# Conceptual question regarding special relativity

I was watching a video lecture on Lorentz Transformation where the lecturer was solving a problem. The statement of the problem was:

Suppose we have two inertial frames- stationary one $$S$$ and another one $$S'$$ which is moving at $$0.6c$$ where $$c$$ is the speed of light. At $$t=0$$, origins of both the frames coincide. Let a light pulse be emitted from the origin at $$t=0$$ making an angle $$\arctan(3/4)$$ with positive $$x$$ axis.

The lecturer was trying to show that the speed of light is measured same in both reference frame by measuring position of light pulse in both frames after $$2 \times 10^{-6}\,\mathrm{s}$$.

Now, he first calculated $$\gamma$$ using $$\gamma = \frac {1}{\sqrt{1-\frac {v^2}{c^2}}}$$ and got $$\gamma = 1.25$$. Also he got the coordinates in $$S$$ refrence frame as $$(480\,\mathrm{m}, 360\,\mathrm{m}, 0\,\mathrm{m}, 2 \times 10^{-6}\,\mathrm{s})$$

Using Lorentz transformations he calculated for $$x' = \gamma \left (x - vt \right) \tag {1}$$ and got $$x' = 150\,\mathrm{m}$$. Now, I noticed that he used $$t= 2\,\mu\mathrm{s}$$ here

Then, then he calculated $$t' = \gamma \left ( t - \frac {vx}{c^2} \right)$$ to get $$t' = 1.3\,\mu\mathrm{s}$$

Now my question is: If all primed quantities are those measured by observer in $$S'$$ reference frame, why did he use $$t= 2\,\mu\mathrm{s}$$ for $$(1)$$? Shouldn't he have calculated $$t'$$ first and then substituted $$t'= 1.3\,\mu\mathrm{s}$$ in $$(1)$$?

• The time in (1) is t and not t'. Why do you think he should have used t'? That equation relates prime coordinate to unprimed coordinates. – nasu Nov 1 '19 at 14:53
• Aren't $(x',y',z',t')$ the quantities as measured by observer in $S'$ frame? For an observer in moving frame, $t' = 1.3 \mu s$ when $t= 2 \mu s$ for an observer in $S$ frame. So how can the observer in $S'$ measure x' at $t= 2 \mu s$? – Nirmal Padwal Nov 1 '19 at 15:14
• Please give a proper attribution to the lecturer. – Ben Crowell Nov 1 '19 at 15:15
• Or does $x' =150$ implies that this will be the x' coordinate measured by observer after $t' = 1.3 \mu s$ – Nirmal Padwal Nov 1 '19 at 15:16
• @Ben Crowell nptel.ac.in/courses/115101011 Lecture 5: Lorentz Transformation. The problem starts at 37:35 – Nirmal Padwal Nov 1 '19 at 15:25

I think conceptual problem is that you are thinking of 2 views of one worldline of the light pulse, $$x(t)$$ and $$x'(t')$$, in which (un)primed position is a function of (un)primed time.
$$x' = f(x, t)$$ $$t' = g(x, t)$$
where of course $$f$$ and $$g$$ are the Lorentz transformation.