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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) $?

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  • $\begingroup$ 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. $\endgroup$ – nasu Nov 1 '19 at 14:53
  • $\begingroup$ 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 $t= 2 \mu s$? $\endgroup$ – Nirmal Padwal Nov 1 '19 at 15:14
  • $\begingroup$ Please give a proper attribution to the lecturer. $\endgroup$ – Ben Crowell Nov 1 '19 at 15:15
  • $\begingroup$ Or does $x' =150$ implies that this will be the x' coordinate measured by observer after $t' = 1.3 \mu s $ $\endgroup$ – Nirmal Padwal Nov 1 '19 at 15:16
  • $\begingroup$ @Ben Crowell nptel.ac.in/courses/115101011 Lecture 5: Lorentz Transformation. The problem starts at 37:35 $\endgroup$ – Nirmal Padwal Nov 1 '19 at 15:25
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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.

As pointed out in the comments, we're just Lorentz transforming coordinates, so that all the primed coordinates are functions of unprimed coordinates:

$$ x' = f(x, t) $$ $$ t' = g(x, t) $$

where of course $f$ and $g$ are the Lorentz transformation.

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