Why don't photons experience time?

• la6ki
In summary, Einstein tried to imagine what an electromagnetic wave would look like from the point of view of a motorcyclist riding alongside it, but found that it didn't make sense. There are two ways that the velocity of the system's center of mass could be: it could be moving at the speed of light, or it couldn't. If V=c, then all the particles are moving along parallel lines, and therefore they can't interact, can't perform computations, and can't be conscious. If V is less than c, then the observer's frame of reference isn't moving at the speed of light.
la6ki
I am asking this question in order to clarify something which I thought I had understood.

First of all, let me say that I understand talking about the perspective (or frame of reference) of a photon doesn't make sense. Yet, after many searches in the web, I feel like there is a consensus among posters who seem to be the experts that for a photon there is no passage of time. Well, I don't understand that.

Let me clarify. When we're talking about, say, a spaceship moving at .5c, we say that an outside observer will 'accuse' the clocks in that spaceship as running slower. But for people on the spaceship, their clocks will still be moving at their regular speed. Is this correct?

If yes, then why can't we extend the same logic to a photon? It is moving at 100% of c and if it... had a clock attached to it, we would say that the clock is stopped. But won't the photon still perceive the clock as ticking at its regular rate?

To repeat, I understand that the question about a photon's perspective doesn't really make sense, but I'm only asking it because in the past 2 hours I saw the answer "a photon doesn't experience time" many times.

la6ki said:
If yes, then why can't we extend the same logic to a photon? It is moving at 100% of c and if it... had a clock attached to it, we would say that the clock is stopped.
You can't make a clock move at c.

la6ki said:
But won't the photon still perceive the clock as ticking at its regular rate?
Photons aren't observers, so they don't perceive anything.

FAQ: What does the world look like in a frame of reference moving at the speed of light?

This question has a long and honorable history. As a young student, Einstein tried to imagine what an electromagnetic wave would look like from the point of view of a motorcyclist riding alongside it. But we now know, thanks to Einstein himself, that it really doesn't make sense to talk about such observers.

The most straightforward argument is based on the positivist idea that concepts only mean something if you can define how to measure them operationally. If we accept this philosophical stance (which is by no means compatible with every concept we ever discuss in physics), then we need to be able to physically realize this frame in terms of an observer and measuring devices. But we can't. It would take an infinite amount of energy to accelerate Einstein and his motorcycle to the speed of light.

Since arguments from positivism can often kill off perfectly interesting and reasonable concepts, we might ask whether there are other reasons not to allow such frames. There are. One of the most basic geometrical ideas is intersection. In relativity, we expect that even if different observers disagree about many things, they agree about intersections of world-lines. Either the particles collided or they didn't. The arrow either hit the bull's-eye or it didn't. So although general relativity is far more permissive than Newtonian mechanics about changes of coordinates, there is a restriction that they should be smooth, one-to-one functions. If there was something like a Lorentz transformation for v=c, it wouldn't be one-to-one, so it wouldn't be mathematically compatible with the structure of relativity. (An easy way to see that it can't be one-to-one is that the length contraction would reduce a finite distance to a point.)

What if a system of interacting, massless particles was conscious, and could make observations? The argument given in the preceding paragraph proves that this isn't possible, but let's be more explicit. There are two possibilities. The velocity V of the system's center of mass either moves at c, or it doesn't. If V=c, then all the particles are moving along parallel lines, and therefore they aren't interacting, can't perform computations, and can't be conscious. (This is also consistent with the fact that the proper time s of a particle moving at c is constant, ds=0.) If V is less than c, then the observer's frame of reference isn't moving at c. Either way, we don't get an observer moving at c.

Well, nobody knows for sure what photons experience.

The idea that photons 'don't experience time' comes from that fact that we know faster moving massive particles experience a slower passage of time than slow moving ones according to the laws of special relativity. That's been convincingly confirmed experimentally. So it's 'easy' [in some people's minds] to extrapolate that to massless photons [light] and figure they must "not experience any passage of time" since they move at the 'ultimate speed'...c. Exactly what that might mean nobody really knows. It doesn't make much sense as the prior post explains.

Was it Einstein who said " Eternity is no time at all for a photon."?? Well, somebody important said something like like that and it captures the idea.

Whatever the exact meaning, I hope eventually some part of the FAQ explanation above will be found incorrect. If we ever do figure it out perhaps we can use it to our advantage in ways not even understood now. Crazier things have happened.

After all, Einstein imagined catching up to light to observe what it would look like, so far understood to be an impossibility as explained above, yet years later emerged general relativity.

Last edited:
Naty1 said:
faster moving massive particles experience a slower passage of time than slow moving ones according to the laws of special relativity.

I think this is a misleading way of putting it, because it is frame-dependent; to the "faster moving massive particles", *we* are the ones whose time is "moving slower".

Naty1 said:
That's been convincingly confirmed experimentally.

Only in the frame-dependent sense given above.

Naty1 said:
So it's 'easy' to extrapolate that to massless photons [light] and figure they must "not experience any passage of time" since they move at the 'ultimate speed'...c.

It's "easy", but that doesn't mean it's correct.

Naty1 said:
Exactly what that might mean nobody really knows.

We know quite well what it means: it means that the concept of "passage of time" doesn't apply to photons. It means that there is a fundamental physical difference between objects that move on timelike worldlines, and objects that move on null worldlines. That's because "timelike" and "null" are two fundamentally different kinds of spacetime intervals.

Naty1 said:
After all, Einstein imagined catching up to light to observe what it would look like, so far understood to be an impossibility as explained above

And that's exactly the conclusion Einstein reached from that thought experiment: what he imagined was not possible. *That* is what led him to relativity (and it was SR, not GR; the insights that led Einstein to GR had nothing to do with imagining catching up to light).

Naty1 said:
Well, nobody knows for sure what photons experience.
I disagree, for the reasons given in my #2. Which part of my argument in #2 do you disagree with?

Naty1 said:
Exactly what that might mean nobody really knows. It doesn't make much sense as the prior post explains.
I disagree with this statement. Does "prior post" refer to my #2? What I said was the opposite of your characterization. I claim that there is nothing at all mysterious or unknown about these issues. They've been well understood for literally 100 years.

Naty1 said:
Whatever the exact meaning, I hope eventually some part of the FAQ explanation above will be found incorrect. If we ever do figure it out perhaps we can use it to our advantage in ways not even understood now. Crazier things have happened.
I disagree that there is any real chance of our current understanding being overturned in this area. Science doesn't progress by overturning theories within the domains where they have already been verified by experiment. It progresses by modifying them to deal with new circumstances under which they had never been tested or were already known or expected to fail.

Naty1 said:
After all, Einstein imagined catching up to light to observe what it would look like, so far understood to be an impossibility as explained above, yet years later emerged general relativity.
I don't understand what you mean by this. I don't think general relativity has anything to do with this issue.

Hi PeterDonis, bcrowell...

I do not disagree with anything you posted...I was attempting to provide some perspective on how theories and understanding evolve...maybe my language is unhelpful.

Regarding my comments on Einstein imagining catching up to light: I was thinking about the fact that it was the issue of the constant speed of light and 'ether'... that got him started on relativity...[according to the accounts I have read] and that it was his 'crazy thought experiment' that eventually led to such an overall revolution in understanding.

Quote by Naty1
So it's 'easy' to extrapolate that to massless photons [light] and figure they must "not experience any passage of time" since they move at the 'ultimate speed'...c.
It's "easy", but that doesn't mean it's correct.

yeah, thanks I did not mean it that way...I edited the earlier post to read...

'easier [in some people's minds]"

referring to those who claim 'photons don't experience the passage of time.

Naty1 said:
I was thinking about the fact that it was the issue of the constant speed of light and 'ether'... that got him started on relativity...[according to the accounts I have read] and that it was his 'crazy thought experiment' that eventually led to such an overall revolution in understanding.

I think this is true as far as it goes, but it's important to understand exactly what it was about the thought experiment that led to the "revolution". Einstein imagined moving at the same speed as an electromagnetic wave, and asked himself what he would observe if that were the case. The answer was that he would observe an electromagnetic wave that was stationary in space--i.e., oscillating in space but not in time. But such a wave is not a solution to Maxwell's Equations: they only allow waves that oscillate in *both* space and time. That made Einstein realize that there was something wrong with the premise of his thought experiment, and *that* led him to SR and its different view of space and time.

I started, then stopped, a search in Google for
"who said 'Eternity is no time at all for a photon'...because I have forgotten...
and what turns up...THIS THREAD>> OMG We ARE being watched!

isn't photon already experience time it self, by having internal clock, the photon frequency?
according to photon, it has frequency, according to us, we do not know the photon exist, up until we hit it. then after that, we assume, more or less the momentum and the frequency of it.

let say the photon sending graviton to earth. according to photon, Earth is the one moving in accelerate, therefore it's getting smaller, while according to Earth photon periodic is increased (T get bigger, f get smaller). Now imagine that the gravity is so big, it will make the frequency so high, than it is the actual meaning the Photon stop in time.

wait, am I wrong in here?

I googled and found this extremely intelligent remark, which I haven't made yet:

bcrowell said:
If CTCs are going to turn up on PF, the relativity subforum would be the logical place.

Naty1 said:
I started, then stopped, a search in Google for
"who said 'Eternity is no time at all for a photon'...because I have forgotten...
and what turns up...THIS THREAD>> OMG We ARE being watched!

If CTCs are going to turn up on PF, the relativity subforum would be the logical place.

isn't photon already experience time it self, by having internal clock, the photon frequency?

No, because this is the frequency measured by some observer who is not at rest relative to the photon, and it can have any value whatsoever depending on the observer. What makes a clock a clock is that its frequency has a specific, special value when you measure it in a frame at rest relative to the clock.

Correct me if I'm wrong here.

Not experiencing time is mean the periodic ticking of the clock is so high, it require eternity to the clock second hand reaching the next mark.

so it's not moving in space that stop, but moving in time. we perceive the clock is not moving, but actually it is moving. but the periodic so high the wave length become near to zero.

Thanks for the discussion guys. I couldn't really follow everything, tbh, but I think I more or less get it. Can you confirm for me if the following is true?

There is a qualitative jump between moving at .999999999999999999c and c. So, a particle with mass moving with the former will experience the same time it experiences if it were stationary, but it would perceive all other clocks as running slower. Similarly, all other observers will perceive the clock of the particle as running slower. However, a photon (moving at c) will not experience time at all, due to the qualitative difference I started with.

Is this more or less true?

What makes a clock a clock is that its frequency has a specific, special value when you measure it in a frame at rest relative to the clock.
If I'm moving in space 0.8c, I will eat for 10 ticking of my clock. But for Earth observer, I eat, I dunno, 13 ticking of their clock. So according to them, my clock ticking is very slow compare to their ticking

So when I move 0.999...c, they will see my clock stop ticking. Isn't it?

Last edited:
One of my motivations for posting here was that sometimes I find the FAQ's difficult to understand...when I first began here in the forums and read a few, I often gave up... Now I am beginning, maybe, to understand them...usually only parts.

I post the following hoping the OP will find answers instructive:

From the FAQ..in post #2:

(This is also consistent with the fact that the proper time s of a particle moving at c is constant, ds=0.)

Could one of you experts explain what this means...maybe put this in the context of the OP's question... do you think it a source of the common refrain the OP encounters...'why don't photons experience time'?? [I am not sure what to make of it.]

and how do we square this
If V=c, then all the particles are moving along parallel lines, and therefore they aren't interacting, can't perform computations, and can't be conscious.

Does this apply to only parallel or also to anti-parallel [opposite directions] light waves?? In other words, how do anti-parallel light waves interact if their frame time doesn't make sense?? [I can't answer this either.]

Also, on behalf of the OP, maybe I could ask:

If photons don't experience any time, how do they interact with a gravitational field? Doesn't that take some time?? Is that a refutation that their time can't be zero??
[I can't answer this one either.]

There is a qualitative jump between moving at .999999999999999999c and c. So, a particle with mass moving with the former will experience the same time it experiences if it were stationary, but it would perceive all other clocks as running slower. Similarly, all other observers will perceive the clock of the particle as running slower. However, a photon (moving at c) will not experience time at all, due to the qualitative difference I started with.

Is this more or less true?

more...yes, until the last sentence...less, the last sentence...
the experts here will not like that I don't think...let's see...maybe the answers to the questions I just posted can shed more 'light' [aggghhhhh,what an awful pun!] on this last part...I sure don't understand it the way I'd like...

a particle with mass moving with the former will experience the same time it experiences if it were stationary, but it would perceive all other clocks as running slower. Similarly, all other observers will perceive the clock of the particle as running slower. However, a photon (moving at c) will not experience time at all, due to the qualitative difference I started with

A particle with mass moving with the former will experience the same time it experiences if it were stationary, but it would perceive all other clocks as running slower. Similarly, all other observers will perceive the clock of the particle as running slower. However, a photon (moving at c) will perceive all other clocks stop (not experience time at all) while the clock of it self (the frequency of the proton?),according to the photon, stay the same.
?

Last edited:
la6ki said:
There is a qualitative jump between moving at .999999999999999999c and c.

Yes. Here's why: suppose you are moving at .999999999999999999c relative to me, and a light beam is traveling in the same direction as you, relative to me. I see the light moving at c, so you appear to me to be moving almost as fast as the light is.

Now I boost myself to move with you, so I am now moving at .999999999999999999c also with respect to the frame in which I started out at rest. I see the same light beam *still* moving at c--boosting myself to .999999999999999999c hasn't changed the speed of the light beam at all, even though it changed your speed relative to me from .999999999999999999c to zero.

la6ki said:
a photon (moving at c) will not experience time at all, due to the qualitative difference I started with.

No. As Naty1 suspected, this is not a valid deduction from the above. The valid deduction from the above is that the concept of "passage of time" does not apply to a photon. Here's why that's true:

For an ordinary object, we can define a "4-velocity" vector, which is a unit vector in spacetime that points along the object's worldline. The components of the 4-velocity in a given frame give the object's "rate of time flow" and ordinary spatial velocity in that frame; but this interpretation depends on the 4-velocity being a unit vector.

For a photon, however, we can't define a unit vector that points along its worldline, because its worldline is null--any vector that points along the worldline has length zero, so there can't be a vector with length one (or any other nonzero length) pointing in that direction. That means we can't even define concepts like "rate of time flow" for a photon.

Naty1 said:
From the FAQ..in post #2:

My recommendation would be to remove the term "proper time" from the FAQ in post #2, and use some other term. Would that help?

a photon (moving at c) will perceive all other clocks stop (not experience time at all) while the clock of it self (the frequency of the proton?),according to the photon, stay the same.

No, this won't work, because you can't make a clock solely out of photons, for the reasons bcrowell gave in an earlier post.

Naty1 said:
how do anti-parallel light waves interact if their frame time doesn't make sense??

Classically, light waves don't interact with each other, period. When you figure in quantum effects, there are very small interactions (due to virtual particle-antiparticle pairs), but I don't think we need to open that can of worms here.

Naty1 said:
If photons don't experience any time, how do they interact with a gravitational field?

Photons don't have to "experience time" to interact, either with a gravitational field or with anything else. Photons have energy, and anything with energy interacts with a gravitational field.

PeterDonis said:
Yes. Here's why: suppose you are moving at .999999999999999999c relative to me, and a light beam is traveling in the same direction as you, relative to me. I see the light moving at c, so you appear to me to be moving almost as fast as the light is.

Now I boost myself to move with you, so I am now moving at .999999999999999999c also with respect to the frame in which I started out at rest. I see the same light beam *still* moving at c--boosting myself to .999999999999999999c hasn't changed the speed of the light beam at all, even though it changed your speed relative to me from .999999999999999999c to zero.

That's a really good explanation, thanks!

No. As Naty1 suspected, this is not a valid deduction from the above. The valid deduction from the above is that the concept of "passage of time" does not apply to a photon.

Fair enough. Well then, can we just say that all those threads in which I read that a photon experiences zero time are just wrong? If the concept of "passage of time" doesn't apply to a photon, then the question of what time a photon experiences are meaningless to begin with, no?

And for the same reason, we can also say that a photon can't experience velocities as well. And hence, the follow-up question which I was going to have, namely, "will two photons moving together perceive each other to have zero velocity (the way you perceived me as having zero velocity when you boosted yours to match my .999999999c)" is also meaningless, correct? (I know that would violate the second postulate of SR as well, but I'm kind of trying to get the same answer from a different point of view).

By the way, I hope you won't mind if I ask a question which is not directly related to the thread title, but one for which I don't want to start a new thread:

Imagine a stationary particle (let's say an electron) sitting next to an electrically neutral wire. It will obviously not experience any force. But now imagine that the electrons inside the wire start moving to the right. From the frame of reference of our stationary electron the density of the electrons in the wire will increase due to length contraction and hence it will look like the wire is negatively charged. So, the electron must be repelled by the wire.

I kept investigating the issue and couldn't find a confirmation to this. Every single link I found was just talking about the force being q*v x B and since v was zero v x B would also be zero. I'm not quite sure what to make of this.

I suspect that this question has a trivial answer, hence my reluctance to start a new thread.

For an ordinary object, we can define a "4-velocity" vector, which is a unit vector in spacetime that points along the object's worldline. The components of the 4-velocity in a given frame give the object's "rate of time flow" and ordinary spatial velocity in that frame; but this interpretation depends on the 4-velocity being a unit vector.

For a photon, however, we can't define a unit vector that points along its worldline, because its worldline is null--any vector that points along the worldline has length zero, so there can't be a vector with length one (or any other nonzero length) pointing in that direction. That means we can't even define concepts like "rate of time flow" for a photon.

But photon has wave length and frequency, at least that "2-velocity" vector there. If we see photon move with speed c, isn't photon will perceive the opposite, that photon think "oh I'm just stay put here, let all other move with crazy speed c". At least the photon will perceive it self as two dimensional thing, complete with the direction of it in this 2D space.

So, a photon will never see it self stop, it will see it self moving, confine in the space of lambda with the frequency f. Same thing at least for two photons moving together. The photon will see the other photon, vibrate at confine space lambda_2 with f_2. The question is, will it perceive that lambda_2, f_2 photon stay at the same distance?

?

Also, Is Special Relativity only work for linear movement, that is there is no angle between the observer? What is the explanation for three photon moving in triangular configuration. Each photon must perceive other photon move toward them with the speed of C, but at the same time it must perceive the other two photon getting close each other with the speed of C. Will it wreck wreck the euclidean space continuum?

I just don't get it,
can't make a clock solely out of photons, for the reasons bcrowell gave in an earlier post.

Since the space where the photon moving is having nearly zero movement in the direction perpendicular with the photon move, sure the photon can measured it vibration in that direction. Like my question before, if the SR apply for the same angle angle movement, then height of the object (not length) must be measure with the relativity in height direction also. In other word, in my opinion, photon move with the speed of C in two direction. Transverse in one direction, with vibration perpendicular to its traversal movement.

Or tell, me, what is the actual meaning that photon is transverse wave?

I'm sorry if my understanding is so weird and at the same time too long to read.

Last edited:
PeterDonis said:
For an ordinary object, we can define a "4-velocity" vector, which is a unit vector in spacetime that points along the object's worldline. The components of the 4-velocity in a given frame give the object's "rate of time flow" and ordinary spatial velocity in that frame; but this interpretation depends on the 4-velocity being a unit vector.

For a photon, however, we can't define a unit vector that points along its worldline, because its worldline is null--any vector that points along the worldline has length zero, so there can't be a vector with length one (or any other nonzero length) pointing in that direction. That means we can't even define concepts like "rate of time flow" for a photon.

No. This is not a valid deduction from the above. The valid deduction from the above is that we can't define rate of time flow for a photon, based on 4-velocity in Minkowski space.

But there is no obligation to define rate of time flow for a photon based on some specific geometrical interpretation of SR. When we go back to the basics of SR we have:

$\Delta \tau = \Delta t \sqrt{1 - \frac{v^2}{c^2}}$

For a photon this gives:

$\Delta \tau = 0$

A.T. said:
When we go back to the basics of SR, the Lorentz transformation we have:

$\Delta \tau = \Delta t \sqrt{1 - \frac{v^2}{c^2}}$

For a photon this gives:

$\Delta \tau = 0$

But the question is not about the length of the photon's $\tau$; the question is whether $\tau$ can be validly interpreted as telling us the "rate of time flow" for a photon. I think the fact that we get into endless threads on this topic shows that that interpretation is not a very good one; maybe that doesn't make it "invalid" (since the question of "validity" of an interpretation has a subjective element), but it does make it a huge pain, pedagogically speaking.

But the Lorentz transformation tells you what observers outside of the frame of reference of the photon will measure as time, not what the photon will measure as time.

la6ki said:
can we just say that all those threads in which I read that a photon experiences zero time are just wrong? If the concept of "passage of time" doesn't apply to a photon, then the question of what time a photon experiences are meaningless to begin with, no?

That would be my position, yes. I think that if an interpretation leads to endless questions that don't really have answers, and having to continuously explain to people why deductions that seem obvious from the interpretation are not valid, that's a good reason not to use that interpretation. But as you can see from other responses in this thread, there are different opinions on this.

la6ki said:
And for the same reason, we can also say that a photon can't experience velocities as well. And hence, the follow-up question which I was going to have, namely, "will two photons moving together perceive each other to have zero velocity (the way you perceived me as having zero velocity when you boosted yours to match my .999999999c)" is also meaningless, correct?

Again, yes, that would be my position. More precisely, we can't define a meaningful concept of "velocity of one photon relative to another photon".

la6ki said:
Imagine a stationary particle (let's say an electron) sitting next to an electrically neutral wire. It will obviously not experience any force. But now imagine that the electrons inside the wire start moving to the right. From the frame of reference of our stationary electron the density of the electrons in the wire will increase due to length contraction and hence it will look like the wire is negatively charged. So, the electron must be repelled by the wire.

I kept investigating the issue and couldn't find a confirmation to this. Every single link I found was just talking about the force being q*v x B and since v was zero v x B would also be zero. I'm not quite sure what to make of this.

Concepts of Modern Physics - Arthur Beiser, chapter 1

I could give you the link for the pdf, but for a moment the extract page regarding SR and Electricity and Magnetism. If you have read the book and still not quite give the answer...I don't know.

Attachments

• Concepts of Modern Physics_ Arthur Beiser 35.pdf
46.2 KB · Views: 181
PeterDonis said:
But the question is not about the length of the photon's $\tau$; the question is whether $\tau$ can be validly interpreted as telling us the "rate of time flow" for a photon.
It is the rate of time flow of a photon for any valid observer. Of course photons themselves are not valid observers, which seems to be the main confusion here. But for any other observer the proper time of a photon has a valid value of zero.

Particles don't actually experience anything, for obvious reasons. So before we can make statements about what a particle "experiences", we must define what it means for a particle to experience something. The idea is simple: All statements about what a particle "experiences" are really statements about the coordinates assigned by an inertial coordinate system that's comoving with the particle. For example, if we say that the particle experiences that B occurs five seconds after event A, it means that the comoving inertial coordinate systems S assigns a 4-tuple of coordinates ##(S^0(E),S^1(E),S^2(E),S^3(E))## to each event E, and the difference ##S^0(B)-S^0(A)## is five seconds.

Now, there there are no inertial coordinate systems that are comoving with a massless particle. Therefore, the definition of "experience" doesn't apply. That's all there is to it.

It really is.

Yes, you can choose to use some other coordinate system that isn't comoving and/or isn't an inertial coordinate system, but which one do you choose, and why? There's no choice that really stands out the way the comoving inertial coordinate systems do for massive particles.

A.T. said:
But for any other observer the proper time of a photon has a valid value of zero.
Proper time is a coordinate-independent property of a curve.

Replies
29
Views
2K
Replies
17
Views
2K
Replies
55
Views
2K
Replies
25
Views
798
Replies
14
Views
1K
Replies
12
Views
1K
Replies
16
Views
1K
Replies
4
Views
898
Replies
27
Views
4K
Replies
83
Views
4K