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


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

The period of a round trip is then:

T=\frac{L}{c \boxminus v}+\frac{L}{c \boxplus v}=\frac{2L}{c}\; \; \; (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.

The Lorentz Adjustment

Expand the mistaken Equation (1) to get

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

Move one γ factor under T to get

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

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

T'=\gamma T\; \textup{and} \; 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 T}{\gamma}=\frac{2}{c}L' \gamma =\frac{2}{c} \frac{L}{\gamma} \gamma

That is,


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