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I'm having a hard time understanding the following problem regarding special relativity:

Let's say we have a light beam moving in the +x direction with a speed of c, and a rocket is moving in the -x direction with a speed of 0.8c. Let's say they are separated by a distance d = 10 light minutes at t = 0 s. Now we want to calculate at what time $t_m$ the light beam will meet the rocket. In order to do this, we can calculate the relative speed $v_{relative}$ of the light beam with respect to the rocket (or vice versa), and then use $t = \frac{d}{v_{relative}}$, where d is 10 light minutes, and $v_{relative}$ is the relative speed. We know that we can't add velocities directly in relativity ($v_{relative} \neq c + 0.8c = 1.8c$), otherwise you break the light barrier, so you have to use Einstein's velocity addition formula. But this formula will return a maximum of 'c' as relative speed? But that would imply that it doesn't matter whether the rocket is stationary, or moving at v = 0.8c towards the light beam? What am I missing exactly? How would you go about solving this problem, using the relative speed?

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    $\begingroup$ You need to think more clearly about what are the inertial frames implicit in the problem, what the relevant events, and how do they transform. $\endgroup$
    – JEB
    Commented Jun 27, 2022 at 14:15

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But that would imply that it doesn't matter whether the rocket is stationary, or moving at v = 0.8c towards the light beam? What am I missing exactly?

You are not missing anything. That is correct. Regardless of the velocity of the rocket in a separate frame, in the rocket’s frame the light is traveling at c.

This is not a problem to be fixed, it is a necessary feature. This is the second postulate in action: the speed of light is c in all inertial reference frames. In order to fulfill this criterion it is necessary that the relativistic velocity addition formula be structured so that the relative velocity between c and anything else is c.

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  • $\begingroup$ I think I already know what my problem was, because I didn't take into account length contraction and time dilation in order to calculate the time that they meet, using relative velocity. $\endgroup$
    – Stallmp
    Commented Jun 27, 2022 at 12:41
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It is useful to draw a diagram $t \times x$. Then calculate the equations for the straight lines of the beam and of the rocket trajectory. The interception is the event $t,x$ where the light reach the rocket. The expression relative velocity can be confusing, because it is used normally for different frames, and there is no frame for a light beam.

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