# I Converting between λ and ν for Blackbody Radiation?!

1. Apr 18, 2016

Forgive me for this stupid question, but how do I convert between

and

I tried c = νλ but that doesn't work. This is the Rayleigh Jeans Law by the way.

Last edited: Apr 18, 2016
2. Apr 18, 2016

### drvrm

pl. give the full expression of the quoted equation and where these two are being used.

3. Apr 18, 2016

4. Apr 18, 2016

### drvrm

The Rayleigh–Jeans law agrees with experimental results at large wavelengths (low frequencies) but strongly disagrees at short wavelengths (high frequencies). This inconsistency between observations and the predictions of classical physics is commonly known as the ultraviolet catastrophe,

see the full expressions
associated Rayleigh–Jeans limits are given by

or

now you can see the approximations- actually they are not exact expressions

5. Apr 18, 2016

### mfig

6. Apr 18, 2016

### vanhees71

You only have to remember that these are probability distributions $\mathrm{d} N/\mathrm{d} \nu$ or $\mathrm{d} N/\mathrm{d} \lambda$. Now you have $\nu=c/\lambda$. This implies
$$B_{\nu}=\frac{\mathrm{d} N}{\mathrm{d} \nu}=\frac{\mathrm{d} N}{\mathrm{d} \lambda} \left|\frac{\mathrm{d} \lambda}{\mathrm{d} \nu}\right| = B_{\lambda} \frac{c}{\nu^2}.$$
Now with
$$B_{\lambda}=\frac{2 c k_B T}{\lambda^4}=\frac{2 k_B T \nu^4}{c^3} \; \Rightarrow\; B_{\nu}=\frac{2 k_B T \nu^2}{c^2},$$
and this was to be shown.

7. Apr 18, 2016

Thank you guys.

For

and

the peak intensities occur at different wavelengths or frequencies.

How do scientists measure the spectral radiance of blackbodies? Are there TWO types of equipment, one for $$B_\lambda$$ and the other for $$B_\nu$$, such that each device yields a peak at a different frequency?