## rayleigh jeans formula

I am to derive the incorrect Rayleigh-Jeans formula from the correct Planck formula to show why plank's constant does not appear in the Rayleigh-Jeans formula. I should also recall the Stefen-Boltmann Law

here's what I have but I'm stuck...

Rayleigh-Jeans formula: $$u(\lambda)=\frac{8 \pi k T }{\lambda^{4}}$$

Planks formula: $$u(\lambda)=\frac{8 \pi k T }{\lambda^{4}}\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1}$$

so I am thinking I am somehow supposed to get: $$\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1} = 1$$ but I dont know how to even begin. any ideas?

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 Recognitions: Homework Help Science Advisor First, the Planck formula is $$u(\lambda) = \frac{8 \pi}{\lambda^4} \frac{ h c / \lambda}{e^{hc/\lambda k T} - 1}$$. You have an extra factor of $$k T$$ in your Planck formula which is the source of the confusion. It should now be a simple matter to obtain the Rayleigh-Jeans formula by taking the appropriate limit.
 so that should lead me to prove that... $$\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1} = kT$$ I am sorry but it is still not apparent how I should go about solving this... as T nears infinity, i get infinity on both sides, so does that prove that $$\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1} = kT$$? and how does the Stefan-Boltzmann Law come into play?

Mentor

## rayleigh jeans formula

For Rayleigh-Jeans, you need to look at what happens the $\lambda$ becomes large.

For Stephan-Boltzmann, you need to look at the total contribution from all wavelengths, i.e., you need to look at

$$\int_{0}^{\infty} u \left( \lambda \right) d \lambda.$$

Make a change of integration variable so that $T$ does not appear explicitly in the integrand. Evaluating the resulting integral requires some specialized knowledge of special function. If you only need to show proportionality to $T^4$, then the integral need not be evaluated. If you need the proportionality constant, use software (e.g., Maple) or tables to evaluate the integral

Regards,
George

 $$\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1} = kT$$ when lambda --> infinity, the left hand side becomes 0/0 so I applied hospital's rule to get: $$\frac{hc(-\frac{1}{ \lambda^2})}{\frac{hc}{kT} e^{hc/ \lambda K T}} = kT$$ $$\lambda \rightarrow \inf$$ $$\frac{hc}{\frac{hc}{kT}}=kT$$ $$kT=kT$$ I did not use the stefen-boltzmann law, did I do this correctly?
 Mentor I guess I mistunderstood - I thought you wanted to derive Rayleigh-Jeans and Stefan-Boltzmann from Planck. To derive Rayleigh-Jeans, expand as a series the exponential in Physics Monkey's expression, and find what happens when $\lambda$ becomes large, but not infinite. Regards, George
 the Rayleigh-Jeans formula is: $$u(\lambda)=\frac{8 \pi k T }{\lambda^{4}}$$ while Planck's formula consists of Rayleigh-Jeans but includes $$\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1}$$ instead of $$kT$$ so what I did was set $$\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1} = kT$$ and solved for when the wavelength was really big, which shows that indeed $$\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1} = kT$$ plugging this into Planck's equation, I will get the Rayleigh-Jeans formula: $$u(\lambda)=\frac{8 \pi }{\lambda^{4}}\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1}$$ $$\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1} = kT$$ Rayleigh-Jeans formula: $$u(\lambda)=\frac{8 \pi k T }{\lambda^{4}}$$ did it do this right? was I supposed to include the stefen-boltzmann law somewhere in there?
 Mentor I don't think so - it appears that you've just gone round in a circle. You need to start with $$u \left( \lambda \right) = \frac{8 \pi hc}{\lambda^{5} \left( e^{hc/ \lambda K T} - 1 \right)},$$ do what I suggested in my previous post, and arrive at $$u \left( \lambda \right) = \frac{8 \pi k T }{\lambda^{4}}.$$ Regards, George
 Blog Entries: 9 Recognitions: Homework Help Science Advisor If T is large, then the exponent in $e^{hc/ \lambda kT}$ is small and you can use Bernoulli's formula $$e^{x}=1+x$$ valid for "x" very small. Daniel.