# Difference between two derivations of length contraction?

I'm working through SR along with a lecture/class series. Last week, we worked through an elementary proof of length contraction after already deriving that $t = \gamma t_0$ with $\gamma$ the familiar Lorentz factor.

This is the proof in the lecture notes:

Consider an observer B and a mirror. They are at rest with respect to each other, separated by a distance $L_0$. The observer reflects a beam of light onto the mirror to measure its distance.

$$L_0 = \frac{ct_0}{2}$$

Now consider an observer A. Observer B is moving away with velocity $v$ with respect to A. A also measures the length with B's experiment:

$$L = \frac{c}{2} (\Delta t_1 + \Delta t_2)$$

Where the terms in brackets represent the motion of the light to the mirror, and the motion of the light from the mirror back to B.

$$\Delta t_1 = \frac{L}{c} + \frac{\Delta t_1 v}{c} \text{ and } \Delta t_1 = \frac{L}{c - v}$$

Analogously follows $\Delta t_2 = \frac{L}{c+v}$.

We now obtain:

$$\Delta t_1 + \Delta t_2 = \frac{2L/c}{1-v^2/c^2}$$

Using that $\Delta t_1 + \Delta t_2 = t_A = \gamma t_0$ we derive

$$\gamma t_0 = \frac{2L/c}{1-v^2/c^2}$$

$$t_0 = \gamma \frac{2L}{c}$$ $$\frac{ct_0}{2} = \gamma L$$ $$L_0 = \gamma L$$

Which is the familiar equation for length contraction.

Here my issue starts.

I was recreating the derivation when I figured "hey, why not make the substitution for gamma earlier." So, I took

$$\Delta t_1 + \Delta t_2 = t_A = \gamma t_0$$

And substituted it into

$$L = \frac{c}{2} (\Delta t_1 + \Delta t_2)$$

This gives

$$L = \frac{c}{2} \gamma t_0$$ $$L = \frac{ct_0}{2}\gamma$$ $$L = L_0 \gamma$$

The reverse of the equation I derived and should have derived. Thus, there must be an illegal operation somewhere. But where?

$$L≠\frac{c}{2}(\Delta t_{1}+\Delta t_{2})$$ See your third equation and the analogous one involving $\Delta t_{2}$, or indeed your fifth equation.
The idea here is that you make the measurement in a rest frame $$S$$, where the length of the rod is measured instantaneously. By this, the only thing that should concern you is that in your frame $$S^{'}$$, $$\Delta t=0$$, and not the measurement of time in the moving frame.