Why don't photons experience time?

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.
  • #36
SysAdmin said:
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?
 
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  • #37
Fredrik said:
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.
 
  • #38
A.T. said:
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).
 
  • #39
PeterDonis said:
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"?
 
  • #40
A.T. said:
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.

A.T. said:
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.
 
  • #41
Fredrik said:
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.
 
  • #42
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?
 
  • #43
SysAdmin said:
mass-less in here is assuming the particle also a point particle, right?

arindamsinha said:
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.

arindamsinha said:
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.
 
  • #44
Fredrik said:
...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?
 
  • #45
Why don't apples experience love?
 
  • #46
arindamsinha said:
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.

arindamsinha said:
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.
 
  • #47
Fredrik said:
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.
 
  • #48
Fredrik said:
And in all "invalid" reference frames.
A.T. said:
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.
 
  • #49
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.
 
  • #50
A.T. said:
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?
 
  • #51
andrien said:
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.
 
  • #52
Fredrik said:
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.
 
  • #53
A.T. said:
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.

A.T. said:
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.
 
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  • #54
PeterDonis said:
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.
 
  • #55
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.
 
  • #56
Naty1 said:
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
 
  • #57
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
 
  • #58
PeterDonis said:
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.
 
  • #59
A.T. said:
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.
 
  • #61
Can somebody tell me what "CTC"s are [from post #26]:

Quote by 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!

BCrowell:

If CTCs are going to turn up on PF, the relativity subforum would be the logical place.
 
  • #62
CTC = Closed Timelike Curve.

If there's a CTC in spacetime, you could in principle move as described by it and meet a younger version of yourself. BCrowell was joking. It got a LOL out of me, so the joke works on some nerds at least. I wouldn't try it as a pick-up line though.
 
  • #63
It was my understanding that the logic (put in laymen's terms, perhaps) that the reason a neutrino cannot be massless is because it can undergo neutrino oscillations, which was the solution to the solar neutrino problem. Neutrino Oscillations are a time-dependent phenomenon and since a massless particle does not experience the passage of time, it would not be able to experience these oscillations. Since it has been shown neutrinos do undergo the oscillations they therefore cannot be massless.

There was a show on the science channel that said as much too.

Also, when you analyze the time dilation formula in the limit of v=c, while the equations are still valid, as you get infinitesimally close to v=c, deltaT moves infinitesimally close to zero.
 
  • #64
Minor Notes:
1]Closed Timelike Curve...haven't seen that one in a long time!

2]It is a lot more relaxing to read you experts picking each other apart than reading when you pick ME apart!

3] la6ki: You lucked out getting all these experts to offer perspectives! Great discussion.la6ki: I hope after reading the posts from a number of the most knowledgeable people here you have a perspective now on WHY I posted early on:

Well, nobody knows for sure what photons experience...blah,blah...Whatever the exact meaning, I hope eventually some part of the FAQ explanation above will be found incorrect.

Now I have been in these forums enough to know better...I should have known THAT wording would get some riled up...far better had I said something like "I hope eventually some part of the FAQ can be revised as a result of new discoveries."

Some of the ways the experts phrase it in this discussion:

"...there are no inertial coordinate systems that are comoving with a massless particle"

That means we can't even define concepts like "rate of time flow" for a photon.

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

...we can't define a meaningful concept of "velocity of one photon relative to another photon".

... We are talking about classical point particles whose world lines are null geodesics in Minkowski spacetime

... I would say that SR says there is no such thing as "the rest frame of a photon",

You can decide for yourself if you think those are conclusive answers to your original question about time for a photon. Such answers, which I think ARE completely accurate within GR, make me SUSPECT we have more to learn. They just seem inadequate to me. I say "we can do better.' Now if quantum theory offered more precise answers I'd be more comfortable...but that is another bag of worms worse than this one!

Another way to express my concern is that I think most of these posters would agree QM and GR have some problems at what we call singularities, apparent infinities...like the center of a black hole and at the big bang. Most probably don't think we have the full answers at those points; my question is whether we should consider that maybe we don't have a full understanding at v = c.

You can decide for yourself what you make of all this.
 
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  • #65
Naty1 said:
Another way to express my concern is that I think most of these posters would agree QM and GR have some problems at what we call singularities, apparent infinities...like the center of a black hole and at the big bang. Most probably don't think we have the full answers at those points; my question is whether we should consider that maybe we don't have a full understanding at v = c.

I'm not sure exactly what aspects of QM you're referring to, but the singularities at the center of a black hole and at the Big Bang have nothing to do with any understanding (or lack thereof) at "v = c".

IMO we understand perfectly well what happens at v = c; the behavior of null curves, and how it differs from the behavior of timelike curves, is well understood. The fact that it's difficult to describe this behavior to lay people in English is because English is not well suited to describing physics, not because the physics is not well understood.
 
  • #66
dm4b said:
(put in laymen's terms, perhaps)

Exactly: that's the point. Putting things in laymen's terms distorts them.

dm4b said:
There was a show on the science channel that said as much too.

Which show? I'd be interested to see if it is on the list of "usual suspects" that tend to generate these PF threads. :wink:

dm4b said:
when you analyze the time dilation formula in the limit of v=c, while the equations are still valid, as you get infinitesimally close to v=c, deltaT moves infinitesimally close to zero.

The Lorentz transformation is *not* valid at v = c; the factor that goes to zero is in the denominator, and you can't divide by zero.
 
  • #67
I'm not sure exactly what aspects of QM you're referring to, but the singularities at the center of a black hole and at the Big Bang have nothing to do with any understanding (or lack thereof) at "v = c".

not yet! [I have high hopes!]

You seem to think relativity is more complete than I..you may be right. 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...

Perhaps I have an inflated hope for science?
 
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  • #68
Naty1 said:
You seem to think relativity is more complete than I..you may be right.

I think my comments were more in the nature of clarifying exactly where the incompleteness is.

Naty1 said:
We haven't any experimental evidence I can think of at either [the 'infinities',

True.

Naty1 said:
nor at v= c]

Here I disagree: we've studied the behavior of light in great detail. That counts as evidence of "v = c".
 
  • #69
PeterDonis said:
Exactly: that's the point. Putting things in laymen's terms distorts them.

I think it boils down to the same thing. Massless particles do not sense the passage of time, or however else one may prefer to say that.
PeterDonis said:
The Lorentz transformation is *not* valid at v = c; the factor that goes to zero is in the denominator, and you can't divide by zero.
Um, that's why I said analyzed in the limit as v=c. Perhaps better wording would have been as v goes to c.
 
  • #70
dm4b said:
I think it boils down to the same thing. Massless particles do not sense the passage of time, or however else one may prefer to say that.

But how one prefers to say it has a huge effect on what inferences lay people draw from it. Say that massless particles are fundamentally different physically from massive ones, so the concept of "passage of time" doesn't even apply to massless particles, and you get questions about why that is, which leads to a fruitful discussion about the behavior of timelike vs. null vectors or worldlines and the way that Lorentz transformations separately take each of those subspaces of Minkowski spacetime into itself.

But say that massless particles do not sense the passage of time, and you get interminable threads about how this means photons don't move in time at all, only in space, how a photon can see the entire Universe all at once, etc., etc., leading to all sorts of further inferences that are just false. Then you have to patiently go back and explain how, when you said massless particles do not sense the passage of time, you didn't really mean that, but something else.

dm4b said:
Um, that's why I said analyzed in the limit as v=c. Perhaps better wording would have been as v goes to c.

But that doesn't cover the case v = c, only v < c but getting closer and closer. Also, the statement as you gave it is frame-dependent: an object can be moving at v = .9999999999999999c in one frame but be at rest in another, and its "deltaT" changes in concert with that. But an object that is moving at v = c in one frame is moving at v = c in every frame. The two kinds of objects (timelike vs. lightlike) are fundamentally different.
 

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