Lorentz Transformation

If an observer is placed in a coordinate system S' that travels with the speed $v$ along the $x$-axis of another system S, then the Galilei transformation can be used to transform one system into the other: $$x' = x - vt$$ $$x = x' + vt$$ We have already seen that the Galilei transformations do not preserve the speed of light in different inertial systems. In order to avoid this problem and still not change too many things in our equations, we just want to introduce an additional factor $\gamma$ which is multiplied to the right sides of the equations. $$\boxed{x' = \gamma(x - vt)}$$ $$\boxed{x = \gamma(x' + vt')}$$ In the next step, a solution for $\gamma$ has to be found. For that purpose, the time $t$ was additionally replaced by $t'$ in the second equation. Otherwise, it would not be possible to find a solution for this equation system. Space and time are therefore not independent of each other anymore, and the measured time changes for the observer of another system. The coordinate of a light pulse can be calculated according to $$x = c t$$ $$x' = ct'$$ Inserting these two equations leads to $$ct' = \gamma(ct - vt)$$ $$ct = \gamma(ct' + vt')$$ To find a solution for $\gamma$, both equations can be multiplied with each other: $$ctct' = \gamma^2(ct-vt)(ct'+vt')$$ Solving this equation for $\gamma$ leads to te so-called Lorentz factor (or $\gamma$-factor): $$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ The term $v/c$ is usually replaced by $\beta$: $$\boxed{\beta = \frac{v}{c}}$$ In this case, the Lorentz factor can be written as $$\boxed{\gamma = \frac{1}{\sqrt{1-\beta}}}$$ The Lorentz transformations for $x$ and $x'$ can then be obtained by inserting the Lorentz factor into the modified Galileo equations. In order to find the transformation of the time, the equation $$t' = \frac{x'}{c}$$ can be used. Inserting $x'$ gives $$t' = \frac{\gamma(x-vt)}{c}$$ The right side can be rearranged according to $$t' = \gamma\left(\frac{x}{c}-\frac{vt}{c}\right) = \gamma\left(\frac{x}{c}-\frac{vtc}{c^2}\right)$$ Since we have already stated that $t$ is equal to $x/c$, it follows: $$\boxed{t' = \gamma\left(t-\frac{v}{c^2}x\right)}$$ and for the back-transformation $$\boxed{t = \gamma\left(t'+\frac{v}{c^2}x'\right)}$$ In order to prove the correctness of the previous derivations, we can go back to the example of a person resting in the system S' and switching on a flashlight in the direction of motion. The speed of the light can be described in S by applying the above-mentioned equations: $$\frac{\Delta x}{\Delta t} = \frac{\gamma(\Delta x' + v\Delta t')}{\gamma(\Delta t' + \frac{v}{c^2}\Delta x')}$$ After canceling the Lorentz factor, the denominator can be rearranged as follows: $$\frac{\Delta x}{\Delta t} = \frac{\Delta x' + v\Delta t'}{\Delta t'\left(1+\frac{v}{c^2}c\right)}$$ With $\Delta x'/\Delta t' = c$ it then follows: $$\frac{\Delta x}{\Delta t} = \frac{c+v}{1+\frac{v}{c}} = \frac{c + v}{\frac{c+v}{c}}$$ Finally, this results in $$\frac{\Delta x}{\Delta t} = c$$ which is exactly the speed of light as measured in the system S'. This example shows that the Lorentz transformation is indeed suitable for transforming the coordinates from one system into another one by keeping the speed of light constant.