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Transcript
Molecular dynamics: Basic ideas 1 Newton’s equations of motion Molecular dynamics (MD) can be discussed in very general form – here, the discussion is kept simple to stress its central ideas. The equations of motion are the central equations in MD. The following equations apply to each molecule i: 2 Newton’s equations of motion Molecular dynamics (MD) can be discussed in very general form – here, the discussion is kept simple to stress its central ideas. The equations of motion are the central equations in MD. The following equations apply to each molecule i: d ri vi dt 3 Newton’s equations of motion Molecular dynamics (MD) can be discussed in very general form – here, the discussion is kept simple to stress its central ideas. The equations of motion are the central equations in MD. The following equations apply to each molecule i: d ri vi dt d vi dt F i ,m m mi 4 Newton’s equations of motion Molecular dynamics (MD) can be discussed in very general form – here, the discussion is kept simple to stress its central ideas. The equations of motion are the central equations in MD. The following equations apply to each molecule i: d ri vi dt d vi dt F m mi i ,m F m i ,m Summation of all forces acting on particle i. 5 The numerical problem Assuming the molecules are spherical: Each position vector has 3 components. Each velocity vector has 3 components. In a system with n molecules, there are 6n ordinary differential equations that need to be solved simultaneously. It is an initial value problem: the positions and velocities need to be known at the initial time to proceed with integration. 6 Future concerns We will need to take precautions if we want all the states to be of same temperature. We will discuss this in the future. If the particles are not spherical, things get more complicated. There are numerical methods tailored to this type of application for maximum numerical efficiency. If the number of molecules is large, many will be too far away from one another – how to reduce the computational load in a meaningful way. How to avoid the effect of borders on simulation results? 7 Equations of motion d ri vi dt d vi dt F i ,m m mi We will assume a system of identical molecules whose energy of interaction is given by the LennardJones potential: U rij 4 rij rij , : energy and size parameters 12 6 rij : intermolecular distance 8 How particles interact Lennard-Jones potential 9 Intermolecular forces The force acting in particle i as result of its interaction with particle j is given by: F ij U rij Fij dU rij drij rij r ij r i r j rij r ix rjx riy rjy riz rjz 2 2 (in Cartesian coordinates) 2 10 Intermolecular forces rij rix r ij rij riy rij riz rij r ix rjx riy rjy riz rjz r 2 2 2 2 12 ij 11 Intermolecular forces 2 12 2 r r rij ij 1 2 1 2 ij rij rix rix rix 2 1 2r r 2 ij 2 12 ij rix 1 r 2rij rix 2 ij 12 Intermolecular forces rij2 rix rix rjx riy rjy riz rjz 2 2 rix 2 2r r ix jx Then: rij rix 1 r 1 rix rjx rij 2rij rix 2 ij 13 Intermolecular forces rij 1 rix rjx rix rij rij rix rij r ij rij riy rij rij riz 1 dU rij Fij r ij rij drij 14 Example: Lennard-Jones in one-dimension Assume all molecules are along a straight horizontal line. Also, assume left-to-right is the positive direction for coordinates and forces. 1 dU rij Fij r ij rij drij 15 Example: Lennard-Jones in one-dimension 1 dU rij Fij r ij rij drij Assume r is such that: ij • there is attraction between molecules i and j: • molecule i is to the right of particle j: dU rij drij 0 r ij ri rj 0 • Then: Fij 0 , i.e., attraction pulls particle i to the left. 16 Example: Lennard-Jones in one-dimension 1 dU rij Fij r ij rij drij Similar analysis shows that: • Attraction pulls particle j to the right. 17 Summary By now, we know: •The equations we need to solve: Newton’s equation of motion; •How to evaluate intermolecular forces using LennardJones potential. 18 Practical difficulty The forces, distances, and time scales in molecular simulations are very small numbers when S.I. units are used. Such small numbers are a potential source of difficulty in molecular simulations because they may give origin to “underflows” or “overflows”: numbers that either too small or too large to be represented in a computer program. It is common to rewrite the equations in terms of dimensionless quantities. 19 Equations of motion in dimensionless form Dimensionless quantities are defined as follows: t t t m t m U r U r U U r r 20 Equations of motion in dimensionless form Equations of motion: i i i mi d ri dr dr dr vi vi vi vi dt mi dt dt dt i dv mi dt F m mi i ,m i dv dt m i dv F F i ,m i ,m dt m d vi d vi mi d vi ai Equivalently: ai ai mi dt dt dt 21 Equations of motion in dimensionless form Lennard-Jones potential and force: U rij 4 rij 12 rij 6 U r 1 12 1 6 ij U 4 rij rij U rij d r 1 1 dU ij F ij F ij F ij r ij rij drij rij d rij 22 Equations of motion in dimensionless form Lennard-Jones potential and force: d r 4 F ij rij r 12 ij 6 ij ij dr r ij 4 13 7 12 rij 6 rij r ij rij 2r r 24 2 ij 12 ij r 6 ij r ij 23 Look again… i i dv F i ,m dt m dr vi dt 1 12 1 6 U 4 rij rij ij F 2r r 24 2 ij 12 ij r 6 ij The Lennard-Jones parameters of a specific substance do not appear. Integration of these equations gives general results, applicable to any substance modeled by the Lennard-Jones potential: an additional advantage of the dimensionless form. r 24 ij Summary The equations inside colored boxes in the previous slide are the practical equations to implement and solve. 25 Homework questions 1) Obtain the dimensionless form of the kinetic energy of a molecule. 2) Using dimensionless quantities, do we get meaningful results by adding the dimensionless potential intermolecular energies and dimensionless kinetic energies? 3) In a simulation with argon, the dimensionless simulated time is 100 units. What is the simulated time in seconds? Argon data: kB 119.8K 2 kg . m k B 1.3806503 1023 2 s .K 3.405 1010 m M 0.03994 kg mol 26 How does the simulation work? 1) Assign initial position and initial velocity to each particle; 2) Call numerical procedure that will integrate the differential equations of motion: a. Compute resulting force acting in each particle; b. Compute the new position and velocity of each particle; 3) After integration from the initial to the final simulated time, analyze the results (plotting, energy conservation, etc…). 27 Future concerns We will need to take precautions if we want all the states to be of the same temperature. We will discuss this in the future If the particles are not spherical, things get more complicated There are numerical methods tailored to this type of application for maximum numerical efficiency If the number of molecules is large, many will be too far away from one another – how to reduce the computational load in a meaningful way How to avoid the effect of borders on simulation results? 28 Numerical integration may be troublesome Many algorithms for integrating ordinary differential equations are not suitable: not accurate enough This is a system of constant total energy – it has to be preserved during the simulation – this is a way of checking the simulation is proceeding well A good algorithm is called velocity Verlet 29 Velocity Verlet algorithm Using the current state of the system: Compute the intermolecular distances, forces, and accelerations; The positions at t+dt are: 1 2 r t t r t t v t t a t 2 Using the new positions, compute the forces and accelerations: 1 v t t v t t a t a t t 2 The new position becomes the current position 30 Future concerns We will need to take precautions if we want all the states to be of same temperature. We will discuss this in the future. If the particles are not spherical, things get more complicated. There are numerical methods tailored to this type of application for maximum numerical efficiency. If the number of molecules is large, many will be too far away from one another – how to reduce the computational load in a meaningful way. How to avoid the effect of borders on simulation results? 31 Periodic boundary conditions What to do if a particle leaves the box as it moves? 32 Periodic boundary conditions To avoid the effect of borders on simulation results, surround the system by replicas of itself 33 Periodic boundary conditions If a molecule leaves the system, another enters from the opposite side 34 Periodic boundary conditions Each particle interacts with the nearest replica of its neighbors 35 Radial distribution function Molecular simulations provide a lot of information that we will uncover as we discuss more of the fundamentals of statistical thermodynamics One type of information – the radial distribution function – has to do with the structure of the substance being simulated Formal definitions apart, the radial distribution function gives information about the average population of neighboring molecules as function of the intermolecular distance to a central molecule 36 Radial distribution function To evaluate it via molecular simulation, one splits the space around each central molecule in spherical concentric shells and counts how many neighboring particles have their center there. Image source: http://rheneas.eng.buffalo.edu/w/images/e/e0/LJ_Rdf_shell_t.GIF 37 Radial distribution function The radial distribution function is the probability of finding the center of a neighboring molecule at a certain distance from the central molecule divided by the same probability for an ideal gas of same density 38 Radial distribution function T Lennard-Jones fluid. Image source: Wikipedia k BT 0.71 3 0.844 39