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 - v2/c2] = 1/√[1 - 0.00012] = 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 Sn 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 distances are the same and the signal is assumed to need 1.0000 sec in both cases, we must conclude that for Sn 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 Sn to Sn+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 Sn precedes the local time of Sn+1 by 0.0001 seconds (i.e. 0.0001/c Δx). 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.


No rights reserved Wolfgang G. Gasser, 1988-2016