lec27 - 2 E 2 ( t ) 2 H 2 2 0 V 0 = E 2 ( t )+( nc/L ) 2 E...

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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
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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 cu ( ν, T ) e ( T )= 1 e ( ν, T )= 1 4 cu ( 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
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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 ² ² ² 1 ² ² r or t average ] 2 2 µ 0 · · V = 1 ± 0 E 2 ( t )+ 1 1 ( nπc 2 /L )
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Unformatted text preview: 2 E 2 ( t ) 2 H 2 2 0 V 0 = E 2 ( t )+( nc/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 j sin( n x x/L ) sin( n y y/L ) sin( n z z/L ) e it | The unit polarization vector j has 2 possible orthog-onal directions and n i = 1 , 2 , 3 . 2 E 2 2 2 2 2 E = 0 2 = c 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 < n 2 2 2 L R = x + n y + n z = c # modes (freq. < ) = 2 1 4 8 3 R 3 3 = L 3 3 c 8.044 L27B7 3 d # V 2 D ( ) = = L 2 = 2 c 3 d c D( ) 8.044 L27B8...
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This note was uploaded on 11/08/2011 for the course PHYSICS 8.004 taught by Professor Staff during the Spring '08 term at MIT.

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lec27 - 2 E 2 ( t ) 2 H 2 2 0 V 0 = E 2 ( t )+( nc/L ) 2 E...

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