Why don't photons experience 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?

 

[WannabeNewton]

photons always travel at speed c.After entering a medium it does not change

This is false and the lack of it traveling at c in generality after entering a medium is a very well known fact. You can start here: http://en.wikipedia.org/wiki/Speed_of_light#In_a_medium and go to textbook sources etc. if you want to learn more.

 

[A.T.]

In each case, the global inertial coordinate systems are especially important, but it seems pointless to label all other coordinate systems "invalid".

By "invalid reference frames" I meant something like "the rest frame of a photon". SR doesn't make any statements about the rest frame of a photon. For all other frames it states that the proper time of a photon is zero.

 

[PeterDonis]

SR doesn't make any statements about the rest frame of a photon.

I'm not sure I agree; I would say that SR says there is no such thing as "the rest frame of a photon", not that it says nothing at all about it.

For all other frames it states that the proper time of a photon is zero.

Once again, this presumes that the zero interval associated with a photon's worldline is appropriately described as "proper time", which is precisely the thing that causes so much confusion about the "time experienced by a photon". I think the term "proper time" should be reserved for timelike intervals only.

 

[WannabeNewton]

I think the term "proper time" should be reserved for timelike intervals only.

I would also agree because one tends to use proper time as an affine parameter along time - like geodesics because we can without running into contradictions but for light - like paths this is not possible and since we can't even use proper time to parametrize the path of light I wouldn't say it holds any physical meaning in the way proper time holds meaning along world lines of massive particles.

 

[Naty1]

Quote by andrien

photons always travel at speed c.After entering a medium it does not change

This is false and the lack of it traveling at c in generality after entering a medium is a very well known fact.

The individual photons DO travel at c; but as they progress thru the material, delays are encountered so the overall, effective transmission rate is slower than c:

From the wiki reference above:

In exotic materials like Bose–Einstein condensates near absolute zero, the effective speed of light may be only a few meters per second. However, this represents absorption and re-radiation delay between atoms, as do all slower-than-c speeds in material substances.

 

[WannabeNewton]

The individual photons DO travel at c; but as they progress thru the material, delays are encountered so the overall, effective transmission rate is slower than c:

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/questions/1909/how-does-a-photon-travel-through-glass

 

[Naty1]

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: ... 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. ...

Short answer: yes, that apparently works...

there have been discussions in these forums about it...search if you want more.
I had not seen such before the discussions here and found the concepts worthwhile.

Check out here: http://en.wikipedia.org/wiki/Relativistic_electromagnetism


I searched...and got sidetracked...found BenCrowell posted about a text he likes on the subject here:

https://www.physicsforums.com/showthread.php?t=665597&highlight=relativistic+electromagnetism

 

[A.T.]

I would say that SR says there is no such thing as "the rest frame of a photon", not that it says nothing at all about it.

It simply doesn't make any predictions about physics in the rest frame of a photon.

 

[PeterDonis]

It simply doesn't make any predictions about physics in the rest frame of a photon.

I hate to keep nit-picking about language, but the way this is phrased implies (at least, I expect it will imply to a lot of newbies) that there *is* something called "the rest frame of the photon", when the whole point is that there isn't. SR says there is no such thing as "the rest frame of the photon". IMO that's the way to phrase it.

 

[Naty1]

la6ki:
In case you haven't seen it, there are some closely related perspectives here:

Light doesn't travel through time?
https://www.physicsforums.com/showthread.php?t=665516&highlight=photon+time

where you should check the posts from Fredrik and Dalespam...and follow the few links provided.