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md2011-05-note_Part_4

# md2011-05-note_Part_4 - Solving for the T(t leads to T dT(t...

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I Solving for the T ( t ) leads to τ T d T ( t ) d t + T = T 0 , T ( t ) = const . × e - t T + T 0 (22) where the constant is T i - T 0 , 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 0 - T ) , (23) we get λ 2 = 2 C V Δ t 3 k B N τ T T 0 T - 1 + 1 . (24) Notes

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I Assuming the Dulong-Petit law ( C V = 3 Nk B ), we finally arrive at λ 2 = 2 Δ t τ T T 0 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 0 - P ( t )] . (26) I The scaling is done by adjusting the volume of the system accordingly V μ 3 V . Notes
I By using the definition of compressibility β - 1 V V P P V = - 1 V β (don’t mix this with the temperature- β earlier in the lecture) we can write d P d t

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