homework and exercises - How to show using Lorentz transformation the $vec{v}$ of one frame to another in either frame is $vertvec{v}vert$

I am trying to show that in the case of one frame at rest, S, and another frame moving in the +x direction, S', that:

1. The S frame see's the S' frame moving with $+\vec{v}$


2. The S' frame see's the S frame moving with $-\vec{v}$

First, we start in the S frame which see's the S' frame moving in the +x direction. We record two spacetime events of the location of the S' frame's origin and find the velocity:

$$Event 1: (t_1,x_1,0,0) $$

$$Event 2: (t_2,x_2,0,0) $$

$$\frac{x_2 - x_1}{t_2 - t_1}=\frac{\Delta x}{\Delta t}=+\vec{v}$$

Now, for the S' frame, they see the S frame's origin moving away in the -x direction. We can transform the events using time dialation and length contraction arguments:

1. 
   

   The length $\Delta x$ reported by the S frame will be longer than the S' frames length (because S' is at rest now), so $\Delta x'=- \frac{\Delta x}{\gamma}$.

   
   


2. 
   

   The time interval $\Delta t$ reported by the S frame will be shorter than the S' frames time interval, so $\Delta t'= \Delta t\gamma$.

   
   

$$\frac{\Delta x'}{\Delta t'}=\frac{\frac{-\Delta x}{\gamma}}{\Delta t\gamma} =\frac{-v}{\gamma^2}$$

Which is obviously wrong, but I am really stuck on how to set up this problem/system.

Answer

To measure length contraction you need to look at two parallel world-lines; you are looking at one individual world-line and therefore you've not been clear about what you're studying.

The coordinate transformation that you want is (with $w=ct$ and $\beta=v/c$ to show a certain symmetry of the equations)$$\begin{align} w'&=\gamma~(w - \beta~x),\\ x'&=\gamma~(x - \beta~w),\\ y'&=y,\\ z'&=z.\end{align}$$ Ignoring the $y$ and $z$ directions as trivial, length contraction comes when we consider the world-lines $(w, x) = (s, 0)$ for all $s$ and $(w, x)=(s, L)$ for all $s$. These both transform to:$$w'=\gamma~(s - \beta~L),\\ x'=\gamma~(L-\beta~s).$$Substituting for $s=w'/\gamma + \beta~L$ we can find the equation $$x'=\gamma~(L-\beta~(w'/\gamma+\beta~L) = -\beta~w' + \gamma~L~(1 - \beta^2).$$

Note that $\gamma~(1-\beta^2) = \gamma / \gamma^2 = 1/\gamma,$ giving our familiar length contraction factor; note also that this line which was stationary is now moving backwards with speed $-\beta~c=-v$.

This effect that when you accelerate, clocks in front of you tick faster the further they are in front of you, and clocks behind you tick slower the further they are behind you, is called the "relativity of simultaneity." It is a very important effect: in fact I have argued before that it is the most important effect, in that for small $\beta\ll 1$ it is the only effect which survives: and we can reconstruct the general coordinate transformation above from this $\gamma\approx 1$ form.