, we could figure
out which of the two predictions was correct.
Well, experiments to measure such bending can and have been performed
using the sun as the source of gravity and using light from particular
stars--light which passes near the sun on its way to us--as the light that
gets bent (it was Einstein who suggested this test, by the way). Normally,
of course, the sun would be too bright to see stars who's light passes near
the sun on its way to us. However, during a solar eclipse, the stars can be
seen. When one compares the positions of such stars which one sees during a
solar eclipse to the positions where the stars should actually be, one finds
that the difference can be attributed to the bending of the light as
predicted by GR, while the SR-Newtonian prediction was incorrect by a factor
of 2.
The third experiment we will look at involves using highly sensitive
atomic clocks taken aboard jets. When one compares the reading on such
clocks to clocks which remained on the ground, one finds that the difference
(though quite small) can only be accounted for completely if one includes
calculations for SR effects and acceleration along with the GR effects of
having the jet fly at high altitudes where the gravitational field is not as
strong as it is on the surface of the earth.
These are a few examples of experimental evidence that exists in favor
of GR. In many cases, more data and more precise measurements would be
needed to rule out all theories other than GR; however, all the evidence we
do have supports the theory.

ve supports the theory.

```

## User Contributions: Part1 - Part2 - Part3 - Part4

[ Usenet FAQs | Web FAQs | Documents | RFC Index ]

Send corrections/additions to the FAQ Maintainer:
Jason W. Hinson <jason@physicsguy.com>

Last Update March 27 2014 @ 02:12 PM