# Obtaining the Rayleigh-Jeans formula - what am I doing wrong?

by quasar_4
Tags: formula, obtaining, rayleighjeans
 P: 290 1. The problem statement, all variables and given/known data Spectral energy density = $$u(\nu, T) = {$\displaystyle \,8\,\pi \,h{\nu}^{3}{c}^{-3} \left( {{\rm e}^{{\frac {h\nu}{kT}}}}-1 \right) ^{-1}$}$$ where h is Planck's constant and k is Bolzmann's constant. Using the relation $$\lambda = \frac{c}{\nu}$$ express the spectral energy density as a function of the wavelength and determine the energy du of a blackbody radiation per unit volume, in a narrow range of wavelength $$\lambda + d\lambda$$. Using an expansion of the exponential factor obtain the Rayleigh-Jeans formula, $$du = \frac{8 \pi k T}{\lambda^4} d\lambda$$ 2. Relevant equations Given some function $$f(x_1, x_2, ..., x_n)$$, $$df = \left(\frac{\partial{f}}{\partial{x_1}}\right) dx_1 + ... + \left(\frac{\partial{f}}{\partial{x_n}}\right) dx_n$$. Also, $$e^{x} \approx 1+x+\frac{x^2}{2!} + ...$$ 3. The attempt at a solution It seems like it should be easy, but I can't get the equation to come out right. Substituting in lambda, I get $${$\displaystyle {\it u}\, = \,8\,\pi \,h{\lambda}^{-3} \left( {{\rm e}^{{\frac {hc}{\lambda\,kT}}}}-1 \right) ^{-1}$}$$ Then from my equation in part b, we should have that $$du = \frac{\partial{u}}{\partial{\lambda}} d\lambda$$ But I get $$\frac{\partial{u}}{\partial{\lambda}} = {$\displaystyle \,-8\,\pi \,h \left( 3\,\lambda\,kT{{\rm e}^{{\frac {hc}{\lambda\,kT}}}}-3\,\lambda\,kT-hc{{\rm e}^{{\frac {hc}{\lambda\,kT}}}} \right) {\lambda}^{-5} \left( {{\rm e}^{{\frac {hc}{\lambda\,kT}}}}-1 \right) ^{-2}{k}^{-1}{T}^{-1}$}$$ Upon using my expansion for the exp. term (expanding around nu = 0, then substituting in nu = c/lambda) the most I can simplify this is to: $$du ={$\displaystyle \,8\,\pi \, \left( -2\,ckT+{\frac {h{c}^{2}}{\lambda}} \right) {\lambda}^{-3}{c}^{-2}$}d\lambda$$ so... what am I doing wrong?? It could have something to do with the $$\lambda + d\lambda$$ (though I just thought that meant I could neglect higher than 2nd order terms in my expansion)...
 HW Helper P: 5,003 A quick check of the units of your original equation, $$u(\nu,T)=8h\pi\nu^3 c^{-3} \left(e^{\frac{h\nu}{kT} }-1\right)^{-1}$$ suggests that the problem lies there, not with your method. If you are having doubts about whether or not using $du=\frac{\partial u}{\partial\lambda}d\lambda$ makes sense, another way to look at it is to realize that the spread in energy over the interval $\lambda$ to $\lambda+ d\lambda$ is defined as $du=u(\lambda+d\lambda)-u(\lambda)$. And for infinitesimal $d\lambda$, a quick Taylor expansion gives $u(\lambda+d\lambda)=u(\lambda)+\frac{\partial u}{\partial \lambda}d\lambda$, from which the result follows.
 HW Helper P: 5,003 Obtaining the Rayleigh-Jeans formula - what am I doing wrong? I must apologize, I was thinking $u$ was just an energy density (units of Joules/m3), not a spectral energy density. Your original equation is just fine. However, you can't simply substitute $\nu=\frac{c}{\lambda}$ into your frequency spectral energy density equation and get a wavelength spectral energy density out of it. In terms of frequency, the spectral energy density has units of Joule-seconds/m3. While in terms of wavelength, the spectral energy density has units of Joules/m4. The first is defined as an energy density per unit frequency, while the latter is defined as an energy density per unit wavelength. For this reason, $u(\lambda,T)d\lambda=u(\nu,T)d\nu$. Or, $u(\lambda,T)=u(\nu,T)\frac{d\nu}{d\lambda}=u(\nu,T)\times\frac{-c}{\lambda^2}$
 P: 1 When we substitute d$\nu$/d$\lambda$ = - c / $\lambda$^2 in Plancks law of Black body radiation (in frequency), we would get an equation that is same as Plancks law of black body radiation (in wavelength) EXCEPT for a negative sign. Because it is an ENERGY equation, the negative sign would be omitted. Is this correct