# Comparing Wein's formula with Rayleigh-Jeans' formula for energy density of blackbody radiation

In our Higher Secondary textbook, it is clearly stated that Wein provided with the formula of Blackbody radiation based on the thermodynamics' law below $$E_{\lambda} d\lambda = 8 \pi hc\lambda^{-5}e^{-\frac {hc}{k \lambda T}} d\lambda$$

Later on, Rayleigh-Jeans gave us the followed equation within the range of $$\lambda + d\lambda$$ $$E_{\lambda} d\lambda = \frac {8\pi kT}{\lambda^{4}} d \lambda$$

For any shorter wavelength, Wein's formula matches with the experimental result but not consistent for longer wavelength whereas formula given by Rayleigh-Jeans makes a good performance for longer wavelength but no longer works for the experimental data used for analyzing energy density for comparably shorter wavelength.

I did somewhat a research for my curiosity. Now if I merge both equations by showing them equal for any optimized range of wavelength with its precise scale for any particular temperature '$$T$$' $$8 \pi hc\lambda^{-5}e^{-\frac {hc}{k \lambda T}} d\lambda = \frac {8\pi kT}{\lambda^{4}} d \lambda$$ $$\implies kT \lambda = hc e^{-\frac {hc}{k \lambda T}}$$ $$\implies \lambda = \frac {hc}{kT}e^{-\frac {hc}{k \lambda T}}$$ $$\implies \ln \lambda = \ln (\frac {hc}{kT}) - \frac {hc}{k \lambda T} \ln e$$ $$\implies \frac{d}{d \lambda} (\ln \lambda) = \frac {d}{d \lambda} {[ \ln (\frac {hc}{kT})]} - \frac {d}{d \lambda} (\frac {hc}{k \lambda T})$$ $$\implies \frac {1}{\lambda} = -\frac {hc}{kT}(-\frac {1}{ \lambda^{2}})$$ $$\therefore \lambda = \frac {hc}{kT}$$

And if we apply the value of $$\lambda$$ into Wein's formula, then we get Rayleigh-Jeans' formula and vice versa. So My Question is:

"Does it mean that in some certain points, both formula works the same for a certain range of wavelength equal to the numerical value of $$\frac{hc}{kT}$$ in such a way that experimental result can predict it with lots of clearance?"

This is a hard question, I cannot provide a full answer but maybe this mathematical insight will be helpful.

When you equate the two formulas and solve for $$\lambda$$. you are actually finding the points where two curves intersect. Now the formula $$\lambda = \frac {hc}{kT}e^{-\frac {hc}{k \lambda T}}$$ defines the intersection implicitly and by the plots, there will be one (edit: at most one) numerical $$\lambda$$ where this holds.

Therefore, this is a single point defining formula and not a function of $$\lambda$$ that can be differentiated.

Perhaps if $$\lambda$$ was also a function of another parameter $$s$$, the formula would define a differentiable function $$\lambda(s)$$ via the implicit function theorem, and then differentiation would be possible.

Just my 2c.