Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Violation of thermodynamics law

  1. Sep 21, 2013 #1
    Black body is a body which absorbs every material and doesn't reflect anything.
    But in thermal equilibrium the radiation absorbed by i t,has to be emitted to maintain thermal equilibrum.
    Which means the radiation falling on it is emitted.ie. it's reflecting the radiation doesn't it?

    When we see at atomic scale the difference between reflection and emission is nothing because a ray is emitted when electron jumps from higher orbital to lower orbital and the incident rays interact with atoms to produce reflected rays.

    So if the black body reflects every radiation, doesn't the thermal equilibrium will be absorbed.
     
  2. jcsd
  3. Sep 21, 2013 #2
    Yes
    No. A blackbody will absorb the incident light and re-emit it, but a blackbody does not directly reflect the incident light. For a blackbody, the emitted light follows the blackbody spectrum, which is in general made up of a wide range of frequencies. For example, we could shine a very high frequency light--say a blue light--onto the blackbody, and all the energy of the blue light would get absorbed by the blackbody. When the blackbody absorbs this energy, it heats up slightly, and it will re-emit blackbody radiation with frequencies depending on its temperature. At room temperature, the majority of the blackbody radiation emitted will be in the infrared range. So at room temperature, if we shine a blue light on the blackbody, it still looks black--nothing is reflected. The absorbed energy gets reradiated isotropically with no memory of what the color of the incident light was. So it is not a reflection.

    This is not a good way to look at why reflections occur. Without any sort of chemistry or quantum mechanics, reflection of electromagnetic waves appears in classical electromagnetism. For example, the surface of a perfect conductor is a perfect reflector in classical EM.

    Real metal conductors (like an aluminum mirror) are not well described the atomic absorbtion-reemission model you suggest. The conduction electrons are "free"--they are shared between many of the metal lattice sites--and light interacts with the "electron fluid" in ways quite dissimilar to the interaction with a hydrogen atom.
     
    Last edited: Sep 21, 2013
  4. Sep 21, 2013 #3
    Emitted rays have the characteristic of the substance emitted rays, without having any memory of the l8get which gave them energy initially, whereas reflected rays have the same frequency of the incident rays.

    But how Can we distinguish the difference between reflection and emission in microscale? Is there any theory to explain?

    Thanks in advance.
     
  5. Sep 21, 2013 #4
    Reflected radiation will preserve properties related to the original source such as wavelength. re-emitted radiation doesn't. For instance, if you reflect the sun's visible radiation with a mirror, you get visible radiation. If you absorb it and re-emit it, you get radiation related to your own temperature - not the sun's - by Wien's law. That will be Infrared radiation for a room temperature black body.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook