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Does a photon know how fast to leave its emitter?

  1. Dec 16, 2011 #1
    Does a photon "know" how fast to leave its emitter?

    Does a photon "know" how fast to leave its emitter, which is moving relative to the photon's receiver, so as to be traveling at c when it reaches the receiver?
     
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  3. Dec 16, 2011 #2

    Bill_K

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    Re: Does a photon "know" how fast to leave its emitter?

    The photon leaves at a speed of c in the emitter's frame. This is also c in the receiver's frame, or in any other frame. The speed of a beam of light is the same in every frame.
     
  4. Dec 16, 2011 #3

    atyy

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    Re: Does a photon "know" how fast to leave its emitter?

    The photon knows by travelling not relative to the emitter or receiver, but by travelling on the "null geodesic" of spacetime.
     
  5. Dec 16, 2011 #4
    Re: Does a photon "know" how fast to leave its emitter?

    c is the speed at which the electromagnetic field propagates through the vacuum. The quanta of this field are individually entirely ignorant of this fact (I presume).

    [On the other hand, every chemical and electron signal in our brains is based entirely on the electromagnetic force, so the processes that make us conscious are entirely due to the interactions of real and virtual photons. Fortunately they all travel at the right speed without us worrying about it.]
     
  6. Dec 16, 2011 #5
    Re: Does a photon "know" how fast to leave its emitter?

    the photons also knows to take the shortest path when it crosses the boundary between matters with different densities. it's as if it knows the path it's going to travel is the most efficient one and less time-consuming, perfect example of effect before cause
     
  7. Dec 16, 2011 #6

    phinds

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    Re: Does a photon "know" how fast to leave its emitter?

    Hm ... I don't believe simple refraction is effect before cause. I'd be interested to hear comments from folks who know more about this than I do. Any optics gurus out there?
     
  8. Dec 16, 2011 #7
    Re: Does a photon "know" how fast to leave its emitter?

    Unfortunately I don't know what a null geodesic of spacetime is. I googled it but it looks like it will take a while to study it.

    Basically I am looking for some explanation of how light speed gets "adjusted" to the motion of emitters or receivers so as to remain the same.

    I am confused by the possibility that the one way speed of light, which cannot be measured, may vary while the round trip is at c. If one way speed does vary, it seems a little strange that light could be traveling at different speeds in different directions (for example within the same object).

    Another problem is that I see no explanation in the slowing of time and contraction of length of the moving object, as the two mechanisms (slowing and contraction) affect equally both oncoming light and catching up light which, if they have different speeds, would continue to have them.
     
  9. Dec 16, 2011 #8

    phinds

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    Re: Does a photon "know" how fast to leave its emitter?

    Light does not adjust itself to emitters and receivers. It is utterly indifferent to them. When a photon is created, it immediately travels at c and is seen to do so in all reference frames.

    The problem is NOT that the one-way speed of light is different than any other speed of light; there is only one speed of light in a vacuum, but what IS a problem is that we can't properly MEASURE the one-way speed of light.

    These effects do not affect light. Light travels at c. Period.
     
  10. Dec 16, 2011 #9
    Re: Does a photon "know" how fast to leave its emitter?

    I read on Wikipedia that the Lorentz theory of relativity uses one-way speeds of light that vary, and that Lorentz's theory is indistinguishable from Einstein's otherwise.
     
  11. Dec 16, 2011 #10

    atyy

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    Re: Does a photon "know" how fast to leave its emitter?

    It's very unclear what Lorentz's theory of relativity was, and probably of interest only from a historical viewpoint.

    There is a view of special relativity which is Lorentzian in spirit, eg. Bell's "Lorentzian pedagogy". It is special relativity described in any particular Lorentz inertial frame. It is equivalent to special relativity.

    Special relativity is a theory that spacetime and all matter have a particular symmetry called Lorentz symmetry (more strictly, Poincare symmetry). Lorentz symmetry means that spacetime at each event is divided into past, future, and elsewhere. A null geodesic forms the boundary between elsewhere and the past or future. A wave travelling on a null geodesic will be seen to have the same "privileged speed" in all Lorentz inertial frames. Lorentz symmetry and the fact that electromagnetic fields have an additional symmetry called conformal symmetry imply that electromagnetic waves travel on null geodesics, which is the same as saying that light travels at the "privileged speed" in all Lorentz inertial frames. The "privileged speed" is therefore usually called the speed of light. In contrast to light, the fields describing a massive emitter and massive receiver have Lorentz symmetry, but not conformal symmetry.
     
    Last edited: Dec 16, 2011
  12. Dec 16, 2011 #11
    Re: Does a photon "know" how fast to leave its emitter?

    I find it is made overly complicated for a layman like myself.

    Know that c is the fastest speed, anything that travels this fast, i.e. 299,792,458 meters in one second is following a "null" geodesic path.


    In the image below the yellow line is a null geodesic path, this is the path a photon would travel since they travel at c.
    any coordinate on the graph that is on the right side of the null path is spacelike, timelike is on the left side.

    This is a very generalized description, but is enough to give the term a context. I came across the null geodesic term and googled it like you did. It wasn't until I started reading about intervals that it became clear what a "null geodesic path" is. Said differently reading about (and asking questions here) intervals made clear what the terms spacelike, timelike and null mean in this context.

    for a mathless approach check out spacetime diagrams.


    c is always the same speed;
    So something travelling at 0.5c emits a stream of photons, those photons are traveling at the same speed the always do, they're not "adjusted" in anyway.

    When the dude travelling at 0.5c calculates the speed of those photons it will be the same as everyone elses calculated value of c.

    It is the measurements that are "adjusted".

    Specifically time dilation and length contraction "adjust" accordingly so that the calculated value of c is always the same, no matter your relative speed to those photons.

    (image below is from wikipedia)
    240px-Minkowski_diagram_-_photon.png
     
    Last edited: Dec 16, 2011
  13. Dec 16, 2011 #12

    PAllen

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    Re: Does a photon "know" how fast to leave its emitter?

    Not at all an optics guru, but a simple mathematical observation may demystify this. When one minimizes an integral quantity (by path variation), the result (Euler-Lagrange equation) is a strictly local differential criterion. This criterion amounts to following the local 'path of least resistance' or 'straightest possible path - in a possibly abstract phase space'. It happens that this strictly local condition is a necessary (but not sufficient) condition for an overall extremum.

    In my view, it is better to think about the light, system, or whatever directly following this local condition rather than 'magically' discovering a global extremum.
     
  14. Dec 16, 2011 #13

    phinds

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    Re: Does a photon "know" how fast to leave its emitter?

    I don't follow that, but let me ask a slightly different question which may already be answered by your statement. Would it be fair to say that refraction is caused ENTIRELY because of what happens at the BOUNDARY between materials (a very local effect) and NOT because of anything outside the boundary (which would be a global effect and COULD imply effect before cause). This seems very reasonable to me because once the light leaves the boundary, it does not change direction again.
     
  15. Dec 16, 2011 #14

    PAllen

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    Re: Does a photon "know" how fast to leave its emitter?

    Yes, you could say that, and I meant to imply that mathematically. Further, if you consider a material with continuously varying index of refraction, the light is really (oops, there's that problem word again..) responding only to local conditions.
     
  16. Dec 16, 2011 #15

    phinds

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    Re: Does a photon "know" how fast to leave its emitter?

    Thanks
     
  17. Dec 16, 2011 #16
    Re: Does a photon "know" how fast to leave its emitter?

    I think if Feynman was asked this question, he would say that a photon takes every path simultaneously, including those that are much faster or slower than the speed of light. Then all the amplitudes for all the paths are added up (a huge integration). If you do this calculation, the only paths that get left with amplitudes different to zero are the equivalent of straight lines in curved spacetime at a speed of c. This is truly a wonderful thing, and may be a hint why the Universe behaves the way it does.
     
  18. Dec 16, 2011 #17

    phinds

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    Re: Does a photon "know" how fast to leave its emitter?

    I think you are overextrapolating Feynman. I don't think he believed in FTL.
     
  19. Dec 16, 2011 #18
    Re: Does a photon "know" how fast to leave its emitter?

    I dont know what you mean by "know" photons know nothing.
     
  20. Dec 16, 2011 #19

    phinds

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    Re: Does a photon "know" how fast to leave its emitter?

    The word was put in quotes and in context the meaning seemed quite clear.
     
  21. Dec 16, 2011 #20
    Re: Does a photon "know" how fast to leave its emitter?

    Well, define SpaceTime as one wave equation, define interactions as the wave equation 'falling out' locally, relative interference as in the Feynman definition and you have one answer :) maybe. Or .. .. ..

    Refraction is a index defined relative 1, as I understands it. Where one is 'space' always being 1. Defining 'space' globaly as 'densities', for example relative 'gravity', should then invalidate the index of refraction as it no longer have a valid definition to be measured relative.

    A photon 'exists' in two ways to us. Due to the 'recoil' observed in a 'source' as demanded by the conservation laws, and in its annihilation at your retina. There is nowhere in between it can be said to 'exist' without a measurement.

    But it also has to do with what you 'think'. Do you expect a propagation? Then you expect a 'path'.
     
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