If you "ride" a photon, in your frame of reference, time is still (that is, no time passes with respect to the outside world, you see the outside world still).
However, suppose you are indeed "riding" on a light beam, and are shining from Alpha Centauri, bounce off a mirror on Earth, then bounce back to Alpha Centauri.
For you, time is still, so nothing happens (everything you watch outside is non-moving in time). However, for an observer on Alpha Centauri, your arrival happens after 8 years (if I'm not messing up the distances), so they are 8 years older.
How can this be explained? Am I missing some general relativity concepts, for example, acceleration of a massless particle when bouncing off a mirror? The question is probably trivial for a cosmologist, but I can't find a satisfactory answer.
Thanks.
First of all, the photon has no (rest) frame of reference. That is, the Lorentz transformation that corresponds to the speed of light is degenerate.
But, I think, your question can be rephrased using an ultrarelativistic massive particle that travels at almost the speed of light, and instead of an instant turnaround, undergoes some tremendous but finite acceleration at the location of the mirror.
And indeed, in the frame of reference of this particle, very little time will have passed during the full voyage. And by taking the limit of its velocity going to c, and its acceleration going to infinity at the turnaround, the total proper time along that particle's path can be made arbitrarily close to zero.
Another way of looking at it: this ultrarelativistic particle will travel (almost) on a null geodesic (where no time passes) for (almost) the entire duration of its flight; it will only travel on a non null-geodesic during the brief period of its turnaround while it accelerates.
First of all, the photon has no (rest) frame of reference. That is, the Lorentz transformation that corresponds to the speed of light is degenerate.
But, I think, your question can be rephrased using an ultrarelativistic massive particle that travels at almost the speed of light, and instead of an instant turnaround, undergoes some tremendous but finite acceleration at the location of the mirror.
And indeed, in the frame of reference of this particle, very little time will have passed during the full voyage. And by taking the limit of its velocity going to c, and its acceleration going to infinity at the turnaround, the total proper time along that particle's path can be made arbitrarily close to zero.
Another way of looking at it: this ultrarelativistic particle will travel (almost) on a null geodesic (where no time passes) for (almost) the entire duration of its flight; it will only travel on a non null-geodesic during the brief period of its turnaround while it accelerates.
@V. T. Toth, so the "make-up" time (in the limit) will be during the acceleration period? And is the question itself meaningless for a photon (truly massless particle)? Although I find the answer to the first question a bit weird, as the acceleration period should be independent of the distance the particle traveled.
I am not sure what it is that you call "make up" time. The ultrarelativistic particle that bounces off a mirror will age very little. Your question suggests that you may be assuming that in this version of the twin "paradox", a traveling clock should show the same amount of time elapsed as the stationary clock. This is not so, because the traveling clock changed direction, so its entire trajectory is not inertial. At the moment when it bounces off the mirror, there is a drastic change from one inertial frame of reference (moving away from the point of origin) to another inertial frame of reference (moving toward the point of origin). It's not so much the duration and magnitude of the acceleration but the magnitude of the change between the "before" and "after" frames of reference (and the amount of time the traveler spends on these two distinct frames of reference) that matters.
I am attaching a diagram that I used in an earlier answer about the twin paradox. It shows three worldlines, one of which (the black line) is stationary, whereas the other two are fast moving (though not quite at the speed of light) with "instant" turnaround. Dashed lines show surfaces of constant time as seen from the three frames of reference; dots along the worldlines show equal amounts of proper time (i.e., "ticks" measured by identical clocks.) Perhaps this diagram helps visualize the problem a little better.
Ok, I think I got it, my question was basically how would you "see the world" when bouncing off that mirror. Before and after the bouncing, everything should be almost still (as you're travelling with 99.9999...9% speed of light), however during the bouncing, it looks like you should see things moving up in time a lot, and this movement should depend on how far away you came up from (if not, the other party that emitted you (the photon, or the ultra-relativistic particle) should see your aging independent of the distance, which is not so). Again, sorry for what it may seem a stupid question, I'm just trying to visualize what happens, and my GR really sucks at this time.
Dear Vlad , one point has been missed in your question, respect to modern optics a photon does not bounce off the mirror, it first absorbed and then another photon emitted from the mirror, Therefore the reflected photon is not the first one. However your question as has been mentioned by Toth , is comparable with the Twin paradox, an old paradox that debated and solved by Von Laue etc, and explained carefully by requirement of a change of reference frames of twin, which shifts the problem from especial to general relativity.
Two events E1 and E2 connected by photons or any other zero rest mass particle have zero proper time interval between the events. One way to see that is if one is travelling with speed c, due to Lorentz contraction, the spatial distance between E1 and E2 in this frame is zero.
A person or a clock riding on a beam of light is an absurd idea. The person or the clock will have to disintegrate and become light in order to acquire the speed of light. Time is a periodic phenomenon. It does not flow in one direction. You can only compare one period of time with another. Smallest unit of time is the period of the wave. Time does not have a physical existence. The associated frequency has the physical existence. T=1/f is the only fundamental relation between physics and imagination.
Thanks all, I got it, it's basically the twin paradox. What about if the universe is finite and has a curvature (so its topology is that of a torus), so if you shine a light beam it will (eventually) come back? And I agree that "riding" a beam of light is non-sensical, however suppose we deal with a very light particle that moves 99.9999...99% of the speed of light.
All experimental evidence suggest that the universe is flat and curvature is 0.
Manouchehr Amiri got it right. Photons either move with the speed c or they do not exist. Even when light moves through a material with speed less than c, the motion is a large number of photon absorptions and re-emissions by the material. As for a reference frame moving with c, this is unattainable for particles with mass, so no experiments can be done -or even conceived- in such a frame. So there are no answers to this non-existent situation.
@ Clifford Miller , just type " quantum, reflection" in google search to find out the answer and rational for my comment. of course you need some insight more than high school physics, up to a level at least to understand QED or QFT.
- Badges
- Science topic