Why does the horizontal distance L contract in the Michelson-Morley experiment?

[Karol]

Homework Statement

Lorentz suggested that L parallel shortens the amount:

And inserting it back cancels the time difference:

$$\frac{2L/C}{\sqrt{1-u^2/c^2}}>\frac{2L/C}{1-u^2/c^2}~\rightarrow~t_1+t_2<2t_3$$
Logic says i have to increase L in order to increase t₁+t₂ so it will equal 2t₃

Homework Equations

The times t₁+t₂ parallel to the velocity u, and 2t₃ vertical, to mirror C are:

The Attempt at a Solution

I understand the mathematical trick, but logically the horizontal distance should grow, no?

 

[kuruman]

This problem statement is ill-defined and so is your "attempt at a solution". I suspect that you will not get many responses until you are clear about what kind of help you are seeking and why.

 

[Karol]

Well i don't know what to write in the Relevant Equations, so i spread the problem statement in both, the Problem Statement and the Relevant Equations.
But i ask why does the horizontal distance L contract instead of lengthen.

 

[MikeLizzi]

Well i don't know what to write in the Relevant Equations, so i spread the problem statement in both, the Problem Statement and the Relevant Equations.
But i ask why does the horizontal distance L contract instead of lengthen.

Seems to me you are trying to compare/contrast the output of the interferometer when it is at rest with respect to the observer and when it is moving with respect to the observer. But that's not what MMX was about. For the original Michelson-Morley experiment, the interferometer, light source and observer were always at rest with respect to each other.