Hi,(adsbygoogle = window.adsbygoogle || []).push({});

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 u_{f}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 Vu_{f}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.

**Physics Forums - The Fusion of Science and Community**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Integrating a differential (Problem in thermodynamic derivation)

Loading...

Similar Threads - Integrating differential Problem | Date |
---|---|

A Integration by parts of a differential | Jul 28, 2017 |

B Some help understanding integrals and calculus in general | May 22, 2017 |

A Time differentiation of fluid line integrals | Apr 7, 2017 |

I Solving a definite integral by differentiation under the integral | Mar 23, 2017 |

Integral for a differential equation problem | Apr 10, 2008 |

**Physics Forums - The Fusion of Science and Community**