I am just working through an argument from Halliday Resnick to derive the Lorentz contraction (see quote below).
Some paragraphs before this, the authors note that:
If the rod is moving, however, you must note the positions of the end points simultaneously (in your reference frame) or your measurement cannot be called a length.
A paragraph later they invoke the following argument:
Length contraction is a direct consequence of time dilation. Consider once more our two observers. This time, both Sally, seated on a train moving through a station, and Sam, again on the station platform, want to measure the length of the platform. Sam, using a tape measure, finds the length to be $L_0$, a proper length because the platform is at rest with respect to him. Sam also notes that Sally, on the train, moves through this length in a time $\Delta t = L_0/v$ where $v$ is the speed of the train; that is, $$ L_0 = v \Delta t \quad \text{(Sam)} $$ This time interval $\Delta t$ is not a proper time interval because the two events that define it (Sally passes the back of the platform and Sally passes the front of the platform) occur at two different places, and therefore Sam must use two synchronized clocks to measure the time interval $\Delta t$.
For Sally, however, the platform is moving past her. She finds that the two events measured by Sam occur at the same place in her reference frame. She can time them with a single stationary clock, and so the interval $\Delta t_0$ that she measures is a proper time interval. To her, the length $L$ of the platform is given by $$ L = v \Delta t_0 \quad \text{(Sally)}. $$
Then they conclude by dividing the two equations above:
$$ \frac{L}{L_0} = \frac{v\Delta t_0}{v \Delta t} = \frac{1}{\gamma}$$ or $$ L = \frac{L_0}{\gamma} $$
which is the length contraction equation.
However I don't see in what sense the length was measured simultanous in the derivation above, how is the detailed connection between the statement that the length measurement has to be simulanous and the quoted derivation?
