Assuming we have 3 objects d,e,f moving on the same line while e is stationary, d is moving to the left at 'almost speed of light' and f is moving to the right at 'almost speed of light'. What is the speed at which f is moving from the perspective of d?

If its 0.9999999% speed of light, what does it say about their course? To my understanding, the perspectives of d,e and f are as followed:

  • from the perspective of d, e and f are both moving away at the sub speed of light
  • from the perspective of e, d and f are both moving away at the sub speed of light
  • from the perspective of f, e and d are both moving away at the sub speed of light

In order to keep all 3 statements true, all 3 objects (d, e and f) are supposed to be moving away from each other at a speed close to speed of light, so they should be virtually located on the vertices of an "invisible" equally-edged expanding triangle.

So my question is - what happened to the original straight line d,e and f were placed on during the acceleration of d and f to 'almost speed of light'?

  • 4
    $\begingroup$ There is no triangle. Nothing happened to the straight line. You simply don’t understand how relative velocities work relativistically. (See en.wikipedia.org/wiki/Velocity-addition_formula for the formula.) And none of them see either of the others moving “at the speed of light”. $\endgroup$ – G. Smith May 6 '19 at 1:55
  • $\begingroup$ If you like this question you may also enjoy reading this Phys.SE post. $\endgroup$ – Qmechanic May 7 '19 at 12:41

To avoid a proliferation of nines, let’s take the leftward velocity of D, and the rightward velocity of F, relative to E to be 0.9 $c$. Then, using relativitic addition of velocities as explained here, we find that D sees F move rightward at

$$\frac{0.9+0.9}{1+0.9\times 0.9}c = 0.9945 c.$$

Similarly F sees D move leftward at this same speed.

There is no triangle, just straight-line relative motion with various relative velocities which are all sub-luminal.

| cite | improve this answer | |
  • $\begingroup$ Thank you. If they move on the same line which appears to be straight to all of them it would have to mean that one of the sentences I wrote would not be true or what am I missing? $\endgroup$ – Uri Abramson May 6 '19 at 6:29
  • $\begingroup$ Each of your three bullet points is wrong. None of the objects are moving away at the speed of light. From the perspective of D, objects E and F are not even moving at the same sub-light speed. $\endgroup$ – G. Smith May 6 '19 at 15:39
  • $\begingroup$ at near speed of light I mean. editing $\endgroup$ – Uri Abramson May 7 '19 at 12:34
  • $\begingroup$ That doesn’t change anything. Your conclusion that there must be triangle is still a non sequitur. There is no triangle. $\endgroup$ – G. Smith May 7 '19 at 15:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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