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Why photons are timeless?

  1. Mar 11, 2015 #1
    As we know light travels very very large distances from far far away galaxy reaches our eyes, why it is timeless, I mean particles like muon disintegrates in time but photon does not why?
     
  2. jcsd
  3. Mar 11, 2015 #2

    BiGyElLoWhAt

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    Because Relativity =P
     
  4. Mar 11, 2015 #3

    BiGyElLoWhAt

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    If something moves at the speed of light, it has infinite time dilation, and doesn't see time.
     
  5. Mar 11, 2015 #4

    mathman

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    Some things are stable (photons, electrons, neutrinos}. Some things are not (muons, tauons, neutrons).
     
  6. Mar 11, 2015 #5
    I've been thinking about this.
    So, yes, that. But also there is length contraction. So despite the particle standing still in time, it has no distance to travel... I may be misunderstanding this horribly.
    Am I?
    Please correct me if I am. Thanks!
     
  7. Mar 11, 2015 #6

    sophiecentaur

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    The problem is that, when you are trying to take your reference frame as one that is travelling at c (relative to what?), any conclusions that your reasoning might produce are likely to be suspect.
    Trying to characterise a quantum particle in classical terms is doomed, I think.
     
  8. Mar 11, 2015 #7
    Ooo, good point.
     
  9. Mar 12, 2015 #8

    sophiecentaur

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    HAHA. No one ever said this was going to be easy. :smile:
     
  10. Mar 14, 2015 #9
    Nothing to disintegrate into!
     
  11. Mar 14, 2015 #10

    PeterDonis

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    Short answer: because there's no other particle a photon can decay into.

    Somewhat longer answer: for a particle to decay into something else, it has to have at least as much energy as the something else it's going to decay into, and the decay process must conserve quantum numbers like electric charge. For example, muons decay into electrons, which are lighter (less massive, hence less energy required to make one) than muons. But electrons can't decay into anything else, because there's nothing lighter than an electron that also has a charge of -1.

    So for a photon to decay into something else, the something else would have to have less energy than the photon. But unless the photon is extremely energetic, such as a gamma ray from a nuclear reaction, there won't be anything else with less energy that it can decay into. Sufficiently energetic photons, like those produced in reactors or particle accelerators, can in fact "decay" into electron-positron pairs (it has to be a pair so that charge is conserved). But photons of visible light have far too little energy to decay into anything else.

    That's not really correct. The correct statement is that the concept of "proper time" does not apply to a photon, or any other particle with zero invariant mass. But there are still distinct events on a photon's worldline; in other words, photons can still have things "happen" to them.

    This is not nearly strong enough. The correct statement is that the concept of "a reference frame traveling at c" is meaningless; there is no such thing. So any reasoning based on such a concept is not just "likely to be suspect"; it's invalid, period.
     
  12. Mar 15, 2015 #11

    DrGreg

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    I'm a little confused by this explanation, because the energy of a photon is observer-dependent. What one observer describes as a high-energy gamma ray could be described by another observer as a low-energy ultra-low frequency radio wave (due to doppler shift). When you say "decay", do you really mean "collide with another photon"?
     
  13. Mar 15, 2015 #12
    Mass implies inertia for motion - since the rest mass of photon is zero - it does not have inertia
     
  14. Mar 15, 2015 #13
    I was also confused by this part. How can we discriminate between low and high energy photons when talking about pair production when being low or high energy depends on our frame of reference? Is it the energy of the photon in the frame of reference in which it was emitted that actually matters?
     
  15. Mar 15, 2015 #14

    Matterwave

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    I think Peter must have meant a photon collision with another photon can produce electron/positron pairs. With one photon, there's really no way to conserve momentum in every frame. In the COM frame of the electron-positron pair there is no momentum afterwards, but the photon surely had SOME momentum before the "decay" in this frame.
     
  16. Mar 15, 2015 #15

    Nugatory

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    Single-photon pair production reactions require a heavy nucleus as well - otherwise there's no way of conserving momentum. The reaction is nucleus plus photon in, nucleus plus electron plus antielectron out. You can choose a reference frame in which the photon is redshifted down to some very small energy level and the nucleus is moving with high energy, or a frame in which the nucleus is at rest and the photon is more energetic, or anything in between. If there's enough energy for the reaction to happen in the frame in which the nucleus is at rest, then there will be enough energy to allow the reaction to happen in any of the frames as well.
     
  17. Mar 15, 2015 #16

    PeterDonis

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    Yes, it is. Matterwave is correct, I was describing how things look in the center of mass frame of the electron-positron pair that gets produced. Also, as Nugatory pointed out, a single photon can't produce an electron-positron pair in free space, there has to be something else present to allow momentum conservation. However, a pair of photons can produce an electron-positron pair in free space, and the description I was giving would be correct in the center of momentum frame of the system (which is the frame in which the photons have zero net momentum before the collision, and the electron-positron pair has zero net momentum after the collision).
     
  18. Mar 15, 2015 #17

    PeterDonis

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    This isn't quite correct. It's true that you can't accelerate a photon the way you accelerate an object with nonzero invariant mass, so you can't measure the inertia of a single photon in the ordinary way. However, it is perfectly possible for a system containing multiple photons to have a nonzero invariant mass, and the energy of the photons will contribute to the inertia of such a system.
     
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