# Why don't photons experience time?

by la6ki
Tags: experience, photons, time
 C. Spirit Sci Advisor Thanks P: 5,637 It seems like the term "passage of time" is being thrown around so loosely I can't even ascertain how it is being defined in this context. If you want to ascribe a quantity / notion of time that is frame independent then you could talk about $\int_{\gamma } d\tau$ (where $\gamma$ is the time - like curve the massive particle is traveling on). What would "passage of time" even mean for light when you can't use proper time as an affine parameter along a null - like path? Are you wanting to use coordinate time? Coordinate time isn't frame independent so what kind of physical significance of "passage of time" can you even define for that?
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 Quote by dm4b Just because something leads to confusion doesn't necessarily mean it is fundamentally incorrect.
Yes, and I wasn't necessarily saying that "photons don't experience the passage of time" is incorrect. Pedagogy always involves judgment calls, which different people can make in different ways; no argument there.

 Quote by dm4b A more technical and exact discussion can alleviate the chances of that and be more fruitful, but that doesn't mean the same kind of confusion can't happen there too.
No argument here either.

 Quote by dm4b Exactly, that's the point of a limit. Plot that up and tell me the trend you see. Exactly, combine that with the trend above and what does that suggest.
That a null interval is exactly zero. Which we already knew since you can plug a null interval directly into the Minkowski interval formula $ds^2 = dt^2 - dx^2 - dy^2 - dz^2$ and get zero.

 Quote by dm4b Combine that with the fact that neutrinos would not able to undergo neutrino oscillations if they had zero mass and what does that suggest. It all suggests that "massless particles do not sense the passage of time"
The general fact that lightlike intervals are zero suggests to me that timelike and lightlike objects are fundamentally different. However, since you mention neutrino oscillations specifically, we can go into more detail for that specific case.

Neutrinos come in three "flavors", electron, muon, and tau, corresponding to the three kinds of "electron-like" leptons. Neutrino oscillation means that a neutrino that starts out as one flavor can change to a different flavor--more precisely, the quantum mechanical mixture of flavors of neutrinos changes over time: the amplitudes for the different flavor eigenstates oscillate.

Oscillating amplitudes in themselves don't require timelike objects: photon amplitudes can oscillate and photons are massless. The point is that the flavor eigenstates of neutrinos--the states in which only one flavor amplitude is nonzero--are *different* than the mass eigenstates--the states in which a neutrino has a definite invariant mass. But for this to lead to neutrino oscillations as defined above, there must be more than one mass eigenstate, so that the amplitudes for different mass eigenstates can oscillate with different frequencies, which in turn means that the amplitudes for each flavor eigenstate (which are just different linear combinations of the mass eigenstates) also oscillate. That means at least one neutrino mass eigenstate must have a nonzero mass. It does *not* require that *all* of the neutrino mass eigenstates have nonzero mass; there could still be one such state with zero mass. AFAIK the current belief is that all of the mass eigenstates have nonzero mass, but that's based on experimental data, not theoretical requirements.

So I would say that the statement "neutrino oscillation requires neutrinos to have non-zero invariant mass" is, while technically correct, a little misleading since it invites the false implication that *any* kind of "oscillation" requires a non-zero invariant mass.
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 Quote by PeterDonis So I would say that the statement "neutrino oscillation requires neutrinos to have non-zero invariant mass" is, while technically correct, a little misleading since it invites the false implication that *any* kind of "oscillation" requires a non-zero invariant mass.
After reading your post, I think we're in agreement on pretty much everything and I enjoyed your summary of neutrino oscillations. I guess I'm just known to not be very picky about some of the laymen's descriptions - but, then again, I'm not in a position where I have to explain away the confusions they create. ;-)

There is one exception - virtual particles. I really wish they invented a different way to talk about those guys! Even graduate level QFT physics texts could do a better job here.
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 Quote by dm4b virtual particles. I really wish they invented a different way to talk about those guys! Even graduate level QFT physics texts could do a better job here.
I'm not familiar with enough QFT texts to comment on them, but I remember having to make a large mental adjustment when I found out about non-perturbative phenomena in QFT. A. Zee's book, Quantum Field Theory in a Nutshell, has a good treatment--at least it made the basics clear to me--and he comments at one point that it took a long time for many QFT theorists to admit that there was more to QFT than Feynman diagrams and perturbation theory, which is where the concept of virtual particles comes from.
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 Quote by WannabeNewton I was talking about light as a wave traveling through the medium. If you want to talk about the individual photons then it is much more subtle than that. This is not related to the thread so for now take a look at: http://physics.stackexchange.com/que...-through-glass
there is no subtlety of individual photon speed.It is always c.The refractive index concept applies to phase speed of light which has nothing to do with photon's speed.refractive index was used when one has no picture of electrons etc.Also in more modern treatment classical theory of refractive index does agree with quantum explanations.Also the retarding of light in a medium of refractive index n is overall written with a factor c/n.But still it is wrong to say that light in a medium light is retarded at speed c/n.
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 Quote by Naty1 We haven't any experimental evidence I can think of at either [the 'infinities', nor at v= c] yet, so a discussion seems moot,maybe that's your point, and that's ok by me....
I don't understand what you mean by saying that we don't have experimental evidence at v=c. SR says massless particles always move at c and massive ones never do. We observe that massless particles always move at c, and we never observe a massive one to.

This is like saying that biology has no empirical evidence about whether humans can reproduce by fission. Biology says that bacteria can reproduce by fission and humans can't. We observe that bacteria reproduce by fission, and we never observe a human to do so.

What experiment would satisfy you, even in principle, that massive particles *can't* move at c? If the only experiment you'll accept is one in which we accelerate a massive particle to c and see what happens, then there is no experiment, even in principle, that would convince you that motion at v=c doesn't exist. This would be like saying that you want to see a human to reproduce by fission so that you can test whether humans can reproduce by fission.
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bcrowell
 I don't understand what you mean by saying that we don't have experimental evidence at v=c.
ben..thanks for the interest. [Look, this could be worse, much worse: just imagine if I were a student of yours with all these crazy perspectives!! ]

I seem to be making things worse rather than better....[That's what my wife always claims!!]

yeah, we seem to have good evidence massive particles can't get to v =c.....
I have never considered that an issue.

This below seems to be one example which I had not seen before....I just stumbled across it....but it conveys the concept I am attempting to describe already:

 The description of event horizons given by general relativity is thought to be incomplete. When the conditions under which event horizons occur are modeled using a more comprehensive picture of the way the universe works, that includes both relativity and quantum mechanics, event horizons are expected to have properties that are different from those predicted using general relativity alone.

I'll start a new thread...that may enable you guys to help me understand "what happens when a null like path [a photon] intersects a null like surface [an event horizon]. [just a first thought as a problem statement]

let's do that separately after [if] I collect my feeble thoughts!!
P: 29
 In a universe full of particles that can only move at lightspeed (i.e. gauge boson) there should be no possibility of interaction from the particles' point of view because time has stopped for them, according to SR
Is it true according current physic knowledge?

If two photon travel parallel in empty universe, what will happen?

Gosh, don't give me warning because of this.
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 Quote by SysAdmin Is it true according current physic knowledge?
No. Photons don't interact with each other, but that isn't because they're massless; see below for further comment on that. There are massless particles that do interact with each other: gluons, for example.

 Quote by SysAdmin If two photon travel parallel in empty universe, what will happen?
Nothing. But that's not because they "don't experience time". It's because (a) photons don't interact with each other period; photons only interact with particles carrying electric charge, and photons don't carry any electric charge; and (b) the two photons are moving in the same direction at the same speed, so their worldlines will never intersect, so even if they could interact in principle, they wouldn't.
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 Quote by Naty1 This below seems to be one example which I had not seen before....I just stumbled across it....but it conveys the concept I am attempting to describe already:
This doesn't have anything to do with photons specifically; it has to do with quantum gravity vs. classical gravity. If quantum effects change the properties of event horizons from what classical GR models them as, that affects *everything* that comes into that region of spacetime, not just photons.
P: 29
 Quote by PeterDonis No. Photons don't interact with each other, but that isn't because they're massless; see below for further comment on that. There are massless particles that do interact with each other: gluons, for example. Nothing. But that's not because they "don't experience time". It's because (a) photons don't interact with each other period; photons only interact with particles carrying electric charge, and photons don't carry any electric charge; and (b) the two photons are moving in the same direction at the same speed, so their worldlines will never intersect, so even if they could interact in principle, they wouldn't.
Photon live in a instant, it's emitted than re-absorb instantly (according to itself), it doesn't decay, not even at the Schwartzschild Horizon and not interact each other in gravitational force. Does all gauge boson behave like this? Does gluon emitted and re-absorb instantly?

Now I understand time dilation is 0 for v=c under SR. Will it be also 0 under GR?
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 Quote by SysAdmin Photon live in a instant, it's emitted than re-absorb instantly (according to itself),
No, this is not correct. You are saying that a photon's worldline contains only a single event; that's not correct, photon worldlines contain multiple events. You can't use proper time to label the events, but you can use other affine parameters; and the fact that you can't use proper time to label the events does *not* mean that "they all happen at the same time".

 Quote by SysAdmin it doesn't decay, not even at the Schwartzschild Horizon
Photons don't "decay", exactly, but they can be absorbed, and this can happen anywhere, including at or inside a black hole's horizon.

 Quote by SysAdmin and not interact each other in gravitational force.
Huh? Photons do interact with gravity, like anything that has energy. That means that beams of photons *can* interact with each other gravitationally. (When you do the math, it turns out that antiparallel beams attract each other, but parallel beams don't; that's due to the way the photons' spin affects the interaction.)

 Quote by SysAdmin Does all boson behave like this?
No. None of them do, including photons.

 Quote by SysAdmin Now I understand time dilation is 0 for v=c under SR. Will it be also 0 under GR?
It's true that null worldlines have a zero spacetime "length" in GR just as they do in SR. But it's not IMO a good description to say that that means "time dilation is 0". The reason it's not a good description is that it leads to invalid inferences like the ones you made in the quotes above.
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 Quote by PeterDonis You are saying that a photon's worldline contains only a single event; that's not correct, photon worldlines contain multiple events.
What are the multiple events on a photon's worldline?
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 Quote by nitsuj What are the multiple events on a photon's worldline?
Draw one on a spacetime diagram, and it will be obvious; a photon's worldline is a *line* on the diagram, containing multiple points, just like other lines. The photon's worldline happens to have a Minkowski length of zero, but that's not the right measurement to use for "number of events on the line"; "what it looks like when you draw it on a spacetime diagram" is a much better measurement (there are still some technicalities, but they're minor for this case).

Another way of looking at it is to ask: a photon gets emitted, and it gets absorbed. Are those two events the same event? Obviously not; they might be light-years apart. So the photon's worldline, which contains both of those events, can't be just a single event; it must contain multiple events (the two endpoints, plus all the ones in between).
P: 29
 Quote by PeterDonis No, this is not correct. You are saying that a photon's worldline contains only a single event; that's not correct, photon worldlines contain multiple events. You can't use proper time to label the events, but you can use other affine parameters; and the fact that you can't use proper time to label the events does *not* mean that "they all happen at the same time".
My writing is not exactly what I understand, since I'm just ordinary people, using English instead of math to talk Physics.

The question about interaction is just because you said
 (a) photons don't interact with each other period; photons only interact with particles carrying electric charge, and photons don't carry any electric charge
Lot of people interpreted τ as the "rate of time flow" for a photon, but in the popular media, they always refer to SR as for explanation. That is why i wonder, have they consider GR before saying that "photon don't experienced time"?
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 Quote by SysAdmin in the popular media, they always refer to SR as for explanation. That is why i wonder, have they consider GR before saying that "photon don't experienced time"?
Yes, these things are essentially the same in GR. (The main difference is that the coordinate systems we're talking about are defined on proper subsets of spacetime instead of on spacetime). Note that the correct conclusion (in both SR and GR) isn't that photons experience zero time. It's that there's no natural way to assign a meaning to statements about what a massless particle "experiences".
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 Quote by SysAdmin The question about interaction is just because you said
Yes, I realized that on reading back; when I said photons don't interact with anything that doesn't have an electric charge, I should have clarified that I was talking about scenarios where there is no gravity (or at least where gravity can be ignored, or handled without having to treat it as an interaction alongside the others). Such scenarios include almost all scenarios where photons are actually studied in practice; experiments which actually study the effects of gravity on photons (as in bending of light by the Sun) are rare.

When gravity is included in the interactions, then yes, photons interact gravitationally. Sorry for the mixup on my part.
P: 29
 Quote by Fredrik It's that there's no natural way to assign a meaning to statements about what a massless particle "experiences".
And if I'm not mistakenly, there is two particle that has massless, photon and gluon. At least two of the neutrino is suspected has mass. Since gluons are never observed as free particles, it left us using photon to define time "experiences" for massless particle. In other word, it doesn't have comparison for other particle. Is it?

Also to clarifying, since this is phenomena for massless particle (that is v=c), it's analyzed using SR and GR. What other theory that can be used to analyzed a massless particle?

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