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Photon's Perspective of Time

  1. Jan 23, 2006 #1
    I'm trying to ask a simple question, which is probably a fatal mistake but...

    According to accepted Einsteinian relativity, say I'm travelling at the speed of light. I understand I can't get to that speed. Suppose I was born at that speed, I'm a photon, whatever. From my "photonic perspective" does time pass by? Is it all one big "now", so that from my perspective I'm eternal, or does time merely slow to some finite crawl?

    If time does change, even if very slowly, I can understand and have no need of the next question.

    If time from the photon's perspective does not change, then how is the perspective of the photon not eternal? In other words, photons are created and destroyed (changed) all the time. Yet if from their perspective time does not change then something does not make sense to me. A particle for which time does not change should have no beginning nor any end.

    I'm confused somewhere...

    JR
     
  2. jcsd
  3. Jan 23, 2006 #2

    pervect

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    Here is your mistake. Your are supposing impossible things, and then asking us to give you an answer.

    This leads to mathematical nonsense.

    Lewis Caroll (who was a mathemetician in real life) has a very amusing and prefectly valid mathematical proof that illustrates the perils of impossible assumptions.

    Given 2+2=5, prove that I am the Queen of England.

    proof:

    2+2=5, but 2+2=4. Therefore 5=4. Therfore 2=1.

    Now, me and the Queen are 2. But 2 = 1. Therefore me and the Queen are 1. Therfore I am the Queen of England.

    (Or was it the King? It doesn't matter, the proof works either way :-)).

    Anyway, the moral of the story is:

    Please don't assume that you can move at the speed of light. You can't.
     
  4. Jan 23, 2006 #3

    robphy

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  5. Jan 23, 2006 #4

    DaveC426913

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    The issue here is not one of whether or not he can move at the speed of light. The issue is one of whether or not a photon can "experience".


    No. Yes.

    I believe I see your point. A given photon exists only from the point where it is emitted from one atom until it is absorbed by another. That means that, even though a photon does not experience time, it does nonetheless, experience a "lifetime". While every point in its lifetime is experienced simultaneously, it is a discreet, and possibly quite small, set of points.
     
    Last edited: Jan 23, 2006
  6. Jan 23, 2006 #5

    pervect

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    It is definitely a mistake to import the baggage of frames of reference applicable to a massive observer to a photon - 3 space + 1 time does not work.

    This is a frequently asked question, and the frequently given answer is basically that it's not a good question.

    For more info, there's always the sci.physics.faq entry

    http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/headlights.html
     
  7. Feb 2, 2006 #6
    The best way of thinking about it is to say that a photon does not experience time from the instant of its creation in some interaction to the instant of its destruction in another
     
  8. Feb 2, 2006 #7
    Thank you for the responses, (the links were useful). This makes better sense to me now, and as FAQ link says, I don't at all feel stupid for asking the question.

    Typically, a good set of answers generate further questions. The link explains how when the equation for relative velocities is used about an object travelling at the speed of light a "meaningless" answer is obtained, signified by an attempt to divide by zero. Some here take that to mean the question itself is somehow meaningless.

    However there is an underlying assumption: is mathematics sufficient to contain philosophy? Though no expert I believe that question received a resounding no some time back. That is to say, mathematics does not wholly encompass methods of thinking or valid logical expressions. I believe that while Godel's theorems set a clear limit in any system that involves counting (mathematics), the much older form of logic (sometimes called "term logic") is not so afflicted.

    Thus, it may be that some questions while meaningless in mathematical terms, may in fact have a sensible construction nonetheless in other systems, such as natural languages. I find the quote above and that about a photon having no life, but nonethelss a measurable lifespan satisfying answers, and again thank the respondents.

    Put in other ways: there are valid questions, and answers, that can not be set in a mathematical framework.

    Or, abusing a cliche: "There is more in Heaven and earth than CAN be dreamt of in mathematical philosophy."

    Again admitting I'm a total neophyte, I await comments.

    JR
     
  9. Feb 2, 2006 #8

    ZapperZ

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    All you need to do is find something in physics that is devoid of an underlying mathematical description. You'll discover that there's none. So here, it is the assumption that there's something that can't be described mathematically that has no validity.

    http://nedwww.ipac.caltech.edu/level5/March02/Wigner/Wigner.html

    Zz.
     
  10. Feb 2, 2006 #9
    Well heck, if we're going to bring philosophy and reasoning into it then things can get a lot more interesting.

    Mathematically describe categorizations.

    It's a matter of opinion whether or not concepts that Math cannot describe are valuable or not. Zapper ascribes to the "not valuable" when it comes to physics. However I can argue that physics is a categorization which is somewhat subjective.

    Zero is a concept that I'm sure Zapper would find valuable but it is still a concept and does not exist in reality. That's why you can't divide by it.

    Hand me 100 nothings...

    So non-physical concepts obviously have some small place in physics as well and the usefullness of a concept is subjective. So I assert that it is not an incontrovertible fact that the question is meaningless. It is instead one of the places that philosophy and reasoning touch physics and the answer is subjective.

    I also postulate that reasoning that requires concepts that cannot be described mathematically is more valuable than math alone. I'll use the difference between computers and humans as my proof. I believe humans are better overall problem solvers than computers even though computers far exceed humans in speed of calculation and errorless math.

    :surprised :surprised :surprised :tongue2: :rolleyes:
     
  11. Feb 2, 2006 #10

    ZapperZ

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    If you think mathematics is analogous to number crunching, then you have mathematics all wrong.

    Again, just show me a physical phenomenon that defies mathematical description. If not, all of this are idle speculation without basis. Now THAT certainly defies mathematical description!

    Zz.
     
  12. Feb 2, 2006 #11

    Elegance, a concept that Einstein claimed was an underlying assumption to his theoretical musings. Please translate elegance into a mathematical formula.

    JR
     
  13. Feb 2, 2006 #12

    ZapperZ

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    Elegance is not a physical concept. It's a quality that human gives, like "beauty", something Einstein also described as what a physical equation should be.

    Furthermore, ask 10 people the meaning of elegance, and you get 10 different description. Ask 10 people if an object is "elegant", and you get 10 very subjective answer.

    Is this what you would consider as a valid quality to describe a physical world, that it simply depends on TASTES? How would you like it that your electronics work simply based on someone's mood and perceived human qualities?

    Try again.

    Zz.
     
  14. Feb 2, 2006 #13

    Hurkyl

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    :grumpy: Zero is no less "real" than one. Or, if you prefer, one is no more real than zero.

    No, that's not why you can't divide by it.

    Already have, and I tossed in 250 extra for free.
     
  15. Feb 2, 2006 #14
    Also the photon always has zero energy in it's reference. So I guess there isn't anything there *to* pass with time.
     
  16. Feb 2, 2006 #15

    jtbell

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    A photon doesn't have a reference frame in which it is stationary, at least not one that you can meaningfully apply the equations of relativity to. At best, "reference frame of a photon" is purely a figurative or metaphorical term, with no physical content (in the context of relativity theory).
     
  17. Feb 3, 2006 #16
    You people are so pedantic. You know what I mean. In the limit of accelerating in the direction of a photon the photon will lose all energy. In other words the reference of the photon is where it's proper energy is zero (since it's proper mass is zero). Of course I'm not worring about the quantum problem here of actually measuring the photon to make sure you're chasing it as it would be impossible for any real observer to even know where it is going.

    Since energy is the rate of change of a system, there is not rate of anything in the reference of a photon. So if the absense of a passage of time concerns you any, it kind of makes sense since there would be no change anyway.
     
    Last edited: Feb 3, 2006
  18. Feb 3, 2006 #17

    robphy

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    Just a few comments to illuminate some issues that seem to be often glossed over... (for which I've begun to realize for myself).

    First, all terms (e.g. "reference frame", "photonic perspective", "simultaneously", etc...) need to be defined precisely. This is one important role of mathematics (namely, a mathematical model of the physics one wishes to describe).

    Here are some "reasonable" properties of a "reference frame" of a massive particle in SR (whose worldline has an everywhere timelike tangent vector). [I am going to emphasize the geometric structures to avoid dealing with and trying to interpret certain algebraic formulas that break down when [probably inappropriately] applied to a massless particle.]
    • If A and B are distinct events on that worldline, either A is in the causal past of B (so that A can influence B), or vice versa.
    • Its Minkowski-arc-length along the worldline is nonzero and can be associated with a clock carried by the particle... this clock marks the particle's "proper time".
    • The hyperplane Minkowski-orthogonal to that tangent vector does not contain that tangent vector... and can be called the particle's "space at an instant" (a "moment of time").
    • The events on this hyperplane are regarded to be "simultaneous" according to this particle since:
      • these events are assigned the same time coordinate as read off by the particle's wristwatch/proper-time (e.g., by a radar method [at least for nearby events]: send off a light signal at wristwatch time t1, receive its echo off the distant event at wristwatch time t2, assign to that distant event the time-coordinate (t1+t2)/2)
      • these events are spacelike-related (and therefore not causally-related) to each other
    • For an inertial massive particle in SR, the entire Minkowski spacetime is foliated by these hyperplanes... meaning that the entire spacetime is sliced into nonintersecting hyperplanes, so that every event in spacetime is assigned a (but certainly not all the same) time-coordinate.
    I think the list above seems reasonable.

    What are the analogous statements for a photon (a massless particle) in SR (whose worldline has an everywhere lightlike [a.k.a. null] tangent vector)?
    • (Does "time stop" for a photon?)
      If A and B are distinct events on that worldline, either A is in the causal past of B (so that A can influence B), or vice versa.
      [still TRUE... so (to me) it DOES NOT make sense to say that "time stops" or that all of its events occur "simultaneously" since there is certainly a sense of causal-sequence of the photon's events.]
    • (Does it make sense to define "proper time" for the photon?)
      Its Minkowski-arc-length along the worldline is ZERO. So, there may be a problem here. Maybe one can define it...but is it useful? Don't ignore the previous point.
    • (Does it make sense to call this hyperplane "space" for the photon?)
      The hyperplane Minkowski-orthogonal to that tangent vector DOES contain that tangent vector [joining two causally-related events]... this is a feature of the Minkowskian geometry of SR.
    • The events on this hyperplane are NOT all spacelike-related to each other... There are events on this plane (namely along the tangent vector) that are causally-related to each other.
    • For an inertial massless particle in SR, the entire Minkowski spacetime is NOT foliated by these hyperplanes. So, there are many events that are not assigned a full set of coordinates. [One might argue here that one really needs a congrunce of worldlines.]
    So, it seems to me that there are some problems trying to define a reference frame for a photon.

    My $0.03
     
    Last edited: Feb 3, 2006
  19. Feb 3, 2006 #18

    ZapperZ

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    I believe that what robphy just did it EXTREMELY valuable. This is because there are many people here who are using these terms without understanding that there are precise definitions and properties as defined in physics. And since we ARE here discussing physics, when you use such terms in ridiculous ways, and the rest of us starts to scratch our heads, do NOT simply assume that we are being "pedantic".

    Look at the term you are using. If you do not have a clue on the underlying mathematical descriptions of it, chances are, you don't know what it is. So leave open the possibility that MAYBE, you used it in an inconsistent fashion, or maybe even in ways that it wasn't meant to be used.

    Zz.
     
  20. Feb 3, 2006 #19

    Tom Mattson

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    Longstreet, the people here are not being pedantic, you're just being sloppy. The reference you refer to here simply does not exist, and it has nothing to do with anything in quantum theory. And your statement that energy is the 'rate of change of a system' is completely devoid of meaning. If you talk rubbish around these parts, you can fully expect to be called out on the carpet for it by the memebership here.
     
    Last edited: Feb 3, 2006
  21. Feb 3, 2006 #20
    It doesn't make sense to talk about a photon's perspective/frame of reference/whatever. But let's pretend that a photon did have a reference frame. Obviously it'd be at rest relative to itself, so [itex]d\tau = 0[/itex].

    If we try to take the norm of the four-velocity we end up with [itex]\frac{dx^{\alpha}}{d\tau} \frac{dx^{\beta}}{d\tau} = \frac{0}{0}[/itex], which as we all know is undefined.
     
    Last edited: Feb 3, 2006
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