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# lec27 - 2 E ˙ 2 t 2 H 2 2 µ 0 V ± 0 ± ² = E 2 t nπc/L...

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MIT OpenCourseWare http://ocw.mit.edu 8.044 Statistical Physics I Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .

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Thermodynamic Approach u ( T ) u ( ν, T ) 0 Then E ( T, V ) = u ( T ) V 1 P ( T, V ) = 3 u ( T ) 8.044 L27B1
This is enough to allow us to find u ( T ). dE = TdS PdV ∂E T ∂S T P T ∂P = = V P ∂V T ∂V ∂T 1 = 1 Tu ( T ) 3 u ( T ) 3 also = u ( T ) 4 u ( T ) = u ( T ) T u ( T ) = AT 4 8.044 L27B2

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Emissive Power of a Black ( α = 1) Body 4 c u ( ν, T ) e ( T ) = 1 e ( ν, T ) = 1 4 c u ( T ) = 1 AcT 4 4 e ( T ) σT 4 This is known as the STEFAN-BOLTZMANN LAW. σ = 56 . 7 × 10 9 watts/ m 2 K 4 8.044 L27B3
Statistical Mechanical Approach Single normal mode (plane standing wave) in a H ? rectangular conducting cavity. E x z 0 L ( 1 x r, t ) = E ( t ) sin( nπz/L ) E 0 , 0 ,n, 1 x 1 y ( 1 y B 0 , 0 ,n, r, t ) = ( nπc 2 /L ) 1 E ˙ ( t ) cos( nπz/L ) 8.044 L27B4

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Energy density = 1 0 E E + 1 B B [ no 1 r or t average ] 2 2 µ 0 · · V = 1 0 E 2 ( t )+

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Unformatted text preview: 2 E ˙ 2 ( t ) 2 H 2 2 µ 0 V ± 0 ± ² = E 2 ( t )+( nπc/L ) − 2 E ˙ 2 ( t ) 2 2 ⇒ Each mode corresponds to a harmonic oscillator. 8.044 L27B5 ± Count the modes. E n x ,n y ,n z = | E ± ² j sin( n x πx/L ) sin( n y πy/L ) sin( n z πz/L ) e iωt | The unit polarization vector ± ² j has 2 possible orthog-onal directions and n i = 1 , 2 , 3 . ··· ∂ 2 ± E 2 2 2 2 ± 2 E = 0 ω 2 = ± πc ² 2 ( n + n y + n z ) L x ∂t 2 − c ∇ ⇒ 8.044 L27B6 ± n z n x n y GRID SPACING 1 UNIT R If the radian frequency < ω ² n 2 2 2 L R = x + n y + n z = ω πc # modes (freq. < ω ) = 2 1 4 × 8 ³ × 3 πR 3 3 = π L ω 3 3 πc 8.044 L27B7 ± 3 d # V ω 2 D ( ω ) = = π ² L ω 2 = π 2 c 3 dω πc D( ω ) ω 8.044 L27B8...
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lec27 - 2 E ˙ 2 t 2 H 2 2 µ 0 V ± 0 ± ² = E 2 t nπc/L...

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