At 56:59 of this lecture by prof. Shankar presents a proof on the lorentz transformation with one observer at rest and other moving a speed $u$ in the following way:

Let $(x,t)$ be the coordinates of observer at rest and $(x',t')$ be the coordinates of the moving observer. Considering the postulates of relativity, it was motivated that: $$x'= \gamma(x-ut) \tag{1}$$ $$x= (x' + ut')\gamma \tag{2}$$ Multiply (1) and (2) and with some algebra: $$ 1= \gamma^2 \left[1+ u \frac{t'}{x'} - u \frac{t}{x}-u^2 \frac{t}{x} \frac{t'}{x'} \right]$$ At 55:38, the professor states that $x=ct$ and $x'=ct'$ , hence, $$\gamma = \frac{1}{\sqrt{ 1- \frac{u^2}{c^2}}}$$

The last step is where I am stuck i.e: $x=ct$ and $x'=ct'$. I am a bit confused here because on why the coordinate functions must equal the speed of light times the time. Hoping for an answer which can explains these two equalities with more clarity.


1 Answer 1


The professor is simply stating that by the postulates or relativity, in both frames the speed of a beam/pulse of light must be equal to the speed of light. He is considering the trajectory of a small beam/pulse of light. In the rest frame its trajectory shall be $x=ct$ (travelling at the speed of light), but also in the moving frame its trajectory shall be $x'=ct'$ (travelling at the speed of light). This condition allows him to derive the formula for $\gamma$.

  • $\begingroup$ Right, but $x$ is a coordinate function . If it was speed of light it should be change in distance by change in time $\endgroup$ May 14, 2021 at 19:31
  • $\begingroup$ @Buraian I don't understand what you're saying. If I was in the rest frame, I would describe the trajectory of such a pulse of light by the following parameterized coordinates: $(t,x(t)=ct,y(t)=0,z(t)=0)$. The velocity would be $d\textbf{x}(t)/dt$. $\endgroup$ May 14, 2021 at 19:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.