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EM radiation becoming heat

  1. May 4, 2009 #1
    Hello everyone! Let me first say how cool it is that this forum exists, I wish I'd thought of coming here years ago.

    I am trying to understand how EM radiation works. From the high school physics that I've learned I have been under the impression that reflection was due to electrical absorption. Furthermore the physics that I've read have implied that light will enter and exit a material with the same frequency.

    So my question is, when the atoms increase in kinetic energy due to exposure to light, what happens to the energy of the light? My idea is that the frequency should degenerate proportionally to the increase in thermal radiation from the body that is associated with the increase in temperature.

    As I indicated before I am also confused as to whether matter only interacts with certain frequencies of EMR. Would a beam of light at a wavelength that does not excite a certain material make any change in the material's thermal energy?

    Edit: And just after posting I of course stumble upon http://en.wikipedia.org/wiki/Raman_spectroscopy" [Broken]. Typical. I will continue studying from there, but I would still appreciate any thoughts on this general subject.
     
    Last edited by a moderator: May 4, 2017
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  3. May 4, 2009 #2

    jtbell

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    The energy carried by a classical electromagnetic wave is associated with the wave's amplitude, not its frequency.
     
  4. May 4, 2009 #3
    In that case I am truly confused. My textbooks and internet sources including Wikipedia often refer to shorter wavelength light as having more energy, but perhaps that designation is no more than the legacy of mechanical waves.

    I thought amplitude represented intensity of the light?
     
  5. May 4, 2009 #4

    jtbell

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    They are probably referring to photons (the units of energy in an electromagnetic wave). Individual photons do have more energy when the wavelength is shorter (and the frequency is higher).

    For a fixed frequency and wavelength, increasing the intensity of the light increases the amplitude of the wave, the total energy in the wave, and the number of photons.

    For a fixed intensity, the energy in the wave stays constant. In that case, increasing the frequency reduces the number of photons (because each photon has more energy), but the amplitude of the wave stays the same because the intensity is the same.
     
  6. May 4, 2009 #5

    Dale

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    To follow up on jtbell's comment: in classical EM the energy flux is given by the http://en.wikipedia.org/wiki/Poynting_vector" [Broken] is proportional to the amplitude squared, not the frequency.

    On the other hand, in QM a http://en.wikipedia.org/wiki/Photon#Physical_properties" is only dependent on the frequency.

    In order to link the QM and classical EM ideas you can essentially express the amplitude of a wave in terms of the number of photons. It is kind of sloppy, but in the limit of a large number of photons you essentially recover classical EM except for things like the photoelectric effect that have no classical EM explanation regardless of the number of photons.
     
    Last edited by a moderator: May 4, 2017
  7. May 5, 2009 #6
    I should have been clearer that it is the photon energy that I am interested in. More specifically I am interested in phenomena that is associated with heating of a material due to exposure to light (without actually exciting any electrons).

    Thank you for your comments about classical EM waves, I wasn't fully aware of the distinction, and I realize now that the difference is significant.
     
  8. May 5, 2009 #7

    jtbell

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    Posting in the Classical Physics forum contributed to the confusion, because photons of course are not part of classical physics. :smile:
     
  9. May 6, 2009 #8
    I was quite unsure of where to post it. First I thought of Quantum and Atomic physics. Finally I decided thermodynamics was my main concern and reasoned that classical physics might be the best place for that discussion.
     
  10. May 6, 2009 #9

    ZapperZ

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    There are two separate parts to your question, and that is creating this confusion.

    When you are asking about "heating of a material due to exposure of light", you have to not only consider the heat source (i.e. light), but also the receiver of that heat source (i.e. the material!).

    For example, if all we care about is a light source having photons with higher energy, then why do we heat food using microwaves and not simply using visible light? After all, microwaves have longer wavelengths (and lower energy per photon) than visible light?

    That observation in itself means that there's something about that frequency within the microwave region that can efficiently transfer its energy to the material. In this case, the vibrational frequency of molecules (especially water or O-H bonds) that matches the microwave frequency.

    The same can be said about the heating of material. The vibrational modes that are available in a solid will dictate what range of frequencies it can absorb and thus, causes that energy to be converted into heat. The IR range is very effective in most material. So the properties of the material itself cannot be ignored in such energy transfer consideration.

    Zz.
     
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