How does a photon experience time at light speed?

[Demystifier]

This isn't a clock made out of light, it's a clock made out of light and matter.

That is true. But the point is that light might be able to perceive the number of its own breaks. Therefore, it might correspond to time perceived by light itself.

Besides, most real clocks are not made of matter, they are made of matter and light. But this does not make such clocks inappropriate.

 

[DaveC426913]

I can picture a photon in that way, but I can also picture a slower-than-light particle in that way. So if your argument were correct, then it would imply that a slower-than-light particle also does not have an experience of time. Consequently, your argument is not viable.

Viewing the universe this way shows that both are possible (particles that do experience time as well as particles that do not are both representable).

I'm simply showing how an object that we see seeming to experience events in time not not necessarily equate to the object actually experiencing events in time.

 

[DaveC426913]

Light can "experience" the number of breaks on the zig-zag trajectory, which is an invariant quantity. Thus, the number of breaks can be interpreted as time "experienced" by light.

Sorry. Light does not experience these zig-zags as time any more than a spring experiences its coils as an oscillation. The light's path is static, occurring in zero time.

One way to reconcile this is that it is we who have an inherent limitation to experience one of the four dimensions only a slice at a time.

 

[bobc2]

If a photon possessed consciousness, does it possesses knowledge of ALL its experiences--past, present, and future--simultaneously? And futhermore, what does it "see" as it whizzes through space?

I'm not sure whether this sequence of spacetime diagrams for observers with increasing speeds will help any at all (someone could explain spacetime diagrams if you need that).

 

[GrayGhost]

First, you are absolutely right that proper time along the photon trajectory is zero. Yet, proper time is not the only meaningful definition of "time". Therefore, we don't know whether "time" experienced by the photon is equal to the proper time.

Yes we do. By definition, the motion of a clock's hands is what signifies the passage of time per the clock. If the hands do not move, then the clock experiences no passage of time.

Second, when one says that photon "cannot be held in a state of rest", one actually means that there are no Lorentz coordinates with respect to which the photon is at rest. Yet, Lorentz coordinates are not the only meaningful type of coordinates. In particular, there exist light-cone coordinates, with respect to which photon is at rest.

If you never heard about light-cone coordinates, see http://en.wikipedia.org/wiki/Light_cone_coordinates

Well, I would say that "it also means" that there are no Lorentz coordinates with respect to which the photon is at rest. When one says a photon cannot be held in a state of rest, it means ... it cannot be stationary per any inertial POV, whatever. What we do know is that all inertial POVs experience the passage of time, that which we call proper time. Wrt the photon's POV, we can only predict what must be the case in that respect, per the LTs ... and that is, the photon exists at all points along its propagational path AT-ONCE within the cosmos per itself. The photon travels a finite distance in zero time (per material entity), and so a speed c plays the role of an infinite velocity in the special theory. It simply cannot experience the passage of time. I'd add that the light-cone coordinates do not suggest that time passes per the photon. It suggests only that the photon is mappable in spacetime.

GrayGhost

 

[DaveC426913]

Yes we do. By definition, the motion of a clock's hands is what signifies the passage of time per the clock. If the hands do not move, then the clock experiences no passage of time.

Or the batteries are dead...

 

[zebing]

Let's take a particle emitted from somewhere far away that travels at 99.999% of the speed of light and that is absorbed by the eyeball of a human observer on earth. If the particle was emitted in what the observer calls Jan 1, 1960 and absorbed by his eyeball in what he thinks is 2008, the particle will think that everything in the universe is almost frozen in Jan 1, 1960 ("almost frozen" because its speed is slightly less than the speed of light). If the particle's velocity is constant (i.e. no acceleration), instead of seeing the human that its about to bash into, it'll see whatever was occupying the point in space where the observer's standing back in Jan 1, 1960 (of course, once it actually bashes into the guy's eyeball, it will decelerate to zero very quickly, thus screwing up the picture).

I like that this analogy was at least attempted, but it has a flaw in its logic.

I've found that most apparent contradictions relating to the theory of relativity relate to issues of simultaneity. The first thing to understand is that two events that may appear to be simultaneous to a stationary observer will not appear to be simultaneous to someone moving near the speed of light. Think of the classic flashing lights on the train thought experiment. Say I were traveling on a train near the speed of light. A light flashes in front of the train and behind the train. To a stationary observer, the flashes appear simultaneous, but to me, the event that occurs in the direction of my motion appears to happen first. Google up the pole and the barn door paradox for another great example.

With a traveling photon, it's important to remember that events in its direction of motion may appear simultaneous to the photon which are not simultaneous to us. For light, it is a 1:1 ratio. In other words, an event that happens one year in the future from our point of view, but which is one light year away (remember that a light year is a measurement of distance) is simultaneous from the photon's point of view if the photon is traveling from where we are and toward that event which is in our future.

You said this: "If the particle's velocity is constant (i.e. no acceleration), instead of seeing the human that its about to bash into, it'll see whatever was occupying the point in space where the observer's standing back in Jan 1, 1960 (of course, once it actually bashes into the guy's eyeball, it will decelerate to zero very quickly, thus screwing up the picture)."

This statement actually isn't true. From a photon's point of view (I'm just going to go with something that travels the full speed of light here), the moment it is emitted and the moment it collides with a human eyeball are the very same moment, even though the two moments may appear to be separated by several decades from our point of view.

This makes perfect sense, because as an object with mass approaches the speed of light, the universe contracts along it's line of motion, and that contraction approaches zero as the speed approaches the speed of light. For a photon moving at the speed of light, the distance to travel between where it was emitted and where it collides with your eyeball is zero. How long does it take to travel zero distance? It takes zero time.

That makes sense because the speed of light is constant for all observers, as stipulated by the theory of relativity. If speed or velocity is expressed as v=d/t, and the distance is zero, then time must also decrease to zero in order for the velocity to remain constant.

So from the photon's point of view, it travels zero distance, and zero time elapses. This is very different from saying that time appears to stop from a photon's point of view when it travels at the speed of light. Remember, there is no time. This is hard to conceptualize in our three-dimensional world that is bound by time, but light operates under different rules than we do. Rules that we can't quite visualize in our sphere.

From our point of view, the photon is very small, traveling very fast through space, and eventually collides with something. From it's point of view, no time or space separate it's emission and its collision.

An interesting thought is to imagine a photon that travels through space and then reflects off of a mirror. It travels no time and through no space to hit the mirror. The sequence of events still remains intact, however, because it changes directions upon hitting the mirror and the universe is then contracted in the line of its new motion.

 

[DrAL]

While I don't quite follow all of the nuances of this discussion (I am not a physicist and I only did a quick read), I am convinced that the issue is valid and extremely important. Why I feel this way is a long story. I do believe that this issue directly impacts in a big way some cutting edge findings in another field of research.

I once heard that if a photon were conscious, it would experience being "everywhere at the same time". A physicist once told me that was silly because a conscious photon would would experience the world (i.e. time/space) just as we do, except the rest of the universe would be wizzing by at c. Well, OK, but what difference does that make? How would anything/anybody experience a world wizzing by at c? What's the difference?

While I am not a physicist, I am rather well versed in Relativity Theory. I understand that the Bern clock would appear to stop to Einstein if his bus was traveling directly away from it at the speed of light. But, how would people on the street (at an angle) appear to Einstein? And, how would someone on the street experience the bus? Please help. Thanks.

 

[pervect]

If you consider the worldline of piece of matter - let's assume it's a piece of matter in free fall for the moment - you can make regularly repeating marks on the worldline (for instance , some repetitive cycle), and the marks are separated by time - or more formally, a timelike interval. So you can define a proper time between said intervals on the worldline of a piece of matter in free fall.

Now consider the worldline of a photon, which we will also consider to be in free fall, furthermore won't be using any of the quantum properties, so we might better call it an electromagnetic wave, except I'm too lazy to edit all this...

We can also make regularly repeating marks along this worldline. For instance, an electromagnetic wave might have points where the electromagnetic field is entirely electric followed by points where the field is entirely magnetic - and these points occur at regular intervals along the line of propagation of the wave.

However, these regularly repeating intervals are not timelike, you cannot assign a "period" or "proper time" to them. They are null intervals, of zero length - different observers will agree that the equally spaced marks are equally spaced, but they'll disagree about the magnitude of the interval. All will agree that the spatial separation is c times the time separation, but the magnitude of the separations can have any value whatsoever, depending on the observer. For instance, a photon that is part of aradio wave with a wavelength measured in meters for one obsever, might be a gamma ray with a wavelength measured in angstroms for another. They will all agree that the photon has _some_ wavelength, but not on its value.

There is a word to describe the sort of geometry that's described when you can measure equal intervals, but can't assign a length to said intervals. This sort of geometry is known as "affine geometry".

"Marking equal intervals" along a freely-falling worldline is most usually called "an affine parameterization of geodesics". The different affine parameterizations are related by affine transformations.

These are usueful keywords for finding more information about the subject, for instance
http://en.wikipedia.org/w/index.php?title=Affine_geometry&oldid=512480357
http://en.wikipedia.org/w/index.php?title=Geodesics_in_general_relativity&oldid=499720552

 

[Mike Holland]

Try going halfway to c to illustrate this. You are coming home from Alpha Centuri, 4 LY away. You travel at 0.87c (I think that is the figure) and get 50% time dilation from our point of view. So we see you take 4/0.87 years for the journey, but your clock only advances half this, 2/0.87 years.

From your point of view, the distance from Alpha Centuri to here is halved by Lorentz contraction to 2 LY, so at 0.87c you take 2/0.87 years to get here.

Time dilation from our point of view, Lorentz contraction from your point of view, and we both see your clock giving the same reading when you get here.

Now extend this example to approach light speed. Then photon time -> zero from our point of view, and distance traveled -> zero from the photons point of view.

The only problem I have is, how does a photon have a frequency when it is frozen in time? Or a wavelength for that matter. But I don't plan to lose any sleep over it.

Mike

 

[harrylin]

[..]
The only problem I have is, how does a photon have a frequency when it is frozen in time? Or a wavelength for that matter. But I don't plan to lose any sleep over it.

Mike

No problem there: traveling with a wave, you stay next to a wave front. Effectively the photon vanishes.

In such a reference frame (which can only be approached for c-light in SR but is perfectly reachable for c-sound), the observed frequency would be zero if one had a reference clock that is unaffected by the motion; and one can find* that for the speed of light the frequency goes faster to zero than that time dilation compensates for it. Einstein found a Doppler equation according to which not only, as he explained, f'=∞ for v=-c; but also f'=0 for v=+c.
- section 7 of http://www.fourmilab.ch/etexts/einstein/specrel/www/

*It's easy to show this with a simple derivation.