Approximation of \hbar\omega << k_{B}T for Proving Formula

[spaghetti3451]

Prove

\hbar\omega &lt;&lt; k_{B}T \Rightarrow \frac{\hbar\omega}{e^{\frac{\hbar\omega}{k_{B}T} - 1}.

 

[spaghetti3451]

I meant that

\hbar\omega &lt;&lt; k_{B}T \Rightarrow \frac{ \hbar \omega}{e^{ \frac{ \hbar \omega}{ k_{B} T} - 1} = k_{B} T.

 

[uart]

I assume you're trying to ask about the black body radiation equation,

I(f, T) =\frac{ 2 h f^{3}}{c^2}\frac{1}{ e^{\frac{h f}{kT}}-1}

and why when hf &lt;&lt; kT this equation approximates to the Rayleigh–Jeans law,

I(f, T) =\frac{ 2 k T f^{2}}{c^2}

 

[uart]

The simple reason is that

e^x - 1 \simeq x

for small x.

This is actually just e^x replaced with the first two terms of it's Taylor series.

 

[spaghetti3451]

Thank you for your answer.

Actually, this was my original expression.

u(\omega) =\frac{ \hbar \omega^{3}}{ \pi^{2} c^{3}}\frac{1}{ e^{\frac{ \hbar \omega}{kT}}-1}.

How do I derive the intensity of a black body (as above) from this expression?

 

[HallsofIvy]

Do as uart suggested- replace
e^{\frac{\hbar\omega}{kT}}-1
with
\frac{\hbar\omega}{kT}