2
$\begingroup$

If we positioned a mirror 1 light year away from earth and shot a particle of light at the mirror so that it would reflect and come back to earth, how long would it take for us to receive that particle of light back to earth and if I jumped on this particle of light for the trip how would it effect how much time has gone by for me and the people waiting for me back on earth

$\endgroup$
6
  • $\begingroup$ The answer to your first question is 2 years obviously! $\endgroup$
    – David H
    Nov 9, 2013 at 7:48
  • $\begingroup$ I ask because it is my understanding that as velocity increases time slows down. So if it took 2 years for us to receive the particle back at earth, it must have seemed like less time for the particle( or me riding the particle). In which case it would have taken the particle less than 2 years to travel 2 light years which I think is impossible. So would that just mean that light does not experience time at all $\endgroup$
    – user32436
    Nov 9, 2013 at 8:42
  • $\begingroup$ ...and that's how I understand it to be. Eigentime (the time measured by someone travelling along) is not defined for lightlike curves. $\endgroup$
    – Danu
    Nov 9, 2013 at 9:09
  • $\begingroup$ This question (v1) appears to be off-topic because it is about fictional physics, i.e. a massive object jumped on this particle of light. If the question can be formulated without using fictional elements, I will be happy to consider re-opening it. $\endgroup$
    – Qmechanic
    Nov 9, 2013 at 9:51
  • 1
    $\begingroup$ Sorry I didn't realize this forum was so uptight about such things. I simply formed the question in my mind and found this as a forum in which to pose it. Thank you for telling me that my question is so terrible you had to completely shut it down. I will not bother this forum with such fantastical, fictional physics questions in the future. $\endgroup$
    – user32436
    Nov 9, 2013 at 15:30

2 Answers 2

4
$\begingroup$

The basic postulate of special relativity is that the speed of light $c$ is equal in any inertial frame. But in the rest frame of a photon, the photon would per definition be at rest, so $c=0$ in that frame. Since the speed of light is then different from all the other reference frames, the rest frame of a photon is not an inertial frame, and the equations of special relativity breaks down; including the equations for time dilation and length contraction. (This may be interpreted as that the photon has no rest frame.)

So as far as I know, there is no proper answer to your question. But it's still fun to think about it though: if you were to assume that a photon is well described as a massive particle with $v \rightarrow c$, then length contraction would cause all of space to contract into a plane. So effectively, the universe would just have two sides, in front of the photon and behind the photon. If we observe the photon, time dilation would make it appear that time is standing still for the photon. But remember that we would also appear to be moving at speed $c$ relative to the photon, so to the photon, the rest of the universe would appear to be frozen in time. So overall, the photon would see the universe as a flat and static place, where nothing happens.

These things are fun to think about, but keep in mind that we're now applying the equations of special relativity well outside their domain of validity; the photon doesn't really have any rest frame, so there is no real answer.

$\endgroup$
2
$\begingroup$

at Light speed change in Proper Time will be zero: $$d\tau=dt\sqrt{1-(\beta^2=1)}=0$$ that means light observer do not experience time.

$\endgroup$

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