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."
(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 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 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?
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. 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.