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.
Thermodynamics – Boltzmann’s Original Derivation of the Stefan–Boltzmann Law
specific-referencethermal-radiationthermodynamics
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Thermal output radiation from the small sphere is given by the Stefan-Boltzman law,
$Power = A σ T^4 = σ 4π r^2 T^4$
but of course the NET radiated power is more interesting; output minus input
$Net\space{power} = σ 4π r^2 (T_{inner} ^4 - T_{outer}^4)$
The important thing to remember, is that the area of that small sphere is part of the scale of its thermal radiation, but the enclosing sphere area is not (because the enclosing sphere radiates to itself as well as to the small sphere, but the small sphere faces only the larger one).
For the thermal equilibrium to occur at equal temperature, there can be no different emitting area and absorbing area, by the zeroth law of thermodynamics: when two items are in contact, heat flows from hotter to cooler.
Yes, it applies, and it's not really related to the Stefan-Boltzmann law.
The energy radiated from a blackbody at temperature $T$ does indeed scale like $T^4$. Any object (blackbody or not) can absorb radiated energy, and that is the part which increases the temperature.
The inverse square law is a statement about the density of radiation (or intensity, in units of $W/m^2$) from a point source, not about either the source or receiving blackbody itself. If at a distance of one meter from a point source an object receives 1 $W/m^2$ of radiative energy, then at a distance of 2 meters the same object will receive 0.25 $W/m^2$ of energy. That's true for a monochromatic point source as well as for a blackbody, and comes exclusively from geometry.
I won't say that it maps directly on to temperature rise of a receiving object (just because that also depends on heat capacity and re-radiation and such) but a rigorous statement is that the rate of thermal energy absorption follows the inverse square law.
Regarding the question of determining the temperature of a blackbody heat source by the inverse square law, the answer is "not really". If you know the surface area of the heater, then measuring the radiated intensity at one distance is all you need. If you don't know the area, measuring the radiated intensity is only going to give you a number proportional to $AT^4$, where $A$ is the heater surface area. You can get more information about temperature by looking at the spectrum of the emitted radiation and using Wien's law: $\lambda_{max} = b/T$, where $b\approx$ 2.9 mm K and $\lambda_{max}$ is the most intense wavelength.
Best 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.