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: [tex]u(\lambda)=\frac{8 \pi k T }{\lambda^{4}}[/tex]

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

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

Physics Monkey

First, the Planck formula is [tex] u(\lambda) = \frac{8 \pi}{\lambda^4} \frac{ h c / \lambda}{e^{hc/\lambda k T} - 1} [/tex]. You have an extra factor of [tex] k T [/tex] 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...

[tex]\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1} = kT[/tex]

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 [tex]\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1} = kT[/tex]?

and how does the Stefan-Boltzmann Law come into play?

George Jones

For Rayleigh-Jeans, you need to look at what happens the [itex]\lambda[/itex] becomes large.

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

[tex]\int_{0}^{\infty} u \left( \lambda \right) d \lambda.[/tex]

Make a change of integration variable so that [itex]T[/itex] 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 [itex]T^4[/itex], 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

[tex]\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1} = kT[/tex]

when lambda --> infinity, the left hand side becomes 0/0 so I applied hospital's rule to get:

[tex]\frac{hc(-\frac{1}{ \lambda^2})}{\frac{hc}{kT} e^{hc/ \lambda K T}} = kT[/tex]

[tex]\lambda \rightarrow \inf[/tex]

[tex]\frac{hc}{\frac{hc}{kT}}=kT[/tex]

[tex]kT=kT[/tex]

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 [itex]\lambda[/itex] becomes large, but not infinite.

Regards,
George

UrbanXrisis

the Rayleigh-Jeans formula is: [tex]u(\lambda)=\frac{8 \pi k T }{\lambda^{4}}[/tex]

while Planck's formula consists of Rayleigh-Jeans but includes [tex]\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1}[/tex] instead of [tex]kT[/tex]

so what I did was set [tex]\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1} = kT[/tex] and solved for when the wavelength was really big, which shows that indeed [tex]\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1} = kT[/tex]

plugging this into Planck's equation, I will get the Rayleigh-Jeans formula:

[tex]u(\lambda)=\frac{8 \pi }{\lambda^{4}}\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1}[/tex]

[tex]\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1} = kT[/tex]

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

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

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

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

[tex]u \left( \lambda \right) = \frac{8 \pi k T }{\lambda^{4}}.[/tex]

Regards,
George

dextercioby

If T is large, then the exponent in [itex] e^{hc/ \lambda kT}[/itex] is small and you can use Bernoulli's formula

[tex] e^{x}=1+x [/tex] valid for "x" very small.

Daniel.