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Black Body Radiation & how it works?

  1. Jul 13, 2011 #1
    This is just for grasping the concept of the Black Body curve and the UV catastrophe. I don't understand how ideal Black Bodies work? or how they absorb and re-emit the energy which they absorbed? Also, why does radiation bounce around the walls of a cavity of a Black Body? shouldn't the black-coated walls absorb the incoming radiation and not let it bounce?

    Thank you anyone who helps! :)
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  3. Jul 13, 2011 #2


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    By bounce they mean that it is continually absorbed and emitted. If the black body is heated and the inside is a hollow cavity with no way for light to escape, radiation passes back and forth between the walls continuously.

    A "perfect" black body is one that absorbs ALL EM radiation that passes through it. In reality there are no perfect black bodies. It turns out that ALL objects will emit EM radiation if they are at a temperature above absolute zero. This radiation is emitted by the charged particles in the atoms that make up the object. It is effectively the reverse of the object absorbing radiation. In the cavity above, if all sides of the black body are equal in temperature, the radiation emitted by AND absorbed by each wall is equal in energy.
  4. Jul 13, 2011 #3
    Thank you for your reply, Drakkith!
    You've given a ton of info that was looking to verify what I thought.

    Anyways, I understand that in reality there is no "perfect" black body and that it is a theoretical but ideal make for phsyics. However, I'm having trouble understanding how this black body is able to absorb and emit equal amounts of energy from wall to wall, are the walls in thermal equilibrium with each other? And what happens to the energy in the black body as it cools down? is the equilibrium distrubed?

    I'm sorry but thank you for your reply!
  5. Jul 13, 2011 #4


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    Well, let's assume a few things here. Let's assume that the entire black body is in thermal equilibrium. Let's also assume that no heat is lost on the outside as well. Now, if all the walls are equal in temperature then they emit, on average, the same amount of radiation over time as they absorb. This is because each wall emits the same amount as every other wall. And assuming that it is a perfect black body, then each wall absorbs all the radiation that hits it. So, each wall is the same temperature on average, it emits the same amount of energy and absorbs the same as every other wall.

    If we cool down one side of this black body, then the corrosponding wall inside the black body will emit less energy thanks to its lower temperature until everything equalizes again.
    Last edited: Jul 13, 2011
  6. Jul 13, 2011 #5
    Oh, it is in thermal equilibrium because the temperature of the walls are constant throughout, hence, causing them to emit, on average, the same amount of radiation and because it is a perfect black body, it will absorb all the radiation falling on the walls. This absorption and emission is done at a constant rate between the atom's electrons (as they lose and gain energy) and the energy avliable inside the hollow cavity?

    So, if we drill a hole inside that cavity and cause radiation to sip out, the energy circulation inside the cavity is decreasing, so would that mean that the electrons are also losing energy and converting it to the pool of energy bouncing around the hollow cavity to adjust the thermal equilibrium, thus cooling down?

    That was just my train of thought.
  7. Jul 13, 2011 #6


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    Yep. That is correct.
  8. Jul 13, 2011 #7
    Oh sweet!
    Thank you for solving my problems! I appreciate the help I got!

  9. Jul 13, 2011 #8
    I'm sorry, could I ask one more thing?
    In the Black Body Radiation curve that Planck theoretical proved, why is only a small amount of UV radiation or high frequency radiation given out of the ideal balck body? If we keep increasing the temperature high enough, above the temperatures we are able to produce on earth, would we see more UV radiation given out?

    Is the amount of UV photons given out up to the probablity of the electrons falling back to their ground state? Is it more probable to be giving out photons of light than of UV or higher?
  10. Jul 13, 2011 #9
    Yes the higher the temperature the higher the energy of the most probable photon. Hot enough and the most probable photon will be 100 Gev gamma rays.
  11. Jul 13, 2011 #10


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    The full version requires a description of Plancks Law, which I am not familiar enough to give accurately. See here for more though: http://en.wikipedia.org/wiki/Planck's_law

    But I can say that yes, when you increase the temperature you have more energy available to make more photons with higher energies.
  12. Jul 13, 2011 #11
    hmm, okay.
    I'll look into it but thank you both for you're help!
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