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The Stefan Boltzmann Law gives a relation between the total energy radiated per unit area and the temperature of a blackbody. Specifically it states that, $$ j= \sigma {T}^4$$ Now using the thermodynamic derivation of the energy radiated we can derive the above relation, which leads to $T^4$. But is there any intuitive reason for the $T^4$ term?

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    $\begingroup$ How about this, physics.stackexchange.com/q/319861 $\endgroup$ May 24 '17 at 14:53
  • $\begingroup$ The fact that it's proportional to $T^4$ is purely classical. The value of the proportionality constant $\sigma$ requires quantum mechanics. For a derivation in sort of the original thermodynamic style, see section 4.6.1 of my special relativity book: lightandmatter.com/sr $\endgroup$
    – user4552
    Sep 12 '17 at 21:57
  • $\begingroup$ $\tau$ has to be power of four, if you includes $\hbar$, and the speed of light $c$ $\endgroup$
    – Shing
    Feb 18 '18 at 10:51
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There's roughly $kT$ energy in each active mode. The active modes are characterized by momenta which live inside a sphere of radius proportional to $kT$, which has volume proportional to $T^3$. Multiplying these factors gives $T^4$, and the result clearly generalizes to $T^{d+1}$ in general dimension.

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  • $\begingroup$ Sorry But I am not familiar to these terms . What do you mean by an active mode $\endgroup$ May 24 '17 at 15:05
  • $\begingroup$ If the energy cost of exciting a mode is much more than $kT$, the energy stored in that mode is exponentially suppressed (see the Boltzmann distribution). If it's much less, then the total energy is about $kT$ by the equipartition theorem. When the latter is true we say the mode is 'active'. $\endgroup$
    – knzhou
    May 24 '17 at 15:06
  • $\begingroup$ This is indeed the outline of the derivation of the S-B law in any good Statistical Mechanics textbook, and is the most direct way to get at the proportionality. Whether it is 'intuitive' or not is a personal matter... $\endgroup$
    – Jon Custer
    May 24 '17 at 15:24
  • $\begingroup$ And the derivative gives the specific heat of solids at low temperatures in the Debye model proportional to $T^3$. $\endgroup$
    – user137289
    Sep 12 '17 at 20:48
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If you know Quantum Mechanics, you know that you can set length to have dimensions of the inverse of energy. This means that $j$ must have dimensions of energy to the four. If you consider that the only variable with energy units is the temperature, then the energy density must be proportional to $T^4$.

If you consider additional constants with energy dimensions, like a mass, then the derivation is no longer valid.

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  • $\begingroup$ This seems a little misleading to me, because it makes it sound like $j\propto T^4$ is quantum mechanical, but it's not. It's purely classical. Only the value of the proportionality constant $\sigma$ is quantum mechanical. $\endgroup$
    – user4552
    Sep 12 '17 at 22:00
  • $\begingroup$ @BenCrowell Uhm. Well, don't you need QM to make thia reasoning. I could replace QM by saying that there is a fundamental action scale $h$, so it sounds more generic. But then maybe you can argue why not introduce a temperature scale too? $\endgroup$
    – jinawee
    Sep 13 '17 at 3:28
  • $\begingroup$ knzhou's argument is classical. $\endgroup$
    – user4552
    Sep 9 '18 at 16:42
  • $\begingroup$ @BenCrowell knzhou's argument is also quantum mechanical because it requires one to say that there are a certain number of modes in a sphere of momentum space (classically there are infinitely many). Indeed, the Stefan-Boltzmann constant is universal, proportional to $k^4/(h^3 c^2)$, which matches the configuration space argument as well since there are 3 factors of $h$. $\endgroup$ Sep 9 '18 at 19:25
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The most basic argument you can make is dimensional. You have three constants: $k$ (energy per Kelvin), $h$ (energy per Hertz), as $c$ (meters per second), and one parameter, $T$ (Kelvin), and need to come up with power per unit area, or energy per second per meter-squared:

Dimensionally:

Energy $\rightarrow E = kT$

Time $\rightarrow h/E = \frac{h}{kT}$, which gives:

Power = Energy / Time $\rightarrow \frac{k^2T^2}{h}$

Length = speed $\times$ Time $\rightarrow c\frac{h}{kT}$

Area = $\frac{c^2h^2}{k^2T^2}$

Combine that into Power per unit Area:

$ j \propto \frac{k^4}{h^3c^2}T^4 $.

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  • $\begingroup$ I have always wondered how dimensionless factors [Variable]/[Particular value] are ruled out in such reasonings. Here $T/T_0$: that could give you any power you like, wouldn't it? $\endgroup$
    – user154997
    Sep 12 '17 at 20:20
  • $\begingroup$ @LucJ.Bourhis: But there is no physically relevant temperature to pick for $T_0$. $\endgroup$
    – user4552
    Sep 12 '17 at 21:55
  • $\begingroup$ The trouble with these dimensional arguments is that they make it sound as if $j\propto T^4$ is quantum-mechanical, but in fact it's classical. It also obscures the fact that the exponent is the number of dimensions of spacetime. $\endgroup$
    – user4552
    Sep 9 '18 at 16:44
  • $\begingroup$ @BenCrowell I don't understand your complaint. I don't know why a dimensional argument is inherently quantum, nor do I understand how counting dimension obscures the number of dimensions; it enters in the definition of a surface (area in the 3+1 case). $\endgroup$
    – JEB
    Sep 10 '18 at 15:38
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Classical radiation energy density is infinite, Rayleigh-Jeans law. How can the thermodynamic derivation, which relays on the relation $$ P=\frac{E_{dens}}{3} , $$ be right? If it works ignoring RJ law it means that the thermodynamic relations used here are more general than classical physics. It looks like that the only needed assumption is that a $P=P(T)$ finite relationship exists which is what Boltzmann must have assumed (before Plank's time) thanks to an article by Adolfo Bartoli, who "demonstrated" that radiation has a pressure. This article is cited in Boltzmann paper, he could read and speak italian as a good austrian. In conclusion the SB law can't be derived classically; it was just derived before QM.

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  • $\begingroup$ This isn't really an answer to the question, but it is interesting. A couple of papers on this topic are Paul, iopscience.iop.org/article/10.1088/0031-8949/2015/T165/014027 , and Montambaux, arxiv.org/abs/1610.05940 . The connection between the classical and quantum issues seems logically complicated, but the Montambaux paper seems to show that the dependence on quantum mechanics is weak, in the sense that the result is completely robust regardless of the specific assumptions made about the statistics. $\endgroup$
    – user4552
    Sep 9 '18 at 17:08

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