If I am not mistaken the entropy for a blackbody per unit area is given by: $$S(T)=\frac{4}{3}\sigma T^3.$$

The volume of a sphere is given by: $$ V(r) =\frac{4}{3}\pi r^3. $$ Is this coincidental? I can't really imagine a hypothetical sphere with volume 'entropy' and radius temperature. It could be that I am misunderstanding the formula.

Where $\sigma$ is $$ \sigma = \frac{2\pi^5k_{\rm B}^4}{15h^3c^2} = \frac{\pi^2k_{\rm B}^4}{60\hbar^3c^2}\,,$$

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    $\begingroup$ I do not know if it is coincidental or not, but it's definitely curious. Nice observation. $\endgroup$
    – user331392
    May 8 at 18:51
  • $\begingroup$ @EdV Sigma could be transcendental or even Pi in the correct units of measurement. $\endgroup$
    – Swike
    May 8 at 19:05
  • $\begingroup$ Sigma is transcendental right? Because it is expressed in $\pi$ $\endgroup$ May 8 at 19:14
  • $\begingroup$ Totally a coincidence. Since $\sigma$ is not dimensionless. $\endgroup$
    – Hosein
    May 8 at 19:19
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    $\begingroup$ It's a nice visualization. This visualization allows one to think of the number of available microstates as the volume of a sphere whose radius is the temperature. Thanks for your observation. Sorry to put this as an answer. I'm too new to comment. $\endgroup$ May 8 at 19:41

1 Answer 1


It's a coincidence, as the lack of $\pi$ indicates. The entropy per surface of a blackbody in $D$-dimensional space is $\frac{D+1}{D}\sigma T^D$. (You can deduce it e.g. by generalizing this.) By contrast, the unit $D$-ball has volume $\frac{\pi^{D/2}}{\Gamma(D/2+1)}$, which decays superexponentially at large $D$.

  • $\begingroup$ Thank you for your comment. It is not entirely lacking $\pi$ right, isn't it hidden in the $\sigma$? I am not proficient enough in math to understand that material in the link. $\endgroup$ May 8 at 19:18
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    $\begingroup$ @bananenheld No, I wouldn't say that. We define $\sigma$ as the proportionality constant in $P=\sigma AT^{D+1}$, where $A$ is the $(D-1)$-dimensional "area". The entropy is then $\frac{D+1}{D}\sigma AT^D$. (Of course, for a spherical blackbody $A$ has a formula including $\pi$.) $\endgroup$
    – J.G.
    May 8 at 19:20

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