Gamma factor between means

Consider the mean between two quantities:

$c+v\; \mathrm{and}\; c-v.$

The arithmetic mean is

$\frac{c+v+c-v}{2}=c.$

The harmonic mean is

$\left (\frac{(c+v)^{-1}+(c-v)^{-1}}{2} \right )^{-1}=\frac{c^2-v^2}{c}.$

Consider a factor γ² that transforms a harmonic mean into an arithmetic mean:

$\gamma^2 \left ( \frac{c^2-v^2}{c} \right )=c$

so that

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

which equals the gamma factor of relativity theory.

Or consider the mean between these two:

$\frac{a}{c+v}\; \mathrm{and}\; \frac{a}{c-v}.$

The arithmetic mean is

$\frac{1}{2}\left ( \frac{a}{c+v}+\frac{a}{c-v} \right )= \frac{ac}{c^2-v^2}.$

The harmonic mean is

$\left (\frac{1}{2}\left (\left (\frac{a}{c+v} \right )^{-1}+\left (\frac{a}{c-v} \right )^{-1} \right ) \right )^{-1}=\frac{a}{c}.$

Consider a factor γ² that transforms a harmonic mean into an arithmetic mean:

$\gamma^2 \left (\frac{a}{c} \right )=\frac{ac}{c^2-v^2}$

so that

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

which equals the gamma factor of relativity theory.

If the harmonic mean was mistakenly used where the arithmetic mean would have been correct, the gamma factor fixes the problem. Or if the arithmetic mean was mistakenly used where the harmonic mean would have been correct, the inverse gamma factor fixes the problem. Hmm …

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