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Disclaimer, not a physicist. When I look at the sky I can see the star Rho Cassiopeiae. In my frame of reference, the photon hitting my retina has traveled for 4000 years and 3.78x10^16 km. In the photon's frame of reference, there's no time elapsed and my retina and the surface of the star are at the same place (no distance traveled). Is that a correct understanding of time dilatation and space compression at relativistic speed?

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  • $\begingroup$ Change subject. The question has nothing to do with the title. $\endgroup$ – David Jonsson Nov 8 at 19:13
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    $\begingroup$ A photon does not have a frame of reference, nor indeed does any massless particle, $\endgroup$ – John Rennie Nov 8 at 19:13
  • $\begingroup$ For info on time dilation see What is time dilation really?. for Lorentz contractin see “Reality” of length contraction in SR. $\endgroup$ – John Rennie Nov 8 at 19:16
  • $\begingroup$ @John, thanks for the links. So, if a object was somehow launched from the star towards Earth at 0.999999999999999999999999c, the distance between the departure and Earth's ground would be about 50km, covered in 180ms, in the object frame of reference. And still 4000 years and 3.78x10^16 km for me on Earth. $\endgroup$ – Moohbear Nov 8 at 20:04
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    $\begingroup$ This is Special Relativity, not General Relativity. $\endgroup$ – G. Smith Nov 8 at 20:59
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Photons are elementary particles, with no rest mass, traveling at the speed of light in vacuum, when measured locally.

Now photons do not have a reference frame, because they do not have rest mass. It does not make sense to say, in the frame of the photon. We cannot say what would I see if I would travel with the photon.

Though, you might try to say that photons experience time in a different way, because they do not have rest mass, traveling at speed c, we could say they do not experience time at all, or that they see the whole time between emission and absorption in just one.

Now you could say that a lightlike worldline, that is a photon moving is, between emission and absorption, the spacetime distance is 0, that is, for the photon, the emission and absorption are causally linked.

Now if you would like to see SR time dilation and length contraction, you could try a neutrino, that is as close as you get to the speed of light. From the neutrino's frame, yes, you are correct, it would be just a very short distance, and short time. From our frame (on Earth) it would be much longer and much more time. This is due to SR time dilation and length contraction.

Time dilation is a difference in the elapsed time measured by two clocks, either due to them having a velocity relative to each other

https://en.wikipedia.org/wiki/Time_dilation

Length contraction is the phenomenon that a moving object's length is measured to be shorter than its proper length, which is the length as measured in the object's own rest frame.[1]

https://en.wikipedia.org/wiki/Length_contraction

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  • $\begingroup$ why the downvote? $\endgroup$ – Árpád Szendrei Nov 9 at 0:51
  • $\begingroup$ @safesphere thank you I edited. $\endgroup$ – Árpád Szendrei Nov 9 at 4:41
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The calculation you performed in your reply to John, which related to an object moving at 0.99999999999999999999999c (or thereabouts!), has given you the right idea about the effects of relative speed when distances and elapsed times are viewed from a moving reference frame.

What you might like to think about is the fact that the effects are entirely symmetrical. Your fictitious object is at rest it its frame of reference, and sees us hurtling toward it at 0.99999999999999999999c (Apologies- I might not have the right number of 9s), so for that object it is our time which has dilated.

If you do decide to consider the reciprocity of the phenomenon, you will also need to take into account the effect known as the relativity of simultaneity. I won't try to explain that here, but it means that we and the fictitious object will have different ideas of when our respective journeys started!

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