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I am not very familiar with the quirkiness of relativity, and I was wondering how to explain this situation.

If a beam of light is shining at some object at some distance from the origin of the beam, does relativity imply that photons in the beam lack time flow and, thereby, never actually experience contact with the object? Wouldn't the photon observe no time passing in the surroundings, immediately at the source of the beam? Could it tell it was moving?

However, we see it make contact, so does that mean that what is observed might never actually ever happen in another reference frame?

Also, when we see the beam of light traveling at c, how can we record the beam's movement? Shouldn't the beam's photons appear to be in some timeless state?

Qmechanic
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John McEnroe
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2 Answers2

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The crucial point here is, that a reference frame where photons are at rest simply cannot be defined in special relativity: There is no Lorentz transformation which transforms you from a given inertial system into a reference frame or a space-time where some photon is at rest.

This, however, means that concepts such as travelled distance and proper time do not make sense when observed from a photon. Even the words "observed by a photon" do not make sense: To be an observer one needs to live in a space-time, means there has to exist a reference frame one can get to via a Lorentz transformation. Since a reference frame where a photon is at rest is not definable within the framework of special (and general) relativity, there is no point in discussing how the world looks like as seen from a photon.

No wonder that nobody can travel at the speed of light. He kind of would have to leave the range of possible space-times to do it :-).

On the other hand, you still can check what happens, when you approach a photon from a given inertial system via a Lorentztransformation: Guess you have a series of Lorentztransformations which bring you nearer and nearer to a fictious "rest system" of the photon. How do events tend to transform within this series?

Events which lie on the route of the photon all approach one point in space-time: This means, space distance and time distance between these events approaches zero. The event of creation of the photon and the event of its absorption (its contact, as you phrase it) thus tend to fall together on the same point in space-time. The other events, which do not lie on the route of the photon) tend to wander away to the infinite future or past (in such a way that the distance in space and the distance in time approaches infinity while the space-time distance stays the same as required for Lorentztransformations).

So, it is mathematically not very clean, but since people are not very satisfied when you simply tell them, things are not defined, lets take a step further and just dare to figure out how such a "photon spacetime" possibly would look like when we consider the series of Lorentztransformations described above in the limit:

It would consist out of events which happen in the here and now and out of events which happen in the infinite future and past infinitely far away. It does not really make sense to travel or to move in such a world, does it?

willyW
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Photons are not considered as observers which would be able to observe their proper time. But if you would do so, their hypothetical proper time would be zero. You obtain this result by multiplying the time observed by any real observer with reciprocal gamma (which is 0 for v=c), see http://en.wikipedia.org/wiki/Time_dilation.

However, you may not forget that not only their hypothetical proper time, but also the hypothetical proper distance of their path would be zero, see http://en.wikipedia.org/wiki/Length_contraction

The result is, that from the hypothetical point of view of photons there would be no movement at all: Time and path would be reduced to zero, and you may calculate the speed dividing 0 km by 0 sec. (the proper distance divided by proper time), finding that speed would be not defined. This seems logic because if you observe a running car for zero seconds you cannot determine its speed.

Moonraker
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  • "if something is at rest you cannot measure any speed". Sure you can, it's speed is 0. Maybe you could say, "if something doesn't have proper time". – rodrigo Aug 13 '14 at 07:42
  • @rodrigo: You're right, I am not satisfied with my formulation. I edited. – Moonraker Aug 13 '14 at 07:45
  • Think twice, you use the same equation for calculating proper time and proper distance. Or to put it differently, proper distance can only be a distance which is not in the direction of movement of the observer. There is thus no meaning to a proper length of a time-like path. – Void Aug 13 '14 at 08:05
  • @Void: No, reciprocal gamma is the common factor for calculating proper time and the proper distance. Length contraction happens only in longitudinal direction of the movement. – Moonraker Aug 13 '14 at 08:57
  • @Moonraker Okay, one wrong statement, ma' bad. Still, write down the explicit integral formula for the proper length of a general (not-linear) path in terms of the four-velocity and the metric. Now write down a similar integral formula for the proper time of a general path. Consider a time-like path. Compare. – Void Aug 14 '14 at 10:49