md2011-05-note_Part_4

Md2011-05-note_Part_ - I Solving for the T t leads to τ T d T t d t T = T ⇒ T t = const × e t/τ T T(22 where the constant is T i T T i is the

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Unformatted text preview: I Solving for the T ( t ) leads to τ T d T ( t ) d t + T = T , ⇒ T ( t ) = const . × e- t /τ T + T (22) where the constant is T i- T , T i is the initial temperature. I By scaling v → λ v , the energy changes by Δ E = ( λ 2- 1 ) 3 2 Nk B T . I Using this and the rate equation in the definition of the heat capacity C V ≡ ∂ E ∂ T = ( λ 2- 1 ) 3 2 Nk B T ∂ t τ P ( T- T ) , (23) we get λ 2 = 2 C V Δ t 3 k B N τ T T T- 1 + 1 . (24) Notes I Assuming the Dulong-Petit law ( C V = 3 Nk B ), we finally arrive at λ 2 = 2 Δ t τ T T T- 1 + 1 . (25) I ( Note that the original article misses a factor of two. ) I With a τ T > 100 Δ t , the temperature fluctuations are natural. I For pressure, the suggested scaling factor (from the same article) can be similarly derived from a rate equation d P d t = 1 τ P [ P- P ( t )] . (26) I The scaling is done by adjusting the volume of the system accordingly V → μ 3 V ....
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This note was uploaded on 02/14/2012 for the course CSE 6590 taught by Professor Kotakoski during the Winter '12 term at York University.

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Md2011-05-note_Part_ - I Solving for the T t leads to τ T d T t d t T = T ⇒ T t = const × e t/τ T T(22 where the constant is T i T T i is the

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