# Michelson-Morley Experiment

I've looked everywhere and I cannot find a complete derivation that includes the step I'm looking for lol...hopefully this will add another link for google.

So the full time for light to travel both directions with the ether then against as follows:

$$T = \frac{L}{(c+v)}+\frac{L}{(c-v)}\tag 1$$ then you get by adding fractions: $$T = \frac{2Lc}{(c^2-v^2)}\tag 2$$

Then everything I read online and my textbook says that equals: $$T=\frac{2L}{c}\frac{1}{1-\frac{v^2}{c^2}}\tag 3$$

then of course: $$T=\frac{2L}{c(1-\frac{v^2}{c^2})}\tag 4$$

I'm not understanding how they get from $(2)$ to $(3)$.

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Hint: $c=\frac{1}{\frac{1}{c}}$ – John Douma Jun 12 '13 at 20:35

You just divide numerator and denominator by $c^2$:
$$T = \frac{2Lc}{(c^2-v^2)}\tag 2=\frac{\frac{2Lc}{c^2}}{\frac{(c^2-v^2)}{c^2}}=\frac{2L}{c}\frac{1}{1-\frac{v^2}{c^2}}$$
That's a very common strategy, specially in special relativity, you try to get that $v/c$ coefficient as much as you can so you're constantly dividing by $c$ :) – MyUserIsThis Jun 12 '13 at 20:47