Why don't photons experience time?

[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.

 

[bcrowell]

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.

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.

 

[Naty1]

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.

 

[PeterDonis]

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".

That's been convincingly confirmed experimentally.

Only in the frame-dependent sense given above.

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.

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.

 

[PeterDonis]

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).

 

[bcrowell]

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?

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.

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.

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.

 

[Naty1]

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.

Hopefully your comments will aid the OP.


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.

 

[Naty1]

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.

 

[PeterDonis]

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.

 

[Naty1]

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!

 

[SysAdmin]

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?

 

[bcrowell]

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

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

 

[bcrowell]

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.

 

[bcrowell]

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.

 

[SysAdmin]

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.

 

[la6ki]

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?

 

[SysAdmin]

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?

 

[Naty1]

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.]

 

[Naty1]

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...

 

[SysAdmin]

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.
?

 

[PeterDonis]

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.

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.

 

[PeterDonis]

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?

 

[PeterDonis]

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.

 

[PeterDonis]

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.

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.

 

[la6ki]

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).

 

[la6ki]

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.

 

[SysAdmin]

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.

 

[A.T.]

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

 

[PeterDonis]

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.

 

[la6ki]

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.

 

[PeterDonis]

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.

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".

 

[SysAdmin]

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.

 

[A.T.]

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.

 

[Fredrik]

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.

 

[Fredrik]

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.

 

[la6ki]

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.

No, I already knew that. But this is about two currents in the same direction. My question is when you have one current in a particular direction but instead of a second current you only have a stationary particle. If the positive and negative charge densities of the wire were exactly equal when there was no current, won't the charge experience a force when a current appears?

 

[A.T.]

Proper time is a coordinate-independent property of a curve.

Yes, its the same in all valid reference frames. For photons it is zero in all valid reference frames.

 

[PeterDonis]

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.

I don't follow this; proper time is not an observer-dependent quantity. I agree that the invariant length of a photon's worldline between any two events on it is zero; AFAIK nobody in this thread is questioning that. But calling that the "proper time of a photon" is precisely the point at issue. I think doing that causes far more confusion than it solves (if it solves any).

 

[A.T.]

I don't follow this; proper time is not an observer-dependent quantity.

Is there a contradiction between "not frame dependent" and "zero in any frame"?

 

[Fredrik]

Yes, its the same in all valid reference frames. For photons it is zero in all valid reference frames.

And in all "invalid" reference frames. It's the same in all coordinate systems, and can be defined without mentioning any coordinate systems at all.

Is there a contradiction between "not frame dependent" and "zero in any frame"?

There is no contradiction there. Compare e.g. to the statements "the vector $x\in\mathbb R^3$ has components (0,0,0) in every basis" and "x is the 0 vector in $\mathbb R^3$". They both say the same thing, but the first statement expresses it in terms of components and bases, and the other one doesn't mention components or bases.

 

[SysAdmin]

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.

mass-less in here is assuming the particle also a point particle, right? wave-length is assumed as quantum value of space in which probability the point particle can be found in this length is 1. photon as 0D particle, moving in space, creating 1D phenomena. in other word, photon is 1D wave.

 

[arindamsinha]

Is it really appropriate to talk about 'photons' in this context? What about photon's traveling through water then? We could potentially catch up with or even overtake such photons in a blaze of Cherenkov radiation.

The question then is - while photons traveling in vacuum do not experience passage of time, photons in denser mediums do? That seems a bit weird.

Any suggestions?

 

[Fredrik]

mass-less in here is assuming the particle also a point particle, right?

Is it really appropriate to talk about 'photons' in this context?

I'm deliberately avoiding the term "photon" in these discussions, because to me that's a term that's only defined in some relativistic quantum field theories (QED, and theories that are like QED but involve additional fields), and the arguments we use are entirely classical. We are talking about classical point particles whose world lines are null geodesics in Minkowski spacetime.

What about photon's traveling through water then? We could potentially catch up with or even overtake such photons in a blaze of Cherenkov radiation.

The question then is - while photons traveling in vacuum do not experience passage of time, photons in denser mediums do? That seems a bit weird.

When you say that they do not experience the passage of time, it sounds like you're saying that they do have an experience, which is that the time that has passed is always zero. But "experience" isn't even defined for particles moving as described by null geodesics.

Photons in a medium do not move at the invariant speed, so we don't have the same problem with them.

 

[arindamsinha]

...Photons in a medium do not move at the invariant speed, so we don't have the same problem with them.

In other words, photons in mediums behave more like massive particles, and they do experience the passage of time? (I am using the same somewhat inadequate terminology here, but I hope my meaning is clear).

What changes about photons when they enter a medium to make this possible?

 

[georgir]

Why don't apples experience love?

 

[Fredrik]

In other words, photons in mediums behave more like massive particles, and they do experience the passage of time? (I am using the same somewhat inadequate terminology here, but I hope my meaning is clear).

Yes.

What changes about photons when they enter a medium to make this possible?

They get absorbed and re-emitted, or at least, a QED calculation to predict the arrival time at a detector must include absorption/emission processes in the calculation.

 

[A.T.]

And in all "invalid" reference frames.

SR doesn't make statements about invalid reference frames. SR applies per definition only to those reference frames which are valid under SR.

 

[Nugatory]

And in all "invalid" reference frames.

SR doesn't make statements about invalid reference frames. SR applies per definition only to those reference frames which are valid under SR.

I think you may have missed Fredrik's point - he's pointing out that the proper interval between two lightlike-separated events is zero no matter what you choose as a reference frame, or even if you don't choose one at all. The validity or invalidity of any particular reference frame is irrelevant.

Now, does special relativity make that statement? This may come down to agreeing on exactly what "The Theory of Special Relativity" is - is it Einstein's early formulation based on Lorentz transformations between one inertial frame and another; or is it the more modern Minkowski-inspired formulation? The former is doesn't make statements independent of reference frame, but the latter does.

 

[andrien]

photons always travel at speed c.After entering a medium it does not change.They don't become massive,may be you should call them some effective mass type thing.

 

[Fredrik]

SR applies per definition only to those reference frames which are valid under SR.

I disagree. I wouldn't define SR that way, and I think most physicists wouldn't.

To me, SR is the idea that spacetime is Minkowski spacetime, and GR is the idea that spacetime is a a Lorentzian manifold with a metric that satisfies Einstein's equation.

There are a few different structure that it would make sense to call Minkowski spacetime (because they can be used to define theories of physics that make identical predictions). It can be defined as a vector space, an affine space, or a smooth manifold. In each case, the global inertial coordinate systems are especially important, but it seems pointless to label all other coordinate systems "invalid".

If we instead define GR as the idea that other coordinate systems are OK too, then what should we call the idea that the metric is to be determined from an equation?