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Is the following proof of Kirchoff's law correct?

  1. Jun 11, 2007 #1
    The standard book used in our university proves the Kirchoff's law of radiation in an elaborate manner...I understand the proof.But I think using the same principles,this can be proved in a much better way:

    Consider an enclosure is at temperature T and the enclosure is filled with temperature radiation emitted by the wall.

    Let a body A has been placed inside an enclosure along with a blackbody B of IDENTICAL geometry.The enclosure have wall opaque to all wavelengths and is insulated thermally from the surroundings.In equilibrium state when A and B attain the enclosure temperature T,the total energy aborbed by them will be equal to the total energy emitted by them.

    Let emissive power of absorptive power be given by e_λ and a_λ.
    If the body A is of abnormal geometry different parts of the body may have different a_λ and e_λ...Suppose an amount of ∑∆U(λ) of radiation falls on the body A in a given time ∆t.As A and B are of identical geometrical shape the radiation falling on the blackbody B is also ∑∆U(λ) in the same time ∆t.

    Then for the blackbody we have ∑∆U(λ)=k*∑(e_bλ) where k is a constant and ∑(e_bλ) is the radiation from the blackbody

    For the body A,we have ∑a_λ*∆U(λ)=k*∑e_λ where ∑e_λ is the radiation from the body A.

    The same proportionality constant has been used because the two bodies are of identical geometrical shapes and radiation emitted in the same time ∆t has been considered.

    Hence from the above relations,

    [∑a_λ*∆U(λ)/∑∆U(λ)]=[∑e_λ/∑(e_bλ)]

    Since this equality should hold for each portion of the spectrum,we must have
    a_λ=(e_λ/e_bλ) Which proves Kirchoff's law.

    Please give me feedback.Any incorrect area or somewhere the logic should be more solid...
     
    Last edited: Jun 11, 2007
  2. jcsd
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