The two-way velocity of light has been measured with the greatest position, and has always been found to be c, the constant speed of light in vacuum. This seems to justify Einstein's postulate that the speed of light is isotropic and constant.

Yet, it has been known for over a century that the Sagnac effect defies special relativity in which a phase shift results for light beams sent in opposite directions around a closed path on a rotating platform. It was Silvertooth's idea to let the area enclosed by the light beams to tend to zero to obtain a positive result for the linear Michelson-Morley experiment. Marett has contended that temperature variations intervence to destroy the Silvertooth experiment. Yet, the question still remains is it true that in the limit of vanishing area enclosed by the light path that the phase shift disappears.

In the Michelson-Morley experiment, the outward and return velocities are c(1-v/c) and c(1+v/c), respectively, where v is the speed of the ether wind. It is clear that the arithmetic average speed is c so that one could not tell whether the return speed was c(1+v/c) or just c,

since v vanishes in the arithmetic average. And since the arithmetic average is greater than, or at least equal to the geometric and harmonic averages there is no reason to doubt that the one-way speed of light is anything but c. And so it has remained since the dawn of special relativity over one hundred and fifteen years ago.

Yet, one is at a loss to explain the time difference of the Sagnac effect being equal to

t"-t'=2Lv/c^2(1-v^2/c^2)

where t" and t' are the times that the light signal that was emitted at time t come back to it after having followed closed paths, L, the circumference of the disc, in opposite directions. The above time difference occurs at the same point on the disc. This time difference,

(t"-t')/L=1/c"-1/c',

would clearly be zero if Einstein's postulate held: c'=c".

Moreover, if c'=c(1-v/c) and c'=c(1+v/c), the arithmetic mean velocity would still be c. That is the time difference propagating in one direction over a semi-circumference, L/2,

t'-t=L/2c(1-v/c)

and in the other direction,

t"-t=L/2c(1+v/c).

Their difference is

t"-t'=L/v(1-v^2/c^2).

Now, let us change our perspective and consider a light signal emitted at time t, reflected at time t' and returns to the point at where it was emitted at time t". In this case, we would encounter length contraction by an amount L=L'(1-v^2/c^2)^(1/2). Or, we could consider the time read on a clock rotating with the disc with respect to one near the disc in the laboratory frame. That time would appear to go slower by an amount t/(1-v^2/c^2)^(1/2). We can apply one or the other, or---choose not to apply either (!) because we are interest in the ratio:

e =(t'-t)/(t"-t)=(1-v^2/c^2)/2(1-v/c)=(1+v/c)/2.

Solving for t' we find

t'=t+e(t"-t),

and Einstein's convention is upheld if we choose e=1/2, in which case the outward journey is the arithmetic mean of the time of emission and the time the light pulse takes to return to the point of emission. Clearly this implies v=0. However, for values of e not equal to one-half, there will be different speeds of light in different directions.

Call c' the speed of light out and c" the speed of return. We have

t'-t=L/2c' and t"-t'=L/2c".

Necessarily, the two-way speed of light must be c; that is, the sum of the two must be:

t"-t=L/2(1/c'+1/c")=L/c.

Their ratio must be e:

e=(t'-t)/(t"-t)=c/2c'=(1+v/c)/2,

where we used the value of e found above. Now, the ratio of the two-speeds is

c"/c'=(1-v/c)/(1+v/c)

which is a Doppler shift under reflection, like that of a light beam striking a mirror. The above equality is

c'=c/(1+v/c)

so the condition on the two-way speed being c gives

c"=c/(1-v/c).

If one speed is greater than c, the other must be less than c, unless v=0 in which case all means coincide. By inverting the speeds, the arithmetic and harmonic means swap places.

The condition that their average is c now applies to the harmonic mean---and not the arithmetic mean! And from the arithmetic (A)-geometric (G)-harmonic (H) mean inequalities we know that the geometric and arithmetic mean speeds are both greater than c! Yet, we have more: In the case of two independent variables--which is our case of the outward and return journeys--the three means satisfy:

A x H = G^2.

Since H=c, A=c/(1-v^2/c^2), and G=c/(1-v^2/c^2)^(1/2).

In contrast, the Michelson-Morley experiment assumed the speeds were c'=c(1-v/c) and c"=c(1+v/c) so that their arithmetic average was simply c. This is the highest average speed attainable and the geometric and harmonic means are c(1-v/c)^(1/2) and c(1-v^2/c^2), respectively.

In the words of Silvertooth, "we are suggesting here an alternative explanation for the original Michelson-Morley null result; namely, that the variation in the speed of light could have had a functional form different from what they expected and designed that experiment to detect." This statement shows their desired to reduce the Sagnac effect to the linear Michelson interferometer, by letting the area of closed path tend to zero. Here, we have shown that it definitely applies to the Sagnac effect which differs from the Michelson-Morley experiment in that the average speed is the harmonic--and not the arithmetic--mean. And since the harmonic mean is the smaller of the three, it proves that the one-way speed of light must be greater than c.

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