Does Time Exist for a Photon Traveling from Point A to Point B?

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SUMMARY

The discussion centers on the concept of time from the perspective of a photon traveling at the speed of light (c). It is established that a photon does not experience the passage of time, as indicated by the proper time equation, which results in zero for light paths. Observers outside a photon perceive time as standing still for it, while the photon itself does not measure time or distance due to Lorentz contraction. The conversation also touches on the implications of Planck time and the distinction between Special Relativity (SR) and Quantum Electrodynamics (QED), emphasizing that the photon exists outside the conventional framework of relativity.

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kernelpenguin
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1. Time seems to have a 'smallest step' -- the Planck time unit.
2. A photon travels at c.
3. When you get closer to c, time dilates causing your passage in time to get slower compared to the rest frame.

And now...

If a photon is emitted at point A and absorbed at point B, then from the point of view of that photon, how much time has passed? An eternity? None at all? It all happened at the same time?

This one's been bugging me for a long time now.
 
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From the photon's point of view? I'd have to say no time at all. From a photon's point of view time might as well not exist.
 
A photon does not experience the passage of time, from the moment it is created to the moment it is absorbed/destroyed.

FYI, item (1) on your list is hypothetical - no one knows what happens at the Plank scale. However, that doesn't change anything. Items (2) and (3), as well as the consequence that photons don't experience time, remain valid.
 
If you travel in your rocket at speed near c, than you would experience time as going normal. However, when an observer outside the rocket looks at your clock in the rocket, time is going slower. So if you're a photon traveling at c, time passes by normally. The photon "experiences" the passage of a year when traveling the distance of a lightyear. The observer, on the outside looking at the clock of the photon, sees time standing still.
 
JV said:
If you travel in your rocket at speed near c, than you would experience time as going normal. However, when an observer outside the rocket looks at your clock in the rocket, time is going slower. So if you're a photon traveling at c, time passes by normally. The photon "experiences" the passage of a year when traveling the distance of a lightyear. The observer, on the outside looking at the clock of the photon, sees time standing still.
This is incorrect. The photon sees neither distance (all space is Lorentz contracted to zero length) nor time. Formally, this is because proper time is defined as
\tau = - \int_{x_1}^{x_2} ds
The integral is to be carried out over the path in spacetime. On paths traveled by light, this is always zero. Informally, suppose the photon could somehow keep track of time. We may ask "how long until an external observer measures any change in the photon's clock". Since that observer measures time as standing still for the photon, he concludes the clock never advances. Otherwise, there would be a contradiction within SR.
 
The - sign corresponds to (-,+,+,+) metric.
I just wanted to clarify a possible dummy source of confusion.
 
But if a photon is created and absorbed without any time passing for it in between, would it not demonstrate that the Planck time is not the smallest unit of time possible?
 
The fact that a photon is incapable of measuring any finite time doesn't mean that time may not be quantized for something that actually can measure time.
 
Actually, ds=o for the photon at all times. This is said to imply that it experiences space and time equally.

You can't really use any of the formulas for SR on the photon except for ds=0. It stands outside of the relativity formulation since its velocity is the same for all observers.

juju
 
  • #10
juju said:
It stands outside of the relativity formulation since its velocity is the same for all observers.
Is this the worst misunderstanding possible !? The photon constant speed is the basis for a need of SR. It is the fundamental observation that there is a maximum speed that requires us to quit the simple world of Galilean relativity.
 
  • #11
Yes, c is the basis of relativity because it is the same for all observers, but stands outside the formalism. c is the least upper bound for the velocity but the interval is open, not closed. Gamma goes to infinity for v=c.

juju
 
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  • #12
kernelpenguin: Your confusion might result from the fact that you're considering SR and QED (or just quantum ideas in general) simultaneously. QED doesn't actually have a 'Planck Time', it's a number that's come out of broader (but less complete) unification theories, and incidently, gets put in a lot of popular science books. Recall that in QED, photons travel all paths, and to find out the probability of it getting from A to B you add up the probabilities of it traveling along every path, and the 'expected' path (if you really want one) looks like the classical straight line between two points. Thus the notion of a photon's path (in spacetime, not just space) isn't really valid in the first place and it's a whole different kettle of fish to be asking what the hell is going on.

The 'classical' SR result is that light travels along paths of no 'proper' length in time or space - this of course might change if someone comes up with a self-consistent and usable theory of quantum gravity.

Kane
 

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