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jader201 4 days ago

Tangential comment, but it’s crazy to think about how, when we look up at the stars in the sky, we’re seeing light in wildly varying degrees of age.

For example, when we look at the sun, that’s 8-minutes-old light. When we look at Polaris (the North Star), that light is 447 years old.

When we look at Andromeda?

Yeah, that light is 2.5 million years old.

SoftTalker 4 days ago | parent | next [-]

Light doesn’t age. From its perspective it hit your retina the moment it left the star.

sfink 4 days ago | parent | next [-]

Yep. From a light emitter's perpective, it is directly embedded in all of the places surrounding it that its light would eventually reach. Your eyeball, a distant dust spec 4 million light years away, and a black hole are all directly adjacent and it tosses photons onto the shell around itself, painting it with light. The photons arrive at the same instant that they are emitted, if you don't count the millions or billions of years in between. And the photons don't.

Rover222 4 days ago | parent [-]

It’s interesting that from the star’s perspective the light is immediately hitting the objects, yes those objects might have 200 million years of random chance and (possibly) free will determining their positions when the light hits

sfink 3 days ago | parent | next [-]

Not to mention that I'm not sure if all photons are guaranteed to hit anything. I have no intuition for calculating how likely that is for things in our neighborhood, but if you think of the star that is closest to the edge of the universe (as in, the farthest light could have traveled from the Big Bang, taking into account expansion of space itself), it seems unlikely that there would be guaranteed to be a chunk of matter between it and the universe's edge in every direction.

So a photon emitted in such a direction takes no time (from the photon's point of view) to be absorbed by its destination... only it has no destination and never will. It gives "what if a tree falls in a forest with no one to hear it?" vibes. Does the universe sort of "lose" the energy of that photon in a way that it doesn't for photons that are absorbed? Is it like The Great Memory Leak of the Universe or something? Is our existence leaking out between the fingers of the hand of God?

SoftTalker 3 days ago | parent | prev [-]

When you're moving at light speed there is no past or future. Everything is compressed into "now." I'm not sure how "random chance" works in such a situation, it's certainly not intuitive at least to me.

jb1991 4 days ago | parent | prev | next [-]

You are saying, from the perspective of light, whether it travels 1 mile or a trillion miles, that journey takes the same amount of time?

sh-run 4 days ago | parent | next [-]

I'm not a physicist, but I believe that's the exact insight that led to special relativity. It goes something like: If your moving at 1,000kmh next to a jet moving at 1,100kmh then the jet is moving at 100kmh relative to you. Eventually people realized those wasn't the case with light. No matter how fast the observer is, light still moves at 299,792,458m/s. Einstein figured out that if the speed of light is fixed despite relative motion, then time must slow down as you move faster. So from the perspective of a photon no time has passed since its departure.

cryptonector 4 days ago | parent | prev | next [-]

Time does not pass for photons. So yes, that is exactly what GP is saying.

There's also a different thing that GP might be hinting at, which is that by convention we assume that the speed of light is the same in all directions, but there are other conventions we can use as long as the round-trip speed of light agrees with that which we've measured (and yes, we can only ever measure the round-trip speed of light, FYI). Another convention is that all the light we see takes zero time to get to us but the light we emit goes out at half the speed one would expect with the standard convention (known as the Einstein synchronization convention). So instead of "light we see from Alpha Centauri is 4 years old" or "we see Alpha Centauri as it was 4 years ago" we can say that we see it as it is right now, but this is not a very commonly used convention.

jb1991 3 days ago | parent [-]

interesting, why would this be?

> we can only ever measure the round-trip speed of light, FYI

cryptonector 3 days ago | parent [-]

Because say you fire a photon from an emitter to some target and you want to know how much time elapsed for that flight, but how would you find out? The target will have to communicate to the observer (you in this case) when it received the photon, but that communication will require.. more photons, and a trip back to you. If you're colocated with the emitter then it's a round-trip, else you need a photon from the emitter and the target, as well as the one from the emitter to the target, and this amounts to a round-trip anyways.

Therefore you can't measure the speed of light in any one direction. You can only measure the round-trip time of flight (e.g., if you have the detector at the emitter and use a mirror).

seabass-labrax 4 days ago | parent | prev | next [-]

I'd really highly recommend the Uncle Albert series of novels by Russell Stannard:

https://booksforkeeps.co.uk/article/visiting-uncle-albert/

The intuition you can develop about special and general relativity from these books is pretty amazing!

oneshtein 4 days ago | parent | prev [-]

Yep, this is what he saying, but this is not what photon does. Photon must perform different amount of wave cycles to reach 1 meter or 1 trillion metters. These cycles can be counted.

cryptonector 4 days ago | parent [-]

> These cycles can be counted.

In a lab setting, yes, but across such distances, no. Photons don't have a cycle counter on them, so they don't keep a cycle count and can't reveal that cycle count. All we can do is measure frequency/wavelength (spectrum, really, since we're going to see lots of photons, not really onesie/twosies) and intensity, and we can use the astrophysical distance ladder to figure out roughly where the emitter must have been.

oneshtein 3 days ago | parent [-]

Red shift allows us to roughly calculate distance and time, so we can multiple time by frequency of light to calculate number of oscillations or cycles and then calculate loss of energy per oscillation at average.

oneshtein 4 days ago | parent | prev [-]

Light is a wave, so it waves something to propagate itself in space and time. Physical photon does countless cycles to leave the star and hit an eye.

It's not an easy task from the prospective of a photon, which can be easyly proven with just two little slits.

BreakfastB0b 4 days ago | parent [-]

Except that's not from the frame of reference of the photon. At the speed of light, the Lorentz transform shows that 1) Time stops completely, 2) All distances in the direction of travel collapse to zero. So in a very real sense, "from the photons perspective" it never exists and the point it is emitted from and the point it is absorbed at, are the same point.

Experiencing time and having mass are linked in a very deep way. Objects that experience time, i.e. have some kind of state evolution, must have mass, this is how we know the neutrino has mass even though it's smaller than we can measure, because we measure them oscillating between the various flavours of nutrinos.

This is also how the Higgs mechanism gives rise to "rest mass" in most particles, by constantly exchanging weak hypercharge with them. This oscillation back and forth gives them mass.

oneshtein 3 days ago | parent [-]

Photon is a wave. It oscillates. It's proven. Math model doesn't account for this to make calculation simpler for human beings.

thisoneisreal 4 days ago | parent | prev | next [-]

That thought randomly hits me all the time when I'm taking out the trash or whatever and just happen to look up. That and the fact that the Bootes Void and Phoenix A* exist out there.

HardCodedBias 4 days ago | parent | prev [-]

Nit: I think that the light from the sun is about 100k years old. Wild.

qnleigh 4 days ago | parent | next [-]

Nit of a nit; the energy might take that long, but the photons that reach us on Earth are not directly created by the nuclear fusion reactions in the sun's core. Fusion creates high-velocity nucliei and other particles, but not visible light. The resulting heat creates photons which are rapidly destroyed by absorption. Only photon emission from the outer most layers of the sun reach Earth.

greenbit 4 days ago | parent | next [-]

I.e., that bit they refer to as the photosphere, effectively the radiating 'surface' of the sun, is the source of the solar photons that strike us here. That trip takes about 8 minutes.

HardCodedBias 4 days ago | parent | prev [-]

You are totally right!

I knew HN would nit my nit, well done!

aplummer 4 days ago | parent | prev [-]

How can that make sense, the photons are emitted and fly straight at us

siavosh 4 days ago | parent | next [-]

The photons were created a long time ago in the core. It takes thousand of years for it to reach the surface, and THEN it takes 8 minutes to get to us.

greenbit 4 days ago | parent [-]

The photons created in the core are some seriously energetic gama rays. Sure, gama rays are very penetrating, but the solar core is dense, and it's about half a million miles to the surface, so these mostly get absorbed right there in the core, making stupendous amounts of heat. At any given depth that means that matter is going to re-emit photons, but never any more energetic than the original ones that are absorbed, but that radiation will be reabsorbed as well. That process of emission and reabsorption means that energy travels to the surface a lot slower than light in a vacuum, and sure, it takes a long time for that energy to reach the surface, but the photons that reach the earth are only the ones created close enough to the 'surface' to escape into space.

stevenwoo 4 days ago | parent | prev [-]

Photons are not created on the surface but in the core where the environment has the higher pressure needed for the physical creation of the photon and the photon takes about that long to work its way out.

Rover222 4 days ago | parent | next [-]

Is this in any sense hydrogen being converted to photons? Photons are massless, but… the mass of the elements in the star are converted to pure energy?

Ekaros 4 days ago | parent | prev [-]

What is the ratio between those and well heat due to nuclear reactions and well pressure. Hot stuff generates visible photons. Say like incandescent light bulb.

So there must be a range in age. As some closer to still hot surface don't need to travel through parts of the sun.

stevenwoo a day ago | parent [-]

So I am not an expert and recalling what I have read in several books and articles, but the conditions of fusion necessary to create photons only exists in the core of the sun. It was a mystery to us and we did not know until scientists were able to use quantum mechanics was able to explain the mechanism, it requires enough temperature and gravitational pressure to force subatomic particles close enough to overcome the forces that ordinarily keep them apart, and this only happens to a small percentage of meeting nuclei with quantum tunneling explaining how they overcome the forces that want to keep the nuclei apart - there's just so many particles squeezed close together that a small percentage that meet (possibly easier to visualize as the quantum wave function describing the position) fuse. This is also why we cannot use this method of fusion on earth - it's impossible to do on earth barring some sci fi artificial gravity invention. If this were not true and fusion could take place anywhere on the sun, the sun could rapidly use up all of the fuel of hydrogen. I am simply repeating what I have read - each photon has to make its way to the surface of the sun after many collisions, since the direction is random and not always outwards, it is theorized they just ping pong back and forth for x years where x can be hundreds of thousands of years. Here's the clearest explanation I found though they only use one slide on the photon's drunken walk https://svs.gsfc.nasa.gov/11084