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Cavity black-body radiation aboard a relativistic rocket

  1. Jan 9, 2007 #1
    1. The problem statement, all variables and given/known data
    If we have a cavity filled with black-body radiation, an this is placed onboard a relativistic rocket (uniform linear motion).
    I want to yield the transformation laws (from the proper frame S' joint to the rocket and moving with constant velocity V in respect to a coordinate frame S) for energy flux phi, energy density epsilon, energy E, pressure p, heat Q, and entropy S.

    2. Relevant equations
    I know these equations expressed in the proper frame:

    phi' = sigma'*T'^4

    epsilon' = (4/c)*phi' = (4/c)*sigma'*T'^4

    E' = epsilon'*psi' = (4/c)*sigma'*T'^4*psi'

    p' = epsilon'/3 = (4/c)*(sigma'*T'^4)/3

    dQ' = (16*sigma'*T'^4*dpsi')/(3*c)

    S' = (16*sigma'*T'^3*dpsi')/(3*c)


    sigma' = Stefan-Boltzmann constant in the proper frame
    T' = temperature in the proper frame
    psi' = volume of the cavity in the proper frame

    3. The attempt at a solution
    Being (it is a known formula):

    E = (E’+p’*psi’*beta^2)/sqrt(1-beta^2)

    by subsituting, i yield:

    E = [(1+(beta^2)/3)/sqrt(1-beta^2)]*E’

    And, if I want to transform pressure? I know in general that:

    p = p’ ----> (4/c)*(sigma*T^4)/3 = (4/c)*(sigma'*T'^4)/3 ----> sigma*T^4 = sigma'*T'^4


    sigma = sigma’*gamma^4

    but this implies also:

    epsilon = epsilon’

    and for the Boltzmann constant k:

    k = k’*gamma (by the definition of sigma, if we suppose invariant the Planck constant, is it right?)

    The heat transforms as follows:

    dQ = (16*sigma'*T'^4*dpsi')/(3*c*gamma)

    Now, there is a problem. In fact, if a calculate E directly from epsilon (considering right its transformation):

    E = epsilon*psi = epsilon’*(psi’/gamma) = E’/gamma

    But this is in disagree with the transformation known of the energy.

    Where is my error?
    Last edited: Jan 9, 2007
  2. jcsd
  3. Jan 10, 2007 #2


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    Science Advisor
    Homework Helper

    First, get a book on relativistic thermodynamics. Second, i hope you don't insinuate that sigma' is any different from sigma.

  4. Jan 10, 2007 #3
    i don't say that sigma' is different from sigma, my calculations said this. in fact at the end of my post i wrote: "where is my error" ? :)
  5. Sep 24, 2009 #4
    Where do you have this relationship?

    dQ' = (16*sigma'*T'^4*dpsi')/(3*c)

    It is saying that the change of heat in the BB is related to the change of volume?
  6. Sep 24, 2009 #5
    I think this is not the right way of using the formula:
    E = (E’+p’*psi’*beta^2)/sqrt(1-beta^2)
    Where ever you got the second term from, for sure in the rest frame E' there is no division by sqrt(1-beta^2).
    the Energy-Momentum states: E^2(moving frame) = E'^2(rest frame) + c^2.p^2.
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