Saturday, 25 July 2015

thermodynamics - 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.



Answer



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.


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