*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 (relativity of simultaneity), which is in the case of small velocities a first order effect, whereas time dilation and length contraction are only second order effects.

Assume
that the earth's orbit, which is around 940 million km long, is exactly
circular. Apart from the earth, 3130 artificial satellites move on this orbit
in such a way that the distances between two neighboring satellites remain
exactly 1 light second (around 300'000 km). The velocities of all satellites
are 0.0001 c (around 30 km/s). We further assume that all relative deviations
from these values are less than 10^{-8}.

If Δx, Δt, etc. denote coordinate differences, the Lorentz transformation for a relative speed of 0.0001 c is:

`[1a] Δx' = γ ∙ (Δx - 0.0001∙c Δt)`

`[1b] Δt' = γ ∙ (Δt - 0.0001/c Δx)`

`[2a] Δx = γ ∙ (Δx' + 0.0001∙c Δt')`

`[2b] Δt = γ ∙ (Δt' + 0.0001/c Δx')`

`γ = 1/√[1 - v`^{2}/c^{2}] = 1/√[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 Lorentz-factor
(gamma, γ) is less than 10^{-8} and lies in the range of distance
and speed deviations. The Lorentz-factor can therefore be ignored:

[4a] Δx' = Δx - 0.0001∙c Δt

`[4b] Δt' = Δt - 0.0001/c Δx`

`[5a] Δx = Δx' + 0.0001∙c Δt'`

`[5b] Δt = Δt' + 0.0001/c Δx'`

All
satellites on the idealized orbit have the same distance from the sun. Their clocks
can exactly be synchronized 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 neighboring 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

The
differences in time and distance result from the fact that the Earth has moved
forward 0.3131 ± 10^{-4} light second during 3131 sec = 52 min + 11
sec.

What
about light speed constancy for observers on the satellites? Two neighboring
satellites move almost in the same direction. According to Special Relativity
the transmission time of electromagnetic 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 a 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 S_{n+1} at global time t + 1.0001 and at S_{n-1}
at t + 0.9999. Because distances are the same and the signal is assumed to need
1.0000 sec in both cases, we must conclude that for S_{n} the following
global times are simultaneous (equation [4b] leads to the same result):

`Sat.n-1: t + 0.9999 +- 10^-7 sec`

`Sat.n : t + 1.0000 +- 10^-7 sec`

`Sat.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 neighboring satellites
(frame S'). Let us assume that each satellite sends at the same global time t a
ray to the satellite in front. If Δx denotes the distance from S_{n}
to S_{n+1} and Δt the needed global time, we get:

`Δx = 1.0001 LS`

`Δt = 1.0001 sec`

Two neighboring 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:

`[6a] Δx' = Δx - 0.0001 c ∙ Δt = 1 LS`

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 = Δx' / Δt = 0.9999 c`

In order to guarantee 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:

`[6b] Δt' = Δt - 0.0001/c ∙ Δx = 1 sec`

With
respect to global time, the local time of satellite S_{n} precedes the
local time of S_{n+1} by 0.0001 seconds (i.e. 0.0001/c Δx). This
relation is valid between the earth and S_{1}, between S_{1}
and S_{2}, ..., S_{3129} and S_{3130}, and between S_{3130}
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.