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Naively, one can attempt to consider the (impossible) light-speed inertial frame. From there you arrive at nonsense conclusions like 'the universe is flattened in the direction of travel' which must 'occur in 0 time' as the former implies '0 travel distance'.

Confusion arises with respect to polarization - I'm not familiar with the details but as I understand it a photon / electromagnetic wave is the solution one arrives at when considering a situation where changes in the electric and magnetic fields influence eachother, the interaction itself propagating as a result in an effective direction with some energy carried (frequency of the wave oscillations). The relation between E and B fields entirely determines the polarization nature of such waves, as I understand it.

Put simply - For any change to occur, time is required; EM waves such as photons require coupled, propagating changes in respective EM fields, which implies that this occurs over some distance. The reasoning being that to note changes in a given space, time must pass, and to note changes over a given time, space must change - governing this is the speed of light, a local ratio of temporal to spatial distances.

Given that a photon is, for lack of better terms, a chain reaction of events in the EM fields, why do we assume that:

  1. A photon will never decay?
  2. A photon does not 'experience' time generally?

This isn't to contest that a photon is (as far as is known) stable so much as to understand how times and distances 'work' for massless propagating entities, and the temporal / casual nature of such propagations.

If these aren't actually assumptions in use today, that hasn't been made very clear to me before. However if one assumes #2 to be true, it would necessitate #1 to be true (no time experienced = no internal changes), so I feel this needs clarification.

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  • $\begingroup$ The "inertial system" of any one photon is degenerate: a photon can not go "sideways", if you like, it can only "see" what's ahead of its path and all of that is "the same" space-time event (if we forget, for a moment, that photons still do count time by counting wavelengths, which is an invariant) for that one photon. But take four photons that have linear independent wave-vectors and you have a unique way of characterizing space-time events. That's exactly what the GPS system does. $\endgroup$
    – CuriousOne
    Commented Jan 2, 2015 at 5:35
  • $\begingroup$ @CuriousOne: I'm a little confused by your use of 'degenerate' (new to QM terminology), could you please clarify? $\endgroup$
    – Xeren Narcy
    Commented Jan 5, 2015 at 0:07
  • $\begingroup$ Oh, I simply meant it in a geometric way. A photon in our picture is a particle moving at the speed of light in every coordinate system (that's wrong for the purposes of QM but suffices here). The only possible way for that is a straight line. If we wanted the photon to deviate from that straight line it would have to violate the speed of light limit. Classical photons don't do that, so they are limited to "rays". One ray can't select a particular spacetime point, but four can. $\endgroup$
    – CuriousOne
    Commented Jan 5, 2015 at 0:13
  • $\begingroup$ Ah, makes sense. And regarding photons 'counting' wavelengths, is it the count or wavelength (or both) that you meant is invariant? $\endgroup$
    – Xeren Narcy
    Commented Jan 5, 2015 at 0:20
  • $\begingroup$ The wavelength depends on the observer but the number of wavelengths that fits into a given physical length is always the same, otherwise one observer would see e.g. ten fringes in an interference experiment, another would see fifteen. That doesn't happen, they both see ten fringes, even though to one observer the color of the light may be red and to the other it may be green. $\endgroup$
    – CuriousOne
    Commented Jan 5, 2015 at 0:25

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Photons certainly decay, they are not more and not less than a momentum that is transmitted from one place to another. Photons do not experience any proper time, and from the hypothetic point of view of particles moving at c length contraction is reducing any distance to zero, and velocity is not defined (= zero meters in zero seconds, that means the photon did not move). This is the result of the analogue application of the reciprocal Lorentz factor to phenomena moving at c.

The electromagnetic waves are moving through spacetime at c, and thus they exist only for us as observers, as a product of spacetime we are living in. Reciprocal gamma of special relativity reduces them to zero. From their hypothetical point of view, photons are outside of spacetime, and their wave in spacetime is replaced by a direct momentum transmission from A to B without need of any wave outside of spacetime.

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  • $\begingroup$ I'm not too sure what you're trying to say - if photons do decay and they are exactly momentum being transmitted, isn't that violation of conservation of momentum? Could you please clarify how photons decay in your view? Also I'm not sure what you mean with your last sentence regarding anything 'outside of spacetime'? $\endgroup$
    – Xeren Narcy
    Commented Jan 5, 2015 at 0:55
  • $\begingroup$ I simply wanted to say that every photon is decaying at the moment of its absorption - certainly not on their travel. --- Spacetime is a Minkowski coordinate system. The hypothetical degenerate coordinate system of photons would be just a dimensionless point - they do not take part in spacetime (instant momentum transmission from A to B) although they are observed by any observer in spacetime in form of a wave. - Even if this "hypothetical" pointlike transmission cannot be observed by anybody, it is the real nature of the photon, the other one being mere observation. $\endgroup$
    – Moonraker
    Commented Jan 5, 2015 at 6:57
  • $\begingroup$ I can see what you mean, although I don't think decay is correct to describe absorption of a photon. To be specific, by photon decay I meant a photon breaking down or changing in some way between emission and absorption - that a photon 'propagates' (at least to an inertial observer) gives rise to questioning why a photon remains or should remain 'intact' wave-to-wave (or peak-to-peak) on its 'path'. $\endgroup$
    – Xeren Narcy
    Commented Jan 6, 2015 at 0:31
  • $\begingroup$ With view to the last sentence of your initial question this should be not a problem - and a mere observation does not concern properties of the photon but rather a property of spacetime, one of its properties being the speed limit of c. $\endgroup$
    – Moonraker
    Commented Jan 6, 2015 at 6:22

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