# Rayleigh jeans formula

1. Feb 13, 2006

### UrbanXrisis

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?

2. Feb 13, 2006

### Physics Monkey

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.

3. Feb 14, 2006

### UrbanXrisis

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?

4. Feb 14, 2006

### George Jones

Staff Emeritus
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

Last edited: Feb 14, 2006
5. Feb 14, 2006

### UrbanXrisis

$$\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1} = kT$$

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?

Last edited: Feb 14, 2006
6. Feb 14, 2006

### George Jones

Staff Emeritus
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

7. Feb 14, 2006

### UrbanXrisis

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?

8. Feb 14, 2006

### George Jones

Staff Emeritus
I don't think so - it appears that you've just gone round in a circle.

$$u \left( \lambda \right) = \frac{8 \pi hc}{\lambda^{5} \left( e^{hc/ \lambda K T} - 1 \right)},$$

do what I suggested in my previous post, and arrive at

$$u \left( \lambda \right) = \frac{8 \pi k T }{\lambda^{4}}.$$

Regards,
George

9. Feb 16, 2006

### dextercioby

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.