Thorntone E. Murray

July 20, 2014

The Lorentz Transformation Contradiction

The Lorentz transformation equations:

1. x’=x-vt/sqrt(1-(v²/c²))

2. y’=y

3. z’=z

4. t’=t-(vx/c²)/sqrt(1-(v²/c²))

Let v represent the speed with which inertial frame K’ is in motion relative to inertial frame K.

In accordance with the Lorentz transformation, judged from frame K, a clock in frame K’ goes more slowly than a clock at rest in frame K. Also in accordance with the Lorentz transformation, a measuring rod perpendicular to the direction of motion in frame K’ and an identical measuring rod perpendicular to the direction of motion in frame K are of equal length relative to each other. As the speed of light is constant and the same for all observers, judged from frame K the time for light to travel the length of each rod is identical in K and K’. Therefore, judged from frame K, the clock in frame K’ goes at the same rate as a clock at rest in frame K, not more slowly.

Then, in accordance with the Lorentz transformation, judged from frame K, the clock in frame K’ goes more slowly than the clock at rest in frame K and, concurrently, that same clock in frame K’ goes at the same rate as the clock at rest in frame K. Following is mathematical confirmation of that contradiction.

Parallel to the Direction of Motion:

The following is a quote from Dr. Albert Einstein’s 1916 Relativity – The Special and General Theory, Section 12 – The Behavior of Measuring Rods and Clocks in Motion; translated by Robert W. Lawson.

“Let us now consider a seconds-clock which is permanently situated at the origin (x’=0) of K’. t’=0 and t’=1 are two successive ticks of this clock. The first and fourth equations of the Lorentz transformation give for these two ticks:

t=0

and

t=1/sqrt(1-(v²/c²))

As judged from K, the clock is moving with the velocity v; as judged from this reference-body, the time which elapses between two strokes of the clock is not one second, but

1/sqrt(1-(v²/c²))

seconds, i.e. a somewhat larger time. As a consequence of its motion the clock goes more slowly than when at rest.” End quote.

Then, judged from K, one second in K’ equals 1/sqrt(1-(v²/c²)) seconds in K. The operative equation is t’=t*1/sqrt(1-(v²/c²)).

Perpendicular to the Direction of Motion:

Judged from K with speed v=0, in frame K light at speed c travels the length L of a measuring rod perpendicular to the direction of motion in the time t.

L=ct or t=L/c with speed v=0

Judged from K with speed v=0, light at speed c travels the length L’ of an identical measuring rod perpendicular to the direction of motion in frame K’ in the time t’.

L’=ct’ or t’=L’/c with speed v=0

Judged from K with speed v=0, length L’ of the measuring rod in frame K’ is equal to length L of the identical measuring rod in frame K.

L’=L with speed v=0

With speed v=0, frame K’ and frame K are at rest relative to each other. With speed v=0, time t’ in frame K’ is equal to time t in frame K.

t’=t with speed v=0

In accordance with the second (and third) equation(s) of the Lorentz transformation, with speed v=0 and speed v>0, the length of a measuring rod perpendicular to the direction of motion in frame K and the length of an identical measuring rod perpendicular to the direction of motion in frame K’ are of equal length relative to each other.

y’=y (z’=z) with speed v>0

Then, judged from K with speed v>0, length L’ of the measuring rod in frame K’ and length L of an identical measuring rod in frame K are equal.

L’=L with speed v>0

As the speed of light is constant and the same for all observers, then:

L’/c=L/c with speed v>0

Shown previously, t=L/c and t’=L’/c. As L/c=L’/c with speed v>0, then:

t’=t with speed v>0

Then, judged from K, one second in K’ equals one second in K. The operative equation is t’=t.

Shown previously:

Judged from K, one second in K’ equals 1/sqrt(1-(v²/c²)) seconds in K. The operative equation is t’=t*1/sqrt(1-(v²/c²)).

The Contradiction:

As a consequence of the Lorentz transformation, judged from frame K, t’=t*1/sqrt(1-(v²/c²)) and t’=t concurrently.

Therefore, the Lorentz transformation is invalid.

Thorntone E. Murray