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Photon question

  1. Apr 14, 2005 #1

    Pengwuino

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    If light is transmitted as photons... and all light is is electromagnetic waves... is it accurate to say all electromagnetic waves travel as photons?
     
  2. jcsd
  3. Apr 15, 2005 #2
    Eh... logically, no: if all light is electromagnetic waves, that doesn't mean all EM waves are light.

    It's a bit hard to interpret your question though; sure, all EM waves propagate as waves, with a certain amplitude and a certain phase... so does light, because of its wave effects... I don't really see your point though.

    I can tell you the inverse: no, I don't see how it can be possible that EM waves have a different sort of propagation than the kind of light waves... all waves are more or less similar.
     
    Last edited: Apr 15, 2005
  4. Apr 15, 2005 #3

    jtbell

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    Sure. There are gamma-ray photons, X-ray photons, microwave photons, radio-wave photons,... The only difference is the energy of the individual photons, as per

    [tex]E_{photon} = hf = \frac {hc}{\lambda} [/tex]
     
  5. Apr 15, 2005 #4
    Light is just a part of the entire EM-spectrum and since photons are the particles that correspond to EM-waves, the answer to your question is YES.

    Beware that when we speak about particles, don't think of objects with clear finite spatial boundaries. This is NOT the case because the position is uncertain in QM due to the HUP. When we talk about particles in the particle wave duality we mean : particle like behaviour : photons are 'little' finite pieces of energy.

    regards
    marlon
     
  6. Apr 15, 2005 #5

    Pengwuino

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    Ok so is "light" defined only as IR, UEV, visible, etc and not x-ray, gamma, short wave, etc?

    I know all of those represents a portion of the RF spectrum, but im asking if say, x-ray is accurate in being called "light" just like visible light is "light".
     
  7. Apr 16, 2005 #6
    Tsunami's right, logically speaking. All light is em, all em "packets" are photons, therefore all light "packets" are photons. Not quite as you stated it, but close enough. Honestly speaking, though some may differ decidedly, in general if
    [tex]E_{photon} = \frac {hc}{\lambda} [/tex]
    holds, call it light to your heart's content. Whether or not it's visible to the human eye is irrelevant.
     
  8. Apr 16, 2005 #7
    I see the photon as an imaginary particle to fulfil momentum and energy preservation. Therefore the photon quantifies the EM radiation. i.e I see it as the power or ampplitude squared of the EM wave integrated over a certain time period. It is convenient solution.

    Look at it this way, your GSM mobile telephone operating at 800MHz and transmitting say at 0,5W needs to emmit 9.43 x 10^23 photons every second when transmitting. No wonder your ear gets cooked by all those photons.
     
  9. Apr 16, 2005 #8
    No, what we call light is just the visible part of the EM-spectrum. IR, UV, X-radiation are other parts of the EM-spectrum (other energies or wavelengths)...So EM-radiation is the general name, if you will.

    marlon
     
  10. Apr 16, 2005 #9
    Somone correct me if I am in any way wrong, I'm not a physicist after all, but the important thing is that these bundles, packets, quanta, have an energy cracterized by
    [tex]E_{photon} = \frac {hc}{\lambda} [/tex],
    which shows that these discreet energies are dependent upon their wavelengths. The photons travel with a velocity c, held to be a constant. Then, by,
    [tex]c = {f\lambda} [/tex]
    there are many wavelengths which satisfy the relation. While light technically only refers to those wavelengths in the visible part of the spectrum, it is not uncommon to use it as a geneal term for all em radiation, anything which satisfies the above relation.
     
  11. Apr 16, 2005 #10

    Pengwuino

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    Well after looking at it, what you answered was closer to what i was thinking and what i was thinken wasnt realy what i wrote down accurately.
     
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