Why Special Relativity does not explain Michelson and Morley's experiment
I'll show that the Lorentz transformation cannot explain the constancy of c in all directions on earth. I'll use only relativistic time displacement, which is in the case of small velocities a first order effect whereas time dilation and lenght contraction are only second order effects.
Assume that the earth's orbit, which is around 940 million km long, is exacly circular. Apart from the earth 3130 artificial satellites move on this orbit in such a way that the distances between two neighbouring satellites remain exactly 1 light second (around 300'000 km). The velocities of all satellites are 0.0001c (around 30 km/s). We further assume that all relative deviations from these values are less than 10^-8.
If Dx, Dt, etc. denote coordinate differences, the Lorentz transformation for a relative speed of 0.0001c is:
Dx' = gamma [Dx - 0.0001 c Dt] [ 1a ]
Dt' = gamma [Dt - 0.0001/c Dx] [ 1b ]
Dx = gamma [Dx'+ 0.0001 c Dt'] [ 2a ]
Dt = gamma [Dt'+ 0.0001/c Dx'] [ 2b ]
gamma = 1 / sqrt[1 - v2/c2] = 1 / sqrt[1 - 0.0001^2] = 1.000'000'005
Whereas time displacement depending on position (see [1b] an [2b]) is 10^-4, the effect of the gamma factor is less than 10^-8 and lies in the range of distance and speed deviations. The gamma factor can therefore be ignored:
Dx' = [Dx - 0.0001 c Dt] [ 4a ]
Dt' = [Dt - 0.0001/c Dx] [ 4b ]
Dx = [Dx'+ 0.0001 c Dt'] [ 5a ]
Dt = [Dt'+ 0.0001/c Dx'] [ 5b ]
All satellites on the idealized orbit have the same distance from the sun. Their clocks can exactly be sychronized from the sun. In principle, time must pass in the same way on all satellites because they are equivalent. This time relevant to all satellites shall be called global time.
Starting from the earth, light rays are sent around the whole orbit in direction of revolution and in opposite direction. The rays are deflected by each satellite to the next satellite. Because all satellites have synchronized watches (global time), the exact temporal course of the ray propagation can be determined.
Two neighbouring satellites move almost in the same direction. The difference of angle is only 1/3131 of the whole circle and can be ignored in the following reasonings. If the speed of light refers to the sun, propagation from one satellite to the next in front needs 1.001 seconds. A ray starting at 0:00:00.0000 in direction of revolution leads to this table:
POSITION GLOBAL TIME COVERED DISTANCE MAX.DEVIATION ----------------------------------------------------------------- Sat. Earth 0:00:00.0000 Sat. 1 0:00:01.0001 1.000l LS 10^-7 sec and LS Sat. 2 0:00:02.0002 2.0002 LS 10^-7 sec and LS ... Sat. 3130 0:52:10.3130 3130.3130 LS 10^-4 sec and LS Sat. Earth 0:52:11.3131 3131.3131 LS 10^-4 sec and LS
A ray in opposite direction leads to this table:
POSITION GLOBAL TIME COVERED DISTANCE MAX.DEVIATION ----------------------------------------------------------------- Sat. Earth 0:00:00.0000 Sat. 1 0:00:00.9999 0.9999 LS 10^-7 sec and LS ... Sat. 3130 0:52:09.6870 3129.6870 LS 10^-4 sec and LS Sat. Earth 0:52:10.6869 3130.6869 LS 10^-4 sec and LS
What about light speed constancy for observers on the satellites? Two neighbouring satellites move almost in the same direction. According to SR the transmission time of e.m. waves between two such satellites must be 1.0000 sec. With respect to the synchronized clocks (global time) however, a ray needs 1.0001 sec to reach the satellite in front and 0.9999 sec to reach the one behind, despite (almost) the same distance in both cases. We must conclude that global time cannot be valid for the satellites. There must be a local time differing from global time.
If satellite S
n sends at global time t a signal in all directions, the signal arrives at satellite Sn+1 at global time t + 1.0001 and at Sn-1 at t + 0.9999. Because both distances are the same and the signal needs 1.0000 sec in both cases, we must conclude that for Sn the following global times are simultanous (equation [4b] leads to the same result): Satellite.n-1 : t + 0.9999 +- 10^-7 sec
Satellite.n : t + 1.0000 +- 10^-7 sec
Satellite.n+1 : t + 1.0001 +- 10^-7 sec
It is instructive to examine how equations [4a, 4b] transform the light speed of the solar frame S into an identical light speed for two neighbouring satellites (frame S'). Let us assume that each satellite sends at the same global time t a ray to the satellite in front. If Dx denotes the distance from S
n to Sn+1 and Dt the needed global time, we get: Dx = 1.0001 LS
Dt = 1.0001 sec
Two neighbouring satellites move at 0.0001 c almost in the same direction and can be approximated by an inertial frame of reference S'. Equation [4a] leads to:
Dx' = Dx - 0.0001 c * Dt = 1 LS [ 6a ]
This result is obvious as relative to the satellites the ray covered only a distance of 1 LS. A classical calculation (global time) of the light speed relative to the satellites leads to:
speed of light = Dx' / Dt = 0.9999 c
In order to garantee light speed invariance also in the moving frames S', time must be transformed in the same way depending on position as position depending on time:
Dt' = Dt - 0.0001/c * Dx = 1 sec [ 6b ]
With respect to global time, the local time of satellite S
n precedes the local time of Sn+1 by 0.0001 seconds (i.e. 0.0001/c Dx). This relation is valid between the earth and S1, between S1 and S2, ..., S3129 and S3130, and between S3130 and the earth. Global time as a common reference leads to transitivity of this relation. It follows that local time on earth must precede the same local time by 0.3131 seconds. This proves that the speed of light cannot be the same for all observers.This is a shortened translation of
Warum die LTG das M.-M.-Experiment nicht erklären können. Comments (on errors) please send to the author.