# Boltzmann’s original derivation of the Stefan–Boltzmann law

Does anyone here know a source where I can find Boltzmann’s original derivation (using primarily thermodynamic arguments) to the Stefan–Boltzmann law (the radiant power emitted by a body in thermal equilibrium is proportional to the fourth power of the absolute temperature)? I have done some research on Google, but I have not found anything much enlightening about this.

• Mar 19, 2017 at 6:39

Boltzmann built on two known properties of cavity radiation.

(1) The energy density, $u$, defined as $u = U/V$, depends only on temperature, $T$.

(2) The radiation pressure, $p$ is given by $p= u/3$. Radiation pressure was given a firm basis c1862 by Maxwell. The factor of 1/3 arises because of the three-dimensionality of the cavity, in which radiation is propagating in all possible directions. [It's easy for us now to derive this equation by considering the cavity as containing a photon gas.]

Boltzmann (1884) used a thought-experiment in which a cavity is fitted with a piston, and we take the radiation inside it through a Carnot cycle. On a $p–V$ diagram the isothermals are just horizontal lines, because $u$ is constant so $p$ is constant. The heat input along the top (temperature $T$) isothermal is $\Delta U + p \Delta V$. This works out to be $4 p \Delta V$. If the lower temperature isothermal is only slightly lower, at temperature ($T - d T$) then the cycle appears as a thin horizontal box, and the net work done during the cycle is simply $dp \Delta V$ in which $dp$ is the infinitesimal pressure difference between the two isothermals. We can then apply the definition of thermodynamic temperature in the form

$$\frac{dT}{T} = \frac{\text{work}}{\text{heat input}} = \frac{dp \Delta V}{4 p \Delta V} = \frac{dp}{4 p}~.$$

This integrates up easily to give the Stefan-Boltzmann law.

A modern thermodynamic derivation would probably start from the fundamental equation (embodying first and second laws of thermodynamics)… $$\text {d}U = T \text{d}S - P \text{d}V.$$ Therefore for an isothermal change $$\left(\frac{\partial U}{\partial V} \right)_T = T \left(\frac{\partial S}{\partial V} \right)_T - p .$$ Using one of the Maxwell relations, this becomes $$\left(\frac{\partial U}{\partial V} \right)_T = T \left(\frac{\partial p}{\partial T} \right)_V - p .$$ But for the radiation, $$\left(\frac{\partial U}{\partial V} \right)_T = \left(\frac{\partial (uV)}{\partial V} \right)_T = u = 3p \ \ \ \ \ \text{and} \ \ \ \ \ \left(\frac{\partial p}{\partial T} \right)_V = \frac {\text{d} p}{\text{d}T}.$$ Making these substitutions and tidying, $$4p = T \frac {\text{d} p}{\text{d}T}.$$ As before, separation of variables and integration gives the Stefan-Boltzmann law.

• How exactly have you written dT/T = Work/Heat input?I get it intuitively, but mathematically? Jan 29, 2022 at 13:03
• @lazearoundall day. Kelvin temperature can be defined by $$\frac {T_1}{T_2}=\frac {Q_1}{Q_2}$$ in which $Q_1$ is the heat input to the Carnot engine at temperature $T_1$ and $Q_2$ is the heat excreted at temperature $T_2$. Subtracting 1 from both sides of the equation, we can write it as $$\frac {T_1-T_2}{T_2}=\frac {Q_1-Q_2}{Q_2}.$$ Using energy conservation (or the first law of thermodynamics, noting that there is no change in internal energy over a complete cycle), $(Q_1-Q_2)$ is the work output. Jan 29, 2022 at 13:19
• @lazearoundallday Please see above comment. Jan 29, 2022 at 14:09
• @PhilipWood Could you please provide me any source containing the derivation of the relation between radiation pressure and energy density of radiatons? Apr 30, 2022 at 8:40
• @AYM Shahriar Rahman Reif: Fundamentals of Statistical and Thermal Physics (End of scection 9.13), Mandl: Statistical Physics (Section 10.4) Apr 30, 2022 at 18:53