1

Your spaceship is traveling at 1/2*c and you approach Bob's spaceship going -1/2*c in the opposite direction. Relative to you, Bob is going the speed of light, and of course he has a giant ticking clock for all to see. From your perspective, Bob's clock looks like it is completely stopped.

1) Can you even see the clock as you approach it? It's coming at you at the same speed light propagates toward you.

2) Do all of Bob's light rays hit you at the same time when Bob passes you so that you would see a long blur/beam of where Bob has been traveling? Could it be more like a big flash because all the photons arrive at the same time, built up like a sound waves before a jet breaks the sound barrier?

3) From Bob's perspective, the light from his clock hit you long ago. Or another way to think about this: say Bob turned on a laser beam pointed at you 10 years before passing you. Bob thinks the laser finally reached you 5 years before you both pass each other. But you still don't see the light until you pass him?! It's tough to wrap my head around this. Idea?

4) Can you see Bob after he passes you? From your perspective, light would have to catch up to him in order to rebound back to you. Perhaps light coming from the side would allow you to see his moving away from you. From Bob's perspective, after you pass him his light will never reach you again because you are traveling away as fast as his light.

5) One step further: if Bob was flying at -2/3*c, is Bob's clock ticking backwards from your perspective? Would you see him pass you before you see him approaching you a mile away?

Qmechanic
  • 201,751
excelCoder321
  • 11
  • 1
    Why do you assert Bob goes at the speed of light relative to you? – QCD_IS_GOOD Nov 06 '15 at 07:21
  • Hi, welcome to Physics SE. It seems that all your confusion is because you think Bob will be coming at the speed of light towards you. But that will not be the case as velocities don't simply add the way they add in classical mechanics. Velocity addition rule in relativity is $\frac{u + v}{1+uv/c^2}$ In this scenario the speed of Bob in your frame will be $\frac{c/2 + c/2}{1+1}$ = $c/2$ –  Nov 06 '15 at 07:24
  • Consider the light that leaves Bob's ship is traveling faster than Bob's 0.5c, so it should reach you before Bob does. Relativistic speed can be counter-intuitive, but there's many explanations of it available. Here's one, similar to, not precisely the same as your question: https://www.youtube.com/watch?v=Mcg4BySwumg – userLTK Nov 06 '15 at 07:33
  • Djiv, to be clear, you're saying Bob is going -c/2, right? negative velocity? does this mean it doesnt matter if we were traveling toward bob or standing still, he's still going -c/2? how can that be! – excelCoder321 Nov 06 '15 at 07:37
  • 1
    thanks for blowing my mind ya'll. my questions are just dumb now. – excelCoder321 Nov 06 '15 at 07:44
  • @Dvij, your math seems off. The relative speed in the spaceship frame is $0.8c$ ($uv = 0.25$, not 1) – BowlOfRed Nov 06 '15 at 09:11
  • @BowlOfRed: Dvij made an error in the denominator $uv/c^2 \ne 1$. – John Rennie Nov 06 '15 at 12:03
  • @BowlOfRed Thanks for pointing out. yeah, $uv/c^2$ is $0.25$. So relative speed will be $0.8c$. –  Nov 06 '15 at 12:06
  • 5
    Possible duplicates: http://physics.stackexchange.com/q/131589/2451 and links therein. – Qmechanic Nov 06 '15 at 13:55

1 Answers1

0

Ok so there is one major problem! You have assumed that because your going at half the speed of light and that bob is going at half the speed of light in the other direction then, relative to you bob goes at the speed of light, but you don't add up velocities in relativity. If your going at velocity $v$ and bob goes at velocity $-u$ then, its velocity relative to you will be $$\frac{v-u}{1+\frac{vu}{c^2}}$$