Register to reply 
Obtaining the RayleighJeans formula  what am I doing wrong? 
Share this thread: 
#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 


Register to reply 
Related Discussions  
Obtaining range from bethebloch formula  High Energy, Nuclear, Particle Physics  2  
What am i doing wrong with this chemical formula?  Biology, Chemistry & Other Homework  3  
Ultraviolet Catastrophe / RayleighJeans Black Body Cavity  Quantum Physics  3  
Rayleigh jeans formula  Advanced Physics Homework  8 