[colechristensen]

Huh? Light has zero lifetime, along with all massless particles. When you travel at c your clock doesn’t tick. From the viewpoint of a photon it is emitted and absorbed at the same instant. You cannot decay if you don’t experience time. I’m not sure exactly how index of refraction works with this.

[ryandamm]

Hoping a physicist can correct me here, but... I believe index of refraction is a function of the photon being absorbed and reemitted by the electrons in the dielectric material, so it's no longer correct to think of a photon moving at a fraction of the speed of light inside the material, it's more like a churning series of them being created (always moving at c) but constantly being absorbed and canceling each other out.

I also seem to recall that the speed of light below c is actually the group velocity, and each individual photon still would move at c. I'm also not entirely sure if photons can be said to exist except at creation and absorption; isn't a photon a phenomenon best described by particle interactions, and moving through free space it's more correctly described as a field? Genuine question, though I somehow doubt I'd understand any good elaboration.

IANAPhysicist, though. I just play with light recreationally.

[Modified3019]

For those curious about what the above poster is talking about, here’s a well done video explaining the topic of the apparent changes of light’s “speed” through materials

3Blue1Brown - But why would light "slow down"? | Optics puzzles 3: https://m.youtube.com/watch?v=KTzGBJPuJwM

[adrian_b]

From the point of view of any observer, a photon has a definite lifetime, between the moments of its emission and its absorption.

The Lorentz transformations are defined only between reference systems where the relative speed between them is less than the speed of light.

It is not possible to attach a reference system to a photon or to any other particle that moves with the speed of light, because there are no conversion rules between the coordinates in such a system and those in a normal reference system.

Therefore it is not correct to say that the lifetime of a photon in a reference system attached to it is zero or infinite or it has any other value.

This lifetime is just undefined, while the lifetimes in any other reference systems are well defined.

The photon does not decay in the absence of interactions with other particles because that would violate several conservation laws. However, when the photons have energies that are high enough, the interaction between themselves can generate other particles, in particle-antiparticle pairs, in order to satisfy all conservation laws.

[akira2501]

If it's travelling at c then isn't the length contraction infinite? Or is that dependent on Lorentz transformations as well?

[adrian_b]

Length contraction and time dilation are words that describe changes that are the consequence of a Lorentz transformation.

Like I have said, the formulae of a Lorentz transformation are defined only when the relative velocity between the two systems is less than the speed of light.

Attempting to pass to a limit when the relative speed approaches the speed of light does not produce any useful result, because at the limit you no longer obtain a reference system, so you no longer get a transformation between reference systems.

Without a reference system, there is no meaning for the concepts of distance and time.

Any reference system for the 4-dimensional space-time must be attached to normal matter made of leptons and quarks, it cannot be attached to photons. In any reference system for the 4-dimensional space-time, the photons are particles that move with equal speeds in space and in time, while the normal matter moves faster in time than in space. The notion of proper time (i.e. the time measured for an object that moves only in time, without moving in space) is not defined for photons, because they always also move in space, not only in time.

This should be obvious from the rule introduced by Einstein that the speed of light is the same in all possible reference systems, from which the Lorentz transformations can be deduced. If a reference system were attached to a photon, in that reference system the speed of light could not have the same value as in the normal reference systems, so within Einstein's theory such a reference system cannot exist.

[animatethrow]

Consider a simpler example from basic math. Is 1/x infinite when x==0? The answer is that 1/x is undefined when x==0. In calculus one can take limits as x "approaches" 0 but x==0 is still undefined. Likewise, the Lorentz length contraction is undefined when traveling at c.

[Nevermark]

> You cannot decay if you don’t experience time.

Interesting take! But if photons couldn't decay due to not experiencing time, they couldn't do anything else either.

The reality is that a photons creation and destruction are not prohibited, but simply "experienced" as two events at different locations at the same time, with the photon being the "thing" that connects those events.

Given that interpretation, it might be reasonable to assume that all photons have beginnings and ends, regardless of the duration we perceive between them, or they wouldn't exist.

Time being no barrier at all for photons.

[nyrikki]

From a photons perspective, emition and absorption are simultaneous.

While from our perspective it is a form of causal connection, that is mearly due to the frame of reference.

While we can infer the connection between each, it is possibly better to consider the speed of light as the speed of causality.

But as there are no privileged reference frames under GR the choice is yours.

But from the photons perspective, it doesn't experience time at all so it can't be a barrier.

But don't confuse the map for the territory. GR is a model, not the system itself.

The fact that almost every test we can figure out has only confirmed it doesn't change that.

Under the 'all models are wrong but some are useful' idea, in GR photons not experiencing time is important to that model.

[pixl97]

>But if photons couldn't decay due to not experiencing time, they couldn't do anything else either.

I mean we are jumping way out of the classical behavior that objects like you and I exist in. To the photon itself is a timeless object. It 'moves' in a null geodesic where t=0. Attempting to apply any classical behavior that occurs in time-like objects just isn't going to work when applying them to massless light-like objects.

[supportengineer]

Would that seem like a fold in spacetime to the photon?

[Nevermark]

I can't top the sibling comment about a summer breeze! But it is an interesting question.

Not only does the photon not experience any delay between its two end points, but it experiences its path between them as a simple shortest-distance straight line segment, even if the same path looks like a curve through gravitationally warped space-time to us.

The photon does experience a form of distance, i.e. the number of wave lengths between its ends. But just the number of cycles, not the actual wave lengths which we would see varying as we experienced dark energy and space stretching the photon's wavelength from our viewpoint.

So a photon "experiences" two spacially separated ends, and a number of wave cycles between them, and that's it? Perhaps.

[nick7376182]

Probably more like a summer breeze

[amelius]

But if your clock doesn't tick, then infinite lifetime means the same as zero lifetime.

From the point of view of the photon, time doesn't even exist. So it is pointless to ask the question from the point of view of the photon.

[kmm]

> You cannot decay if you don’t experience time

That's a common misconception, there's no a priori reason a particle without a restframe can't decay. For all known particles with a finite lifetime we give this lifetime as measured in its restframe (i.e. with the particle standing still), but in principle it is an observer-dependent quantity, faster moving particles will take longer to decay. If we, for example, assume the lifetime of a massless particle is proportional to its energy, we retain the same expected Lorentz covariance.

Of course, if you actually go through the math, the known massless particles in our universe, photons and gluons, turn out to be stable.[1]

1: https://arxiv.org/abs/hep-th/9508018

[TheOtherHobbes]

I wish people would stop repeating this.

If light is emitted and absorbed at the same instant, how does it know when/where in spacetime it's been emitted and absorbed?

You cannot apply SR to light in this naive way. SR works just fine for anything with mass, but without a theory of quantum gravity no one has the first clue how massless particles operate in spacetime, or what they do or don't "experience."

[bequanna]

Right, but doesn’t light travel at less than c in some situations (passing through glass, etc)? Would we say they experience “time” in those cases?

[samus]

In these situation photons are either bouncing off matter or are getting absorbed and emitted by matter. In the former case they don't travel a straight path, in the latter case there is a short time lag between absorption and emission.

[amluto]

I think this is sort of besides the point. If you build a box, paint the walls black, and put a flashlight in the box, then the photons coming from the flashlight are shorter lived than if you shine the flashlight into the sky on a cloudless day or night. Not shorter lived from their own perspective — shorter lived from an outside observer’s perspective. Sure, one could quibble about the choice of observer, but you would he hard-pressed to put an observer in the box who thinks the photons last very long.

[pixl97]

Why is the speed of causality beside the point?

lets take two magical particles that have clocks on it. One is a photon and the other is a neutrino. I send these off towards you in a perfect vacuum. When you receive these particles the clock on the photon will be 0. It will be be the exact same photon that left my emitter, it will not have changed in any way as it did not interact with anything along the way. And as long as you are not moving relative to me, you'll perceive the photon as the same color/wavelength I emitted it at.

Meanwhile that neutrino will arrive billions of a second later (well depending on our distance) and will have oscallated at least trillions of times if not far more. The clock on the neutrino will have ticked the difference between the photon arrival to the neutrino arrival.

Don't apply classical behavior to light-like objects. They play be different sets of rules.

[amluto]

This is all true, but the article isn’t about how long a photon thinks it lives or how much it experiences the passage of tone. It’s about whether the photon keeps going forever from the perspective of someone approximately at rest [0] in the universe (like astronomers on Earth!).

[0] General relativity has no preferred “at rest” frame, but the generally accepted FLRW model of the universe does. You can be at rest with respect to the universe, or you can be moving. If you are moving, distant objects in front of you will appear blue-shifted on average as compared to distant objects behind you.