Chapter3_10

# Chapter3_10 - dots in mode space and dV = 4 r 2 dr Thus dr...

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PHY3063 R. D. Field Department of Physics Chapter3_10.doc University of Florida E&M Modes in a Conducting Cavity For a cube with sides of length L get 2 2 2 2 z y x n n n L + + = λ where n x , n y , n z = all positive integers . This equation describes all possible wavelengths of electromagnetic radiation in the cubical conducting cavity and 2 2 2 2 z y x n n n L c c f + + = = , all possible frequencies. We must count the number of allowed frequencies in the cavity. Let N(f)df = number of allowed frequencies between f and f+df . Now transform to “mode space” ( each dot corresponds to an allowed frequency ) and define r L c f 2 = with 2 2 2 z y x n n n r + + = . Note that N(f)df = N(r)dr = ρ dV , where N(r) is the number of dots ( in mode space ) between concentric shells of radius r and r+dr and ρ = 1 is the density of
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Unformatted text preview: dots ( in mode space ) and dV = 4 r 2 dr . Thus, dr r dr r dV r dr r N 2 2 1 2 8 1 4 ) ( ) ( ) ( = = = and df c L c Lf dr r dr r N df f N ) / 2 ( ) / 2 ( 4 ) ( ) ( ) ( 2 2 1 2 8 1 = = = , where I used r = (2L/c)f and dr = (2L/c)df and where the 1/8 comes from including only the positive integers. Hence, df c Vf df c Vf df f N 3 2 2 3 2 8 2 8 ) ( = y-axis z-axis x-axis O L L L n x n y n z Mode Space r Dot at every positive integer ( i.e. =1) corresponds to an allowed frequency! Must multiply by two for the two polarization state of light! Can show that N(f)df is independent of the shape and depends only on the volume V!...
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