Nature is quirky. Whenever things don't quite match up, She changes them so they will. The results often seem to be bizarre and nonsensical, but the more you study it you realize how profoundly wise Nature is. It all started with a thought experiment that Einstein said he came up with at around the age of 16. The young Einstein wondered what would happen if he chased a light beam and caught up with it. This essay describes two of the most important discoveries in science: The Special Theory of Relativity and the General Theory of Relativity. Both of these discoveries were made by a single man, Albert Einstein, over a period of one decade (1905 – 1915). This essay is directed at an audience of amateur scientists like myself. I will approach these two theories on the basis of their underlying principles, deriving as much as possible using basic geometry and a bit of elementary calculus. I will not go into the depth needed to become a “relativist.” Mastery of general relativity would require a good working knowledge of tensors, which is beyond the scope of this essay. Nevertheless, I think amateur scientists like myself will get something useful out of it.

1. Relativity in Easy Steps
(A Primer for Amateur Scientists)
by John Winders

2. Note to my readers:
You can access and download this essay and my other essays through the Amateur
Scientist Essays website under Direct Downloads at the following URL:
https://sites.google.com/site/amateurscientistessays/
You are free to download and share all of my essays without any restrictions, although it
would be very nice to credit my work when quoting directly from them.

3. This essay describes two of the most important discoveries in science: The Special Theory of
Relativity and the General Theory of Relativity. Both of these discoveries were made by a single man,
Albert Einstein, over a period of one decade (1905 – 1915).
This essay is directed at an audience of amateur scientists like myself. I will approach these two
theories on the basis of their underlying principles, deriving as much as possible using basic geometry
and a bit of elementary calculus. I will not go into the depth needed to become a “relativist.” Mastery
of general relativity would require a good working knowledge of tensors, which I lack.1
Nevertheless, I
think amateur scientists like myself will get something useful out of it, so let's get started.
___________________
Nature is quirky. Whenever things don't quite match up, She changes them so they will. The results
often seem to be bizarre and nonsensical, but the more you study it you realize how profoundly wise
Nature is. It all started with a thought experiment that Einstein said he came up with at around the age
of 16. The young Einstein wondered what would happen if he chased a light beam and caught up with
it. A light beam is electromagnetic, consisting of two waves propagating in space, a magnetic wave and
an electrical wave. The most intuitive answer to Einstein's question is that the light would “freeze” so
that the two components would just hang in space. Maxwell's equations describe the propagation of
electromagnetism through space. The funny thing about those equations is that the speed of
propagation in a vacuum is a constant c = √1/μ0 ε0, where μ0 and ε0 are the permeability and
permittivity of free space, respectively. But c does not refer to any particular fixed reference frame, so
light could never “freeze.” Unlike sound or ocean waves, which travel at a constant speed with respect
to some material medium, light travels at a constant speed in any reference frame, period.
This simple conclusion is the basis of the Special Theory of Relativity, which Einstein published in
1905. Even if there is no fixed reference frame for light, there still must be a way to measure the
motion of any object using a metric that produces the same numerical result for all observers when they
are in relative motion to each other. Einstein examined the special case where all observers and the
object being measured were moving in uniform motion relative to one another, which is why it is called
the special theory of relativity. Einstein himself preferred calling it the theory of invariance.
The metric Einstein came up with is: s2
= c2
t2
– x2
– y2
– z2
A change Δs defines distance in space-time, which is invariant for all observers. For example, suppose
we observe an object moving with constant velocity relative to us, starting from a point (x1, y1, z1) and
ending at a point (x2, y2, z2). If Δt is the time it takes to travel that distance, we compute the distance it
traveled through space-time as:
Δs = √c2
Δt2
– (x2 – x1)2
– (y2 – y1)2
– (z2 – z1)2
Now the $64,000 question is what do we use to measure space and time in this instance? The answer is
that each observer uses his own measuring rod for distance and his own clock for time. It's that simple.
Now suppose you're sitting in your easy chair and not going anywhere. How fast are you traveling in
space-time? The obvious answer is that you're not traveling at all, but that would be wrong. It's true
that you're not traveling along any distance, but look at your watch. Is it moving? If so, then you're
moving through space-time. For each observer in the universe, his/her own space-time travel is
1 It's comforting to note that as brilliant as Einstein was, he still needed help with expressing the principle of general
relativity in mathematical form. The great mathematician David Hilbert was on hand to provide the help he needed.
- 1 -

4. measured simply by using a clock, because (x2 – x1)2
– (y2 – y1)2
– (z2 – z1)2
= 0.
Δs = c Δt light-seconds
In other words, everything in the universe is traveling through space-time at the speed of light,
regardless of how fast they are moving relative to each other!2
That's the special theory of relativity in
a nutshell. Now let's see how this works out in practice.
Suppose Alice marks off a starting point and ending point 0.866 light-seconds apart. Bob, her trusty
partner is attached to a rocket sled that accelerates him to 0.866 times the speed of light as he crosses
the starting point. Alice measures the time it takes him to reach the ending point traveling at that
velocity using a stop watch: Δt = 0.866 light-seconds / 0.866c = 1.0 second. She computes Bob's
travel through space-time as follows:
Δs = c √1.02 – 0.8662
= 0.5 light-second
Bob also measured his own space-time travel using his own stop watch. What do you think it will
read? That's right, it will read 0.5 seconds. How can that be? Well, Nature had to “glitch” space and
time a bit to make Alice's reality match Bob's. Alice and Bob both agree that Bob's speed relative to
Alice and Alice's speed relative to Bob are both 0.866c. However, they clearly don't agree on the time
it took Bob to travel between the two markers. So Nature had to fix that disagreement by reducing the
distance between the markers as seen by Bob. By moving the markers closer together, say 0.433 light-
seconds apart, Bob only takes 0.5 seconds to travel between them at a velocity 0.866c. You see,
whenever Nature is faced with a paradox, She'll just rearrange things – in this case She did it to both
space and time – in order to make things work out okay.
The slowing down of time shown on Bob's watch (as seen by Alice) and the shortening of the distance
traveled (as seen by Bob) are both computed using a factor symbolized by the Greek letter gamma (γ):
γ = 1/√1 – v2
/c2
, where v = the velocity Bob relative to Alice (and vice versa).
t' = γ∙t
d' = d / γ
But there is one more item Nature must tinker with in order to preserve her integrity: Mass. To get an
object moving relative to another object requires energy. When a 0.50 caliber bullet is fired from a
high-powered rifle, we can easily observe that it has lots of energy.3
That energy equals ½ m∙v2
.
However, if we chase after that bullet and catch up to it, we find that all its energy has disappeared. So
the energy an object possesses is relative to an observer and not absolute. Now you might think that
you can make a bullet travel as fast as you want simply by using more gunpowder and putting more
energy into it. But you can't, because γ turns into an imaginary number as soon as v > c, which is
definitely no good.4
So how can you put unlimited amounts of energy into a bullet, while still
maintaining its velocity (relative to you) below the speed of light? Well, Nature came up with a clever
solution: She makes the bullet heavier (relative to you) when you try to make it go faster. Eventually,
She makes the bullet so heavy that you simply cannot make it go any faster. Her trick is to apply the γ
factor to the rest mass, m0: m = γ∙m0. That way, anybody can add as much energy as they want to the
bullet, but its velocity relative to them will never be greater than c.
2 There is one important exception to that rule, which will be revealed shortly.
3 If you don't believe it, try standing in front of a high-speed 0.50 caliber bullet.
4 Nature definitely frowns on imaginary quantities.
- 2 -

5. This leads to an important result: Putting energy into an object increases its mass relative to whoever is
supplying the energy, so mass and energy are equivalent. If two things are equivalent, it means there is
a proportionality constant between them, and the proportionality constant between mass and energy
turns out to be c2
. In other words, e = m∙c2
, which is the final piece of the special-relativity puzzle.
Footnote 2 on the previous page mentioned that there is one exception to the principle that everything
in the universe travels through space-time at the speed of light. That exception is light itself. If you
compute Δs for a particle of light, you'll find that Δs always equals zero no matter which direction it
travels through space. Light is motionless in space-time. That kind of makes sense because if
everything is moving through space-time at the speed of light except light itself, then everything must
be traveling at the speed of light relative to light, which is exactly what is expressed by Maxwell's
equations and what Einstein discovered when he tried to chase a light beam in his imagination.
The fact that motions are relative means that two observers in relative motion see each other's clocks
slow down by the same factor γ, which leads to an apparent contradiction known as the Twins Paradox.
I must point out at this juncture that the Twins Paradox is almost always presented incorrectly in the
literature. The usual description goes something like this:
Alice and Bob are twins. Bob is an astronaut who goes on a journey to a distant Planet X, while Alice
remains earthbound. Bob blasts off in a rocket ship and accelerates to nearly the speed of light toward
Planet X. When Bob nears Planet X, he turns the ship around and fires off his rocket motors,
decelerating to a stop and then accelerating to nearly the speed of light toward Earth. When Bob finally
reaches Earth, he turns his ship around again and fires off his rocket motors one final time to decelerate
and land. His twin sister Alice is there to greet him, but whereas Bob is still a spry young astronaut, he
discovers to his horror that his sister has turned into an old hag. How can that be? If their relative
velocities were equal, why didn't Alice's aging slow down just like Bob's?
At this point, most authors use a lot of hand-waving to describe what is taking place with Alice and
Bob within the context of special relativity. The problem is that you cannot use special relativity to
explain what is happening to Bob and Alice when Bob is accelerating and decelerating like mad.
Special relativity only applies to uniform motion, which is why it's “special.” If you want to use
special relativity to describe The Twins Paradox, you can only have uniform motion – no acceleration
or deceleration is allowed.
The diagram on the following page is the correct version of the Twins Paradox using special relativity.
It requires a second astronaut, whom we will call Charlene. In this scenario, Planet X and Earth are
stationary with respect to each other and are 8.66 light-years apart. Bob happens to be whizzing by
Earth toward Planet X at a speed of 0.866c. As he passes by Earth, he communicates with Alice and
synchronizes his clock with hers. It is 12:00:01 am on January 1, 2017. He then coasts toward Planet
X at a constant speed of 0.866c. As he approaches Planet X, Bob observes Charlene heading in the
opposite direction toward Earth. She is also traveling at a constant speed of 0.866c relative to Earth
and Planet X.5
Bob communicates with Charlene and tells her to synchronize her clock with his, which
was previously synchronized with earthbound Alice's clock. Charlene obliges and according to Bob, it
is January 1, 2022. Bob continues on his way, and he can do whatever he likes from then on. Charlene
continues on her journey to Earth, which is still separated by 8.66 light-years from Planet X. As
Charlene whizzes by Earth, she communicates with Alice and compares her clock with Alice's.
According to Charlene it is January 1, 2027, but according to Alice it is January 1, 2037. Alice
apparently had “aged” ten more years than Bob and Charlene.
5 It so happens that Bob's and Charlene's speed relative to each other is 0.99c, but that's neither here nor there.
- 3 -

6. I think you will see that there really is no paradox at all, and this can easily be explained using special
relativity without the usual hand-waving needed to explain away Bob's acceleration.
The key is that the distance between Earth and Planet X are only 4.33 light-years apart in Bob's and
Charlene's reference frames. From Alice's perspective, it takes Bob ten years to reach Planet X and it
takes Charlene ten years to reach Earth. But from Bob's and Charlene's perspectives, each of their
journeys only take five years. Nature had to reduce the distance between Earth and Planet X for both
Bob and Charlene in order to slow down Alice's clock with respect to them. The distances that Bob and
Charlene traveled through space-time are the same for everybody.
That completes the part of the essay that derives the special theory of relativity, but there are a couple
of concluding remarks I want to add.
- 4 -

7. • Before Einstein published the results of his theory of invariance, a.k.a. special relativity,
scientists believed that since light is a vibrating wave, it needs a medium to vibrate. They
proposed this medium was comprised of a substance they called æther (pronounced ee-ther).6
If
that were true, the speed of light would vary depending on an observer's velocity through the
æther. Since the Earth revolves around the Sun, its relative motion with respect to the æther
should vary seasonally, and they should be able to measure this variation. In 1887, two
scientists, Albert Michelson and Edward Morley, set up a sensitive device called an
interferometer that was designed specifically to measure the Earth's “drift” through the æther.
Repeated attempts failed to uncover any drift at all. Various theories were put forward to
explain this lack of success, including one that proposed that objects somehow become shorter
by the factor γ in the direction of the æther drift. Amazingly, after Einstein published his
landmark theory in 1905, he stated that he had never heard of the Michelson-Morley
experiment, even though his special theory of relativity explained its results perfectly!7
• I think some people refer to space-time as a “four-dimensional space-time continuum” because
they are trying to impress other people by using more words than are necessary. The problem
with using that term is it implies that time is just another ordinary spatial dimension, so we tend
to represent the universe as some sort of four-dimensional Euclidean object. The geometry of
space-time is not Euclidean. You can certainly map points in space-time onto points on a sheet
of paper, but the resulting diagram on paper does not represent any physical object. Distances
in space-time are based on taking the difference between c2
Δt2
and (Δx2
+ Δy2
+ Δz2
) and not
the sum of the squares of all four dimensions, as would be the case if space-time really were
Euclidean. Thus, whenever someone (most often a cosmologist) shows you some sort of
diagram or artist's rendition of the “universe,” be rest assured those pictures are wrong.
• Because light doesn't move through space-time, Δs = 0 leads directly to the following equation:
c Δt2
= Δx2
+ Δy2
+ Δz2
. This is nothing other than the equation of a spherical wave front
expanding at a constant speed, c. In other words, it's the equation of light propagating through
space. If we chop off one of the dimensions, we get an expanding circle. By plotting
expanding circles along a time axis in both the positive and negative directions, we get a pair of
expanding cones.
Now we're ready to proceed with a much trickier proposition: What happens when objects accelerate.
For that we need to dive head-first into the general theory of relativity (GR).
___________________
Einstein said that one of his happiest moments was when he watched a man fall off a ladder. At that
moment he realized that the poor guy didn't feel a thing (until he hit the ground). Actually, I don't think
he really saw a guy fall of a ladder; instead, he imagined what it would be like to be in free fall near the
Earth and he realized it would be the same as being in a place that had zero gravity. Somehow,
accelerating toward the Earth canceled the “force” of gravity.
If you put a charged object in an electric field, it will accelerate in the direction of the field, and an
accelerometer attached to that object will register an acceleration. But if you put a mass in a
gravitational field, it will still “accelerate,” but an accelerometer won't register anything. This led
6 Although pronounced the same, this substance is not the same as ether that puts people to sleep.
7 I guess that shows what happens when you don't have the Internet.
- 5 -

8. Einstein to conclude that gravity wasn't a “force” after all because objects in free fall don't feel it. But
if gravity isn't a force, then what is it? Einstein concluded that gravity is equivalent to acceleration.
This brilliant insight started Einstein on a ten-year quest to form a new theory of gravity based entirely
on objects moving through an altered geometry of space-time.8
In order to do this, he employed
multidimensional mathematical objects known as tensors, which express sets of linked differential
equations that have to be solved together simultaneously. Tensors are way, way, way beyond my
limited mathematical abilities, so I must completely gloss over them.9
But there is still a lot about GR
that we can grasp using simple math based on a few underlying principles.
Suppose Bob is standing on the surface of a planet that has an ultra-strong gravitational field. High up
in a balloon is his assistant Alice. Bob sends a signal to Alice using a laser pointer, but when the signal
reaches her, she notices that the frequency of the light has shifted downward. It's as if time on the
surface is slow. Using a pair of binoculars, Alice observes Bob on the surface of the planet and notices
that everything in Bob's vicinity has indeed slowed down. How can we explain this?
If Bob were communicating with Alice by shooting marbles at her, the marbles would have to lose
kinetic energy as they travel upward against gravity, which they manage to do by slowing down. But
since Bob is communicating using light, the light can only lose energy by slowing down its frequency
since it must travel at the speed of light. A light packet – a photon – has an energy ep = ħ∙f, where ħ is
Planck's constant and f is the light's frequency. This energy is equivalent to mass, per SR:
mp = ep / c2
= ħ∙f / c2
As the light packet ascends in the positive y-direction through a gravitational field g(y), it loses energy:
dep = – g(y) mp dy = – g(y) (ħ∙f / c2
) dy = ħ df
df / f = – g(y) dy / c2
Solving the simple differential equation above is easy:
f (y) = f (0) exp{– ∫0
y
g(y) / c2 dy}
As the observer Alice increases her height above Bob, the frequency of Bob's laser signal decreases
exponentially. Bob is at the bottom of a so-called “gravity well,” and not only does Bob's laser slow
down, but everything else in Bob's world slows down relative to Alice's world.
So by introducing gravity into the picture, Nature apparently decided that She had to tinker with time.
It would seem logical that Nature would tinker with space as well, and that's quite true. However, the
exact formula She uses for that isn't so easy to determine. If it were easy, Einstein wouldn't have spent
ten years of his life struggling to find out what it is. But we can still approach the problem in sort of an
intuitive way, also relying on the information we gleaned from special relativity.
Free falling in a gravitational field is exactly equivalent to floating in space without gravity. Also,
standing still in a gravitational field is exactly equivalent to accelerating in space without gravity. Now
suppose Bob were in an enclosed room where he couldn't tell if he were standing on a planet with
gravity or accelerating in space. There is a laser pointer attached to one of the walls of the room that is
8 At this point I must stress that I believe that while space-time is a convenient frame of reference, it isn't an actual
physical “thing.” Of course, I could be wrong.
9 I'm going to use the excuse that tensors are for nerds.
- 6 -

9. aimed at the opposite wall. When the room is free-floating without any gravitational force, the light
beam hits a certain spot on the wall, but while the room is sitting on the planet, Bob notices that the
beam hits the wall slightly below that spot. He suspects the laser's aim may be a little off, so decides to
fill the room with smoke and trace the beam's path. To his amazement, the beam curves (!), but light
ain't supposed to curve. Then Bob realizes he's either sitting in a gravitational field or the room he's
sitting in is accelerating in empty space. If he were shooting marbles across the room, the marbles
would trace out parabola-shaped paths in either case. Similarly, the light should trace a path that
approximates a parabola in an accelerating room or when sitting in a gravitational field.10
Would Alice also notice the light path bending near the planet? You betcha. There's no way that
Nature could hide such a glaring anomaly from Alice, although She may have to tweak the exact shape
of the path a little before revealing it to Alice.
What would happen if Bob were free falling toward the planet? Would he see light bend then? The
answer is no. A free-falling Bob would not feel any effect from gravity, so as far as he's concerned he
could be a billion miles away from any planet. The light just traces a perfectly straight line and hits the
original spot on the opposite wall, with a constant horizontal velocity. This brings up an interesting
question about Bob's free fall. If the gravitational field were constant and the light path were an exact
parabola with a constant horizontal velocity, then a constant downward acceleration could straighten
out the curved path from Bob's perspective, in compliance with Newtonian physics. But the light's
horizontal velocity is changing, so in order to straighten the light path and restore a constant horizontal
velocity, Bob's free-fall acceleration is not quite constant in a uniform gravitational field. This example
illustrates why the math of GR is so difficult. In Newton's world, you never have to worry about
details like making sure the velocity of light is always constant. In Einstein's world you do.
In summary, an observer looking at another frame of reference where gravity is present would see time
slowing down and light curving. If you could trace a particle of light as it passes near the Sun, you
would definitely see it curving and slowing down. Although all observers must measure the same
speed of light in their own frames of reference, they can observe light traveling at different speeds in
other frames of reference. The Shapiro time delay is a case in point. When a radar signal is bounced
off Mars or Venus and the signal's path almost grazes the Sun, there is an additional time delay
10 The fact that light traces out a curve that approximates a parabola is important. The horizontal velocity of the marbles
is constant, which is why they trace out a parabola. On the other hand, the horizontal velocity of light cannot be
constant, because the total velocity, combining horizontal and vertical velocities, must equal c. Thus, the horizontal
velocity decreases as the vertical velocity increases, so the curve bends a little more than a parabola. This is one of the
ways GR differs from Newtonian physics.
- 7 -

10. compared to radar signals that don't pass close to the Sun. The only explanation for this is that light
slows down in the vicinity of the Sun, which means time itself slows down relative to a distant observer
due to the Sun's gravity.
The important thing to remember is that even though light paths bend in space, light travels in perfectly
straight lines through space-time. In fact any object not being pushed or pulled by any forces (besides
gravity) travels in perfectly straight lines through space-time. Those lines are called a geodesics, and in
space-time, they're the longest possible paths, whereas in normal space, geodesics are the shortest
possible paths. Remember how an observer measures his or her own travel in space-time? Yes, by
using a clock. So the longest possible space-time path for an observer is the path that maximizes the
elapsed time on the observer's own clock. This invokes a very important principle of physics: The
principle of least action, which we'll delve into a little later.
But next, we need to revisit the Twins Paradox. This time, we're letting Bob accelerate.
In this version, Bob blasts off from Earth at a constant acceleration headed toward Planet X. Because
of the equivalency between gravity and acceleration, Bob might think the entire universe is immersed
in a giant gravitational field (marked with the green arrows in the diagram above) pulling the Earth,
Alice, and Planet X in a direction behind his rocket ship. In that case, Bob would find himself at the
bottom of a huge gravity well, making a clock on Planet X (shown in blue) speed up relative to Bob.
Alice is a little farther down in that gravity well, so her clock (shown in red) slows down relative to
Bob. But the cumulative effects of speeding up or slowing down build up exponentially over distance,
so whereas Alice's clock slows down a little, the clock on Planet X speeds up a lot.
When Bob reverses direction at Planet X, he again finds himself at the bottom of another huge gravity
well with Alice far, far away at the top of it. This makes her clock (in blue) speed up a lot relative to
Bob. The combined effects of both gravity wells cause Alice to age much faster than Bob over the
- 8 -

11. course of his trip, although the results are not even close to those you'd get from special relativity. Of
course, the time-dilation and distance-dilation effects from Bob's motion relative to Alice à la special
relativity must be added to the effects of Bob's acceleration, but it would be nearly impossible to
account for both sets of effects without diving into a full-blown GR analysis, which I'm not going to do.
I can assure you that the results of a GR calculation would be far different than the simple case of
special relativity.
The principle of least action is revealed in many different areas of physics, including optics, mechanics,
and relativity. In optics, refraction is explained as light taking the shortest optical path, thereby
minimizing action. Newtonian physics, the trajectory of an object in a gravitational field is the path that
minimizes action integrated over time. Here, action is defined as an object's instantaneous kinetic
energy minus its instantaneous gravitational potential energy. In order to minimize action over time,
kinetic energy should be kept to the smallest possible value, thereby maximizing the potential energy.
The resulting path will depend on the details of how action is computed. Computing kinetic energy is
straightforward; ek = ½ m∙v2. The simplest way to compute gravitational potential energy is ep = m∙g∙h,
which assumes the gravitational acceleration, g, is constant with respect to the height, h, above the
Earth's surface. Minimizing (ek – ep) integrated over time requires a rather difficult technique
minimizing something called the Lagrangian. The resulting trajectory is a parabola – no surprise.
We can refine the computation of gravitational potential energy by using ep = – m∙M∙G / r, where M is
the Earth's mass, G is the gravitational constant and r is the distance between the object and the center
of the Earth. Minimizing (ek – ep) integrated over time then results in a elliptical orbit around the
Earth's center – again no surprise. Over short distances, a parabola very closely approximates an
ellipse near its vertex. Within the confines of a baseball stadium, a simple parabola is accurate enough
to describe the flight of a baseball, but it's not nearly accurate enough for traveling to he Moon.
We saw how Newtonian physics deviates slightly from GR in computing the path of light near a
gravitating planet. GR has its own definition for the gravitational component of action, called the
Einstein-Hilbert action:11
S = 1/2κ ∫ R √- g d4x
This is a fairly innocent-looking expression, but be warned that we're now dealing with four-
dimensional objects12
and the d4x means the integral has to be solved four separate times in four
dimensions over the whole of space-time. After you finish all the horrendous math required to solve
this thing, you get something that's tantalizingly close to an elliptical orbit derived from Newton's laws,
with a small difference. It turns out that Mercury's orbit around the Sun is highly elliptical, and the
perihelion of its orbit shifts forward over time. The measured shift is 43 seconds of arc per century
greater than Newton's laws predict, and that small discrepancy had puzzled astronomers for a very long
time. When Einstein ran the numbers for Mercury's orbit using GR, he found the shift matched the
astrological measurements to a tee. Eureka! Einstein knew then he was on the right track.13
But duplicating existing measurements of the precession of the perihelion of Mercury's orbit wasn't
enough to propel Einstein into fame and celebrity. The general theory of relativity needed to make a
falsifiable prediction involving something that had never been accurately measured before: The
bending of light near the Sun. That confirmation would happen during the May 1919 solar eclipse.
11 It is said that Hilbert came up with this alone, but he graciously allowed Einstein to share the credit with him.
12 R is the Ricci scalar and g is the determinate of the metric tensor matrix (as if I know what those are).
13 Einstein recalled that when he made this discovery, he was so excited he couldn't sleep that night.
- 9 -

12. Appendix A – Chasing Moonshadows
One of the nice things about the Moon is that its apparent size is almost equal to the Sun's apparent size
as seen from Earth's surface. So about two times every year, the Moon totally blocks out the Sun
somewhere on Earth. During these events, called total solar eclipses, the shadow of the Moon passes
over the Earth from west to east,14
turning day into night on the surface. Stars that would ordinarily be
blotted out by the glare of the Sun and a blue sky then become visible. So an eclipse would be a great
opportunity to test out Einstein's prediction from GR that light paths are bent toward the Sun by the
Sun's gravity.15
Having taken a 2-year breather after publishing his special theory of relativity, Einstein began working
on GR in 1907. He published an interim version of GR in a paper entitled “On the Influence of
Gravitation on the Propagation of Light,” appearing in Annalen der Physik in June, 1911. In this
version of GR, he recognized that time slows down in a gravitational field, so the speed of light in one
reference frame doesn't equal c when observed from a different gravitational reference frame. Using
that information and the mass/energy equivalency from special relativity, he predicted the image of a
star near the limb of the Sun would appear to shift toward the Sun's center with respect to images of
stars farther away from the Sun. His formula for the angle of the path's deflection is given below.
α = 2∙G∙M / c2∙Δ, where M is the mass of the Sun and Δ is the distance from the center to the path
The number worked out to be 0.85 seconds of arc, denoted as 0.85''. This is an incredibly small angle,
but a German astronomer named Erwin Finlay-Freundlich thought he could detect it. Freundlich was
an associate and a great admirer of Einstein, and he believed wholeheartedly that gravity bends light.16
So he set about organizing an expedition to observe the August 21, 1914 solar eclipse and measure the
deflection of the images of stars near the Sun. That particular eclipse traversed the following countries
(in chronological order): Greenland, Norway, Sweden, [Latvia, Lithuania, Belarus, and Ukraine, which
were all part of Russia in 1914], [Turkey and Iraq, which were part of the Ottoman Empire in 1914],
Iran, which was called Persia in 1914, and Pakistan, which was part of India in 1914.
It turned out that Freundlich's expedition was star-crossed.17
He decided that a place called Feodoriya
on the Crimean Peninsula would offer the best possibility for making his observations. As fate would
have it, World War I broke out in July, 1914 while he was in Crimea setting up his experiments.
Crimea belonged to Russia, and Germany and Russia were now enemies. Being a German citizen,
Freundlich was detained and interred there for a little while before being allowed to return to Germany.
Tragically, the Russians wouldn't allow him to carry out his experiments, so he missed out on the 1914
eclipse completely.18
There was a silver lining around that particular cloud, however. The angle of deflection that Einstein
had predicted in his 1911 paper, 0.85'', was way off. Had Freundlich been able to do his experiments
with a high degree of accuracy, his measurements would have contradicted Einstein bigly. What would
14 The Moon's shadow moves more than 1,000 mph over the surface, so a supersonic jet might be able to keep up with it.
15 Nowadays, scientists can create an artificial solar eclipse by putting a camera into space and blocking out the Sun's
image with an opaque circular disk.
16 Actually, Newton believed it too. He thought that light consisted of tiny particles called corpuscles, which traveled
really fast. So since gravity attracted everything else, why not light corpuscles too? Ironically, the angle of deflection
using Newton's laws worked out to be the same number as predicted in Einstein's 1911 paper.
17 Pun intended.
18 My research couldn't determine if the Sun was even shining on Feodoriya on August 21, 1914.
- 10 -

13. have happened to Einstein then? Would he have given up working on GR and settled for a long, boring
career at the Swiss Patent Office? This is one of those interesting questions that historians like to
debate, like what if Robert E. Lee had defeated Ulysses S. Grant at the Battle of Gettysburg, or what if
Germany had perfected the atomic bomb in early 1944? Sometimes it's better to be lucky than good,
and it turned out that Einstein was lucky enough to be given a second chance.
While the “War to End All Wars” was still raging across Europe in November, 1915, Einstein submitted
four papers to the Prussian Academy:
1. “Fundamental Ideas of the General Theory of Relativity and the Application of the Theory in
Astronomy”
2. “On the General Theory of Relativity”
3. “Explanation of the Perihelion Motion of Mercury from the General Theory of Relativity”
4. “The Field Equations of Gravitation”
The first paper cleared up some of the misconceptions in the 1911 paper, but the second paper
contained a glaring error. The final and correct version of GR was found in the fourth paper. The later
version of GR raised the predicted angle of deflection of starlight near the Sun to 1.75'', making it
somewhat easier to detect during a solar eclipse.
Sir Arthur Eddington, a British astronomer extraordinaire, was very impressed by Einstein's new theory
and he considered himself as being one of very few experts who understood it.19
After WWI formally
ended on November 11, 1918, European scientists, who were former enemies, could then freely
collaborate on their work. Plans were made to confirm the 1.75'' gravitational deflection predicted by
Einstein during an upcoming solar eclipse on May 29, 1919. Nobody was taking any chances on the
weather, so there were two expeditions – one led by Eddington on the island of Principe off the coast of
Africa, and the other in Brazil led by Sir Frank Watson Dyson.
On Principe, Eddington used the fairly bright Hyades Cluster for his experiment. A lovely photograph
of the cluster is shown below.
The weather turned out to be ideal in both locations. Eddington and Dyson snapped photographs of
stars close to the edge of the Sun, and then returned to England to analyze them. Eddington compared
19 Eddington, to put it politely, was full of himself. It is said when a writer complimented him on being one of three
people in the world who understood GR, Eddington paused in silence. The writer said, “Don't be modest, Eddington.”
Eddington replied, “On the contrary. I am trying to think who the third person is.”
- 11 -

14. this eclipse photographs to photographs of the region around the Hyades Cluster taken in January and
February (at night when the Sun was absent). In 1920, Eddington and Dyson co-authored a paper
claiming the angular deflections of stars near the Sun they measured during the 1919 solar eclipse
perfectly matched Einstein's prediction!
To have some idea of how tiny 1.75 seconds of arc is in comparison to the apparent size of the Sun,
look at the three figures below.
The disk in the first figure represents the Sun, which subtends an angle of 0.52 = 1872'' as seen from
the Earth's surface. A deflection of 1.75'' is less than 0.1% of that angle. The small box over the Sun's
edge is blown up 10x in the second figure. The small box over the Sun's edge in the second figure is
blown up 10x again in the third figure. The two tiny red dots shown in the third figure are stars
separated by 1.75'' drawn to scale. Was it really possible for Dyson and Eddington to measure that
small a separation with the kind of accuracy required to distinguish 1.75'' from, say 0.85''?
There's a phenomenon known as annual aberration of light due to the orbital motion of the Earth
relative to distant stars. The Earth revolves at a velocity of 30 km/sec around the Sun. This changes
the relative motion between the Earth and the stars by 30 km/sec every six months, displacing their
apparent positions toward the direction of the Earth's motion. The figure below illustrates this effect
for stars observed during a May 29 solar eclipse versus observed at night in January/February.
The red stars represent “true” positions, and the blue stars represent apparent positions due to annual
aberration. The maximum annual aberration of a star is on the order of 20''. That's quite a lot
compared to 1.75''. Also, the Earth is rotating on its axis. The Earth's rotation causes an angular shift
called diurnal aberration. So if you're observing a star directly overhead at the equator, your relative
- 12 -

15. motion with respect to it is different than when you're observing the star when it's closer to the horizon.
Were either of these effects taken into account in Dyson's and Eddington's analysis, or would
everything just come out in the wash because the stars in the general direction of the Hyades Cluster
would all have the same annual and diurnal aberrations? I did some research on that question, but I
couldn't find an answer.
While we're on the topic of aberration, it would be worthwhile to explore where it comes from, in light
of what we learned from special relativity. Think of running through a rain shower. If the rain drops
are falling vertically, the front of your body will be soaked while the back of your body will stay dry.
In your moving frame of reference, it appears that the direction from which the rain is falling has
shifted toward the direction you're running. Similarly, if a star is directly overhead and you run in a
horizontal direction, the light rays falling vertically will appear to shift forward a little. That's the
classical explanation of aberration.
Things get a little more complicated with special relativity, because velocities don't add together the
same way as they add in classical physics. It's one of the consequences of light traveling at the same
speed in the frames of reference of all observers. If a booster stage of a rocket is traveling at 0.8c and
the second stage blasts off at 0.6c relative to the booster, the total velocity is not 0.8c + 0.6c = 1.4c. It's
0.946c instead. Taking relativity into account, let θ be the angle of the star's “true” position above or
below the observer's direction of motion, and φ the star's apparent angle above or below the observer's
direction of motion.
tan (φ) = sin (θ) / [γ (cos (θ) + v/c)]
The difference between φ and θ is the aberration. You'll remember that γ =1/√1 – v2
/c2
. When v << c,
the formula above still works using γ = 1.
The visible part of the Sun that's covered by the Moon during a solar eclipse is called the photosphere.
The photosphere is not a solid surface; it's simply a spherical surface from which most of the Sun's
light radiates. The actual “Sun” extends many miles beyond the photosphere, gradually thinning out
into space. The corona is a super-hot region immediately surrounding the visible disk, which is only
visible on Earth during a solar eclipse due to the blinding brightness of the photosphere. In other
words, the Sun has an atmosphere that extends quite a distance from the visible disk.
Feng Xu published a paper in the April 2002 Solar Physics journal, entitled “Light Deflection Near the
Sun's Limb: Refraction by the Solar Atmosphere.” The paper's abstract is quite interesting:
“Light refraction by the Sun's atmosphere is calculated. As detected from the Earth, the refraction can deflect a light ray
emitted from the Sun's limb by 13′′ or a starlight ray grazing the solar limb by 26′′, an effect 15 times larger than the
gravitational deflection.”
I couldn't find out anything about Feng's credentials (the Chinese name Feng Xu is about as common as
the English name John Smith), so I cannot offer any opinions about the author's veracity or the validity
of the claims in the paper. But I couldn't find any evidence that Dyson and Eddington even considered
the effect of refraction by the Sun's atmosphere in their conclusions. Were their conclusions published
in their 1920 paper an example of confirmation bias? Did they simply throw away data from their
photographs that didn't match the 1.75'' deflection they were looking for? Were Dyson's and
Eddington's equipments even technically capable of measuring such small a deflection with accuracy
and precision? Who knows, although these questions do make a good conspiracy theory. Fortunately
for Einstein, there have been many other confirmations of his general theory of relativity using much
more modern and precise techniques. So I think his reputation is still quite secure in history.
- 13 -

16. Appendix B – Más Masa
In the earlier part of this essay, I talked about how the mass of an object increases relative to an
observer when their mutual velocity increases. I said it's Nature's way of storing unlimited amounts of
kinetic energy within an object to avoid raising its speed above c. I'm now going to try and give a
scientific reason why m = γ∙m0 and how that leads directly to e = m∙c2.
The simplest explanation for why m = γ∙m0 is to consider transverse motions of objects in a reference
frame that is moving at relativistic velocities relative to an observer. If a mass, m0, moves at a velocity,
u, in the moving reference frame, it results in a momentum, p = m0∙u in that frame. Here, bold letters
indicate the quantities are vectors, having directions as well as magnitudes. If the moving frame's
velocity is v with respect to an observer and u is in the transverse direction to v, then p is also in the
transverse direction to v. Since the observer sees time slowing down in the moving frame, the
transverse velocity would appear to slow down by the same factor γ: u' = u / γ. In order to conserve
transverse momentum, p' measured in the observer's frame must have the same value as p in the
moving frame. In order to do accomplish this, the apparent mass of the object, m, as seen by the
observer must increase by the same factor as u' decreases:
p' = m∙u' = (γ∙m0) (u/γ) = m0∙u = p → m = γ∙m0.
Now we're ready to explore the relationship between m = γ∙m0 and e = mc2. There have been many
proofs of e = mc2 given over the years. Einstein presented his famous “two blackboard derivation” in
1934. I've explored a number of proofs and derivations, but many of them seem a bit “hand-wavy” to
suit me.20
A very simple derivation involves an approximation to γ = 1/√1 – v2
/c2
when v<<c. It turns out that
when γ ≈ 1, you can substitute (1 + v2 / 2c2) for γ. Combining this with the classical definition of kinetic
energy, ek = ½ m∙v2, we get
m ≈ m0 (1 + v2 / 2c2) = m0 + ½ m0∙v2 / c2 = m0 + ek / c2
ek / c2 ≈ (m – m0)
In other words, boosting the energy of an object by e increases its apparent mass by approximately
e / c2. That's fine, but Nature frowns on approximations.21
She wants exact derivations that don't
depend on any approximations or hand waving. I found a pretty good exact derivation in a video
produced by Physics Reporter, which you can go to here: Video Tutorial on e=mc^2
The proof uses a bit of calculus as follows.
m = γ∙m0 → m2 (1 – v2/c2) = m0
2
m2 c2 – m2 v2 = m0 c2
Taking the derivative d/dm of both sides of the equation:
2m c2 – 2m v2 – 2m2 v dv/dm = 0 (The derivative of the right side = 0 since m0 and c are constants)
20 Somebody actually posted a proof using a radioactive cat traveling at relativistic speeds and emitting gamma rays. I
wonder if it was Schrödinger's cat after accidentally swallowing the radioactive isotope used in the experiment.
21 Actually, this approximation can be resolved by noting that equating kinetic energy to ½ mv2 is itself an approximation
that applies when v << c. Using the relativistic definition of kinetic energy would result in the exact equation e = mc2.
We'll see how that works out in the next proof.
- 14 -

17. Multiplying both sides of the above equation by dm and dividing both sides by 2m, we get
c2 dm – v2 dm – m∙v dv = 0
c2 dm = v2 dm + m∙v dv (Equation 1)
Let dW be incremental work done on the mass, m, by a force, F, applied over an incremental distance,
ds, in the direction m is traveling with respect to the observer.
dW = F∙ds
F = dp/dt = d(m∙v)/dt = m (dv/dt) + v (dm/dt) (This accounts for changes in both v and m)
dW = m (dv/dt) ds + v (dm/dt) ds
Note that ds/dt is just the instantaneous velocity, v. Substituting that into the above equation, we get
dW = m∙v dv + v2 dm (Equation 2)
By comparing Equation 2 with Equation 1, it's easy to see dW = c2 dm. Integrating dW and c2 dm, we
get the following.
e – e0 = ∫e0
e
dW = c2 ∫m0
m
dm = (m – m0) c2
This is the most direct proof I could find for proving e = m∙c2, beginning with m = γ∙m0. It involves
adding energy incrementally to a mass, m, traveling at a relativistic velocity, v, and computing the
resulting change in mass without using any low-velocity approximations or sleights of hand. The trick
was to find the derivative of (m / γ)2 with respect to m and set it equal to zero.
It seems like the root cause of the apparent mass of an object increasing relative to an observer when
the object is moving relative to that observer can be traced to the fact that relative motion slows down
time in the object's reference frame from the observer's point of view. That brought up an interesting
question in my mind: If gravitation slows down time relative to a distant observer, does gravitation
also increase the apparent mass of an object relative to a distant observer? Or to put it another way,
does acceleration increase the inertial mass of an object? I think that if conservation of momentum
holds and apparent velocity decreases, the answer to both those questions would have to be a
resounding yes. But since inertial mass is the same as gravitational mass, wouldn't an increase in
inertial mass of an object also increase the gravitational field around it? Just thinking about an endless
regression like this makes my head hurt. Unfortunately, I wasn't able to find a good reference that
gives a definitive answer. Using Internet search engines entering questions like “does gravitation
increase mass?” mainly generates responses like, “Why yes, mass does produce gravity,” which is
certainly true, but it's not exactly the answer I was looking for.
I suspect that acceleration affects space in the transverse direction as well as in the direction of
acceleration, so the resulting reduction in transverse momentum may not be proportional to the slowing
down of time. In other words, if there is any increase in inertial mass due to gravity, it may not be
proportional to the slowing down of time and it may have to vary with direction.
I did find a paper “Gravity, Not Mass Increases with Velocity” by Eli Peter Manor, published in the
August 2015 edition of Journal of Modern Physics. It is found at this link: Eli Peter Manor Paper.
Manor claims that when an object moves in the direction of a gravitational field, its inertia is different
than when it moves in other directions. That's very interesting. I'll keep you posted.
- 15 -

18. Appendix C – QM Meets SR
Scientists complain a lot about the lack of a single theory that unifies quantum mechanics and gravity
through the general theory of relativity. However, there's a very deep and beautiful connection between
special relativity, SR, and quantum mechanics, QM, which we'll explore in this appendix. Suppose a
laser emits green light and is boosted to some relativistic speed, v, as depicted below.
A box is drawn around the laser that defines a moving reference frame traveling at v. Let's see what
goes on inside the box. The laser emits equal amounts of light (quanta) in two directions. For every
quantum heading in the direction of motion of the laser, there is an identical quantum heading in the
opposite direction. Let the wavelength of the laser light equal λg. According to quantum mechanics,
the energy of each quantum equals h∙c / λg, where h is the Planck constant. A light quantum also carries
a momentum, p, equal to h / λg. After emitting the green quanta, energy exits the reference frame. The
total energy lost is Δe = 2h∙c / λg. This translates into a reduction in mass using m = e / c2:
Δm0 = – 2h / c∙λg
The change in momentum of the system inside the box is zero, because the two photons are going in
opposite directions and carry off equal amounts of momentum. The speed of the laser, v, remains
unchanged, although reducing its mass will reduce its forward momentum by Δm0 ∙v.
Now let's look at what a stationary observer outside the box sees. The green light emitted in the
forward direction of travel will leave the box blue-shifted, shortening its wavelength to λb. The green
light emitted in the backward direction of travel will leave the box be red-shifted, lengthening its
wavelength to λr. The blue and red shifts combine two factors: 1) relativistic time dilation, and 2) the
Doppler shift. Formulas can be derived entirely from classical wave mechanics and Lorentz
transformations used in SR. I won't bother to derive them here, but will simply write them down:
λb = λg √(1 – v/c) / (1 + v/c)
λr = λg √(1 + v/c) / (1 – v/c)
The total energy exiting the box, Δe', as seen by the stationary observer is the energy of the blue
quantum plus the energy of the red quantum.
Δe' = h∙c / λb + h∙c / λr = h∙c [√(1 + v/c) / (1 – v/c) + √(1 – v/c) / (1 + v/c) ] / λg = 2 h∙c∙γ / λg
Δe' = γ Δe
This is what you would expect; i.e., an observer sees more energy emitted from a moving laser than
from a stationary laser; if a fixed mass increases by a factor γ when it's in motion, then relative motion
should increase any loss of mass (or energy) by the very same factor. Now comes the fun part.
- 16 -

19. When we look at the momentum of the light exiting the box, things definitely are not balanced. A blue
light quantum has more momentum than a red light quantum, so there is net change in the momentum
of the moving reference frame as seen by the stationary observer. Specifically,
Δp = pr – pb = h / λr – h / λb = h [√(1 – v/c) / (1 + v/c) – √(1 + v/c) / (1 – v/c) ] / λg = – 2 h∙γ∙v / c∙λg
By definition p ≡ m∙v
Δp = v Δm + m Δv
We saw from examining the system from within the moving frame that Δv = 0, which means that the
relative speed, v, doesn't change in the stationary reference frame either. Therefore, if Δv = 0, the only
way p can change is to change m.
Δm= v Δm' = – 2 γ∙h∙v / c∙λg
Δm = – 2h∙γ / c∙λg = γ Δm0
Here, we derived the change in mass observed from outside the moving reference frame entirely from
the change in momentum, even though the change of momentum measured inside of the moving frame
was zero and there was no change in relative speed! What's so amazing to me is that two seemingly
disconnected theories – special relativity and quantum mechanics – are so interdependent. There is
nothing in special relativity that relates to light quanta and there is nothing in quantum theory that
relates to Lorentz space contraction and time dilation in moving frames of reference. And yet the two
theories blend and work together seamlessly.
In fact, this scenario is another way to prove e = mc2. The equation Δm = – 2h∙γ / c∙λg works even when
v << c and γ = 1. In this case, – 2h / c∙λg = Δe / c2, so QM can be used to actually prove Δe = Δm0∙c2.
Nature has some pretty neat tricks up Her sleeve.
This analysis works for the general case when a body in motion emits light in all directions. The light
emitted ahead of the moving body is blue-shifted while the light emitted behind the moving body is
red-shifted. For each quantum emitted, there is a corresponding p. Combining all p values together
produces a net negative Δp, which reduces the mass of a moving body instead of slowing it down.
Comparing the Δm to the energy emitted, Δe, reveals Δe = Δm0∙c2 → e = m0∙c2.
- 17 -

20. Appendix D – The Amazing Travels of Johnny Photon
Throughout his life, Einstein wondered exactly what a photon is. Is it a wave? A particle? Both?
Neither? Photons are simply quanta or energy packets that can't be subdivided. As a youth, Einstein's
imagined what it would happen if he caught up to a light wave. Instead of catching up to a light wave,
I'd like to imagine what it would be like to actually become a light wave named Johnny Photon. In the
diagram below, Johnny is sitting in a spaceship at rest relative to 12 stars positioned all around him.
The stars emit green light, which is kind of unusual,22
and are they spaced evenly 30º apart around a
circle. Seven of the stars are labeled 1 through 7. Johnny blasts off toward Star 1 and as his speed
approaches 86.6% of light speed, things change drastically for him, as shown in the figure below.
The directions of all the stars have shifted forward from their “true” positions relative to Johnny's
motion, with the exception of Star 7, which is still directly behind Johnny. Actually, the term “true”
22 We don't see green stars in the heavens for a very good reason. Our Sun really isn't a “yellow dwarf” star as stated in
some astronomy books. It's actually a “green dwarf” because the spectrum of sunlight has a peak intensity at green
wavelengths. So why doesn't the Sun look green? Because we evolved on Earth, our eyes have completely adapted to
sunlight and interpret its greenish color as “white,” so red-hot objects look red and blue-hot objects look blue, but all
green-hot objects look white to us.
- 18 -

21. position is a misnomer: “True” positions are where things are observed to be, and not where we think
they “aught” to be based on where we observed them in the past.
The formula that gives the altered positions is the formula for spatial aberration found in Appendix A:
tan (φ) = sin (θ) / [γ (cos (θ) + v/c)]
Two other things occur as the spaceship reaches 86.6% of light speed relative to the stars:
1. Light from stars move toward the front of Johnny's spaceship are blue-shifted according to the
relativistic Doppler effect, which also takes into account time dilation from relative motion.
Stars that are still located behind the ship are red-shifted.
2. Distances between the space ship and the stars shrink in the direction of motion due to Lorentz
contractions. This squashes the circle of stars around the spaceship into an ellipse, and squashes
the universe itself into an ellipsoid.
As starlight reflects off the surface of Johnny's spaceship, he detects a definite “drag” tending to slow
him down. This is due to the fact that blue-shifted photons carry more momentum than red-shifted
ones.23
As these photons bounce off the spaceship, they impart a backward force.
Johnny puts the pedal to the metal and speeds up to 99.999% of light speed. The blue light is first
shifted to purple, then ultraviolet, then X-rays, and finally gamma rays, which penetrate the spaceship
and begin to fry Johnny to a crisp. Everything outside the spaceship has been compressed into a tiny
ultra-blue-shifted cone in front of it, while all the points that were exactly 180º in back of it have
collapsed together into an ever-shortening line, red-shifted into oblivion.
While everything outside the spaceship is shrinking, the ship itself retains its original dimensions, so
Johnny is literally running out of room, and it's becoming impossible to gain more speed. The only
way Johnny can go any faster is by reducing the mass of the spaceship along with everything inside it.
In order to attain the unattainable speed of light, Johnny must shed all of his rest mass and shrink down
to the size of a point. By doing that, he turns into Johnny Photon.24
When this happens the question is, “Where's Johnny Photon at?” The answer is surprising: He's
literally everywhere in space-time. That's right. Johnny Photon fills the universe, or more correctly the
universe has collapsed around him. Remember earlier when it was revealed that all material objects
travel through space-time at the speed of light, except light itself, which is stuck motionless? Well,
Johnny Photon now finds himself stuck in space-time, and every material object in the universe must
now travel with respect to him. The material objects see different facets of Johnny Photon as they
make their way through space-time. Johnny Photon seems to be absent in certain places, like in a
shadow cast behind a brick wall. That simply means that the probability of finding Johnny Photon
behind the wall has been reduced substantially; Johnny still exists there probabilistically.
23 Standard cosmology insists the cosmic microwave background, CMB, is thermal background radiation having an
apparent temperature of around 2.7ºK. If the CMB is of a true cosmological nature, any Doppler shift from motion with
respect to the CMB would make the CMB itself a preferred fixed frame of reference, similar to the long-discredited
æther, which a big no-no according to relativity. On the other hand, we would expect to see a Doppler shift due to
motion with respect to a locally-generated “CMF”; i.e., foreground material radiating microwaves at 2.7ºK.
24 Actually, Johnny could have saved himself the trouble of using a spaceship to attain light speed and being fried by
gamma rays in the process. He could have simply reduced his rest mass to zero right off the bat, and that would have
boosted him to light speed automatically. According to special relativity, objects with rest mass can never travel at light
speed relative to other such objects, but objects without rest mass must always travel at light speed relative to every
object with mass.
- 19 -

22. Appendix E – The Schwarzschild Effect
Almost immediately after Einstein published his four papers on general relativity in November, 1915,
the physicist/astronomer Karl Schwarzschild found and exact solution to the field equations for a non-
rotating spherical mass of mass M. It come in two versions. The first, and most famous one, is the
exterior Schwarzschild metric, shown below. The interior version will be presented later.
c2
dτ2
= (1 – rs/r) c2
dt2
– (1 – rs/r)-1
dr2
– r2
(dθ2
+ sin2
θ dφ2
)
The time parameter τ is the proper time (the time measured on a clock) of an observer near the mass.
Everything to the right of the equal sign are quantities measured by a distant observer, not under the
influence of gravity. The radius r is equal to the circumference along a circle centered over the
spherical mass divided by 2π. The reason r is stated that way is because measuring things in a radial
direction is different than measuring them in a tangential direction, which why the metric is expressed
in polar coordinates instead of (x, y, z) coordinates. This is one of the strange things about general
relativity. The angles θ and φ are latitude and longitude of a point distance r from the center of the
mass.
You'll notice that M doesn't seem to be found anywhere to be found in the metric. However, the radius
rs, called the Schwarzschild radius, and it is equal to 2 G∙M / c2, so M is embedded in there after all. If
M is zero, then the exterior Schwarzschild metric reduces to this:
c2
dτ2
= c2
dt2
– dr2
– r2
(dθ2
+ sin2
θ dφ2
) = c2
dτ2
= c2
dt2
– x2 – y2 – z2
The expression for zero M above should look familiar: It's the metric for empty space we encountered
in special relativity.
As the name implies, the interior Schwarzschild metric describes space-time inside the gravitating
sphere. So if you tunnel through the sphere, you would want to use the interior metric. The underlying
assumption is that the density of the material making up the sphere is constant throughout,25
and the
interior metric is as follows.
c2
dτ2
= ¼ (3 √ 1 – rs/rg – √ 1 – r2
rs/rg3
)2
c2
dt2
– (1 – r2
rs/rg3
)-1
dr2
– r2
(dθ2
+ sin2
θ dφ2
)
The radius rg is the outside radius of the gravitating sphere itself. At r = rg, the interior and exterior
metrics become one and the same. If M → 0, the interior metric is transformed into the metric for
empty space used in special relativity. For the sake of comparison, let us solve rs / rg for a typical large
gravitating “sphere,”26
like the Earth.
rg = 6.371 106 m
G = 6.674 10 –11 m3 kg –1 sec –2
M = 5.972 1024 kg
c = 300 106 m/sec
rs = 2 G∙M / c2 = 0.00886 m = 0.349 inch, or about the size of a marble.
25 This can only be approximately true for weak gravity. Density must increase for strong gravity because all substances
are compressible because the speed of sound in incompressible substances is infinite, which we know is impossible.
26 The Earth isn't perfectly spherical, but it's close enough.
- 20 -

23. The ratio rs / rg = 1.391 10 –9, or around one part per billion
So it seems that space-time in and around the Earth is pretty much like empty space far away from
everything for the most part. What about the Sun? Well, the Schwarzschild radius is proportional to
mass, so since the Sun's mass is 332,965 larger than the Earth's, its Schwarzschild radius is greater by
the same factor. This works out to be rs = 2.95 km for the Sun, which is still pretty small compared to
its radius, rg = 695,700 km. The ratio rs / rg = 4.24 10 –6, or about four parts per million.
What would happen if you drilled a tunnel through the Sun until you got below 2.95 km from the
center? Besides the fact that you'd be burned to a crisp by the Sun's sweltering core temperature of
27,000,000F, well … nothing. According to the interior Schwarzschild metric, there's nothing special
about being inside the Schwarzschild radius, per se. In fact, at the exact center of the Sun (r = 0),
space-time is no different than empty space. Hold that thought for a while, because things get pretty
hairy when r  rs for the exterior Schwarzschild metric. Let's go back to the exterior metric now.
Free-falling objects in empty space – those not influenced by outside forces besides gravity – follow
paths through space-time defined by geodesics. Simply put, a geodesic is the path from point A to
point B that maximizes the proper time, τB – τA. In a way, it's the same thing minimizing the “action”
or effort to get from A to B. So when you're near a large, gravitating sphere, the shortest path through
space is the one that maximizes the distance through space-time. To calculate the geodesic close to
large non-rotating sphere of mass M, you would try to maximize everything to the right of the equal
sign in the exterior Schwarzschild metric. This turns out to be very difficult in many cases, like when
calculating the orbit of Mercury around the Sun, so I'm only going to show a very limited number of
really simple examples.
For example, let's take the case of a clock suspended motionless over the Sun. That's pretty easy,
because dr2
, dθ2
and dφ2
are all zero. This reduces the exterior Schwarzschild metric to
c2
dτ2
= (1 – rs/r) c2
dt2
or dτ = √(1 – rs/r) dt
In other words, time measured by a clock near the Sun slows down by the factor β = √(1 – rs/r). What
happens to the speed of light near the Sun? Well it also slows down, but in a peculiar way. The
Schwarzschild metric for a light path is found by replacing c2
dτ2
with a zero. If a light particle travels
only in the radial direction, dθ2
and dφ2
are both zero, and the light velocity, vλr , is computed below.
0 = (1 – rs/r) c2
dt2
– (1 – rs/r)-1
dr2
vλr  (dr / dt) = c∙(1 – rs/r) = β2 c
But if light is traveling in a tangential direction to the gravitating body, with dr2
and dφ2
both set to
zero,
0 = (1 – rs/r) c2
dt2
– r2
dθ2
vλt  r∙θ / dt = c √(1 – rs/r) = β c
In other words, light slows down more traveling in a radial direction than traveling in a tangential
direction. Bear in mind the speeds above are measured by a distant observer. The speed of light in any
local reference frame, measured as dr / dτ or r∙dθ / dτ, will always be c. This speed asymmetry greatly
adds to the difficulty of using the Schwarzschild metric to calculate things like orbital motion.
Earlier in this essay, I raised the question of what happens to space near a gravitating body. We're now
- 21 -

24. ready to answer that question. Consider Einstein's famous light clock, where light bounces back and
forth between two mirrors. Using this clock, time is measured by counting the number of times a
bundle of light bounces. To a distant observer, time should appear to slow down by the same factor β
regardless of which way the mirrors are oriented. Suppose the clock is very far away from the Sun
with the mirrors oriented in the vertical direction, so the light travels at tangential velocities, vλt. to the
Sun. If the clock is lowered toward the Sun, time slows down by the factor β. But since the speed of
light also slows down by the same factor, the space between the mirrors remains unchanged.
Now suppose the same clock has mirrors oriented in the horizontal direction, so the light travels at
radial velocities, vλr. If the clock is lowered toward the Sun, time slows down by the factor β, but the
speed of light slows down by the factor β2. Therefore, the distance between the mirrors must shrink by
β to compensate for a greater reduction in speed. Distances across space near a gravitating body as
seen by a distant observer become asymmetrical. Objects flatten in the r-direction.
We're now going to use a modified version of the exterior Schwarzschild metric to calculate the
bending of light around the Sun the way Einstein did it back in 1915. He used something called the
“weak gravity” version, where the asymmetry of space is ignored.
c2
dτ2
= (1 – rs/r) c2
dt2
– (1 – rs/r)-1
(dx2
+ dz2
+ dz2
)
With (rs/r) << 1 the term (1 – rs/r)-1
can be approximated as (1 + rs/r). You can see that by doing the
long division 1 / (1 – rs/r) and getting rid of everything after the second term. Since we're dealing with
light, c2
dτ2
= 0, and the “weak gravity” metric is reduced to the following.
0 = (1 – rs/r) c2
dt2
– (1 + rs/r) (dx2
+ dy2
+ dz2
)
(dx2
+ dy2
+ dz2
) = c2
dt2
(1 – rs/r) / (1 + rs/r)
Because space is now symmetrical, the speed of light is the same in any direction:
vλ  √(dx2
+ dy2
+ dz2
) / dt2
= c √(1 – rs/r) / (1 + rs/r)  β2 c
Einstein assumed light's bending angle would be very small, so if light is traveling in the x-direction at
the point where it grazes the Sun, deviations perpendicular to the path, dy, would also be very small. In
other words, y along the path would be very nearly constant, equal to the Sun's radius, R. The figure
below illustrates this, with changes in the y-direction along the light path greatly exaggerated.
- 22 -

25. As the light wavefront moves left to right, a speed differential exists in the y-direction.
dvλ / dy  (dr / dy)∙(dvλ / dr)  (dr / dy)∙(d / dr)∙c∙(1 – rs / r ) = (dr / dy)∙c∙rs / r2
(dr / dy)  y / √ x2 + y2 = R / r
dvλ / dy  R∙c∙rs / r3 = R∙c∙rs / (x2 + R2) 3/2 , substituting the constant R for y
A change in the speed of light in a direction perpendicular to the light's path causes the phenomenon of
refraction to occur, where the direction of the light's path bends by a certain angle toward the direction
where the speed is slower. Over a small interval dx, a positive gradient in speed, dvλ / dy, causes a
small incremental change in the angle, dα, in the negative-y direction.
dα  (dvλ / dy ) / c = R∙rs dx / (x2 + R2) 3/2
To get the total angle of deflection, integrate dα over the interval 0 < x < to the right of the Sun,
then double that angle to account for the interval –  x < 0 to the left.
α = 2∙R∙rs ∫0

dx / (x2 + R2) 3/2 = 2∙(rs / R) {lim (x / √x2 + R2) – 0 } = 2∙rs / R
x→
We saw earlier that rs = 2.95 km for the Sun. If the light from a star just grazes the visible surface of
the Sun at R = 695,700 km, then α = 2 rs / R = 0.00000848 in radians. Converting this into an angle in
degrees, α = 0.00000848 180 / π = 0.0004859°. There are 3,600'' of arc per degree, so α works out to
be 1.749'', Einstein's famous prediction. It turned out that his initial assumption that y  R = constant
over the entire light-bending distance was pretty accurate.
Now lets go back to the question of what happens at r  rs for the exterior Schwarzschild metric.
Suppose – hypothetically – that all the mass of a spherical object could be squashed inside its own
Schwarzschild radius. If that happens, the object becomes – hypothetically speaking – a “black hole.”
The spherical surface surrounding M having a radius rs is called an “event horizon.” The following
statement is repeated over and over in popular science literature.
“Nothing, not even light, can escape a black hole.”
But there's another truth that's seldom mentioned: Nothing, not even light, can infiltrate a black hole
either. As can be seen from the exterior Schwarzschild metric, several things happen at the event
horizon as seen from the outside: a) time stops, b) light freezes, and c) radial distances shrink to zero.
An object observed from the outside falling toward the event horizon could never actually cross it as
seen from the outside. So if nothing can cross over rs from the outside, 1) how did M cross over rs in
first place, and 2) how can M appear to increase as seen from the outside by additional matter crossing
over rs? The answers to those questions are: 1) it didn't, and 2) it can't.
Event horizons cannot form in the physical universe; true black holes are not physical objects.
General relativity itself provides fundamental reasons why M cannot fit within its own Schwarzschild
radius. I'm not alone in this position; an increasing number of professional scientists are also becoming
convinced of it. The purported existence of black holes have produced some very difficult paradoxes,
and even furious hand-waving by theoretical physicists can't paper over some of the more serious
contradictions. I go into more detail about this in another of my essays, entitled “Why There Are No
True Black Holes.” You can discover all the reasons by clicking on This Link.
- 23 -

26. Appendix F – The God of Thieves
Mercury, a.k.a. Hermes, is the god of quite a lot of things: financial gain, commerce, eloquence,
poetry, messages, communication, travel, luck, trickery, and thieves. Mercury is also the only metallic
element in a liquid state at room temperature, as well as the innermost planet of the solar system with
the distinction of having the most eccentric elliptical orbit of all the planets.27
One of the things about
Mercury that puzzled astronomers for a long time is the amount of precession of its perihelion, which is
a fancy term for the fact that the major axis of its elliptical orbit slowly rotates around a circle. The
gravitational influences of Venus and the Earth could explain some of that precession, but even after
taking those into account, the precession was just too much by an angle around 43'' per century.28
As you might suspect, general relativity can account for the exact amount of excess precession.
Unfortunately, its derivation is way more difficult than the bending of light described in Appendix E.
Einstein's original 1915 paper on the subject is about as clear as mud, unless you can handle the tensor
terminology, which I can't. Using Einstein's field equations as a starting point is simply not for an
amateur scientist like me. I looked around for simpler explanations, but they're not easy to find.
Fortunately, I was able to locate an excellent one by Doug Sweetser, called “24 Steps to the Precession
of the Perihelion of Mercury.” Sweetser gives a complete derivation using the exterior Schwarzschild
metric and some calculus, but his 24 steps are a bit tedious, so I'm going to summarize and condense
them in a manner I hope won't be too confusing or boring for my fellow amateur scientists.
We start out with the exterior Schwarzschild metric.
c2
dτ2
= (1 – rs/r) c2
dt2
– (1 – rs/r)-1
dr2
– r2
(dθ2
+ sin2
θ dφ2
)
Since we're dealing with elliptical orbits, we'll need an “r” term and at least one angle term, so let's pick
the angle θ. Setting dφ2
= 0 reduces the ESM to four terms below.
c2
dτ2
= (1 – rs/r) c2
dt2
– (1 – rs/r)–1
dr2
– r2
dθ2
Multiplying all the terms by (1 – rs / r) and dividing them by dτ2,
(1 – rs / r) = (1 – rs / r)2 (dt / dτ)2 – (dr / dτ)2/ c2 – (r / c)2 (1 – rs / r) (dθ / dτ)2
Sweetser then goes into a rather obscure dissertation about things called “killing fields.” These are
used to reconfigure the Schwarzschild metric into something that will ultimately describe an orbit; in
other words, they're simply quantities that are substituted for other quantities in the equation above.
1) E / m∙c2 = (1 – rs / r)∙(dt / dτ)
2) L / m∙c = (r2 / c)∙(dθ / dτ)
3) U = 1 / r
The third substitution allows (dr / dθ) to be replaced by – (1 / U2)∙(dU / dθ), a step that wasn't at all
obvious to me, but turned out to be absolutely essential to the outcome. The quantities E and L actually
do represent energy and angular momentum as their labels imply. After we make the proper
substitutions and do a little algebra – or more correctly quite a lot of algebra – we arrive at the final
version of the exterior Schwarzschild metric with everything moved to the right of the equal sign.
27 That honor used to belong to Pluto, until Neil deGrasse Tyson and others demoted it to the status of a dwarf planet.
However, a group of NASA scientist are pushing to have it re-instated as a full-fledged member of the solar system.
28 This goes to show you what lengths astronomers will go to in measuring angles.
- 24 -

27. 0 = (E / m∙c2 ) 2 – (L2 / m2∙c2 )∙(dU / dθ)2 – 1 – (L2 U2 / m2∙c2) + rs∙U + rs L2 U3 / (m2∙c2)
Now if you're like me, you'd be hard-pressed to see how this is any better than the original equation.
But there's one final trick that can get rid of more of the terms; namely, by taking the derivative of each
of the terms with respect to θ, and replacing rs with the value 2∙G∙M / c2:
0 = d2U / dθ2 + U – G∙M∙m2 / L2 – 3∙G∙M∙U2 / c2 (1)
The part of (1) highlighted in red can be reduced to zero if r (θ) describes an ellipse! Seeing this isn't
obvious at first, but we can test it by setting U to the function that follows. The blue part of (1) is
Einstein's “correction” to the ellipse that also needs to be reduced to zero, which we'll solve later.
U = 1 / r = (G∙M∙m2 / L2)∙(1 + ε∙cos θ ) , which is an equation of an ellipse, flipped upside down
We take the second derivative of U with respect to θ,
d2U / dθ2 = – (G∙M∙m2/L2)∙ε∙(1 + ε∙cos θ )
We plug the above expressions for U and (d2U / dθ2) into the part of (1) in red,
d2U / dθ2 + U – G∙M∙m2 / L2 = – (G∙M∙m2 / L2)∙ε∙cos θ + (G∙M∙m2 / L2)∙(1 + ε∙cos θ) – G∙M∙m2 / L2 = 0 ✓
That's the way mathematicians solve differential equations sometimes – guessing the solution and
plugging it in to see if it works. Flipping U upside down does, in fact, give us an ellipse:
1 / U = r = (L2 / G∙M∙m2 ) / (1 + ε∙cos θ)  r = a∙(1 – ε 2) / (1 + ε∙cos θ) , the general form of an ellipse
Okay, that's great. But where's the precession? Well, the blue part of the equation supplies the
precession part. First we calculate – 3∙G∙M∙U2 / c2 using U2 = [(G∙M∙m2 / L2)∙(1 + ε∙cos θ )]2 and plug it
into (1) to find out exactly what needs to be canceled next: It's shown in blue below.
0 = d2U / dθ2 + U – G∙M∙m2 / L2 – (3∙G∙M / c2)∙(G2 ∙M2∙ m4 / L4)∙(1 + 2ε∙cos θ + ε2∙cos2θ)
We would like to cancel out the entire blue part, but the only term that matters is the cosine term,
because it repeats every cycle. Adding the part in green below to U will cancel all the red terms and
also (approximately) the cosine term: U = (G∙M∙m2/L2)∙[1 + ε∙cos (θ – 3∙G2∙M2∙m2 θ / c2∙L2)].
The Newtonian advancement of 2π per cycle divided by 2π∙(1 – 3∙G2∙M2∙m2 / c2∙L2) yields Δθ.
Δθ = 2π / (1 – 3∙G2∙M2∙m2 / c2∙L2) – 2π  2π (1 + 3∙G2∙M2∙m2 / c2∙L2) – 2π = 6π∙G2∙M2∙m2 / c2∙L2
(L2 / G∙M∙m2 ) / (1 + ε∙cos θ)  a∙(1 – ε 2 ) / (1 + ε∙cos θ)
Using the above relationship, Δθ can now be expressed in terms of Mercury's orbital parameters.
Δθ (a, ε ) = 6π∙G∙M / a∙(1 – ε 2 )∙c2
This is the angle the perihelion shifts every revolution around the Sun. For Mercury, there is a
revolution every 88 days, so there are 415 revolutions per (Earth) century. The semi-major axis, a, is
5.79 1010 m, and the eccentricity, ε, is 0.206. Plugging a, ε, G, M, and c into the formula for Δθ and
multiplying that angle by 415, we get a shift angle of 42.8'' per century, which matches up well with
the “extra precession” that had puzzled astronomers for a long time.
Hats off to Doug Sweetser, who organized the messy math in a fairly comprehensible fashion. I was
able to check everything he did using only a basic level of calculus. You can find his detailed
presentation of all 24 steps by clicking on This Link.
- 25 -

28. Appendix G – Surf's Up
The bending of light and the precession of Mercury's perihelion aren't too hard to fathom when you
start out with the exterior Schwarzshcild metric; just apply a little calculus with a few approximations
to make the job easier. Gravity waves are a different matter (pun intended). The tensor algebra is
extremely difficult for ordinary amateur scientists like me to master, so I'll have to treat this topic with
kid gloves. Ordinary electromagnetic (EM) waves are so simple in comparison. All you have to do is
generate an alternating current in a wire, and Maxwell's equations clearly show how coupled electric
and magnetic fields propagate through space at the speed of light. Gravity waves are much harder to
analyze precisely. In order to generate them, you need something called a time-varying quadrupole
moment. Once you have that, you can derive gravity waves with about 9 ½ pages of impenetrable math
with a lot of approximations. So what is a time-varying quadrupole anyway? Refer to the sketch
below of two bodies, M and m, orbiting around their common center of mass.
The time sequence is from left-to-right. The upper part of the sketch show the orbiting bodies from
above the orbital plane. The yellow object represents the quadrupole moment tensor, ITT, of the system.
Calculating ITT is devilishly complicated, so let's just leave it at that. In the top view, the bodies form a
counter-clockwise rotating dipole moment, which generates something called a strain gradient, h0, that
rotates at the frequency of revolution of the two bodies, fr. The bottom part of the sketch shows the
same bodies viewed along the orbital plane. In that view, the quadrupole moment changes shape
periodically at twice the revolution frequency, generating two strain gradients h+ and h of f = 2 fr. In a
nutshell, transverse gravity waves are time-varying strain gradients h0 or {h+ , h}.
One of the tests of general relativity is to see whether gravity waves actually exist in nature.
Unfortunately, the effect is extremely weak. Only astronomical-sized bodies like orbiting stars are
large enough produce anything close to measurable waves, and even then it is extremely hard to
measure them directly. There is at least one way they could be observed indirectly – measuring the
decaying orbits of rapidly-revolving binary star systems.
- 26 -

29. After many pages of excruciatingly difficult math using a number of assumptions, you can come up
with an approximate formula giving the total power lost by gravity waves radiating from a binary
system having two masses, M and m, in circular orbits around their common center of mass:
P = dE/dt = 32/5 G4 (M m)2 (M + m) / r5 c3 , where r = separation distance between M and m
It turns out that a binary system with elliptical orbits would produce more radiation than one with
circular orbits. An interesting thing to note is that P→∞ as r→0. This is because gravity can supply an
infinite amount of negative energy. Of course, physical bodies have radii, so it would be impossible for
the r between them to ever reach zero in the first place, but you can see that smaller orbits will radiate
significantly more power than larger ones.
The total orbital energy (kinetic + potential) of a two-body system is given by E = – ½ M m / r. If the
bodies radiate energy, then E must become more negative and r must decrease. Let's see how long it
will take r to reach zero, starting with r = R.
dE/dr = ½ M m / r2
dr/dE = 2 r2 / M m
dr/dt = (dr/dE) (dE/dt) = – (2 r2 / M m ) ( 32/5 G4 (M m)2 (M + m) / r5 c3 )
A few of the terms above will cancel, so
dr/dt = – [ 64/5 G4 (M m) (M + m) / r3 c3 ]
r3 dr = – [ 64/5 G4 (M m) (M + m) / c3 ] dt
Integrating both sides of the second equation,
¼ (r1
4 – r2
4) = [ 64/5 G4 (M m) (M + m) / c3 ] (t2 – t1)
t decay = R4 {5 c3 / [256 G4 (M m) (M + m) ]} ,r1 = R, r2 = 0, t1 = 0, t2 = t decay
A normalized orbital decay is obtained by setting {5 c3 / [256 G4 (M m) (M + m) ]} = 1. The plot below
shows the radius shrinking over time with 1.0 ↔R on the y-axis and 1.0 ↔t decay on the x-axis.
- 27 -

30. For the Earth-Sun system, t decay is on the order of 10 25 years, which is good news because it means
orbital decay is one less catastrophe we have to worry about. So what are the chances of us “catching a
wave” so to speak, by actually observing a decaying orbit somewhere else in the universe? There have
been several such observations, one involving the Hulse-Taylor binary star. This is an unusual binary
in that it consists of pulsars, emitting radio signals in synch with the orbital frequency. That frequency
appears to be increasing, meaning the orbital radius is decaying in a manner similar to the normalized
plot above, which may be an indication Hulse-Taylor is emitting gravity waves. Of course, it may be
an indication of something else, such as tidal forces. For example, the orbit of the Moon is getting
larger because it produces tides on Earth, which spins 27 times faster than the Moon's 27-day period.
Those tides boost the Moon's orbital energy by a small amount, but it's many orders of magnitude more
than the energy the Moon radiates away by gravity waves. If the Earth were spinning more slowly than
the Moon's revolution, or if the Earth were spinning in the opposite direction, tidal forces would drain
much more of the the Moon's orbital energy than radiating gravity waves ever could.
Could scientists detect gravity waves directly, using something like a gravity antenna? Well, there's a
problem with this. Three important assumptions are made when developing the gravity-wave
equations. These are summarized in the following three inequalities involving distances.
R universe >> r >> λ >> s
The first distance is the radius of the universe itself, which is expanding. The second distance, r, is the
size of the source of the waves. The third distance is the wavelength, λ = c / f. The fourth distance, s, is
the scale of a system affected by the waves. The first inequality means we don't have to consider the
expansion of the universe affecting the propagation of gravity waves. The second inequality means
that the gravity waves are transverse and not simply static tidal forces due to variations in the proximity
to gravitating bodies. The third inequality means the the gravity waves are not generated by bodies
moving at relativistic speeds, which would produce higher-order effects due to non-linear properties of
the GR field equations. The third inequality is important, because it means that any gravity antenna
system would be much smaller than typical gravity waves it's trying to detect.
A binary star, consisting of a pair of solar masses separated by the radius of Mercury's orbit, will
complete one orbit every 88 days. Because f = 2 fr for h+ and h waves, their wavelength, λ, is 44 light-
days, a distance of 1.14 1012 km. The orbit of Neptune, the outermost “actual” planet in the solar
system, has a diameter of about 9 109 km. Therefore, the λ of radiation from the binary star is 127
times larger than what is currently considered the size of the solar system. I'm not sure if that satisfies
λ >> s, but it's close enough for me. The bottom line is that you'd need a gravity antenna larger than the
solar system to reliably detect gravity radiation emitted by a typical binary star.
According to Einstein's equivalency principle, free-falling objects in uniform gravitational fields don't
feel any effects of gravity. This is much different than EM fields. If you attach accelerometers to
electrically-charged objects and release them in an electric field, their accelerometers will measure
different rates of acceleration depending on the object's mass and electrical charge. If you perform the
same experiment in a uniform gravitational field, the accelerometers all record zero regardless of how
much mass the objects have, and they all move together. Things are a little different when objects are
placed in non-uniform gravitational fields, and when they are exposed to gravity waves.
The next appendix will explore attempts professional scientists are making at detecting gravity waves,
including looking for signs of quadrupole gravity patterns in the so-called CMB (cosmic microwave
background), and listening for the sound of black holes colliding. Stay tuned.
- 28 -

31. Appendix H – Catch a Wave
What's it like to have a gravity wave slam into you? Imagine yourself floating weightless in space
surrounded by other objects, shown as red dots, whose distances and angles to your position can be
measured with a transit equipped with a laser range finder. Getting rid of any notion that you're in an
absolute frame of reference, you might as well consider yourself being at the center of the universe.
The situation you find yourself in is depicted below in the first diagram on the left.
Now suppose an h+ gravity wave arrives at your location traveling in a direction perpendicular to this
page. You notice that distances and angles to the other objects are all changing, although you don't feel
any acceleration or sense of motion at all. The objects around you change their positions as shown in
the second diagram. It seems as if the whole universe has been squashed in the horizontal direction and
stretched in the vertical direction. After a little while, things return back to normal, as shown in the
third diagram above. Then it happens all over again, but this time the objects have shifted to the
configuration shown in the fourth diagram, as if the universe has been stretched in the horizontal
direction and squashed in the vertical direction. Again, you don't feel any sensation of motion while it
is happening. This sequence of events repeats over and over as gravity waves continue to waft over
you and your surroundings.
The funny thing is when you ask your neighbor what she experienced, she says exactly the same thing
happened to her. Taking a survey of everyone around you who encountered the h+ wave, they all tell
the very same improbable story: Every object around them seemed to shift horizontally and vertically
at the same time, yet nobody felt any motion.
The above occurrence describes an encounter with gravity plane wave, based on the the premise that it
wavelength λ >> s, where s is the scale of collection of surrounding objects. So what kind of events
can cause such a gravity wave and what are the chances of observing one directly? A gravity-wave
source must be time-varying and cannot be spherically symmetrical. There are plenty of those sources
in the universe. As we saw earlier, a binary star is an emitter of gravity waves, although it's a very
weak one. A supernova is another one. You might think an exploding star would be spherically
symmetrical, but it's actually not if it spins on its axis, which most stars do. But the best chance we
have in observing gravity waves would be if they came from collision of two extremely massive
objects, commonly called black holes.29
That's because the frequencies of the strains h+ (or h)
29 I'll refer to them as black holes here, even though I'm convinced that no true holes actually exist. Large stars collapse
- 29 -

32. generated by such a collision are supposedly around 100 Hz or greater, which is very high for gravity
waves, so the wavelengths will be 3 million meters or less. It may be possible to build a gravity
antenna that can detect these waves by building something that's not too far below that scale. Scale is
very important because since h is strain, defined as h = ΔL / L, in order for ΔL to be detectable, let
alone measurable, L must be very large, preferably on the order of λ.
The LIGO (Laser Interferometer Gravitational-Wave Observatory) project is one such very elegant
gravity antenna. It's actually an updated version of the famous Michelson-Morley interferometer,
which failed to detect any motion of the Earth through the “æther wind,” caused by the motion of the
Earth through a hypothetical medium that transmits light. LIGO has two perpendicular arms, like the
Michelson-Morley version, with some significant differences. Whereas the M-M interferometer had a
scale of 11 meters, LIGO's arms are four km long and that was stretched to 1,120,000 meters by having
the light reflect back and forth between two mirrors 280 times. A scale that large is getting pretty close
to being in the range of λ. The mirrors in the M-M interferometer were rigidly fixed to a stone slab, but
that won't work with a gravity antenna because the idea is to have the mirrors “float” as if they are in
free fall in space, allowing the distances between them vary, so LIGO's floating mirrors are suspended
from special pendulums in a vacuum. An ideal gravity wave for LIGO would be an h+ wave that came
straight down from the sky with a polarization aligned with the arms. In that configuration, the
effective length of one arm would increase by the maximum amount while the effective length of the
other arm would simultaneously decrease, maximizing the interference signal at the photodetector.
There seems to be a paradox however. If space is stretching and shrinking, wouldn't that also affect the
the measuring rod – light – that is used to measure the length of the arms? It would seem that the exact
same number of light wavelengths would travel to the mirrors and back if the light stretched and shrank
the same amount as the arms. It turns out not to be the case. What LIGO is actually measuring is the
time it takes for the light to be reflected back and forth 280 times, regardless of how the light's
wavelength is affected. A time differential between the arms translates into a phase shift. But how
could LIGO succeed when Michelson and Morley were doomed to failure? It's because the M-M
interferometer was set up to measure a time differential between the arms that simply does not exist –
the speed of light is constant in every direction.
In February 2016, the LIGO team announced that it had detected the long-awaited gravity-wave signal
generated by two colliding black holes. As a retired engineer, I was very interested in finding out how
this feat was accomplished, so I read over LIGO's design specifications on Caltech's Site, and I found
them to be quite extreme and very impressive. LIGO is certainly not your grandpa's interferometer.
LIGO was designed to detect a strain of 10 –21. That's a displacement of only 4  10 –17 m over a 4 km
stretch, compared to the diameter of a proton, 10 –15 m. LIGO's light source is a highly-coherent light
beam from a 200-watt laser.30
The light passes through the beam splitter, where it enters the two arms.
Each arm has a light cavity where it bounces back and forth between two mirrors – one near the beam
splitter and one at the end of the light tube 4 km away. The cavity is designed to accomplish two
objectives: 1) allow the beam to make 280 passes back and forth before returning to the beam splitter,
thus increasing the effective length of the arm by a factor of 280, and 2) amplify the 100 watts coming
into ECOs (eternally collapsing objects), which are very gravitationally compact but do not have true event horizons.
To a distant observer an ECO resembles a black hole in almost every respect because an ECO's gravity will red shift
any light emitted from it into oblivion. The major difference between ECO and a black hole observed from afar is that
an ECO has an immense magnetic field, whereas a true black hole cannot possess any magnetic field at all.
30 Michelson and Morley were stuck with pretty primitive light sources. Thomas Edison patented the electric light bulb in
1879, but M-M actually used oil lamps as light sources in their 1881 and 1887 experiments! Go figure.
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33. from the laser into a mammoth 750 kW monster beam.31
The latter objective sharpens the interference
pattern by reducing “shot noise.” Light consists of tiny energy packets, or quanta, called photons.
Photons act like individual particles like pellets from a shot gun. The pellets, or shot, produce noise
like rain drops falling on a metal roof. By increasing the number of pellets per second, the individual
photons merge into a classical wave that produces a sharp image as opposed to the blotchy image you'd
get from individual light quanta.
Light applies a force to a mirror when it reflects from it. A 750 kW beam produces 0.005 nt or about
½ gram of force on the mirrors. LIGO's mirrors weigh in at 40 kg. A force of 0.005 nt would produce
an acceleration of 0.000125 m/sec2, which isn't much; but it's obvious that even a tiny % variation in
the power of a 750 kW beam could change the force enough to move the mirrors more than the width
of a proton. Thus, extreme care must be taken to keep the cavity beams at constant power levels.
When you look down at the lights of Los Angeles from Griffith Observatory, you'll see a noticeable
twinkle from light quanta colliding with air (and smog) molecules. Light colliding with air molecules
inside the light cavities would wreck any chances of using light as a measuring rod to detect mirror
displacements smaller than a proton, so LIGO's mirrors are suspended in a vacuum that has an air
density 10% of the air density in outer space.32
The mirrors are suspended by pendulums (or pendula?) having a four-stage design, shown below. Four
weights, including the mirror at the lower end, are suspended from
four pendulum arms of varying lengths so that the pendulum won't
resonate at any particular frequency. If the support should move to
the left or right from external vibrations, the pendulum dampens the
motion so the mirror stays aligned with the red line. The picture is
very exaggerated to show the effect. Other optics in the LIGO
system that are required to stay in perfect horizontal alignment are
suspended the same way. In addition, LIGO employs active
damping, which senses vibrations and applies counter forces that
cancel them out. Without these extreme measures, LIGO would be
little more than a very expensive seismograph.
Even with these extreme measures, some noise gets added to the
signal that needs to be dealt with. This is a classic problem
encountered in communication engineering: to reliably detect signals
coming from noisy channels. In the 1940s, Claude Shannon, the father of modern communication
engineering, pointed to techniques that solved that problem. Intuitively, you would think you can pass
the noisy signal through a filter that somehow cancels the noise, but that's not effective because you
don't know a priori what needs to be canceled since noise is random. In fact, reducing the signal's
bandwidth with filtering just makes matters worse. Shannon showed that in order to extract a signal
from a noisy channel, you have to process the signal before it enters the channel by encoding it using a
clever algorithm that adds redundancy to the signal and spreads its frequencies over a wider bandwidth.
At the receiving end, the noisy signal is processed by a decoder that removes the redundancy,
dramatically reducing the noise and making reliable communication through the noise possible.
31 750 kW is enough power to supply the electricity used by 750 American homes.
32 I'm not sure what Cal Tech really means by “outer space.” It could mean space where Earth satellites are in orbit, or
maybe interplanetary space, interstellar space, or even intergalactic space. Who knows? In any case, LIGO's vacuum is
pretty good.
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