md2011-03-note_Part_1

md2011-03-note_Part_1 - will start to fail I Obviously...

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Molecular Dynamics simulations Lecture 03: Solving Equations of Motion Dr. Jani Kotakoski University of Helsinki Fall 2011 Notes
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C±²²³´µ¶ · T´¸¹ Sº¹» I From the last lecture, we know that we have to solve either the 3 N Newtonian second order differential equations or the equivalent set of 6 N Hamiltonian first order equations: m i ¨ r i = f i (1) ˙ r i = p i m i ˙ p i = f i (2) where f i = - r i U I In both cases, we have to numerically integrate f i over some time. I The time increment used for the integration is defined as the time step Δ t . Notes
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I Choosing a proper time step for the problem is of crucial importance in MD: I If Δ t is too small, performing the calculations is not efficient (takes very long). I On the other hand, a large Δ t will also cause problems: I Atoms can get too close to each other, which leads to large sudden changes in kinetic energies, which is of course completely unphysical. I Energy conservation becomes more of a problem. I Reversibility condition (as discussed during the last lecture)
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Unformatted text preview: will start to fail. I Obviously, it’s better to err rather toward too short than too long Δ t . I Conventionally, Δ t has been a constant troughout the simulations, and has been chosen using some rules-of-thumb, based on experience. Notes I For example: “ The distance traveled by the atoms should be limited to 1/20 of the interatomic distance b ( Δ x 6 b / 20 ) ”: I From equipartition theorem ± 1 2 m i v 2 i , α ² = 1 2 k B T (3) I While h v i i = q ∑ α ³ v 2 i , α ´ , this leads to h v i i = s 3 k B T m i (4) Notes I As the distribution goes much beyound the h v i i , we use 5 h v i i as the velocity for the estimation: 5 h v i i = Δ x Δ t ⇔ Δ t = Δ x 5 h v i i 6 b 5 × 20 r m i 3 k B T (5) I Note that this depends on the mass of the atom ( m i ). Notes...
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md2011-03-note_Part_1 - will start to fail I Obviously...

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