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SurprisedPotato t1_ja6pztf wrote

For the photon, no time would have passed at all. What that means in practice is that photons can't change as they travel. The photons we receive are exactly the same as the ones that were sent.

Here's another example. Back in the 80's, it was thought that neutrinos were massless, and traveled at the speed of light. Neutrinos are extremely light particles that are emitted from some nuclear reactions - and they come in three "favours".

Also, back in the 80's, careful measurements had been done of the number of neutrinos coming from the sun, and the figure was only about 1/3 of what it should have been.

There were a few ideas proposed to explain that. One was that the sun had switched off and we would all freeze to death within 10000 years or so, but another was that some of the neutrinos from the sun were transmuting into the other forms, and we were only detecting the 1/3 that stayed in their original form. However, if neutrinos were traveling at c, that couldn't happen - if, for the neutrino, no time had passed, then it couldn't transmute, since change needs a passage of time, and for objects at c, no time passes.

Since then, we have confirmed that, in fact, neutrinos do have mass, and don't travel at c, and so time does pass for them on their journey to us from the sun, so we aren't doomed to an icy future. This was very exciting for the physics world, and probably won (or will win) someone a Nobel Prize.

However, for light itself, we know it travels at c, and therefore no time passes for the photons as they cross the 8 light-minutes between the sun and us. (For them, it also seems like no distance has been traversed).

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Dvorkam t1_ja73a6a wrote

I was planning to write something like this, but you beat me to it. I really like this idea, that from the perspective of photon, there is no such thing as vacuum. Every atom along its trajectory(from the "perspective" of photon) is right next to each other.

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phunkydroid t1_ja8phat wrote

From the perspective of the photon, there is no journey at all. It begins and ends at the same moment with zero time or distance in between. This is why a photon's "point of view" is just not a valid reference frame for any measurement.

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Bensemus t1_ja9rtvx wrote

> For the photon, no time would have passed at all.

This is often said but isn't correct. There is no reference frame for the photon. Without one you can't say how much time has passed for it.

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SurprisedPotato t1_jaa602j wrote

What we can say is that particles travelling at c can not change state, hence the puzzlement in the 20th century about "missing" solar neutrinos.

"Can't experience change" is something we might describe as "time does not pass", without implying they have a reference frame.

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left_lane_camper t1_jaamcrt wrote

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".

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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.

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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.

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Any-Growth8158 t1_ja9arcy wrote

Aye. Two ways of looking at it.

Time is slowed down so much that no time passes.

An equal statement is that the length of the path in front of photon is contracted to down to zero length so it isn't really going anywhere.

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