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Describing the Life of a Photon

  1. Apr 12, 2013 #1
    I heard Neil deGrasse Tyson talk about the life of a photon that was emitted from the Sun and hit oneself. He said that its entire life was an instant. I hope I'm not misquoting or misunderstanding what he said.

    With that in mind, does that instant take into account the thousand of years (relative to us) it spends trapped in the mantle and core of the Sun, trying to get out?

    Also, how would one describe the life of a photon that is emitted from the Sun, or any Star for that matter, and travels indefinitely throughout the Universe, never coming into contact with anything? Is this an indefinite instant?
     
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  3. Apr 12, 2013 #2

    ghwellsjr

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    A photon only exists from its point of emission to its point of contact with something so you don't have to worry about all the other photons that are trapped in the sun with regard to one that escapes.

    I don't know what Neil said but he should have made it clear that time does not apply to photons instead of implying that time for a photon is zero.
     
  4. Apr 12, 2013 #3

    PeterDonis

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    I don't think you're misquoting or misunderstanding; but I do think Tyson is stating things in a way that is likely to cause confusion. (He's not the only one, btw: we often get threads here on topics like this after one of Brian Greene's specials airs on PBS.)

    The correct way to say what Tyson was saying is that a photon's worldline has zero length. In the simplest case, where gravity is negligible so we can use the formulas of special relativity, the length of a worldline is given by the Minkowski metric:

    [tex]\tau^2 = c^2 t^2 - x^2[/tex]

    which is basically a spacetime version of the Pythagorean theorem: [itex]t[/itex] is the travel time of the photon in some inertial frame, and [itex]x[/itex] is the distance the photon covers in that time in the same frame. Zero length for a photon's worldline means [itex]\tau^2 = 0[/itex], which just means that [itex]ct = x[/itex]: the photon moves at the speed of light.

    Really what this is saying is that objects that move at the speed of light, like photons, are fundamentally different, physically, from objects like us, that move slower than light, and for which [itex]\tau^2 > 0[/itex]. In that case, of objects moving slower than light, [itex]\tau[/itex] is just the time elapsed on a clock that moves with the object. By analogy, then, pop science discussions often say that "no time elapses for a photon" because [itex]\tau = 0[/itex] for a photon.

    However, the analogy is flawed because [itex]\tau[/itex] has another meaning for objects that move slower than light (we call the worldlines these objects move on "timelike", by contrast with "null" worldlines that photons move on): each event on a timelike worldline can be labeled by a unique value of [itex]\tau[/itex] (which we normally refer to as the "proper time" of the event according to the observer following the worldline). Interpreting [itex]\tau = 0[/itex] for a photon as meaning the photon "only takes an instant" to go from, say, the Sun to the Earth, would imply that the photon's worldline is not a line but a single point--a single event. Obviously it's not, because we could put a photon detector anywhere between the Sun and the Earth and detect the photon, and each possible location for the detector must be a distinct point on the photon's worldline. So [itex]\tau[/itex] for the photon simply doesn't mean "time" the way it does for a timelike object.

    No; that's a whole separate discussion. The "instant" only refers to the photon traveling from the surface of the Sun to Earth, through vacuum.

    Not really. The condition [itex]\tau = 0[/itex] for the photon's worldline still holds, but that doesn't make it an "instant". See above.
     
  5. Apr 12, 2013 #4

    A.T.

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    I see no problem with "no time elapses for a photon", because it indicates that proper time is meant. I agree that "only takes an instant" is misleading, because it leads to confusion with coordinate time.
     
  6. Apr 12, 2013 #5

    PeterDonis

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    This is fine as long as people don't immediately infer "only takes an instant" from "no time elapses". However, it seems like a lot of people do.
     
  7. Apr 12, 2013 #6

    Dale

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    I still have a problem with that because proper time is only defined along timelike worldlines. I would agree that the spacetime interval is 0, but I wouldn't call that 0 proper time.
     
  8. Apr 12, 2013 #7
    I don't see how this makes sense. Isn't the photon created by nuclear fusion in the core, and thus exists?
     
  9. Apr 12, 2013 #8

    A.T.

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    What's the downside of calling it proper time?
     
  10. Apr 12, 2013 #9
    May I ask about the origin of photons from the sun's nuclear furnace?

    Where do they come from under the standard model?
     
  11. Apr 12, 2013 #10

    Bill_K

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    The photon that escapes from the sun's surface is not the same one as was produced in the core. Photons do not bounce, they only travel in a straight line at velocity c until they hit something. At each collision, one photon A is absorbed and another photon B is created.
     
  12. Apr 12, 2013 #11

    Bill_K

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    The adjective "proper" means "in the particle's own rest frame."
     
  13. Apr 12, 2013 #12
    So, is there math to describe the frame of the photon or not?

    If not why not?
     
  14. Apr 12, 2013 #13

    DrGreg

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    "Frame" usually means "inertial frame", in which case the answer is no: a photon is not at rest in any inertial frame. See our FAQ: Rest frame of a photon.
     
  15. Apr 12, 2013 #14

    Yes, I could not agree with you more.

    A photon does not occupy an inertial frame.

    Now, I did not imply this and am back to my question.

    So, is there math to describe the frame of the photon or not?

    If not why not?

    The frame would be that of the photon, not an inertial frame.
     
  16. Apr 13, 2013 #15

    Nugatory

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    There is no math to describe the frame of a photon because "the frame of <something>" is just a convenient verbal shortcut for "a frame in which that <something> is at rest", and there is no such thing when the <something> is a photon.

    Inertial or not is beside the point; a frame in which something is at rest is not necessarily an inertial frame.
     
  17. Apr 13, 2013 #16

    sophiecentaur

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    This assumes that it has to be the 'same photon' throughout the journey. If you think in terms of the energy propagating from inside the Sun to its surface as a series of interactions between ions (nuclei?) on the way through, you are really talking in terms of a whole series of photons, for each step in the journey. Energy finally arrives at the surface, is emitted, in quanta (photons) which have no individual identity, until the energy arrives at Earth, where a further photon interaction occurs when you 'see' the light. Photons don't actually need to 'exist' at all, on the way - just at each end, with the energy existing in the form of waves in between. The 'little bullet' model is not a good one.
     
  18. Apr 13, 2013 #17

    Dale

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    There are several downsides. First, why should it be proper time and not proper distance? Second, proper time is the time measured by a clock traveling along that worldline and no clock can travel that worldline. Third, a timelike worldline can be parameterized by proper time, but a null worldline cannot, you have to introduce an affine parameter instead. Fourth, for geodesics in flat spacetime the proper time is the coordinate time in the reference frame where the starting and ending events are co-located, there is no such frame for light.
     
  19. Apr 13, 2013 #18

    Dale

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    They come from conservation of energy and momentum in many of the nuclear reactions that take place there.
     
  20. Apr 13, 2013 #19

    Dale

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    There is a subtle distinction between a frame and a coordinate system. You can make a coordinate system where a pulse of light is at rest, but he coordinate basis of such a coordinate system is not orthonormal, so it doesn't form a valid reference frame. Frames must, by definition, be orthonormal, but a pulse of light is null.
     
  21. Apr 13, 2013 #20
    Many thanks for all the responses. Much appreciated.
     
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