# The entropy given by stefan Boltzmann's law looks remarkably similar to the volume of the sphere; $S(T)=\frac{4}{3}\sigma T^3$

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

• I do not know if it is coincidental or not, but it's definitely curious. Nice observation.
– user331392
May 8 at 18:51
• @EdV Sigma could be transcendental or even Pi in the correct units of measurement. May 8 at 19:05
• Sigma is transcendental right? Because it is expressed in $\pi$ May 8 at 19:14
• Totally a coincidence. Since $\sigma$ is not dimensionless. May 8 at 19:19
• 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. May 8 at 19:41

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 $$D$$.
• 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. May 8 at 19:18
• @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 $\pi$.)