Photons in vacuum have no proper time, and they are not considered as observers and not as reference frame. But what about photons travelling through matter? Their velocity is lower than light speed, thus the proper time formula (multiplication with reciprocal gamma) would yield a proper time, but do we consider that they have a proper time/ that they are observers/ that they may be reference frames?

  • $\begingroup$ You surely don't mean a pure state in the sense of an element of the Hilbert space, do you? Every physical system may be in a pure state. A pure state is the right description whenever our information about the system is maximized one allowed by the uncertainty principle. $\endgroup$ Aug 13, 2014 at 6:52
  • $\begingroup$ Sorry yes, careless commenting! $\endgroup$ Aug 13, 2014 at 6:55
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    $\begingroup$ See this answer. In a medium photons are no longer just photons. The interaction with the medium means they need to be described by a wavefunction covering both the photon and the medium. This combined state has a mass and a non-zero proper time. $\endgroup$ Aug 13, 2014 at 6:56

1 Answer 1


As my answer to How do photons know they can or can't excite electrons in atoms? explains, when a photon interacts with matter it is no longer just a photon. The photon/matter system has to be described by a wavefunction that describes both and that isn't separable into a photon bit and a matter bit.

When the interaction is strong, e.g. in Bose-Einstein condensates, the interacting system is described as a quasiparticle called a polariton. For light passing through glass the interaction isn't strong enough for this to be a useful description, but the idea remains the same. The photon/solid system now has a mass and propagates at less than $c$, so it does have a non-zero proper time.

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    $\begingroup$ A great answer, simply incredible! Before validation I wait if there are other opinions. Your link to yesterday's question is very useful. Italian Wikipedia "photon" (it.wikipedia.org/wiki/Fotone#Fotoni_nella_materia) goes to the same sense. $\endgroup$
    – Moonraker
    Aug 13, 2014 at 7:37

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