-1
$\begingroup$

I read this question at another forum but the thread was already closed. Here's the description: Two observers A and B are both moving at a velocity of 0.9 times the speed of light with respect to a stationary object. So, won't A see B moving at a speed greater than light?

$\endgroup$

marked as duplicate by jinawee, John Rennie, Abhimanyu Pallavi Sudhir, user10851, Kyle Kanos Feb 23 '14 at 18:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ Stuff moving at such great speeds don't follow normal everyday physics. Look up "Theory of Relativity". :) $\endgroup$ – mikhailcazi Feb 23 '14 at 6:42
  • $\begingroup$ I know that. I asked this question after reading some part of SR. $\endgroup$ – Yashbhatt Feb 23 '14 at 6:54
2
$\begingroup$

No, A will not see B moving faster than the speed of light due to time dilation. What you are doing is a "Galilean Transformation" which is really just an approximation for objects moving with a velocity much less than the speed of light. The proper equation for velocity transformations in special relativity is:

$$ u'= \frac{u+v}{1+\frac{uv}{c^2}} $$

Where u is B's speed with respect to the stationary observer and v is the speed of A's reference frame which is moving with respect to the stationary observer. A is stationary within its own frame. Plugging in we see that A sees B moving away at $.9945c$ indeed less than the speed of light. Full derivations are available everywhere online like the one mentioned above. Hope this helps.

$\endgroup$
  • $\begingroup$ And what about the formula where time difference between two seconds of stationary observer will be 1/sqrt(1-v^2/c^2) for A where v is the velocity of A? $\endgroup$ – Yashbhatt Feb 23 '14 at 6:28
  • $\begingroup$ I think that you are referring to differences in observed elapsed time. For the "stationary" observer we define proper time $\Delta\tau$ and we define the moving observer's time as $\Delta t$. Then the equation for time dilation is $\Delta t = \gamma \Delta\tau$ where $\gamma$ is the equation you mentioned in the last comment. All of this information can be found by simply searching time dilation on Wikipedia or Google. en.wikipedia.org/wiki/Time_dilation $\endgroup$ – ClassicStyle Feb 23 '14 at 6:44
  • $\begingroup$ Yes, exactly. I understood what you wrote about addition of velocities but can you please also explain what happens when we take into account the equations you just mentioned? $\endgroup$ – Yashbhatt Feb 23 '14 at 6:48
  • $\begingroup$ We have the velocity of B relative to A. So calling proper time the time that elapses in A's reference frame you can plug in the velocity of B relative to A into the time dilation equation and see how time is dilated in B. Fundamentally we say that $c$ is the same in all reference frames. For this to happen time will have to change. $\endgroup$ – ClassicStyle Feb 23 '14 at 7:01
  • $\begingroup$ That's what i did. I replaced v with 0.9c in 1/sqrt(1-v^2/c^2). I got 0.22. So that means that 1s for a stationary object will be 0.22s for A. Right? $\endgroup$ – Yashbhatt Feb 23 '14 at 7:06
0
$\begingroup$

No, you are imagining the Newtonian addition of velocities. For parallel velocities, the sum is $\frac {0.9+0.9}{1+0.9\cdot 0.9} \approx 0.9945 \lt 1$

$\endgroup$
  • $\begingroup$ If observer A is moving at 0.9c then time will be slower for him. According to special relativity 1s for A will be about 5s for a stationary observer. So, in A's 1s b would have moved 0.9*5 = 4.5c. So, won't A see him as travelling 4.5c distance in 1s? $\endgroup$ – Yashbhatt Feb 23 '14 at 6:18
  • $\begingroup$ While it is true that the lab sees A's clock running slower, it is not true that A's clock runs slow in an absolute sense. You are trying to use the lab measurement of B's speed with the lab view of A's clock to compute what A will see. You need to do a proper Lorentz transformation, which (of course) yields a speed less than $c$ $\endgroup$ – Ross Millikan Feb 23 '14 at 6:26
  • $\begingroup$ But suppose A has an atomic clock at his disposal and measures the distance traveled by B between t = 1 and t = 0 according to his atomic clock, then won't he feel that B has moved 4.5c and so is travelling at 4.5c? $\endgroup$ – Yashbhatt Feb 23 '14 at 6:36
  • 1
    $\begingroup$ No @Yashbhatt, he'll feel B has moved $\approx 0.9945c$ $\endgroup$ – Pranav Hosangadi Feb 23 '14 at 15:49
  • $\begingroup$ I understood what everyone said about addition of velocities but I am unable to understand it from the point of view of time dilation. Can you explain it? $\endgroup$ – Yashbhatt Feb 26 '14 at 17:41

Not the answer you're looking for? Browse other questions tagged or ask your own question.