![]() ![]() I dont understand why only long wavelengths matter. It is plausible that these modes are not particularly sensitive to the discrete nature of the solid: i.e., the fact that it is made up of atoms rather than being continuous. This approach is reasonable because the only modes which really matter at low temperatures are the long wavelength modes: i.e., those whose wavelengths greatly exceed the interatomic spacing. The molar heat capacity does not decrease with temperature as rapidly as suggested by Einstein's model because these long wavelength modes are able to make a significant contribution to the heat capacity even at very low temperatures. Long wavelength modes have lower frequencies than short wavelength modes, so the former are much harder to freeze out than the latter (because the spacing between quantum energy levels, $\hbar \,\omega$, is smaller in the former case). In this case, the field decays exponentially along the waveguide axis and the wave is thus evanescent.Dulong and Petit came up with $C_v$ = $3R$ using classical equipartion theorem, which Einstein modified (because it could not hold up for objects other than metals) by using Planck's quantization and assumed that all the atoms in a solid vibrated with the same $\omega$ which made it act like $3N$ independent harmonic oscillators.and it gave the (incorrect) result that the heat capacity tends to $0$ as temperature tends to $0$ K. ![]() The wave equations are also valid below the cutoff frequency, where the longitudinal wave number is imaginary. ![]() As a voltage ratio this is a fall to 1 / 2 ≈ 0.707 ![]() Most frequently this proportion is one half the passband power, also referred to as the 3 dB point since a fall of 3 dB corresponds approximately to half power. In electronics, cutoff frequency or corner frequency is the frequency either above or below which the power output of a circuit, such as a line, amplifier, or electronic filter has fallen to a given proportion of the power in the passband.
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