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Black-body radiation of a parabolic mirror in a vacuum violates thermodynamics?

  1. Jul 18, 2012 #1
    Preface:
    I understand that regardless of any further details, the correct answer is "No, it does not, because..." I'm looking for an answer because clearly I don't understand something about either thermodynamics, mirrors, or black body radiation. In my mind, the scenario I describe follows the rules of black body radiation and mirrors to violate the second law of thermodynamics. This is almost certainly incorrect, so I need someone to tell me what I'm getting wrong.

    Scenario:
    Imagine you have a 1 meter long and 200 cm wide aluminum parabola with a 99% reflective mirrored surface on the inside. Suppose the outside of the parabola has been blackened by a laser. To keep this mentally simple, let's get rid of convection by placing this mirror in a chamber with a near-perfect vacuum, and let's get rid of conduction by placing the chamber and mirror on the international space station so that the mirror can stay in the middle of the chamber without touching anything. Assuming you don't change the temperature of the chamber, the mirror and chamber should (presumably) reach an equilibrium at room temperature.

    Problem:
    The mirror will produce black-body radiation. However, most of the black-body radiation emitted from the inside of the parabola will be reflected in a single direction, towards the chamber. This means that the parabola will be emitting directed energy which heats up one side of the chamber. We should be at thermodynamic equilibrium, but we're not. What's going on here?
     
  2. jcsd
  3. Jul 18, 2012 #2
    How do you know this? By "single direction" you mean a parallel beam?
    And why do you think that the back of the mirror does not emit?
     
  4. Jul 18, 2012 #3

    mfb

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    Staff: Mentor

    A parabola can reflect everything coming from one specific direction to a specific point. Everything else will have a focus elsewhere.

    If the inside is reflecting in the infrared, it will not emit much radiation (it is "white"). Photons from the chamber walls just hit the chamber walls again, some photons stay at the parabola, some photons are emitted by the parabola. There is nothing special about the shape here.
     
  5. Jul 18, 2012 #4
    No, but a somewhat focused beam. If you imagine a line going from the center of the parabola and extending outward, I think most of the energy would be focused to within 60 degrees of that line.

    Parabolas tend to reflect light away from themselves in mostly the same direction, right?

    I think it would emit at least as much as the mirror. The Stefan-Boltzmann law suggests that temperature and surface area are the only variables which determine the energy released by black-body radiation. It might be the case that the blackened exterior is hotter than the reflective interior, but I think that would mean that the system is not in thermodynamic equilibrium.

    The problem is that the mirrored surface is reflective, while the exterior is not. The exterior would be very good at absorbing heat, the interior would be very good at reflecting it. This might create a cycle where heat is absorbed by the exterior of the parabola, then emitted by both the exterior and interior, but focused in a certain direction when it's emitted by the interior.
     
  6. Jul 18, 2012 #5
    Right, it wouldn't be precisely focused. But it would be more focused than ordinary.

    If I understand you correctly, this might be the explanation. Are you saying that reflective surfaces tend to emit less radiation, or tend to emit a different spectrum of radiation?
     
  7. Jul 18, 2012 #6
    If it helps, let's try something that just seems wrong: a perpetual motion device. Imagine you place a blackened cap on the end of the parabola to absorb the heat, and you thoroughly insulate the cap so that it can get quite hot. Then, you use the hot cap to run a heat engine at room temperature without any input of energy. Where would this device fail?

    Edit: Intuitively, I'd say that the cap would not get hot at all, but I can't explain why. The mirror would radiate heat and most of it would be reflected to the cap. Most of the heat radiated from the cap to the mirrors would also be reflected back to the cap. How is it possible that the cap would stay cool, or would the device actually heat it up but fail at some other point?
     
    Last edited: Jul 18, 2012
  8. Jul 18, 2012 #7

    Dale

    Staff: Mentor

    Here is the key mistake. The mirror is (by definition) not a black body and will not produce black-body radiation. In this case, the mirror is 99% reflective, which means that it absorbs only 1% of the energy of a black body at the same temperature and also it emits only 1% of the energy of a black body at the same temperature.
     
  9. Jul 18, 2012 #8
    True, but the back of the mirror is a black body. Heat would be conducted from the blackened exterior to the reflective interior, right?

    Or does the mirror actually emit less energy even if it's at the same temperature?
     
  10. Jul 18, 2012 #9

    Dale

    Staff: Mentor

    Yes.
     
  11. Jul 18, 2012 #10
    Very cool. While I was waiting for a reply I checked some Wikipedia articles regarding non-black body radiation (it didn't even occur to me to check for that!) and found Kirchhoff's law of thermal radiation, Low emissivity, and Thermal radiation which indicate that for any material, the coefficients of emissivity and reflectivity should add together to equal 1.

    Thanks for helping me everyone. :)
     
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