I was reading Brian Greene's book The Elegant Universe and in the special relativity part, I was told that even if I was chasing a beam of light at light speed, it would still appear to be moving at $3*10^8$ metres per second.

He writes that in a moving train when there are two observers sitting in front of each other round a table, if a bulb is lit midway on the table, the two observers would see the bulb light exactly at the same instant but an observer standing on the platform would disagree with this.

If I'm moving with near light-speed and am trying to catch up with a light beam (which still moves with light speed ahead of me), would a stationary observer not find the light beam to be moving with a velocity approximately $2(3*10^8)$ m/s? (Isn't that a velocity greater than the maximum allowed?)

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    $\begingroup$ No, everyone sees $c$. This is one of the many ways that relativity is in conflict with common sense. You just have to get used to it. $\endgroup$ – garyp May 23 '18 at 18:11

No, if the stationary observer set up a measurement device, she would measure the approaching light beam's velocity as exactly c in each and every case, regardless of the state of motion of the light beam's source. In special relativity, velocities like this do not simply add as they do in the galilean picture.

  • $\begingroup$ Thanks, a friend told me the maths too. The addition works as $$v_{AB}=\frac{v_A + v_B}{1 + \frac{v_A + v_B}{c^2}}$$ $\endgroup$ – Ayush Arya May 24 '18 at 3:56

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