**Chapter 12**

** **

**Is the
Speed of Light a Constant?**

"It is impossible to travel faster than the
speed of light, and certainly not desirable, as one's hat keeps blowing
off."

Woody Allen

**Introduction**

(Special thanks to Dr. George Marklin for a key
concept that made this paradox possible.)

Special Relativity ("** SR**") states, as
its second postulate, that the "[

In a sense we are in the same situation we were in when we
discussed the first postulate. In the
1905 version of the SR it essentially talked about "**imaginary
time**." This
"imaginary time" was replaced with "

Let me present a paradox to the second postulate in the
SR. I will call this the "**Three
Space Ship Paradox**."

**The Setup of the Experiment**

Let us assume there are three space ships. Each of these space ships has exactly the
same capabilities. Let us further
assume that all three of these ships are traveling in open space, half-way
between two galaxies, and all velocities are measured relative to Cosmic
Microwave Background Radiation or CMBR.
Let us pick a reference point, relative to CMBR, and call it Point N,
that is a point half-way between two galaxies.
Even though the galaxies are moving relative to CMBR, this point is not
moving relative to CMBR. Point N is
perfectly at rest, relative to CMBR, during this entire experiment.

Let us also define a flat 2D plane that is "at
rest" relative to Point N, 1 light year in radius, with center at Point
N. Let us further draw a "**straight
line**" from Point N to the edge of this flat 2D plane. This line is also perfectly "at
rest," relative to CMBR and Point N, during the entire experiment. Obviously, the straight line forms a visual
radius for the flat 2D plane. Now let
us pick a point that is 1/2 light year from Point N, and is on the
"straight line," and thus on the flat 2D plane, and call it Point
C. From Point C, we will pick a point
1/4 light year "above," and normal to the flat plane at Point C, and
call it Point A.

Now let us define some directions. Using Point A as a reference point and vantage point, and looking
"down" at Point C, and also looking "towards" Point
N," we will define that a space ship that is headed directly towards Point
N, and is on the straight line, is headed "North." Obviously, if a ship were headed away from
Point N, and were on the straight line, it would be headed "South."

Again using Point A as a reference point, any ship traveling
on the plane that is headed perpendicular to the straight line, and is headed
from right to left is headed "West" and any ship traveling in the
opposite direction is headed "East."
This completes our coordinate system.
Point A will be our vantage point for the rest of this experiment.

On the top of each of the three ships is a painted circle
that is 100 meters in diameter. On this
circle is drawn a line, from one end of the circle to the other, passing
through the center point of the circle, such that it is parallel with the
direction the ship moves when it is moving "forward." In other words, as the ship is moving
forward, the line is parallel to the direction vector that the ship is
headed. A second line is drawn on the
circle that is perpendicular to the first line and also crosses the center of
the circle. The two lines form a
"cross," with its center at the center of the circle. At the four endpoints of these two lines is
put a short stick, vertical or normal to the 2D plane formed by the circle
painted on the ship.

We will number these four sticks. The stick that is on the first line, and at the front of the
ship, as it heads forward, is called Stick 1.
The sticks are numbered consecutively, clockwise, around the ship.

Let us now think about how the speed of light is
measured. We first calculate how far
the light travels, call it 'd', for distance, and we measure how much time it
takes to travel that distance, call it 't', for time. We use the following formula: c = d / t

In each of the three ships we will measure how long it takes
photons (assuming the photon theory) to travel between two of the sticks. We will use very accurate atomic clocks to
measure the time, t, it takes the photons to travel between the two sticks on
each ship.

Ship 1 is traveling at exactly 90% of the speed of
light. It is traveling
"Northbound," towards Point N, 1,000 meters ** below** the
flat 2D plane and directly beneath the straight line. Ship 2 is also traveling at exactly 90% of the speed of
light. It is traveling
"Southbound," away from Point N, directly on the flat 2D plane. Ship 3 is traveling "Westbound"
(from the vantage of Point A), 1,000 meters

First of all, let me point out that the atomic clocks on all
3 ships are recording "time" exactly the same. This is because all 3 ships are traveling at
exactly the same velocity relative to Point N, which is stationary relative to
CMBR. This comment requires some
explanation.

With the Hafele-Keating experiment, the direction the jets
headed was significant to their velocity relative to the "at rest"
reference point. However, the direction
of the jets was important only because the jets were carried with the surface
of the earth in its daily rotation. It
was the rotation of the earth that made their direction important.

In the case of the Three Spaceship Paradox, direction is not
important because there is no rotating planet that is affecting their net
velocity relative to Point N. Relative
to Point N, all three ships are traveling at exactly the same velocity. Thus, the atomic clocks, and the people
inside of the ships, are all measuring time identically. We will also synchronize all of the clocks,
meaning not only are all three ships measuring (a change in) time identically,
but all of them are recording exactly the same time.

**The Experiment Begins**

A single pulse of laser light is sent from Point N along the
straight line. It is 4,000 meters in
diameter when it arrives at Point C. At
the exact instant the beam arrives at Point C. the ships are positioned (as
they are in motion at 90% of the speed of light) such that this light
simultaneously hits the "**first stick**" (i.e. the
closest stick of each ship to Point N) of each space ship. In other words, in Ship1, the laser beam
hits Stick 1 first. In Ship 2 it hits
Stick 3 first. In Ship 3 it hits Stick
2 first. Obviously, our concern is how
long it takes this light to hit the "

Now comes the question, will all three ships measure the
same speed of light? This question is
equivalent to the question, will it take the same amount of time for the light
to travel from the "first stick" to the "second stick" for
all three ships?

To answer that question, we must think about the MTLs. Since the beam of light hits the "first
stick" of all three ships at exactly the same instant, and all three
atomic clocks are synchronized, we can use this instant in time to start the
MTLs operating. In other words, the
"second stick" is the "target" and it is a moving target. Thus, we are concerned with how much the
target (i.e. the "second stick") moves while the laser beam is
"in the air," meaning while the beam is traveling from the
"first stick" to the "second stick."

When the MTL is started, in the time it takes the light beam
to travel to the second stick in Ship 1, the second stick is moving ** towards**
that light at 90% of the speed of light.
In the time it takes the light beam to travel to the second stick in
Ship 2, the second stick is moving

When the speed of light is calculated, ** the motion of
the two measuring points, relative to CMBR, is not taken into account**. In other words, if I were measuring the
speed of light in a laboratory in Overland Park, Kansas, USA, I would measure
the speed of light based on how much time it took the light to travel between
two fixed points in the lab. Let us say
the two points were 100 meters apart in the lab. I would not take into account the velocity of the earth towards
Leo, the orbit velocity of our earth around the sun, or any other factor. I would always use "100 meters" as
'd'. Likewise, when the people on all
three ships measure the speed of light, the 'd' that they use will also be 100
meters on all three ships.

However, our concern here is not how far it is between the
two sticks relative to the occupants of the ships, but how far the light has to
travel ** relative to CMBR** before it hits the "second stick

Let's do the math (assuming the speed of light is exactly
300,000 kps):

First, let us calculate how much time it takes the laser
pulse to travel the 100 meters, d, in Ship 1.
Ship 1 is headed "North," directly towards Point N. Thus, with a simple amount of mathematics,
the light only travels 52.63 meters (relative to Point N) before it hits the
second stick, meaning it takes 0.0000001754333 seconds to travel between the
two sticks. The people on Ship 1, who
know nothing about CMBR, will calculate the speed of light, c, as: 570,017.1
kps because from their perspective the light travels 100 meters.

Second, let us calculate how much time it takes the laser
pulse to travel the 100 meters, d, in Ship 2.
Ship 2 is headed "South," directly away from Point N. Thus, with a simple amount of mathematics,
the light travels 900.00 meters (relative to Point N) before it hits the second
stick, and it takes 0.000003 seconds to travel between the two sticks. The people on Ship 2 will calculate the
speed of light, c, as: 33,333.3 kps.

Third, let us calculate how much time it takes the laser
pulse to travel the 100 meters in Ship 3.
Ship 3 is headed "West," perpendicular to the straight line
that emanates from Point N. In this
case the same photons (assuming the photon theory) that directly pass over the
first stick will not hit the second stick.
But this does not matter because other portions of the same laser beam
will hit the second stick. To simplify
things, I will consider the "first stick" and the "second
stick" to be infinitely wide, and perpendicular to the straight line, and
will simply measure the time it takes any of the photons to pass between these
two infinitely wide sticks. By definition
the people in Ship 3 will calculate the speed of light, c, as: 300,000 kps.

Since the atomic clocks on all three ships are synchronized,
and because they all measure time the same way, and because they are all
traveling at the same velocity relative to Point N, and because all of them use
'd=100', the only variable is how long it takes the light to travel between the
two sticks of each ship, which is 't'.
It is clear that the time that it takes the light to travel between the
two sticks will be different for each ship.
Thus we must conclude that ** c will be different for each of the 3
ships**.

The postulates of Special Relativity are assumptions, not
laws. The MTLs are laws. We must give priority to laws that are
proven, rather than to assumptions which are not proven.

Thus, we have no choice but to conclude that the speed of
light cannot be the same for all observers.
In fact, even with the ether theory the same results would be obtained,
since we are measuring c from the perspective of each ship.