# Michelson Morley Experiment Re-examined

The Michelson-Morley experiment, compared the longitudinal and transverse cases of reflected light, expecting to detect an ether wind (Figure 1).

Figure 1. Michelson-Morley apparatus

They explain: “Let sa … be a ray of light which is partly reflected in ab, and partly transmitted in ac, being returned by the mirrors b and c, along ba and ca. [Then] ba is partly transmitted along ad, and ca is partly reflected along ad. If then the paths ab and ac are equal [lengths L], the two rays interfere along ad.”

By rotating the apparatus they expected to detect an ether wind parallel to ba or ca. They calculated the round-trip duration (in the notation here) as

$T=\frac{L}{c-v}+\frac{L}{c+v}\;&space;\;&space;\;&space;(1)$

But since the independent variable is the distance traversed, duration is a dependent variable, and the experiment is in the distance domain. Speeds are distance rate speeds, which add by harmonic addition. So the denominators should be:

$c&space;\boxminus&space;v=\left&space;(\frac{1}{c}-\frac{1}{v}&space;\right&space;)^{-1}$

and

$c&space;\boxplus&space;v=\left&space;(\frac{1}{c}+\frac{1}{v}&space;\right&space;)^{-1}$

The period of a round trip is then:

$T=\frac{L}{c&space;\boxminus&space;v}+\frac{L}{c&space;\boxplus&space;v}=\frac{2L}{c}\;&space;\;&space;\;&space;(2)$

As with a light clock, the round-trip periods would be equal whether or not there was an ether wind, so their null result should have been expected.

Expand the mistaken Equation (1) to get

$T=L\left&space;(&space;\frac{2c}{c^{2}-v^{2}}&space;\right&space;)&space;=\frac{2L}{c}\left&space;(&space;\frac{1}{1-v^{2}/c^{2}}&space;\right&space;)&space;=\frac{2L}{c}&space;\gamma^{2}$

Move one γ factor under T to get

$\frac{T}{\gamma}=\frac{2}{c}L&space;\gamma&space;\;&space;\;&space;\;&space;(3)$

Equation (3) is still mistaken but suggests an adjustment by defining variables T′ and L′ such that

$T'=\gamma&space;T\;&space;\textup{and}&space;\;&space;L'=\frac{L}{\gamma}$

in which time is ‘dilated’ and length is ‘contracted’. Then substitute T′ and L′ into Equation (3):

$\frac{T'}{\gamma}=\frac{\gamma&space;T}{\gamma}=\frac{2}{c}L'&space;\gamma&space;=\frac{2}{c}&space;\frac{L}{\gamma}&space;\gamma$

That is,

$T=\frac{2L}{c},$

which is correct. The Lorentz transformation compensates for the mistake in Equation (1) but doesn’t fix it.