0
$\begingroup$

Special Relativity tells us - the faster things travel their time is slower relative to a stationary observer. Do massless particles, like photons travelling at the speed of light, “experience” zero time and in their frame of reference travel "instantaneously". Putting it in a thought experiment. Imagine travelling to a distant star. On reaching the star, several decades have past as shown on the spaceship's clocks and on our aging bodies. Now consider doing the same journey but incredibly close to the speed of light (not possible in practice but not violating any physical laws) we would see the clocks have changes only by a few minutes and our bodies are unchanged. Wouldn't it be reasonable to extrapolate and predict that if we were to travel at a smidgen less than speed of light, it would appear (to us astronauts) to have taken close to zero time?

$\endgroup$
4
  • 2
    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/29082/2451 and links therein. $\endgroup$ – Qmechanic Feb 14 at 16:53
  • 3
    $\begingroup$ Does this answer your question? How does a photon experience space and time? $\endgroup$ – BioPhysicist Feb 14 at 17:20
  • $\begingroup$ Note that there's no qualitative difference between your two theoretical trips to the distant star. Any trip requires motion, and any motion implies time dilation. The only reason we talk about it as two distinct cases is because, for a macroscopic object like a space ship, it would take more energy than we know how to supply in order to create enough time dilation for us to notice it. $\endgroup$ – Solomon Slow Feb 14 at 19:51
  • $\begingroup$ You can’t do it in minutes because it takes the better part of a year to accelerate to close to c without crushing yourself with G forces. $\endgroup$ – Ben51 Feb 14 at 23:22
1
$\begingroup$

Your reasoning is true. If an observer on, for example, earth were to measure the time passing on the spaceship, they would find that the time that passed for the astronaut is

$$t'=t\sqrt{1-\frac{v^2}{c^2}}=\frac{t}{\gamma}$$

where $t'$ is the time elapsed on the spaceship moving with relative velocity $v$ and $t$ is the time that has passed on earth. One can see as $v\to c$, $t'\to 0$.

One could argue that since for a photon, $v=c$, $t=0$. However, since $c$ must always be constant, there is no frame of reference moving along with a photon $-$ if there were one, the photon would be at rest in it, which would violate the constant speed of light principle.

So it does not really make sense to talk about how much time has elapsed for a photon or a hypothetical observer at the speed of light; but one can accurately describe what happens as the velocity of the observer approaches $c$, where the reasoning that you elaborated on is correct.

$\endgroup$
-2
$\begingroup$

Yes, your conclusion is correct. Time dilation tend to infinity as we approach the speed of light, so does the energy needed to make us move that fast.

$\endgroup$

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