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 "[apparent]

speed of light" is the same for all observers. If the observers have adequate

equipment, we must assume that Einstein meant: the "measured speed of light"

is the same for all observers.

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 "actual time" long before 1971.

Since there is no point in talking about the "imaginary speed of light," we will do

exactly the same thing we did with the first postulate and instead talk about the

"measured speed of light."

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 halfway

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

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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

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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 above the flat 2D plane.

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

"second stick" (i.e. the stick at the opposite end of the same line the "first stick"

is on) of each ship. For Ship 1 the "second stick" is Stick 3. For Ship 2 the

"second stick" is Stick 1. For Ship 3 the second stick is Stick 4. By design, even

though all three ships are traveling at 90% of the speed of light, they all arrive at

Point C, such that, at exactly the same instant the light hits the "first stick" of

each ship.

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?

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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 away from that light at 90% of the speed of

light. In Ship 3, the second stick is moving perpendicular to the light beam path

at 90% of the speed of light, and the light beam will miss the second stick, but

other light beams from the same laser will hit the second stick.

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! The people in the space ships may have no idea

how fast they are traveling relative to CMBR, they are only interested in how

much time, t, it takes the light to travel the 100 meters, d, the distance between

the "first stick" and 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.

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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.

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