Talk:Phy5645

Given that Planck's energy distribution equation is:

$\rho _{Planck}={\frac {2c^{2}}{\lambda ^{5}}}{\frac {h}{e^{\frac {hc}{\lambda kT}}-1}}$

show that in the long wavelength limit this reduces to the Rayleigh-Jeans equation:

$\rho _{Rayleigh}={\frac {2ckT}{\lambda ^{4}}}$

Solution:

Evaluate the limit:

$\displaystyle \lim _{\lambda \to \infty }$

by expanding the exponential in the denominator for first order in lambda:

$e^{\frac {hc}{\lambda kT}}-1\approx (ch)/(k\lambda t)\implies$ ${\frac {2c^{2}}{\lambda ^{5}}}{\frac {h}{e^{\frac {hc}{\lambda kT}}-1}}\approx {\frac {2c^{2}}{\lambda ^{5}}}\left({\frac {h}{(ch)/(k\lambda t)}}\right)$

then

${\frac {2c^{2}}{\lambda ^{5}}}\left({\frac {h}{(ch)/(k\lambda t)}}\right)={\frac {2ckt}{\lambda ^{4}}}$

3.1.2 Parity operator and the symmetry of the wavefunctions and 4.1.6 Commutators & symmetry are talking about the same thing.