Photons, and anything else moving at c, don't really have a perspective.

An object moving at near c would experience almost no time at all during the journey. The distance between Earth and the Sun would, from its perspective, contract such that its near light speed movement allowed it to cross the distance much faster.

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).

To a photon, there's no such thing as time. From its point of view, it would be instant. One way of thinking about this is that in special relativity, distances shrink by a factor of sqrt(1-v^2/c^2). That is zero for a photon where v equals c, so from the photon's point of view, the distance between the Earth and the Sun is zero.

More technically, "age" is a funny concept in relativity. Time is seen differently by different observers. You need to specify both where and when something happened (not just "Alice met Bob at the corner of Main and Elm, but Alice met Bob at the corner of Main and Elm at 4:30 on Friday). If you take two different events at different times and different places, different observers won't agree on how far apart they are or when they happened, but they will always agree on the difference between distance squared and time multiplied by the speed of light squared (dx^2-c^2 dt^2). Since a photon moves at the speed of light the distance it moves dx in a time dt is just speed times dt, or dx=cdt, so dx^2-c^2 dt^2=0. As far as the universe is concerned, the distance between a photon leaving the sun and that photon hitting the earth is exactly zero. In our frame, that means dx is 93 million miles and dt is 8 minutes, but to the photon dx=0 and dt=0. There's no "right" answer for the age of the photon, since every frame is valid, but if you ask the photon, you'll get zero.

NASA: How long does it take light to get out from the inside of the Sun?

According to the famous 'drunkard's walk' problem, the distance a drunk, making random left and right turns, gets from the lamp post is his typical step size times the square root of the number of steps he takes. For the sun, we know how far we want to go to get out....696,000 kilometers, we just need to know how far a photon travels between emission and absorption, and how long this step takes. This requires a bit of physics!

The interior of the sun is a seathing plasma with a central density of over 100 grams/cc. The atoms, mostly hydrogen, are fully stripped of electrons so that the particle density is 10^26 protons per cubic centimeter. That means that the typical distance between protons or electrons is about (10^26)^1/3 = 2 x 10^-9 centimeters. The actual 'mean free path' for radiation is closer to 1 centimeter after electromagnetic effects are included. Light travels this distance in about 3 x 10^-11 seconds. Very approximately, this means that to travel the radius of the Sun, a photon will have to take (696,000 kilometers/1 centimeter)^2 = 5 x 10^21 steps. This will take, 5x10^21 x 3 x10^-11 = 1.5 x 10^11 seconds or since there are 3.1 x 10^7 seconds in a year, you get about 4,000 years. Some textbooks refer to 'hundreds of thousands of years' or even 'several million years' depending on what is assumed for the mean free patch. Also, the interior of the sun is not at constant density so that the steps taken in the outer half of the sun are much larger than in the deep interior where the densities are highest. Note that if you estimate a value for the mean free path that is a factor of three smaller than 1 centimeter, the time increases a factor of 10!

Typical uncertainties based on 'order of magnitude' estimation can lead to travel times 100 times longer or more. Most astronomers are not too interested in this number and forgo trying to pin it down exactly because it does not impact any phenomena we measure with the exception of the properties of the core region right now. These estimates show that the emission of light at the surface can lag the production of light at the core by up to 1 million years.

The point of all this is that it takes a LONG time for light to leave the sun's interior!!

Zero seconds.

The closer you get to light speed, the slower you perceive time. Once you hit light speed, time stops for you.

A trip to the opposite side of the universe would be an instant to a photon, but billions of years to an outside observer.

It doesn't really make sense to ask "how much time has passed from the perspective of a photon", because we can't build something called a "reference frame" for the photon.

Whenever we talk about special relativistic time dilation we must do so by comparing the rate that time is passing in two different reference frames. If I am moving along with you, we are in the same reference frame, but if we are moving relative to each other, then we are different ones. You can think of a reference frame as being a way of describing the universe from your perspective, and when we talk about time dilation or length contraction we have to have two reference frames to compare together. Time always passes normally and distances are always the same in your own reference frame: you won't see your own clocks ticking slow or fast or your own rulers changing length, but you will see that your clocks disagree with those of someone who is moving relative to you and that you and that same person will disagree on how long your rulers are.

Time dilation and length contraction can be thought of as consequences of one basic fact about the universe: all observers, no matter how they are moving, will agree on how fast light is moving. If you're moving away from me at 100 mph and you shine a laser at me, you will measure the light leaving the laser as moving at c. I will measure the light from your laser as moving at c as well, not c-100 mph. In order to agree on how fast light is moving while we are moving relative to each other, we must disagree on how fast our clocks are ticking and how long our rulers are.

But if we were, hypothetically, to be moving relative to each other at c, we would encounter a paradox: light moving parallel to that reference frame would have to be both moving at c (as it must be in all valid reference frames) and be stationary at the same time. This is a contradiction, so we cannot construct a reference frame for the photon, and without a reference frame it doesn't really make sense to talk about time dilation or length contraction.

What we can say, however, is that as something moves arbitrarily close to c that the time it would see pass while traveling between two points would get arbitrarily close to zero and the distance between those points would as well. People often make the slight error of thinking that because the limit goes to zero here that the answer is that it is zero when v = c, but it's more precise to say that the limit approaches zero as v approaches c, but does not exist at v = c. Sort of like how 1/x gets arbitrarily large as x approaches 0 from the positive side, but 1/x does not actually exist when x = 0.

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.

Photons don't experience time since they travel at the speed of light. Time passes slower the closer you get to the speed of light, until it stops at the speed of light. A photon that is part of the cosmic microwave background is indistinguishable from a photon created by my microwave oven unless you see where it's coming from.

If you are looking at the photon from an outside perspective, when fusion happens in the core, a lot of energy is released. The same photons from that reaction dont necessarily make it to the surface, rather get absorbed and remitted before taking 8 minutes to make it to Earth. That process of absorbtion and remitting from the core to the surface, on average takes about 100000 years.

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.

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.

This is the basis of the "trick" question that many people fail to get.

An astronaut takes off on a trip as a twenty year old. What is the farthest that he can he travel before he turns 30?

> 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.

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.

Where can i find more on this? Very confused.

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".

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.

https://www.askamathematician.com/2013/08/q-why-does-it-take-thousands-of-years-for-light-to-escape-the-sun/

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.

Fair enough