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Integrating a differential (Problem in thermodynamic derivation)

  1. Jun 10, 2014 #1
    Hi,

    I'm looking at a derivation of the thermodynamics of black-body radiation. My question is in regards to the math of the derivation.

    Using the first law of thermodynamics and considering an adiabatic expansion of the cavity, it can be stated that

    [itex]dU = -\frac{u}{3}dV [/itex]

    Here small u is the volumetric density of the internal energy U/V.

    Small u can also be described as

    [itex]u = \int_{0}^{\infty}u_fdf[/itex]

    Here uf is the distribution function for how the internal energy density is distributed amongst the various frequencies (wheew!).

    The steps where I'm confused goes like this:

    [itex]u = \frac{U}{V} = \int_{0}^{\infty}u_fdf \Longleftrightarrow[/itex]
    [itex]U = \int_{0}^{\infty}Vu_fdf \Longrightarrow[/itex]
    [itex]dU = \int_{0}^{\infty}d(Vu_f)df [/itex]

    My confusion comes from the last step. I can't see how the differential of U is the same as the integral of the differential of Vuf with respect to the frequency f.

    Can anybody explain the math behind this? It's quite possible that it's just a small thing I just can't see :D But any help is most appreciated :)

    Thank you.
     
  2. jcsd
  3. Jun 11, 2014 #2
    Well I figured it out now, and as I thought, it was quite simple.

    Since

    [itex]U = \int_{0}^{\infty}Vu_fdf = \int_{0}^{\infty}Fdx[/itex]

    F is now just equal to Vuf. I just made a change of variable names to make it more simple.

    [itex]dU = Fdx[/itex]

    [itex]dF = fdx[/itex]

    [itex]\int_{0}^{\infty}dFdx = \int_{0}^{\infty}f dx dx = F dx = dU[/itex]

    This corresponds to the problem I was asking.
     
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