I was so bored on a flight, so I borrowed a pen from my neighbor passenger and wrote a quick derivation of a few results in special relativity:
Suppose an observer is doing a light speed measurement on a train by measuring the time it takes to go upwards and then reflect back and measure the time elapsed, he measures the time elapsed to be $ 2t $ and the distance traveled to be $ 2h $.
But the train is moving, so to an observer outside of the train, he measures the light moved in diagonal manner, let the time spent to be $ t_0 $, and the overall distance the light traveled is $ 2\sqrt{(vt_0)^2 + h^2} $. We assumed the two observers measure the same $ h $ and the same $ v $.
Both measurements are speed of light, so they equals:
$ \frac{2h}{2t} = \frac{2\sqrt{(vt_0)^2 + h^2}}{2t_0} = c $
First, we cancel out all the $ 2 $.
$ \frac{h}{t} = \frac{\sqrt{(vt_0)^2 + h^2}}{t_0} = c $
Suppose an observer is doing a light speed measurement on a train by measuring the time it takes to go upwards and then reflect back and measure the time elapsed, he measures the time elapsed to be $ 2t $ and the distance traveled to be $ 2h $.
But the train is moving, so to an observer outside of the train, he measures the light moved in diagonal manner, let the time spent to be $ t_0 $, and the overall distance the light traveled is $ 2\sqrt{(vt_0)^2 + h^2} $. We assumed the two observers measure the same $ h $ and the same $ v $.
Both measurements are speed of light, so they equals:
$ \frac{2h}{2t} = \frac{2\sqrt{(vt_0)^2 + h^2}}{2t_0} = c $
First, we cancel out all the $ 2 $.
$ \frac{h}{t} = \frac{\sqrt{(vt_0)^2 + h^2}}{t_0} = c $
Square the second equality to get
$ (vt_0)^2 + h^2 = c^2t_0^2 $
Rearranging, get
$ h^2 = (c^2 - v^2)t_0^2 $
Dividing by $ c^2 $ get
$ \frac{h^2}{c^2} = (1 - \frac{v^2}{c^2})t_0^2 $
But $ \frac{h^2}{c^2} $ is simply $ t^2 $ by the first equality, therefore
$ t^2 = (1 - \frac{v^2}{c^2})t_0^2 $
or
$ t = \sqrt{1 - \frac{v^2}{c^2}} t_0 $.
The equation shows a few things:
First, $ v \leq c $, for it does not make sense to have imaginary time. That indicates nothing can run faster than the speed of light.
Second, $ 0 \leq \sqrt{1 - \frac{v^2}{c^2}} \leq 1 $, therefore, the time interval between the same events is shorter for the observers on the train. Suppose $ t_0 = 1 $ second, then $ t $ is less than a second, so if there is a clock there, it doesn't tick yet. So for the observer outside the train, it appears that a moving clock go slower. This is called time dilation.
Third, since the both observers agrees on the relative velocity, the distance traveled, measured by the two observers are $ vt $ and $ vt_0 $ respectively. We know $ t \leq t_0 $, therefore, to the observer on the train, distance appears shorter as well. This is called length contraction.
