The Lorentz Transformation for length gives longer instead of shorter values I'm really trying, but can't seem to understand equations for Lorentz transformations. For example, take the 5.6.2 example from this site.
It uses this equation for Lorentz transformation:
$$
x' = \frac{x - v*t}{\sqrt{1 - \frac{v^2}{c^2}}}
$$
In the step 3. of the example, the equation looks like this:
$$
x'_2 - x'_1 = \frac{x_2 - x_1}{\sqrt{1 - \frac{v^2}{c^2}}}
$$
From what I understand, it is the same as:
$$
L' = \frac{L}{\sqrt{1 - \frac{v^2}{c^2}}}
$$
If I input $L = 100 m$ in there, I get $L' = 102.062 m$. That obviously can't be right as $L$ is the length of that street in a frame where it is stationary, so spaceship flying by can only see that it is shorter. In the step 4 they used this formula:
$$
L' = {L} * {\sqrt{1 - \frac{v^2}{c^2}}}
$$
which gives correct solution of $L' = 97.976 m$.
What I can't understand at all is how the equation from step 3 became formula from step 4? As far as I can tell, that definitely isn't the same equation (in step 3 there is a division, but in step 4 there is multiplication!). It really makes understanding these transformations impossible for me. I understand that there are 4 main equations for Lorentz transformations, but why they give wrong solutions (equation from step 3!) can't get in my head.
 A: $x_2'-x_1'$ is not the length of an object as seen by a moving observer.  This is because $x_2'$ and $x_1'$ are not measured at the same time in the moving observer's reference frame.  The front end of the board is at $x_1'$ at time $t_1'$.  The back end of the board is at $x_2'$ at time $t_2'$.  Length of the board in the moving observer's frame is not just $x_2'-x_1'$.  From the moving observer's point of view, the positions are not measured at the same time.  To get $L'$, you need to find $x_2'-x_1'$ when $t_1'=t_2'$.  This will not happen when $t_1=t_2$.
A: You are transforming the spatial distance between two events from one coordinate system to another.  That is not the same thing as transforming a length from one coordinate system to another.  That's because a length is the distance between two ends of a rod at the same time, and the meaning of "at the same time" changes when you change coordinate systems.
A: I too had this similar doubt, while deriving the expression for Length contraction. I would like to make it as clear as possible.
Note- There will be many times frames $S$ and $S'$ will be used. So bare that with me.
See, say there is a rod in a frame $S'$, where the rod act as a stationary object or not moving, also called as Proper Length, although it is moving for an observer looking, say in frame $S$, such that frame $S'$ is moving in $+x$ direction with velocity say $v$ wrt frame $S$. Now for the observer in frame $S$, will see the rod moving.
A conceptual point- You always measure length of an object when it is still or not moving, by noting one end of the rod and then another end "at the same time". But this is not possible for observer in $S$ frame. So the only logical way to measure the length of the rod in a moving frame, is to measure the rod by noting time of the two end at the rod at the same time.
The major part of confusion. Now by LT, if we keep/maintain, time measured of two ends of the rod same, which is appearing moving in frame $S$, then what about time measured (or if you measure it, do measure it) then in $S'$ frame. You will see, it will be non-zero. So what going on after all at the end then?
The answer is it doesn't matters. Why you asked, this is because the rod is at rest in frame $S'$, and think for a second, does it matter for a rod at rest, we measure one end of the rod, say today and another end after 5 days, as rod length will always be measured accurately in frame $S'$.
So yeah by LT it is necessary to keep time same for moving frame and not for the frame where rod is at rest. I hope it makes clear to you know and in solving problems too.
