What went wrong in the proof of the theory that nothing can travel faster than the speed of light?

In a “normal” size vehicle, let say with the height of the interior of the vehicle (the distance from the floor to the ceiling) is about 3 m (≈3.28 yards), the time it takes the light beam emitted from its source on the floor to hit the ceiling is 10

^{-8} of a second – literally a split bit of a second.

The right triangle used to prove in the theory above is obtained under the assumptions that:

- Light would travel at the constant speed c=3×10
^{8} m/s regardless of the frame of reference. - From the external observer’s perspective, light would hit the ceiling of the vehicle at the point B on the ceiling directly above the point A where the source of light is on the floor.
- Two different frames of reference (the passenger’s perspective and the external observer’s perspective) could be related in one single diagram.

The first assumption would seem to hold in the case the velocity of the vehicle is insignificant compared to the speed of light: v≪c which is the underlying assumption which leads to the same conclusion. But what if v≈c? From the perspective of the passenger, would the light beam still seem to travel at the constant speed c=3×10

^{8} m/s?

The second assumption is made under the presumption that the light beam emitted from its source at the point A on the floor will hit the point B directly above it on the ceiling. Once “airborne,” the light beam does not “know” that it is in a moving vehicle. Thus, it will travel in a straight, vertical path. Again, if v≪c, then it would seem that the light beam would hit the point B on the ceiling directly above the source on the floor, as it would if the vehicle were stationary. Indeed, it would seem insignificant where on the ceiling the light beam hits because it would require nanotechnology to measure the infinitesimal distance between the point C on the ceiling where the light beam actually hits and the point B which is directly above the light source at the point A on the floor. However, this insignificant distance is actually the length of the base of the right triangle used in describing Einstein’s thought experiment. So, as insignificant as it may seem, it must be noted that the light beam will not hit the point directly above its source, but towards west (if the vehicle is traveling in the east direction) and distance x away, where x=vt

_{0} is the distance traveled by the vehicle in time t

_{0} it takes the light beam to hit the ceiling, about 10

^{-8} second, as observed by the external observer.

Assume, for the moment, the third supposition above – that the scenario can be described in a triangle by combining the two different frames of reference – is justifiable. Then the right triangle used in proving the theory above is as follows, where

- z is the distance the light beam traveled in time t
_{0} that it takes for the light beam to hit the ceiling, as observed by the external observer, - x is the distance traveled by the vehicle in time t
_{0}, again, as observed by the external observer, - y is the distance the light beam traveled in time t
_{1}, as observed by the passenger traveling in the vehicle

Note that x≪z is in any realistic situation, since in the best of scenarios, even if manmade, supersonic vehicles attain ten times the speed of sound (Mach 10), this right triangle would be so skinny (y≈z) that it would be almost a vertical line segment, rather than a triangle.

However, there is no reason to presume that v≪c before anything is proven yet. In that case, there is no reason to make the first two assumptions. For the sake of observing accurately, suppose both the passenger and the external observer can observe the scenario in an extremely slow motion, slow enough that the scenario which occurs within 10

^{-8} second can be viewed over, say, 10 seconds. Then, what would be true is that

- From the perspective of the passenger, the light beam would not travel directly from the point A to the point B directly above it, but to the point C which is x meters to the west away from the point B on the ceiling, where x=vt
_{0} is the distance traveled by the vehicle in time t_{0} it takes the light beam to hit the ceiling, about 10^{-8} second, as observed by the external observer. That is, the passenger (traveling in east direction in the vehicle) would see the light beam as traveling northwesterly direction to the point C. - On the other hand, from the external observer’s perspective, the light beam which does not “know” that it is in a moving vehicle, would travel in a straight path vertically and hit the ceiling at the point C, rather than at the point B directly above the point A, since the vehicle would have moved x meters.

Then, again, assuming that the scenario can be described by combining two different frames of reference, the right right-triangle (or the correct right triangle) would be then one with y as the hypotenuse, and x the base and z the height.

Here, this new right triangle (again, very skinny in any practical scenario, almost like a vertical line segment) can be used in two different cases: one in which the distance traveled by the vehicle is from the external observer’s and another in which the distance traveled by the “ground” is from the perspective of the passenger’s (if the vehicle is transparent, say). This is by retaining the third assumption made in proving the theory above.

In the first case, then

- z is the distance the light beam traveled in time t
_{0} , as observed by the external observer, - x is the distance traveled by the vehicle in time t
_{0}, again, as observed by the external observer, - y is the distance the light beam traveled in time t
_{1}, as observed by the passenger traveling in the vehicle

In the second case,

- z is the distance the light beam traveled in time t
_{0} , as observed by the external observer, - x is the distance traveled by the “ground” in time t
_{1} that it takes for the light beam to hit the ceiling, as observed by the passenger, - y is the distance the light beam traveled in time t
_{1}, again, as observed by the passenger traveling in the vehicle

Since we do not presume that the vehicle cannot travel at the luminal or a superluminal speed, we also do not presume that the passenger sees the light beam as traveling at the speed of c=3×10

^{8} m/s. We will label it as s for speed. Then, by applying the Pythagorean Theorem on the right triangle in two different cases above yields the following. In the first case, (vt

_{0})

^{2}+(ct

_{0})

^{2}=(st

_{1})

^{2}, and in the second case, (vt

_{1})

^{2}+(ct

_{0})

^{2}=(st

_{1})

^{2}. Setting the two left-hand-sides equal to each other, we see that t

_{0}=t

_{1}. That is, there is no time dilation. The time that it takes for the light beam to hit the ceiling is the same as observed by the external observer, as well as by the passenger. And the speed at which the light beam would seem to be moving away from the perspective of the passenger is s is the value that satisfies s

^{2}=v

^{2}+c

^{2}. This is not the actual speed of the light beam, but only the perceived speed of the light beam, as observed by the passenger who thinks the light beam is traveling in a diagonal fashion, rather than in a vertical fashion.

This perceived, non-actual speed of the light beam computed is a result of combining two different perspectives in one equation, which is an assumption made in “proving” Einstein’s theory: that the two different frames of reference can be combined to describe the scenario in a single diagram. If we do not retain that assumption, then there is no triangle to describe the scenario, let alone a right triangle. Whether we retain that assumption or not, we do not arrive at a result that precludes any object from traveling at superluminal speeds.

In the correct right triangle, what does the distance y actually mean then, if it is to have any meaning in the diagram above without producing a non-actual, perceived quantity due to combining two difference references of frame?

- z is the distance the light beam traveled in time t
_{0} , as observed by the external observer, - x is the distance traveled by the vehicle in time t
_{0}, as observed by the external observer, - y is the distance between the source of the light beam on the floor and the head-end of the light beam at time t
_{0}, as observed by the external observer

With this diagram which includes only one perspective, that of the external observer in a non-moving frame of reference, there is no non-actual, perceived quantity produced in computations.

This scenario may be more easily visualized if we change it by exchanging a small ball with a light beam. Here is a thought experiment.

A thought experiment: Consider a vehicle with a ceiling of 100 m, traveling at a speed of v=100 m/s. Assume that in the vehicle is vacuum and with no gravity. A ball is emitted vertically upward at a speed of c=1000 m/s (v<c), 100 m/s (v=c), or 10 m/s (v>c). What path of the ball does a passenger in the vehicle observe and an observer outside the vehicle observe? Once “airborne,” the ball does not “know” that it is in a moving vehicle, and it will travel in a vertical fashion as observed by an external observer, though it will seem to the passenger that it is traveling diagonally to where it hits on the ceiling.