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Debate About Photons

  1. Jun 23, 2010 #1
    Last night an EE I was talking to here in San Diego got noticeably upset with me when I asserted that all visible light consists of photons, including light from such sources as the sun, an electric spark, and a lit match. I found myself being treated like a crackpot for asserting what I took to be the most trivial and accepted tenet of Quantum Physics, that all visible light consists of discrete bundles of energy called "photons". His claim was that all the light sources I mentioned emitted "radiant energy", which did not consist of photons. "True" photons, he angrily insisted, were only emitted in the special case where electrons were bouncing in and out of "holes" such as you find in a light emitting diode. Scowling and growling at me, he advised me to google "photonic energy", to be enlightened.

    He's an EE and I'm not, so is there really only an authentic special case "true" photon, as he asserts?
     
  2. jcsd
  3. Jun 23, 2010 #2

    Dale

    Staff: Mentor

    There are some quantum states (e.g. coherent states) where the number of photons is uncertain. But even so I am with you on this. If you don't like the idea of virtual photons then you may not want to consider a near-field antenna like a microwave to be photons, but it is certainly accepted by the QED community.
     
  4. Jun 23, 2010 #3
    Your story makes me ashamed to be a EE. I apologize for his ignorance.

    Of course, the sun emits more than just photons, but I don't know why he would say that the radiation does not consist of photons at all.

    By the way, Googling "photonic energy" did not enlighten me at all. How about you?
     
  5. Jun 23, 2010 #4

    K^2

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    Science Advisor

    All light is EM field. But all linear fields can be quantized. So while I'd leave alone debate on what really the light is made up of to philosophers and holy men, any light can be properly represented and described physically as "bundles of energy" we affectionately call 'photons'.

    To me, as a physicist, that's all the truth I need. If it can always be described as photons, it may as well always be photons, or at any rate, something physically indistinguishable.

    The rest is theology.
     
  6. Jun 23, 2010 #5
    Well, here is where you can easily make him see his error. Since he thinks photons only come from the combination of hole/electron pairs in a semiconductor, how does he explain the reverse process in which solar panels and photodiodes convert visible light to hole/electron pairs, thus producing usable current and electrical power?
     
    Last edited: Jun 23, 2010
  7. Jun 23, 2010 #6
    We weren't even discussing anything like RF or microwaves. He sneered at my assertion the sun itself was emitting photons. He flicked his cigarette lighter on and said "You're trying to tell me that's photons? No."
     
  8. Jun 23, 2010 #7

    K^2

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    I think you had a moral, if not legal, right to hit him over the head with something heavy for being so smug about being an idiot.
     
  9. Jun 23, 2010 #8
    I don't know either. It seems to me Huygens, Newton, Young, Planck and Einstein were working to explain light before the invention of the LED, and that the ultimate explanation; that light had to be discrete bundles of energy, was made to explain all the conventional, pre-technological sources of light.
     
    Last edited: Jun 23, 2010
  10. Jun 23, 2010 #9

    DaveC426913

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    Gold Member

    The guy's a dope.
     
  11. Jun 23, 2010 #10
    Like I said, he's an EE and I'm not. For all I know there could be some technical reasoning I'm not aware of whereby he's right.
     
  12. Jun 23, 2010 #11

    DaveC426913

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    At best, he has an industry-specific understanding of the subject. It may be the case in his world that that's how they look at it, but you would still be well within your rights to correct him.
     
  13. Jun 23, 2010 #12
    That's what I'm thinking. His notion is some misapplication of an EE method of thinking to general physics.
     
  14. Jun 23, 2010 #13
    As an EE, let me set the record strait- its simply a misunderstanding on his part.

    EE's don't need to take quantum effects into account unless:
    1) Wavelengths are extremely short
    2) Very low noise is important
    3) Doing anything in solid state electronics

    You get exposed to these in later years so it would be easy for a budding EE to think that quantization was limited to semiconductors.
     
  15. Jun 23, 2010 #14
    Unfortunately this guy is in his late 40's-early 50's.
     
  16. Jun 23, 2010 #15
    The classical electromagnetic field is described by Maxwell's equations:

    [tex]
    \begin{array}{rcl}
    \nabla \cdot \mathbf{E} & = & 4 \pi \, k_{0} \, \rho \\

    \nabla \cdot \mathbf{H} & = & 0 \\

    \nabla \times \mathbf{E} & = & -\frac{k_{1}}{c} \, \frac{\partial \mathbf{H}}{\partial t} \\

    \nabla \times \mathbf{H} & = & \frac{1}{c \, k_{1}} \, \left( \frac{\partial \mathbf{E}}{\partial t} + 4\pi \, k_{0} \, \mathbf{J} \right)
    [/tex]

    Here [itex]\rho[/itex] is the volume density of electric charges and [itex]\mathbf{J}[/itex] is the surface density of electric current.

    The second equation implies that the magnetic field [itex]\mathbf{H}[/itex], being solenoidal (divergenceless) can always be represented as a curl of another vector field:

    [tex]
    \mathbf{H} = \nabla \times \mathbf{A}
    [/tex]

    where [itex]\mathbf{A}[/itex] is called the vector potential of the field. Since the curl of the gradient of any scalar function is zero, we see that this potential is determined up to a gradient of an arbitrary scalar function:

    [tex]
    \mathbf{A} = \mathbf{A}' + \nabla \Lambda
    [/tex]

    Substituting the expression for [itex]\mathbf{H}[/itex] in terms of [itex]\mathbf{A}[/itex] in the third Maxwell equation, changing the order of derivatives w.r.t. time and coordinates and rearranging, we get:

    [tex]
    \nabla \times \left( \mathbf{E} + \frac{k_{1}}{c} \, \frac{\partial \mathbf{A}}{\partial t}\right) = 0
    [/tex]

    The term in the parentheses is potential (irrotational) and we can write:

    [tex]
    \mathbf{E} = -\frac{k_{1}}{c} \, \frac{\partial \mathbf{A}}{\partial t} - \nabla \Phi
    [/tex]

    Here [itex]\Phi[/itex] is called the scalar electromagnetic potential. The minus sign is for traditional reasons. If we change the vector potential according to the above rule, we need to simultaneously change the scalar potential according to:

    [tex]
    \Phi = \Phi' - \frac{k_{1}}{c} \, \frac{\partial \Lambda}{\partial t}
    [/tex]

    for the electric field [itex]\mathbf{E}[/itex] to remain unchanged.

    We may impose an additional condition to the potentials using the arbitrariness of the gauge function [itex]\Lambda[/itex]. We will choose it to simplify the remaining to equations as much as possible. Substituting the expressions for the fields in terms of the potentials in the first and fourth Maxwell equations and performing some vector calculus, we get:

    [tex]
    \begin{array}{l}
    -\nabla^{2} \Phi - \frac{k_{1}}{c} \, \frac{\partial}{\partial t}\left( \nabla \cdot \mathbf{A} \right) = 4\pi \, k_{0} \, \rho \\

    \frac{1}{c^{2}} \, \frac{\partial^{2} \mathbf{A}}{\partial t^{2}} - \nabla^{2} \mathbf{A} = -\nabla\left((\nabla \cdot \mathbf{A}) + \frac{1}{c \, k_{1}} \, \frac{\partial \Phi}{\partial t}\right) + \frac{4\pi \, k_{0}}{c \, k_{1}} \, \mathbf{J}
    \end{array}
    [/tex]

    to be continued...
     
  17. Jun 23, 2010 #16
    Wow...I understood all the math in Dickfore's post, but I'm not sure what he was getting at. I eagerly wait the continuation!

    Anyway zoobyshoe, you were correct. All light consists of photons. The whole idea of quantum electrodynamics is that you can quantize the EM field as photons. And I can't think of any electromagnetic wave which couldn't be represented as photons. There's no such thing as "radiant energy" in the sense that he's describing. It's just a term that some people (probably in biology and chemistry) fuse for the power transmitted by EM waves.
     
  18. Jun 23, 2010 #17
    wth? photon is a unit measurement of light and sunlight is a source of light. spark itself may not be photon but since it the spark probably excites the gases and then the gases stabilize themselves by emitting light or photon. thats all there is to it.
     
  19. Jun 23, 2010 #18

    alxm

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    Science Advisor

    And in a funny role-reversal, most chemists are perfectly comfortable with quantized states, but may easily be unaware of where those strange continuous "bands" come from.. :tongue2:
     
  20. Jun 24, 2010 #19
    It's been my understanding, not just that it is possible to represent an EM wave as quantized, but that Planck demonstrated that light HAS to be quantized, physically, literally, to avoid the "Violet Catasrophe" that would result if light energy were transmitted in a continuous flow. I have not, myself, bothered to dig into Planck's reasoning in any detail, but this dilemma he faced is cited in every introduction to quantum physics.

    Yes, it has been my understanding the light emitted by a spark is actually an epiphenomenon of the hot, ionized gases the spark causes. Regardless, it's been my understanding the light emitted in this situation consists of photons, like all light.
     
  21. Jun 24, 2010 #20

    berkeman

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    Staff: Mentor

    Zooby, will you see this gentleman again? Hand him a Post-It note with the URL of this thread on it... o:)
     
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