# Planck's Form of Wien's Distribution Law

Can anyone explain how Wien's distribution law seen in Wien's original "On the Division of Energy in the Emission-Spectrum of a Black Body":

$$\phi_\lambda = \frac{C}{\lambda^5} e^{-\frac{c}{\lambda \theta}},$$

where (using Wien's notation) $\phi_\lambda$ is the intensity, $\lambda$ is the wavelength, $C$ and $c$ are constants and $\theta$ is the temperature, becomes in Planck's "On the Law of Distribution of Energy in the Normal Spectrum" (in Planck's notation)

$$E.d\lambda = \theta^5 \psi(\lambda \theta).d\lambda,$$

where $E$ is the energy, $\lambda$ is the wavelength and $\theta$ is the temperature.

SOLUTION: We want to show that Planck's expression $E = T^5 \psi (\lambda T)$ is equivalent to Wien's distribution law, which has a general form $E = F(\lambda)e^{-\frac{f(\lambda)}{T}}$, but which in more concrete terms (as shown by Wien) must be $E = \frac{C}{\lambda^5}e^{-\frac{c_1}{\lambda T}}$

First off, we divide the r.h.s. by $c^4$ which, unlike Planck's case where he deals further with frequency, here, in the wave-length expression, $c^4$ is nowhere to be found. We'll do the division nevertheless, for the sake of the discussion and will get

$$E = C T^5 \psi (\lambda T),$$

where $C = \frac{1}{c^4}$. Then we will multiply and divide the r.h.s. by $\lambda^5$ and will get

$$E = \frac{C}{\lambda^5} (\lambda T)^5 \psi (\lambda T)$$

From the above we get

$$F(\lambda) = \frac{C}{\lambda^5}$$

and

$$e^{-\frac{c_1}{\lambda T}} = (\lambda T)^5 \psi (\lambda T).$$

So, the function $\psi(\lambda T)$ in \S6 of Planck's paper must be

$$\psi (\lambda T) = \frac{e^{-\frac{c_1}{\lambda T}} }{(\lambda T)^5}$$

in order for Planck's and Wien's expressions to be equivalent. The function $\psi$, however, doesn't constitute the whole exponent part, which is $(\lambda T)^5 \frac{e^{-\frac{c_1}{\lambda T}}} {(\lambda T)^5} = e^{-\frac{c_1}{\lambda T}}$, and that, indeed, makes the two expressions equivalent.

## 1 Answer

The two are clearly compatible, as we can take $$\psi(x) = \frac{1}{x^2}e^{-c/x}$$ which gives Wien's law in the way stated by the OP. In the original paper by Wien, the author starts from a version of the law which appears to be incompatible with the one stated by Planck: $$\phi_\lambda = F(\lambda) e^{-f(\lambda)/\theta}$$ where $\theta$ is the temperature, in accordance with the custom at the time. The original article by Planck mentions a specific reference for the formula he's using: M. Thiesen. Ver. Deutsch. Phys. Ges., 1900, 2, 66. The referenced journal can be found here. M. Thiesen fails to provide a derivation but quotes Wien, despite noting that

Die hier gegebene Formulierung des Gesetzes fehlt bei Wien

which means (translation mine):

The formula given here is not present in Wien's work

I could not find the two papers referenced by M. Thiesen. However, I managed to find this letter submitted to Nature in 1948 by E.O. Hercus ("The Proof of Wien's Law"), which is a brief review of Wien's proof of Planck's statement, which apparently involves fairly complicated considerations about light in a cylindrical cavity.

I could not find Wien's referenced paper, but I managed to find another one of the references, the book "The Electron Theory of Matter" by O. Richardson, which gives a proof at pages 339-342. The proof involves a lot of geometrical considerations about light in a cylindrical box, some experimental considerations and Stefan-Boltzmann's law.

EDIT. Perhaps it is useful to understand the historic progression. By considering spherical cavities, Wien (1893-1894) proves that $$\phi_\lambda = \theta^5 \psi(\lambda \theta)$$ Or equivalently, $$\phi_\lambda = \frac{\theta^5 \lambda^5}{\lambda^5} \psi(\lambda \theta)=\frac{1}{\lambda^5} \chi(\lambda \theta)$$ where we have defined a new function $\chi(\lambda\theta) = (\lambda\theta)^5 \psi(\lambda \theta)$. The functions $\chi$ and $\psi$ are unknown. This is emphasised by Richardson (p. 343):

It is probable that the relations (23) and (24) which involve the universal undetermined functions $\phi$ and $\chi$ of the argument $\lambda T$ are as far as we can get, from such very general considerations as have been employed above.

where the author adopts a different notation (his $\phi$ is our $\psi$ and $T$ is temperature). However it should be pointed out that this is enough to establish a very important result, Wien's displacement law. Indeed differentiating $\phi_\lambda$:

$$\frac{\partial \phi_\lambda}{\partial \lambda} = \lambda^{-6} \left[\lambda \theta \chi'(\lambda \theta) -5\chi(\lambda\theta) \right]$$

so the maximum of $\phi_\lambda$ is given by setting the thing inside square brackets to zero. The relevant thing is that this quantity only depends on the product $\lambda \theta$, so if a maximum is attained, it will be at $\lambda \theta=b$ for some $b$. This is Wien's displacement law.

Later Wien (1896) tries to get an explicit formula for the function $\chi$ (or equivalently for $\psi$). And he does this via a completely different path, which is illustrated in the paper referenced by the OP. The result he gets is more specific (he employs a number of assumptions, see the paper itself) but as mentioned at the beginning is compatible with the previous result.

• Nature appears to have published only p.144 of Hercus' letter. Page 143 is missing. Does Richardson's book derive an expression of Wien's law containing $T^5$? – ganzewoort Dec 23 '17 at 1:38
• Yes. He derives the formula which is the one you quoted. I don't think the proof is particularly interesting (it's very nineteenth century) but I can post an outline if you want – John Donne Dec 23 '17 at 1:55
• Yeah, please do. Mind you, I'm not interested in the derivation per se. All I want to see is how the two formulae in the OP are equivalent to each other. If you can show that equivalence that'll be great. Thanks. – ganzewoort Dec 23 '17 at 2:47
• OK, Richardson derives $L\left(\lambda\right) d\lambda = T^5 \phi(T\lambda)d\lambda$ and then multiplying and dividing the r.h.s. by $\lambda^5$ he gets $L\left(\lambda\right) d\lambda = \lambda^{-5} \chi(T\lambda)d\lambda$. But, then, how did $\chi(T\lambda)$ become $C e^{-\frac{c}{T\lambda}}$ given by Wien? – ganzewoort Dec 23 '17 at 15:44
• How did $T^5$ melt into $T$? – ganzewoort Dec 23 '17 at 15:50