# Frequency and wavelength of thermal radiation [duplicate]

Why for a given temperature these forms of Wien's law implies a different maximal wavelength?

• Do they? I haven't done the calculations to verify the exact numbers stated, but remember $\nu$ is the frequency, which is equal to $c/\lambda$. It looks like the equations in terms of the exact constants are correct. Commented Oct 2, 2019 at 3:13
• I carried out a computation fro $T = 6000 \ {\mathrm K}$. It leads to results $\nu = 352 \ {\mathrm THz}$ and $\lambda = 849 \ {\mathrm nm}$ for the equation $\nu_{max}$ and to results $\nu = 620 \ {\mathrm THz}$ and $\lambda = 482 \ {\mathrm nm}$ for the equation $\lambda_{max}$ . So are they two different functions? Commented Oct 2, 2019 at 3:50
• @JackCeroni They’re different. The $x$ in the two formulas is not the same number. See en.wikipedia.org/wiki/… Commented Oct 2, 2019 at 4:22
• @G.Smith I see, my bad, thanks for clarifying. Commented Oct 2, 2019 at 12:52

These are maxima of two different functions. The former, $$\lambda_\text{max}$$, is the maximum of the blackbody radiance per unit wavelength,

$$B_\lambda(\lambda,T)=\frac{2hc^2}{\lambda^5}\frac{1}{e^{hc/\lambda k_BT}-1}.$$

The latter, $$\nu_\text{max}$$, is the maximum of the blackbody radiance per unit frequency,

$$B_\nu(\nu,T)=\frac{2h\nu^3}{c^2}\frac{1}{e^{h\nu/k_BT}-1}.$$

Note that $$B_\lambda$$ and $$B_\nu$$ are not related by $$\lambda\nu=c$$; there is a fifth power of the wavelength but a third power of the frequency. Instead, they are related by

$$B_\lambda d\lambda=-B_\nu d\nu,$$

which expresses the fact that the radiance in a particular spectral interval is the same regardless of whether one is characterizing the spectrum by wavelength or by frequency.

Because they are two different functions, their maxima are not related by $$\lambda\nu=c$$.