10
$\begingroup$

It is often said that photons do not experience time. From what I've read, this is because that when travelling at the speed of light, space is contracted to infinity, so while there is no time to cover any distance, there isn't any distance to cover.

But the fact remains that as the universe expands, the photon's wavelength stretches as well. So from everyone's else perspective, that photon's wavelength is gradually changing.. But since photons don't experience time, how do they account for that change in their own wavelength?

I mean, the photon should exist for at least one plank-time, right? Otherwise it wouldn't really exist, and we couldn't detect it. (I'm assuming things here. Please correct me if I'm wrong).

So if it was "born" as a certain wavelength, and then immediately absorbed as a different wavelength, then couldn't it be said that the photon experienced time?

Also, 2 photons (from the same source) might get absorbed at different times (from our perspective), but from the photon's perspective they should experience the same amount of time (zero). Is there something going on here with different-sized infinities? How is that phenomena explained?

Thanks!

$\endgroup$
  • 12
    $\begingroup$ Photons are notoriously uninquisitive and never seem to ask themselves these questions. $\endgroup$ – WillO Jul 13 '15 at 1:11
  • 5
    $\begingroup$ The frequency of the photons is a property measured only by an observer and thus depends only on the observer's frame of reference. You can take any photon and assign it any frequency you want just by changing your frame of reference (by a boost). The gradual change in wavelength is due to your "stationaty" frame of reference and gradual curvature of the space-time. $\endgroup$ – Alexander Jul 13 '15 at 1:47
  • 1
    $\begingroup$ @Alexander you should make that an answer. $\endgroup$ – ragnar Jul 13 '15 at 14:38
  • 1
    $\begingroup$ I've deleted some comments that contained personal aspersions and were degenerating into a back-n-forth argument. $\endgroup$ – dmckee Jul 13 '15 at 15:26
8
$\begingroup$

We don't really have a good perspective on what a photon "feels" or, indeed, anything about what its universe would look like. We're massive objects; even the idea of "we must travel at the speed of light because we're massless" makes little sense to us. But we can talk, if you like, about what the world looks like as you travel faster and faster: it's just that obviously that doesn't tell us truly what happens "at that point" of masslessness.

One thing that happens, as you go faster and faster, is that everyone else sees your clocks ticking slower and slower. This is the basis for the statement that photons don't "experience time." It's a little more complicated than that: suppose you are emitting light, say, as periodic "flashes": there is a standard Doppler shift which has to be corrected for before you see this "time dilation". In fact, as you get faster and faster, the flashes undergo "relativistic beaming", the intensity of the pulses will point more and more in the direction that you're going, as seen from the stationary observer.

The same effect in reverse happens for you: as you go faster and faster, the stars of the universe all "tilt" further and further into the direction you're going.

By these extrapolations, in some sense a photon experiences no time as seen from the outside world. But in another sense: if the photon had any way to communicate to the rest of the world, it could only communicate to the thing that it's going to hit anyway, and no faster than it itself can travel there. So in some sense it simply "can't" communicate its own state at all.

So a key lesson, I guess, is that we have to think of the particle's frequency as interactive: in some sense the photon's energy that gives it a frequency $f = E / h$ where $h$ is Planck's constant, but in another sense it is changeless, it's not "oscillating."

Quantum electrodynamics actually reifies this notion (makes the idea "solid" in the mathematics) pretty well: the photon's frequency lives in its complex phase, but a quantum system's overall phase factor is not internally observable and can only be observed by its interaction with an outside system with a different phase. In turn, you only observe their phase difference; there is a remaining overall phase for the interacting system which becomes unobservable, and so forth.

$\endgroup$
2
$\begingroup$

In Minkowski spacetime, the spacetime interval of lightlike movements is zero. That means, from the (hypothetical) point of view of a massless particle such as a photon, it does not even exist one Planck time.

At a proper time zero, any wavelength becomes meaningless, even if the physical process is the same that we observe. For the answer you have to take into consideration cause and effect: The photon including all its features has been created by an electron at A, and features have been transmitted to an electron at B. If from the (hypothetical) point of view of the photon it is vanishing, you have to refer to what is remaining, i.e. the electrons A and B.

Electron A is transmitting a momentum to electron B, and (due to space expansion) electron B is receiving a diminished momentum. From the (hypothetical) point of view of the photon, between A and B there is nothing (spacetime interval = zero).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.