How does the spectral distribution of blackbody radiation relate to Wien's law?

[spaghetti3451]

Homework Statement

a) Show that for photons of frequency \nu and wavelength \lambda :

1) d\nu = - c d\lambda / \lambda^{2}
2) u(\lambda)d\lambda = - u(\nu)d\nu
3) u(\lambda)d\lambda = u(\nu) c d\lambda / \lambda^{2}

b) Show that the Rayleigh-Jeans spectral distribution of blackbody radiation, u_{RJ}(\nu), is of the form required by Wien's law, u_{W}(\nu) = \frac{W(\lambda T)}{\lambda ^{5}}

c) Obtain the correct form of Wien's undetermined function W(\lambda T) from Planck's formula.

Homework Equations

The Attempt at a Solution

Solution to a):

1) d\nu = \frac{d\nu}{d\lambda}d\lambda = - \frac{c}{\lambda^{2}}d\lambda
2) can't do
3) substitute d\nu in 1) to d\nu on the RHS of 2)

Solution to b):

u_{RJ}(\nu)d\nu = \frac{8\pi\nu^{2}}{c^{3}}kTd\nu
- u_{RJ}(\lambda)d\lambda = (\frac{8\pi\frac{c^{2}}{\lambda^{2}}}{c^{3}}kT)(-\frac{c}{\lambda^{2}}d\lambda)
u_{RJ}(\lambda)d\lambda = \frac{8\pi k(\lambda T)}{\lambda^{5}}d\lambda
So, W(\lambda T) = 8\pi k(\lambda T)

Solution to c):

u(\nu)d\nu = \frac{8\pi h \nu^{3}}{c^{3}} \frac{d\nu}{e^\frac{h\nu}{kT} - 1}
- u(\lambda)d\lambda = \frac{8 \pi h \frac{c^{3}}{\lambda^{3}}}{c^{3}} \frac{- \frac{c}{\lambda^{2}}d\lambda}{e^{\frac{hc}{\lambda kT}} - 1}
u(\lambda)d\lambda = \frac{8 \pi hc}{\lambda^{5}} \frac{d\lambda}{e^{\frac{hc}{\lambda kT}} - 1}
So, W(\lambda T) = \frac{8\pi hc}{e^{\frac{hc}{k(\lambda T)}} - 1}
Need help with a)2). Also, can you check the rest, please?

 

[clamtrox]

Need help with a)2).

You know what you get when you integrate u(λ) dλ and u(v) dv. You can for example integrate first integral from 0 to some λ, and second one from v to ∞. Then these integrals are equal, right?