1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Conservation of quantized energy

  1. Jul 17, 2014 #1
    The following text is an excerpt from a physics book:

    I find this paragraph a bit hard to grasp, so I have some questions about it and I would appreciate if they are clarified.

    1) What is meant by "the energy carried off by the radiated electromagnetic waves"? As I interpret this it means the energy that are within (travelling) electromagnetic waves. Right? (I think the word "off" confuses me, why not just say "carried BY"?)

    2) What does it mean that the energy is "lost"? As energy can't be destroyed but only transformed, what is it transformed into?

    3) As I understood a a perfect blackbody, it is a body which absorbs AND re-emits all electromagnetic waves that are incident on it. But if the body re-emits everything that is incident on it, how can the object both gain (absorb) this energy and at the same time re-emit it? That would mean it duplicated the energy? Or does it only absorb half the incident energy and re-emits the other half?

    4) Are "the energy carried off by the radiated electromagnetic waves" and "the energy lost by the atomic oscillators" always identical? Or only so for a perfect blackbody?

    5) Is there some other energy form involved in this situation that is not mentioned here?
  2. jcsd
  3. Jul 17, 2014 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    1. Correct. It is the energy of the EM wave.

    2. Lost just means that the oscillator had 3hf energy, and now it has 2hf. An amount of energy equal to hf is no longer possessed by the oscillator. It has been "lost" by the oscillator to the EM wave.

    3. A perfect blackbody absorbs all EM radiation incident on it. It emits EM radiation based on its temperature, not based on the radiation absorbed. In other words, you could shine 100 watts of microwave radiation on it, heating it to 1,000 k and it would be emitting primarily in the infrared range. Heat it to 5,000 k and it would emit primarily in the visible range.

    4. They are always identical since energy is conserved.

    5. Nope.
  4. Jul 17, 2014 #3
    But what does it mean it has been "lost"? Where did it go?

    Ok, I understand the emitted radiation depends on temperature. But then I have some follow-up questions:

    6) Under which circumstances does a perfect blackbody re-emit all incident radiation?

    7) Is the absorbed energy of a perfect blackbody always equal to the energy the same body re-emits?

    8) If we shine 100 wattseconds on a perfect blackbody, will it a) also absorb 100 wattseconds? b) AND re-emit 100 wattseconds? Or how would the energy be distributed?

    Thanks in advance.
  5. Jul 17, 2014 #4


    User Avatar
    Science Advisor
    Gold Member

    6) In thermal equilibrium a blackbody will emit just as much ENERGY as it absorbs. The absorbed radiation is absorbed, it can't in a sense be "re-emitted" in the sense that you seem to be using this word. You seem to be using the word re-emitted like reflected.

    7) see 6 above, in thermal equilibrium yes, otherwise no.

    8) If we had a blackbody at 0K in a complete vacuum (no CMB) and shined a 100w flashlight on it for 1 second, it would absorb 100 watts. It would increase in temperature a little, and then eventually emit 100 joules of radiation. How long it takes to do so depends on its surface area.
  6. Jul 17, 2014 #5
    It's true I think of the "re-emitting" as reflection, are you saying this view is incorrect? How can then one think of it, how is it different from reflection?

    So if I understand it correctly (please correct me if I'm wrong) when we have electromagnetic radiation shining on a perfect blackbody, the body absorbs (all of) the energy of the radiation, which makes the body heat up to some extent. Then the body emits the same amount of energy again. The energy being radiated is equal to the energy being absorbed which is again equal to the energy being emitted. Is that correct?
  7. Jul 17, 2014 #6


    User Avatar
    Staff Emeritus
    Science Advisor

    It went to the emitted EM wave. That's it. It wasn't destroyed, it was moved.

    As Matterwave said, the radiation is absorbed, not reflected.

    In Matterwave's example, the blackbody absorbed 100 joules of energy in 1 second (100 watts). It then re-emits 100 joules of energy over a much longer period of time since the 100 joules only heated it up a little bit. So yes, it emits the same amount of energy that it absorbs.

    100 watts per second is 100 joules of energy, so let's use that. If you shine 100 joules of energy onto the blackbody it will re-emit that same amount of energy as radiation over time. The distribution of frequencies depends on how hot the blackbody is.

    Think back to my earlier example with the microwaves. The blackbody absorbs all of this radiation. If the blackbody is heated up to room temperature, most of the radiation is emitted as infrared radiation, not microwaves. The hotter it is, the higher the average frequency of the emitted radiation.

    That is correct.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook