Why does Planck's constant not appear in the Rayleigh-Jeans formula?

[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 don't know how to even begin. any ideas?

 

[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.

 

[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?

 

[George Jones]

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

 

[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?

 

[George Jones]

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

 

[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?

 

[George Jones]

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

 

[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.