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Unformatted text preview: x µ ∆ x √ 2 πσ 2 e − ( s − <s> ) 2 / 2 σ 2 δ ( s − i ∆ x ) p ( s ) = i ± ²³ ´ comb envelope p(s) s ∆ x 8.044 L8B5 ± Example K.E. of an ideal gas 1 2 p ( v x ) = σ = kT /m √ 2 πσ 2 e − v x / 2 σ 2 1 K.E. x = 2 mv 2 x 1 2 1 mean = 2 m ± ² v ³´ x ² µ = kT 2 σ 2 mean sq.= ( 1 4 2 m ) 2 ± ² v ³´ x ² µ 3 σ 4 3 = 4 ( kT ) 2 p (K.E. x ) variance = 1 2 ( kT ) 2 K.E. x 8.044 L8B6 ± 3 directions, N atoms E = total K.E. E ± = 3 N ² K.E. x ± = 3 NkT 2 U ≡ ² Variance( E ) = 3 N Variance( K.E. x ) = 3 N ( kT ) 2 2 1 exp[ − ( E − (3 / 2) NkT ) 2 p ( E ) = ² 2 π { (3 / 2) N ( kT ) 2 2 { (3 / 2) N ( kT ) 2 ] } } 3 s.d. 2 N kT 1 = = 3 ± mean 2 N kT 3 2 N 8.044 L8B7...
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 Spring '08
 staff
 Physics, Central Limit Theorem, Normal Distribution, Probability theory, 1 m

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