Converting between λ and ν for Blackbody Radiation?

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

Rayleigh-Jeans law written as B_λ(T) = 2ck_B T / λ^4, where B_λ(T) is spectral radiance per wavelength, c is the speed of light, k_B is the Boltzmann constant, T is temperature and λ is wavelength.


and

Equation B_ν(T) = 2ν²k_B T / c², the Rayleigh-Jeans law for spectral radiance as a function of frequency ν and temperature T.


I tried c = νλ but that doesn't work. This is the Rayleigh Jeans Law by the way.
 
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tade said:
Forgive me for this stupid question, but how do I convert between

Equation B_λ(T) = 2ck_BT / λ^4, the Rayleigh-Jeans approximation of spectral radiance per unit wavelength for blackbody radiation.


and

Rayleigh-Jeans approximation B_ν(T) = 2ν²k_BT/c², written as an equation in plain text with B sub nu of T on the left.


I tried c = νλ but that doesn't work.
pl. give the full expression of the quoted equation and where these two are being used.
 
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

Planck's law for spectral radiance B_λ(T) = (2hc²/λ⁵) · 1/(e^(hc/λk_BT) − 1), approximately equal to 2ck_BT/λ⁴ in the long-wavelength limit.

or

Planck's law for spectral radiance B_ν(T) = (2hν³/c²) × 1/(e^(hν/k_BT) − 1), approximately equal to 2k_BTν²/c² in the low-frequency limit.

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

For
Equation B_λ(T) = 2ck_BT / λ^4, the Rayleigh-Jeans approximation of spectral radiance per wavelength for a blackbody at temperature T.


and

Rayleigh-Jeans law equation: B_ν(T) = 2ν²k_B T / c², where B_ν is spectral radiance per unit frequency, k_B is Boltzmann's constant, T is temperature and c is the speed of light.


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 [tex]B_\lambda[/tex] and the other for [tex]B_\nu[/tex], such that each device yields a peak at a different frequency?