All Photons Move at 300,000km/s.... But Don't?

Discussion in 'Pseudoscience' started by TruthSeeker, Jun 12, 2015.

  1. TruthSeeker Fancy Virtual Reality Monkey Valued Senior Member

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    Maybe I'm drunk and not thinking straight, but if all light is always moving at the same speed of "c", how come photons have different levels of energy, frequency and speed?
     
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  3. paddoboy Valued Senior Member

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    The energy of a photon is proportional to its frequency and also the amount of energy inferred on any photon, depends on the observer and his/her frame of reference.
     
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  5. TruthSeeker Fancy Virtual Reality Monkey Valued Senior Member

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    Ok, but if all photons move at the same speed, how can they have greater or lesser amounts of energy relative to each other? Shouldn't they all have the same amount of energy, or the more energetic ones move faster than the less energetic ones?
     
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  7. exchemist Valued Senior Member

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    You are speaking as if you are thinking the energy of a photon is kinetic energy (E=1/2 mv²). However it isn't like that. m (the rest mass) =0 for a photon. The energy it has is by virtue of the fact it is an excitation in the electromagnetic field. If you like, you can think of it as having energy in a similar way to the stored energy in the magnetic field of a solenoid, or something. But because in the photon's case it's stored in undulating fields, the energy turns out to depend on how fast the undulations go, i.e. the frequency. E=hν, where ν is the Greek character "nu" and stands for frequency, while h is Planck's constant.

    So it's really quite different from the kinetic energy that we are all used to in Newtonian mechanics.
     
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  8. rpenner Fully Wired Valued Senior Member

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    For a free particle in quantum field theory, these relations relate energy, momentum, mass, velocity, frequency, wavenumber and wavelength:

    \(E^2 = \left( mc^2 \right)^2 + \left( c \vec{p} \right)^2 \\ E \vec{v} = c^2 \vec{p} \\ E = h f\\ \vec{p} = \frac{h}{2 \pi} \vec{k} \\ | \vec{k} | = \frac{2 \pi}{\lambda} \)


    For a massless phenomenon that carries energy and momentum, these equations say:

    \(E = c | \vec{p} | = \frac{h c}{2 \pi} | \vec{k} | = \frac{h c}{\lambda} = h f \\ | \vec{v} | = c \)

    The conclusion is that Energy, Momentum, Frequency and Wavenumber are all proportional while the magnitude of the velocity is fixed. Massless phenomena which carry energy and momentum travel at the speed of light.

    These equations, and the theories they are based on, are consistent with more than 300 years of detailed observation. They differ from the equations of Newtonian mechanics because Newton didn't conduct experiments on momentum and energy of particles moving faster than one tenth the speed of light. Neither did Newton investigate radioactive nuclei and other measurable mass changes and didn't have a concept of rest energy. (Historically, the concept of energy itself came after Newton.) Neither did Newton investigate the physics of quantum phenomena.

    Since momentum is apparently more central to these equation than velocity, we can rewrite the first equations for massive particles in terms of a new quantity \(x = \frac{| \vec{p} |}{m c} \) as:

    \(| \vec{p} | = m c x \\ E = \left( mc^2 \right) \sqrt{ 1 + x^2 } \\ | \vec{v} | = \frac{x}{\sqrt{1 + x^2}} c \)

    \(x\) has no units associated with it, so it's ideal to use in calculus.

    The Taylor series for energy and velocity are:
    \( E = \left( mc^2 \right) \left( 1+ \frac{x^2}{2} -\frac{x^4}{8} + \frac{x^6}{16} - \frac{5 x^8}{128} + \dots \right) \\ | \vec{v} | = c \left( x - \frac{x^3}{2}+ \frac{3 x^5}{8}- \frac{5 x^7}{16} + \dots \right) \)
    So when \(x = \frac{| \vec{p} |}{m c} << 1 \) then we derive the first order relationships:
    \( E \approx \left( mc^2 \right) \left( 1+ \frac{x^2}{2} \right) = mc^2 + \frac{p^2}{2 m} \\ | \vec{v} | \approx c x = \frac{ | \vec{p} | }{m} \)
    Which gives us the Newtonian relations (valid for massive particles at low speed): \( \vec{p} = m \vec{v}, \; E_{\textrm{kinetic}} = E - E_0 = \frac{1}{2} m v^2\).

    // edit: changed \( \nu \) to \( f\) so it wouldn't be confused with velocity.
     
    Last edited: Jun 12, 2015
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  9. TruthSeeker Fancy Virtual Reality Monkey Valued Senior Member

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    Thank you guys, that was very helpful. I was actually watching Cosmos and Neil deGrasse Tyson said that when light goes through a prism it slows down and splits into all wavelengths. I assumed he was correct, but I realize now that light can't be slowing down...
     
  10. exchemist Valued Senior Member

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    That, actually, is rather a tricky and non-trivial issue. When people talk about the "speed of light" ("c"), they generally mean its speed in a vacuum. In media containing matter, the light interacts with the matter, because matter is, to a greater or lesser degree, depending on what it is, polarisable and thus it reacts to the electric field of the photons passing through, and this in turn influences the passage of the photons through it.

    The effect of this is a bit complicated. A photon can be thought of as a "wave packet", i.e. a pulse of waves. The "velocity" of the overall pulse can in some conditions be different from the velocity of the waveforms that make it up. One speaks in such cases of the "group velocity", for the overall pulse, and the "phase velocity" for the waveforms comprising it, to distinguish them. The difference between them is shown in the simulation here: https://en.wikipedia.org/wiki/Phase_velocity , in which the red dots move with phase velocity while the group velocity is indicated by the green dots. You will see the green dots move more slowly than the red ones.

    With light in a material medium, both can be altered, so while it is fair to say the speed of light can be seen as effectively lower in a material medium than in vacuo, some precision is needed as to what one is talking about. The splitting of white light is due to higher frequency light having its phase velocity reduced by more than lower frequency light. This is called dispersion.

    P.S. I hope I have got this right! It is all too easy to get it muddled. Refraction is notoriously hard to explain simply and clearly - if you want your explanation to be correct, that is.

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  11. TruthSeeker Fancy Virtual Reality Monkey Valued Senior Member

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    I think I get it... Seems to me like photons are like massless beams of pure energy that can move unchanged in the vacuum, and then when they interact with matter they lose part of their energy until being fully absorbed by an electron...
     
  12. rpenner Fully Wired Valued Senior Member

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    In a transparent medium, light does not scatter or lose energy. That's related to conservation of momentum and light's wave-like qualities that I don't wish to address.

    Photons are how electromagnetic energy propagates in vacuum, like the vacuum between electrons and protons. Light is the macroscopic propagation of electromagnetic disturbances. It seems intuitively clear that when light propagates though a transparent medium some of the time the energy is bound to the configuration of the matter and not propagating as photons.

    The crudest possible model is that light spends a fraction of it's time, \(\frac{n - 1}{n}\), as measured in the material's rest frame, bound to the matter and the rest traveling at c.

    Indeed, Hippolyte Fizeau's 1859 experiment is explained in terms of this toy model when light passes through a transparent material with refractive index n > 1. For the same medium moving at speed v, the average propagation speed in this toy model is \(\frac{ c^2 + c v n }{ c n + v}\) which is consistent with the findings of Fizeau and those that followed. But it does not support the toy model, because the consistency of Special Relativity requires any phenomena which averages to a propagation speed of \(\frac{c}{n}\) to obey the same rule.
     
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  13. BrianHarwarespecialist We shall Ionize!i Registered Senior Member

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    Ok here is the interesting part, when photons of different frequencies interact in a scenero of "Quantum Entanglement" they appear to break the light speed barrier. In my personal opinion what I think is happening during this process is the photons or particles merge at a "singularity" at the higher frequency using "space-time" itself as a medium to travel FTL.
    So since nothing can travel FTL the particles photons or electrons dont actually move at all but when they become entangled they merge at a singularity warping space with the help of space-time creating a miniture warmhole causing them to become one particle at the point of singularity in spacetime then expressed as dual particles in the probable reality of space. I also believe this process causes the affect of gravity these are my assumptions. would love to discuss it further...
     
  14. BrianHarwarespecialist We shall Ionize!i Registered Senior Member

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    Wave particle duality and probablity I believe are the same thing expressed differently...
     
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  15. paddoboy Valued Senior Member

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    In actual fact when light enters any medium, it is said to "slow down" according to its refraction index. It is though absorbed and re emitted and has travelled a longer distance upon exiting that medium....hence it has a longer distance to travel.
    Thta's a fairly rough description but adequate I think.

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  16. Q-reeus Valued Senior Member

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  17. exchemist Valued Senior Member

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    Well yes and no. As others have pointed out, they do not lose energy when travelling at a net lower speed in a material medium. Remember, for photons E= hν. ν, the frequency, is NOT changed on entering a material medium. Therefore the energy is not changed. You need to let go entirely of this idea that speed has something to do with energy, where photons is concerned. The interaction with the medium changes the effective, net, speed of phase and group velocity but does not affect energy.

    But you are onto something interesting with the comment about absorption by an electron. The refractive index (the ratio of c to the net phase velocity in the medium, if I have it right) is a function of frequency. This is "dispersion", and is why red and blue light are bent to different degrees in a prism. The interesting thing is that this dependence of refractive index on frequency gets stronger and stronger as the frequency approaches one of the absorption frequencies of there material, i.e one of the "resonant" frequencies at which an electron in the material will absorb the photon. (Of course, at an absorption frequency itself, the material becomes opaque, because the light is no longer transmitted through the material. It is absorbed instead.)

    So the phenomenon of dispersion is most definitely related to the process of absorption and emission. Both are due to the photons "coupling" with the electrons in the medium they are passing through, and at the absorption frequency, the coupling hits resonance and becomes powerful enough to cause absorption.

    I think this is rather cool stuff, personally.
     
  18. Q-reeus Valued Senior Member

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    If the topic is broadened to include lossy interaction with the medium, I agree absorption and emission must be included. However, as the last part of that vid I linked to explains, pure refraction (including dispersion) is best thought of as a collective excitation involving atomic polarizations not atomic absorption/emission processes as discussed earlier in that vid. You have a quibble with that picture?
    [Of course there are phonon-photon interactions too which are dissipative AND collective in nature, but I think it needs to be made clear refraction is NOT about atomic absorption/re-emission of a photon as often envisaged.]
     
    Last edited: Jun 13, 2015
  19. exchemist Valued Senior Member

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    Not at all. Don't get me wrong, like you, I have little time for the "explanation" of lower effective speed in a medium as a rest of absorption and re-emission. Firstly absorption can't occur if the frequency doesn't correspond to an absorption line and secondly, if it did, the re-emitted light would be scattered in random directions, which it manifestly is not. (I have not watched the video so don't know whether this point is made - it certainly is on one of the better videos on the topic that I have seen before).

    No, what I'm suggesting (and of course these are all hand-waving, qualitative pictures, capable of expressing only a part of the QM model) is that absorption is a sort of limiting, "resonant", case of the more general polarisation interaction between light and matter. The behaviour of refractive index, as a function of frequency, as one scans the light frequency up to and through an absorption line makes the link between the two very obvious, it seems to me.

    Are you OK with that or you do think I misrepresent it?
     
    Last edited: Jun 13, 2015
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  20. paddoboy Valued Senior Member

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    Nicely put exchemist and I'm sure most would agree.
     
  21. Q-reeus Valued Senior Member

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    Very ok with that exchemist.

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    You will have much greater in-depth knowledge of the various QFT intricacies, and there are probably other processes possible not even mentioned but then again not really germane to the OP.
     
  22. exchemist Valued Senior Member

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    Actually I know bugger-all about QFT

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    . Whatever I know about this subject comes from the Quantum Chemistry supplementary option I took for my degree in the 1970s, a lot of it courtesy of Peter Atkins, whose inspirational lectures led me to buy his book, which I still have and occasionally dip into. In those days I could understand most of the maths. Now, it is only the concepts I really remember…...…..c'est la vie…...
     
  23. Q-reeus Valued Senior Member

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    And I made an error in saying QFT when just vanilla QED was more appropriate. Whatever - I think we have collectively converged to something useful for OP.

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