Obtaining the RayleighJeans formula  what am I doing wrong??by quasar_4 Tags: formula, obtaining, rayleighjeans 

#1
Jan2610, 09:41 PM

P: 290

1. The problem statement, all variables and given/known data
Spectral energy density = [tex] u(\nu, T) = {\[\displaystyle \,8\,\pi \,h{\nu}^{3}{c}^{3} \left( {{\rm e}^{{\frac {h\nu}{kT}}}}1 \right) ^{1}\]} [/tex] where h is Planck's constant and k is Bolzmann's constant. Using the relation [tex] \lambda = \frac{c}{\nu} [/tex] 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 [tex] \lambda + d\lambda [/tex]. Using an expansion of the exponential factor obtain the RayleighJeans formula, [tex] du = \frac{8 \pi k T}{\lambda^4} d\lambda [/tex] 2. Relevant equations Given some function [tex] f(x_1, x_2, ..., x_n) [/tex], [tex] df = \left(\frac{\partial{f}}{\partial{x_1}}\right) dx_1 + ... + \left(\frac{\partial{f}}{\partial{x_n}}\right) dx_n [/tex]. Also, [tex] e^{x} \approx 1+x+\frac{x^2}{2!} + ... [/tex] 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 [tex]{\[\displaystyle {\it u}\, = \,8\,\pi \,h{\lambda}^{3} \left( {{\rm e}^{{\frac {hc}{\lambda\,kT}}}}1 \right) ^{1}\]} [/tex] Then from my equation in part b, we should have that [tex] du = \frac{\partial{u}}{\partial{\lambda}} d\lambda [/tex] But I get [tex] \frac{\partial{u}}{\partial{\lambda}} = {\[\displaystyle \,8\,\pi \,h \left( 3\,\lambda\,kT{{\rm e}^{{\frac {hc}{\lambda\,kT}}}}3\,\lambda\,kThc{{\rm e}^{{\frac {hc}{\lambda\,kT}}}} \right) {\lambda}^{5} \left( {{\rm e}^{{\frac {hc}{\lambda\,kT}}}}1 \right) ^{2}{k}^{1}{T}^{1}\]} [/tex] 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: [tex] du ={\[\displaystyle \,8\,\pi \, \left( 2\,ckT+{\frac {h{c}^{2}}{\lambda}} \right) {\lambda}^{3}{c}^{2}\]}d\lambda [/tex] so... what am I doing wrong?? It could have something to do with the [tex] \lambda + d\lambda [/tex] (though I just thought that meant I could neglect higher than 2nd order terms in my expansion)... 



#2
Jan2610, 10:46 PM

HW Helper
P: 5,004

A quick check of the units of your original equation, [tex]u(\nu,T)=8h\pi\nu^3 c^{3} \left(e^{\frac{h\nu}{kT} }1\right)^{1}[/tex] suggests that the problem lies there, not with your method.
If you are having doubts about whether or not using [itex]du=\frac{\partial u}{\partial\lambda}d\lambda[/itex] makes sense, another way to look at it is to realize that the spread in energy over the interval [itex]\lambda[/itex] to [itex]\lambda+ d\lambda[/itex] is defined as [itex]du=u(\lambda+d\lambda)u(\lambda)[/itex]. And for infinitesimal [itex]d\lambda[/itex], a quick Taylor expansion gives [itex]u(\lambda+d\lambda)=u(\lambda)+\frac{\partial u}{\partial \lambda}d\lambda[/itex], from which the result follows. 



#3
Jan2610, 10:50 PM

P: 290

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.




#4
Jan2610, 11:09 PM

HW Helper
P: 5,004

Obtaining the RayleighJeans formula  what am I doing wrong??
I must apologize, I was thinking [itex]u[/itex] was just an energy density (units of Joules/m^{3}), not a spectral energy density. Your original equation is just fine. However, you can't simply substitute [itex]\nu=\frac{c}{\lambda}[/itex] 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 Jouleseconds/m^{3}. While in terms of wavelength, the spectral energy density has units of Joules/m^{4}. 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, [itex]u(\lambda,T)d\lambda=u(\nu,T)d\nu[/itex]. Or, [itex]u(\lambda,T)=u(\nu,T)\frac{d\nu}{d\lambda}=u(\nu,T)\times\frac{c}{\lambda^2}[/itex] 



#5
Jul211, 12:47 AM

P: 1

When we substitute d[itex]\nu[/itex]/d[itex]\lambda[/itex] =  c / [itex]\lambda[/itex]^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 


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