Submitted by Opposite-Shoulder260 t3_11d4jwo in explainlikeimfive
left_lane_camper t1_jaamcrt wrote
Reply to comment by SurprisedPotato in Eli5: how old is a photon from the sun when it arrives to the earth? by Opposite-Shoulder260
It's a subtle difference and one we should be careful to distinguish: the lack of a valid reference frame for the photon means that it does not make sense to talk about the passage of time from the photon's perspective. Without such a reference frame, something moving at c (such as the postulated massless neutrino) cannot change, which would also be true if the elapsed time was 0, though that is not the case here.
But lacking a valid reference frame is not the same as saying the elapsed time (or distance) for a massless particle between its creation and some later time (from our perspective) is zero. Talking about elapsed time at all for a massless particle from its perspective doesn't make sense. If you are familiar with coding, it's vaguely analogous to the difference between a value being 0 (a real number that can be found on the number line) and NULL (a value that is not a number at all).
Saying "time does not pass" is kind of ambiguous and I wouldn't describe it as wrong per se, though without clarification it might lead someone to think that it means that "the amount of time is 0" rather than "it does not make sense to ask how much time has passed, as the photon lacks a valid reference frame in which time could pass in".
SurprisedPotato t1_jaan7z1 wrote
I'll think about this. I will note that the Lorentz transformation equations as given in the introduction to this wikipedia page look like they'd work fine, giving non-null answers, in the case v=c. They imply that in the photon frame, only t' = 0 is possible, but that's what you'd expect of a particle not traveling through time. The equations don't say "t' does not exist".
Maybe you've seen a formulation where something is divided by gamma = sqrt(1 - v^2 / c^2 ). Then, certainly, such a formulation might say "such and such a quantity does not exist when v=c". I would suggest that that's a limitation of the formulation, not necessarily a reflection of reality.
left_lane_camper t1_jabvm53 wrote
First, be careful with your Lorentz factor (γ). You are missing a sign on an exponent there, which is pretty critical in this case. We usually write
γ = ( 1 - ( v / c )^2 )^**-**1/2
and so γ diverges in the limit where v->c, and then you can see that the Lorentz transformations are not defined at v=c. But we shouldn't get too hung up on this, as this doesn't really address your real question:
>I would suggest that that's a limitation of the formulation, not necessarily a reflection of reality.
And to do that, we should step back from the math for a second and think carefully about applicability. Even if we have a quantitative description of a phenomena that gives a real, non-divergent answer we must be very careful that it is actually applicable to a given situation so as not to over-extend a model. Not all answers given by an equation are correct: sometimes we're just doing math and not physics.
In this case, we build a Lorentz transform by comparing two valid inertial reference frames. One of the postulates we use to construct one such frame is that the speed of light is invariant for all observers in any frame, which leads to the Lorentz transformations. However, if we try to construct such a frame at v=c we encounter a paradox: light moving parallel to this frame must be moving at c and also must be stationary in the frame. This cannot be, so we cannot construct the frame and without the frame the Lorentz transformations are meaningless (and also undefined as the Lorentz factor is undefined at v=c).
As such, in this case, it is quite the opposite: that the Lorentz factor is undefined at c is not an artifact of the mathematics, but a reflection of something fundamental to relativity.
SurprisedPotato t1_jabxkfe wrote
Fair enough
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