In the perspective of the photon

Discussion in 'Physics & Math' started by Cyperium, Nov 16, 2010.

  1. Cyperium I'm always me Valued Senior Member

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    To itself the photon doesn't travel at the speed of light, and doesn't travel the distance of 1 lightyear per year. At the only moment it can exist, it would exist at all points as all distances shrink to zero and time stops (there is only one everlasting moment as it can only travel in the speed of light - which is in the perspective of the photon infinite speed).

    What would this mean? The photon exists so it has a perspective, but it is infinity, the infinity of only one moment, as time stops the outside world would move through time at a infinite rate in the perspective of the photon, always existing at all points.

    So if this is the perspective of the photon, the true nature (as I would perceive it), then what is the process that determines a certain speed for it, and direction, and all other things that the photon just doesn't have in its own perspective?

    What is more real, the perspective of the photon? or the perspective of the outside observers?
     
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  3. Green Destiny Banned Banned

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    A way to avoid this paradox is by the realization that a photon doesn't even have a frame of reference from it's point of view, so questioning whose frame of reference is preferred becomes a non-problem.
     
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  5. prometheus viva voce! Registered Senior Member

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    As miraculous as it sounds, green destiny is right, more or less. Specifically, you cannot define an inertial frame moving relative to another at the speed of light. You can see this if you try and do this using a Lorentz transformation and plugging in v = c.

    As the OP kind of says, photons do not experience time so every event that a particular photon encounters happens instantly. Also, Lorentz contraction tells us that at c, every length is zero, so the photon perceives the universe as a timeless point.
     
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  7. Cyperium I'm always me Valued Senior Member

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    Yes, still I'm not questioning whos frame of reference is preffered, as both does exist. We can easily see that the photon does exist, so what lifetime of the photon are we observing? As it has no time, and how can we perceive it at a certain place, when it is at all places at once? It has never during its lifetime been at one certain place, or at one time instead of another, as it is at all places (has arrived at every point in no time).


    What I'm questioning is the photon itself, as it shouldn't even be anywhere specific as it moved infinitely fast to everywhere, and how can it have a direction and all other properties when it has only one event (existence itself?) as the only thing it has done is to exist, and instantaniously it is everywhere as all distances shrink to zero.

    So what we perceive is a reality that just shouldn't be and that the photon has never actually experienced.
     
  8. rpenner Fully Wired Valued Senior Member

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    Or you could point out that Lorentz invariance puts more limits on physics than just conformal invariance, and so experiment should be able to distinguish between the photon world view and the world where time and mass matter.
     
  9. arfa brane call me arf Valued Senior Member

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    Does that mean the photon sphere is a null-geodesic surface with no past endpoints?
     
  10. jmpet Valued Senior Member

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    ALL HAIL HYPNOTOAD!

    With a great deal of respect to the moderator of this forum, who has the ability to suspend you or throw your ideas into the trash if he so chooses, I present my humble two cents.

    -----
    I am a believer in "tired light". Case in point- light coming from a galaxy to your telescope's focal point. That light traversed million of light years for million of years to be received by your quantifying telescope.

    In cosmic terms, I see light at a path of photons spaced out in Planck Units from A to B... ball after ball of a photon packet in a row from there to here. Each packet is an interaction between that moment in Spacetime and the ball at that Planck moment in Spacetime.

    With that said, there are a quadrillion septillion quadrillion billion photon packets between points A and B of Spacetime.

    Mathematically, a measurable amount of those Planck moments contain matter or energy..."stuff". The straight line of photons hits this stuff and refracts, reflects or absorbs... in all cases it distorts.

    No wonder the further away a galaxy is, the less we can see of it... a galaxy gives off a quadrillion trillion photons in all directions. A stream of them comes towards us. There is a lot of stuff in empty space between points A and B! Lots and lots of photons and anti-quarks and top spinning subatomic particles and planets and suns. No wonder of the trillion photons sent directly to Earth only one or two actually hit Earth at the right point in Spacetime to be hit by a telescope, which collects photons over time- the rest are absorbed, reflected or refracted by the interstellar medium.

    Photons have their own laws and among them, I believe in the tired light hypothesis.

    ALL HAIL HYPNOTOAD!
     
  11. AlexG Like nailing Jello to a tree Valued Senior Member

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    Wow.

    After reading that, my intelligence has diminished.
     
  12. jmpet Valued Senior Member

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    Try thinking bigger- it produces results. Look past the math ingrained in your head to logic. Put the calculator away and think for yourself.
     
  13. prometheus viva voce! Registered Senior Member

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    Photons and other massless particles follow null geodesics. Where these geodesics end depends on the spacetime that you're in. In flat space there is a distinct timelike infinity in the past and future where the world lines of massive observers end (the bottom and top corners). Light like world lines end at null infinity (the lines linking all the corners)

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    In AdS for example the situation is different because light can reach "infinity" in a finite time and be reflected back.
     
  14. arfa brane call me arf Valued Senior Member

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    Ok, but I'm trying to understand what a Cauchy horizon is. According to Hawking, the Hawking-Penrose theorem says a Cauchy horizon is generated by null-geodesic segments without past endpoints. I assume if a photon is "trapped" in the photon sphere of a black hole, it loses information about its past development. But I also assume that doesn't mean it has no future outside the photon sphere--it could interact with a material particle for instance.
     
  15. chaos1956 Banned Banned

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    It's at an infinite speed because the universe expands at this constant infinite rate. Still the mass that it is attracted to does not slow it down it just allows for it to spirial at different angles due to the spins particles themselves have.

    In theory yes, but we don't have the ability to shrink ourselves down to the level of a photon yet. As you pass that event horizon so to speak the matter begins to rearrange itself in a funny way. Inversly porportional to the way in which it is percieved. Its just how far past the horizon we are able to shrink ourselves to keep up with it as we are moving at a very fast constant speed.
    [qoute]
    So if this is the perspective of the photon, the true nature (as I would perceive it), then what is the process that determines a certain speed for it, and direction, and all other things that the photon just doesn't have in its own perspective?[/quote]
    We calculated the speed using precise tools to determine the speed of the strings that bind the universe the universe together. There is enough of them to say they share a strange perspective when assimilated into larger quanta. Still in space light moves as a beam, through mass no matter how dense the mass light does no such thing. We may not be able to pick up the light escaping from the black hole as a fissionable larger denser Sun, but the radiation that is able to escape its swirling conjectures is only more confirmation of lights infinite speed and presence within the universe. Surely a larger object is still taking advantage of these radiation emissions as they are scattered thorough.

    The perspective of the photon.
     
  16. Green Destiny Banned Banned

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    The terminology is a light-like vector of the Minkowskian Spacetime, and has the signature of \(\mu_{\nu \nu}=0\), and it describes the timelike cone of trajectory, which is null described. The timelike vector is obviously part of the four-vector written in Einstein summation \(v=v^{\mu}e_{\mu}\) which can be written in row vector form \((v^0,v^1,v^2,v^3)\).

    The light does not vanish - time vanishes for the light because the dilation has been stretched if you like, to infinity.
     
  17. AlphaNumeric Fully ionized Registered Senior Member

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    GD, that's precisely the kind of post I have commented about to you, where you present your 'understanding' as if its simply stating well known results when in fact you don't know the material, you're just winging it on stuff you read other people say.

    To illustrate this :
    Signatures are to do with metrics, not trajectories. \(\mu_{\nu\nu}\) is meaningless, its not standard notation for anything. The metric in SR is \(\eta_{\mu\nu}\) but that doesn't equal 0, its a (2,0)-tensor. Its signature is -+++ (up to a choice in convention), which is entirely independent of the trajectories of things moving in the space the metric refers to. A null (or light-like) trajectory satisfies \(v^{\mu}v_{\mu} = 0\) or is formed by paths relating to \(ds^{2} = \eta_{\mu\nu}dx^{\mu}dx^{\nu} = 0\).

    Misuse of the word 'null'. And null trajectories lie on a light cone but the light cone, by definition of the fact its formed by light trajectories, is not time-like, an path on it is null. Vectors and trajectories entirely within the light cone are time-like, not the cone itself.

    Meaningless. A time-like vector will satisfy \(v^{\mu}v_{\mu} = \sum_{\mu}v^{\mu}v_{\mu} < 0\), all you gave is the generic expansion of a general vector in a particular choice of coordinates, nothing of what you said actually relates individually and specifically to time like or null or space like vectors

    There is no such thing as a rest frame in terms of null trajectories, you cannot reduce them to the form \(v^{\mu} = (v^{0},0,0,0)\) by a Lorentz transformation, as such a vector satisfies \(v^{\mu}v_{\mu} < 0\) and Lorentz transformations leave \(v_{\mu}v^{\mu}\) unchanged, by definition.

    If you know you don't grasp something properly don't post. Or at the very least make it clear you're asking a question due to not being sure. When you post as if you're certain about your 'facts' and you're familiar with the subject but in fact you aren't then you can infect other people, who also don't know, with your own misconceptions or ignorance. You said in another thread your presence isn't disrupting but spreading large amounts of misinformation, when you know full well you're not entirely competent at this stuff, is disruptive.
     
  18. Green Destiny Banned Banned

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    None of my statements are technically incorrect. None of my statements said anything of a rest frame for a photon.

    Piss off alphanumeric. My day hasn't even begun and you're pissing me off already.
     
  19. prometheus viva voce! Registered Senior Member

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    Ahh, ok. My understanding is that if you have some spacetime, and you define initial conditions for your wave equation of choice on some spatial slice (a Cauchy surface) and evolve that in time you should be able to predict what happens at any future time. If you find that, when you do that, there are points in your future spacetime that you can't predict what happens from those initial conditions then those points lie behind a Cauchy horizon. Another way of seeing this is that, on one side of the Cauchy horizon you have closed spacelike geodesics and on the other you have closed timelike geodesics. I have never heard of closed null geodesics.

    This breakdown in predictability doesn't mean the photon is trapped or something weird like that. It just means that you have to supply more boundary conditions at the Cauchy horizon to specify the solution. FYI, the simplest solution where this happens is the Reissner Nordstrom Black hole (charged but not rotating).

    Another way to see what a photon "sees" as it passes through some spacetime is to take the Penrose limit, which is something like the first order expansion of a metric around a null geodesic.
     
  20. prometheus viva voce! Registered Senior Member

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    A lot of what you wrote was meaningless gibberish. I suppose you could say it said nothing rather than something wrong, but I wouldn't say that.
     
  21. Green Destiny Banned Banned

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    It certainly was not wrong. If you think it was meaningless, then so be it, but it wasn't wrong.
     
  22. Green Destiny Banned Banned

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    In wiki, they used a different term to me η(v,v) = 0 which they denote as Null. This is obviously what I was trying put forward.

    and it describes the timelike cone of trajectory, which is null described.

    Is correct. timelike cones are the realm of photons, the path is a null path which are all relevantly issued for photons. I don't see anything wrong in the statement.

    He says what I said in the third paragraph that its meaningless. It certainly isn't since we are talking about how its represented using Einstein summation, and the row vector quantity roles off the four vector of spacetime.

    And I never mentioned anything about a rest frame, or even hinted at such.
     
  23. AlphaNumeric Fully ionized Registered Senior Member

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    Yes, they were and I went through them. Why don't you actually respond to them, rather than doing your usual thing of a mass quote and then ignoring what you're actually quoting.

    If I'm wrong, point out where. Saying "Shut up, it was right!" doesn't retort what I said and when done all the time (as you do) it becomes clear its just code for "Crap, I can't retort anything he said but I must still respond". Happens every time I dissect one of your posts.

    That was me explaining something, not just correcting you. You talked about time for the photon. Time for an object is what a clock in the objects rest frame would measure.

    So you're angry because you posted nonsense and I pointed it out? I didn't make you post nonsense and do you seriously expect people to just sit by and watch you spout crap as if you understand it? If you don't want people replying to your posts get a blog. If you come to a discussion forum and spout nonsense why are you surprised when someone says "That's nonsense"?

    You failed to understand the notation and you got confused with the usual metric notation. This demonstrates you don't understand the notation, else you'd never have typed what you did.

    \(\eta(v,v) = \eta_{\mu\nu}v^{\mu}v^{\nu} = v^{\mu}v_{\mu}\), which for null v gives \(\eta(v,v) = v^{\mu}v_{\mu} = 0\). The fact you mashed that into \(\mu_{\nu\nu}\) shows you don't understand it, you're just mindlessly copy and pasting.

    Photons move on null trajectories. A light cone is defined by paths of light rays relating to a particular point in space-time. All trajectories on the light cone are by definition null. If a trajectory is within all light cones associated to all points along the trajectory then it is time-like (or equivalently its tangent vector is always time-like). Light cones separate the time-like and space-like regions, ie \(v^{\mu}v_{\mu} < 0\) is time-like, \(v^{\mu}v_{\mu} = 0\) is null and \(v^{\mu}v_{\mu} > 0\) space-like (in the -+++ signature). You can define time-like cones but they are frame dependent and nothing to do with photons. Your statement is incorrect.

    I said you gave nothing more than the definition of components of a vector. The expressions had nothing specific to do with time-like vectors, yet you were presenting them as if they did. This is not the first time you talk about something specific and then the expressions you give are utterly generic and/or unrelated, you did it with entanglement. You claimed to reach some expression for an entangled system and yet the expression was nothing but the generic expansion for a 2 dimensional state, nothing to do with entanglement or anything specific.

    Go on, explain why that expression is specifically about time-like vectors. Show my comments are wrong, rather than just saying they are but never doing anything about it.
     

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