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

[quasar_4]

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

Homework Statement

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

Homework 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!} + ...

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)...

 

[gabbagabbahey]

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.

 

[quasar_4]

Interesting. This is an exam problem (from a previous year) at my school, so maybe the question has a typo. I can't see what else one would do to solve for this equation.

 

[gabbagabbahey]

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}

 

[bkrishnan39]

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