ChemWiki - User contributions [en]

Rep:Mod:JK21056

2016-11-18T11:56:31Z

Jk2914:

==Task 6: Dynamical Properties and the Diffusion Coefficient==
===Task 1===
====Using data from simulation====
For the following graphs, the series Timestep was replaced by Time, which was achieved by multiplying Timestep series by a tenth of the simulation timestep (0.002).
[[File:JK2105TIMEOwn_simulation_gas.jpg|500px|thumb|center|''Figure 1'': Mean Squared Displacement of the gas as a function of Time. The gradient of the linear section = 1.41]]
[[File:JK2105TIMEOwn_simulation_liquid.jpg|700px|thumb|center|''Figure 2'': Mean Squared Displacement of the liquid as a function of Time. The gradient of the line = 0.0509]]
[[File:JK2105TIMEOwn_simulation_solid.jpg|700px|thumb|center|''Figure 3'': Mean Squared Displacement of the solid as a function of Time. The gradient of the line is assumed to be zero at equilibrium.]]

The trends are as expected. The gas becomes the most displaced as it is the most disordered phase. The rate of change of displacement increases until about timestep 2000. This is probably because the gas has diffused sufficiently to overcome Lennard Jones Potential at this point. It will continue to diffuse.

The liquid becomes more diffuse with time but at a much slower rate than the gas because it has stronger interactions between particles.

The solid rapidly takes its lattice position as this is energetically favourable. It will oscillate slightly before remaining in the lattice. Although there is a slight translation, the scale shows that this is very small so the position varies very little from its average.

Estimation of <math> D </math>, the Diffusion Coefficient:

<math>D = \frac{1}{6}\frac{\partial\left\langle r^2\left(t\right)\right\rangle}{\partial t}</math>

Using the gradients of the linear sections of the graphs, as mentioned above:

Gas: <math> D = \frac {1}{6} \times 1.41 = 0.235 </math> Reduced units

Liquid: <math> D = \frac {1}{6} \times 0.0509 = 0.00848 </math> Reduced units

Solid: <math> D = \frac {1}{6} \times 0 = 0 </math> Reduced units

====Data from one million atom simulations====
Time could not be calculated as the simulation's timestep was not given:

[[File:JK2105Million_gas.jpg|700px|thumb|center|''Figure 4'': Mean Squared Displacement of the gas as a function of Timestep. The gradient of the linear section = 0.0376]]
[[File:JK2105Million_liquid.jpg|700px|thumb|center|''Figure 5'': Mean Squared Displacement of the liquid as a function of Timestep. The gradient of the linear section = 0.00105]]
[[File:JK2105Million_solid.jpg|700px|thumb|center|''Figure 6'': Mean Squared Displacement of the solid as a function of Timestep. The gradient of the line is assumed to be zero at equilibrium.]]

Using the gradients of the linear sections of the graphs, as mentioned above:

Gas: <math> D = \frac {1}{6} \times 0.0376 = 0.00627 </math> Unknown units

Liquid: <math> D = \frac {1}{6} \times 0.00105 = 0.000175 </math> Unknown units

Solid: <math> D = \frac {1}{6} \times 0 = 0 </math> Unknown units

===Task 2===
Evaluating the normalised velocity autocorrelation function for a 1D harmonic oscillator:
[[File:Jk2105Derivation page 1.jpg|700px|thumb|center|''Figure 7'': VACF evaluation]]
[[File:Jk2105Derivation page 2.jpg|700px|thumb|center|''Figure 8'': VACF evaluation part 2]]

[[File:Jk2105VACF Graph.jpg|700px|thumb|center|''Figure 9'': The VACF for a harmonic oscillator, solid and liquid plotted for timesteps 0 to 500.]]

===Task 3===
The trapezium rule was used to evaluate the area beneath the graphs for my own simulation.
This gave an integration of -5.39 for the solid, -5.39 for the liquid and 3499 for the gas.
The diffusion coefficient, <math> D </math> was calculated in each case:
<math> D_{solid} = \frac{1}{3} \times -5.39 = -1.80 </math>

<math> D_{liquid} = \frac{1}{3} \times -5.39 = -1.80 </math>

<math> D_{gas} = \frac{1}{3} \times 3499 = 1166 </math>

[[File:Running integration solid, liquid.jpg|700px|thumb|center|''Figure 9:'' Running integration using the trapezium rule for solid and liquid. The two integrations overlap.]]

[[File:Running integration gas.jpg|700px|thumb|center|''Figure 10:'' Running integration using the trapezium rule for gas.]]

File:Running integration gas.jpg

2016-11-18T11:54:40Z

Jk2914:

File:Running integration solid, liquid.jpg

2016-11-18T11:54:11Z

Jk2914:

Rep:Mod:JK21056

2016-11-18T11:44:37Z

Jk2914:

File:Jk2105VACF Graph.jpg

2016-11-18T11:21:57Z

Jk2914:

Rep:Mod:JK21056

2016-11-18T10:57:18Z

Jk2914:

==Task 6: Dynamical Properties and the Diffusion Coefficient==
===Task 1===
====Using data from simulation====
For the following graphs, the series Timestep was replaced by Time, which was achieved by multiplying Timestep series by a tenth of the simulation timestep (0.002).
[[File:JK2105TIMEOwn_simulation_gas.jpg|500px|thumb|center|''Figure 1'': Mean Squared Displacement of the gas as a function of Time. The gradient of the linear section = 1.41]]
[[File:JK2105TIMEOwn_simulation_liquid.jpg|700px|thumb|center|''Figure 2'': Mean Squared Displacement of the liquid as a function of Time. The gradient of the line = 0.0509]]
[[File:JK2105TIMEOwn_simulation_solid.jpg|700px|thumb|center|''Figure 3'': Mean Squared Displacement of the solid as a function of Time. The gradient of the line is assumed to be zero at equilibrium.]]

The trends are as expected. The gas becomes the most displaced as it is the most disordered phase. The rate of change of displacement increases until about timestep 2000. This is probably because the gas has diffused sufficiently to overcome Lennard Jones Potential at this point. It will continue to diffuse.

The liquid becomes more diffuse with time but at a much slower rate than the gas because it has stronger interactions between particles.

The solid rapidly takes its lattice position as this is energetically favourable. It will oscillate slightly before remaining in the lattice. Although there is a slight translation, the scale shows that this is very small so the position varies very little from its average.

Estimation of <math> D </math>, the Diffusion Coefficient:

<math>D = \frac{1}{6}\frac{\partial\left\langle r^2\left(t\right)\right\rangle}{\partial t}</math>

Using the gradients of the linear sections of the graphs, as mentioned above:

Gas: <math> D = \frac {1}{6} \times 1.41 = 0.235 </math> Reduced units

Liquid: <math> D = \frac {1}{6} \times 0.0509 = 0.00848 </math> Reduced units

Solid: <math> D = \frac {1}{6} \times 0 = 0 </math> Reduced units

====Data from one million atom simulations====
Time could not be calculated as the simulation's timestep was not given:

[[File:JK2105Million_gas.jpg|700px|thumb|center|''Figure 4'': Mean Squared Displacement of the gas as a function of Timestep. The gradient of the linear section = 0.0376]]
[[File:JK2105Million_liquid.jpg|700px|thumb|center|''Figure 5'': Mean Squared Displacement of the liquid as a function of Timestep. The gradient of the linear section = 0.00105]]
[[File:JK2105Million_solid.jpg|700px|thumb|center|''Figure 6'': Mean Squared Displacement of the solid as a function of Timestep. The gradient of the line is assumed to be zero at equilibrium.]]

Using the gradients of the linear sections of the graphs, as mentioned above:

Gas: <math> D = \frac {1}{6} \times 0.0376 = 0.00627 </math> Unknown units

Liquid: <math> D = \frac {1}{6} \times 0.00105 = 0.000175 </math> Unknown units

Solid: <math> D = \frac {1}{6} \times 0 = 0 </math> Unknown units

===Task 2===
Evaluating the normalised velocity autocorrelation function for a 1D harmonic oscillator:
[[File:Jk2105Derivation page 1.jpg|700px|thumb|center|alt text]]
[[File:Jk2105Derivation page 2.jpg|700px|thumb|center|alt text]]

File:Jk2105Derivation page 2.jpg

2016-11-18T10:56:17Z

Jk2914:

File:Jk2105Derivation page 1.jpg

2016-11-18T10:55:07Z

Jk2914:

Rep:Mod:JK21056

2016-11-17T20:57:30Z

Jk2914:

Rep:Mod:JK21056

2016-11-17T20:52:41Z

Jk2914:

Rep:Mod:JK21056

2016-11-17T20:44:55Z

Jk2914:

==Task 6: Dynamical Properties and the Diffusion Coefficient==
===Task 1===
Using data from simulation:
For the following graphs, the series Timestep was replaced by Time, which was achieved by multiplying Timestep series by a tenth of the simulation timestep (0.002).
[[File:JK2105TIMEOwn_simulation_gas.jpg|500px|thumb|center|''Figure 1'': Mean Squared Displacement of the gas as a function of Time. The gradient of the linear section = 0.0281]]
[[File:JK2105TIMEOwn_simulation_liquid.jpg|700px|thumb|center|''Figure 2'': Mean Squared Displacement of the liquid as a function of Time. The gradient of the line = 0.00102]]
[[File:JK2105TIMEOwn_simulation_solid.jpg|700px|thumb|center|''Figure 3'': Mean Squared Displacement of the solid as a function of Time. The gradient of the line is assumed to be zero at equilibrium.]]

The trends are as expected. The gas becomes the most displaced as it is the most disordered phase. The rate of change of displacement increases until about timestep 2000. This is probably because the gas has diffused sufficiently to overcome Lennard Jones Potential at this point. It will continue to diffuse.

The liquid becomes more diffuse with time but at a much slower rate than the gas because it has stronger interactions between particles.

The solid rapidly takes its lattice position as this is energetically favourable. It will oscillate slightly before remaining in the lattice. Although there is a slight translation, the scale shows that this is very small so the position varies very little from its average.

Estimation of <math> D </math>, the Diffusion Coefficient:

<math>D = \frac{1}{6}\frac{\partial\left\langle r^2\left(t\right)\right\rangle}{\partial t}</math>

Using the gradients of the linear sections of the graphs, as mentioned above:

Gas: <math> D = \frac {1}{6} \times 0.0281 = 0.00468 </math>

Liquid: <math> D = \frac {1}{6} \times 0.00102 = 0.00017 </math>

Solid: <math> D = \frac {1}{6} \times 0 = 0 </math>

Data from one million atom simulations:

[[File:JK2105Million_gas.jpg|700px|thumb|center|''Figure 3'':]]
[[File:JK2105Million_liquid.jpg|700px|thumb|center|alt text]]
[[File:JK2105Million_solid.jpg|700px|thumb|center|alt text]]

File:JK2105TIMEOwn simulation solid.jpg

2016-11-17T20:44:02Z

Jk2914:

File:JK2105TIMEOwn simulation liquid.jpg

2016-11-17T20:43:31Z

Jk2914:

File:JK2105TIMEOwn simulation gas.jpg

2016-11-17T20:42:48Z

Jk2914:

File:JK2105Million solid.jpg

2016-11-17T20:13:44Z

Jk2914:

File:JK2105Million liquid.jpg

2016-11-17T20:13:15Z

Jk2914:

File:JK2105Million gas.jpg

2016-11-17T20:12:59Z

Jk2914:

Rep:Mod:JK21056

2016-11-17T20:06:42Z

Jk2914: Created page with "==Task 6: Dynamical Properties and the Diffusion Coefficient== ===Task 1=== File:JK2105Own_simulation_gas.jpg|500px|thumb|center|''Figure 1'': Mean Squared Displacement of..."

==Task 6: Dynamical Properties and the Diffusion Coefficient==
===Task 1===
[[File:JK2105Own_simulation_gas.jpg|500px|thumb|center|''Figure 1'': Mean Squared Displacement of the gas as a function of Timestep. The gradient of the linear section = 0.0281]]
[[File:JK2105Own_simulation_liquid.jpg|700px|thumb|center|''Figure 2'': Mean Squared Displacement of the liquid as a function of Timestep. The gradient of the line = 0.00102]]
[[File:JK2105Own_simulation_solid.jpg|700px|thumb|center|''Figure 3'': Mean Squared Displacement of the solid as a function of Timestep. The gradient of the line is assumed to be zero at equilibrium.]]

The trends are as expected. The gas becomes the most displaced as it is the most disordered phase. The rate of change of displacement increases until about timestep 2000. This is probably because the gas has diffused sufficiently to overcome Lennard Jones Potential at this point. It will continue to diffuse.

The liquid becomes more diffuse with time but at a much slower rate than the gas because it has stronger interactions between particles.

The solid rapidly takes its lattice position as this is energetically favourable. It will oscillate slightly before remaining in the lattice. Although there is a slight translation, the scale shows that this is very small so the position varies very little from its average.

Estimation of <math> D </math>, the Diffusion Coefficient:

<math>D = \frac{1}{6}\frac{\partial\left\langle r^2\left(t\right)\right\rangle}{\partial t}</math>

File:JK2105Own simulation solid.jpg

2016-11-17T19:56:58Z

Jk2914:

File:JK2105Own simulation liquid.jpg

2016-11-17T19:56:32Z

Jk2914:

File:JK2105Own simulation gas.jpg

2016-11-17T19:55:40Z

Jk2914:

Rep:Mod:JK21055

2016-11-17T18:43:50Z

Jk2914:

==Part 6: Structural properties and the radial distribution function==
===Task 1===
[[File:JK2105RDF graph.jpg|700px|thumb|center|''Figure 1'': Radial Distribution Functions for Solid, Liquid and Gas]]
The three different phases show very different distributions. All three show a very low probability until a distance of r=1. This is due to the Lennard-Jones potential being very high. All three phases also show a peak at 1 which is the equilibrium energy, so it is most favourable for neighbours to sit here.

The solid is in a face centred cubic lattice which means that an atom's neighbours are going to be very particular distances away. This explains why there are sudden peaks and troughs in probability.

The liquid is more diffuse and fits a weak, almost lattice-like, pattern while the Lennard Jones potential is favourable. Further away, the distribution becomes random. It is equally likely to find an atom at any distance at high (r).

The gas is very disordered and is distributed completely randomly, though the probability is higher while there is a favourable Lennard Jones potential. The average probability tends to 1 as the Lennard-Jones potential tends to 0.

[[File:JK2105FCC diagram.jpg|500px|thumb|center|''Figure 2'': Diagram showing the three atoms which the first three peaks on the solid RDF diagram correspond to.]]

''Figure 2'' illustrates which are the closest three lattice positions to the central atom. These are therefore the three first peaks in the RDF graph.

The lattice spacing equals 1.375, the x-value of the second peak on the RDF graph, as this is the second closest position to the central atom.

[[File:JK2105Integration of solid RDF.jpg|700px|thumb|center|''Figure 3'': Cumulative integration of Solid RDF, used to calculate the coordination number to each lattice position]]

''Figure 3'' was used to calculate the coordination number for the first three peaks. The central atom coordinates to the closest position (marked 1 in figure 2) 12 times. It coordinates to the second closest position 6 times and to the third closest position 24 times.

Rep:Mod:JK21055

2016-11-17T18:37:09Z

Jk2914:

==Part 6: Structural properties and the radial distribution function==
===Task 1===
[[File:JK2105RDF graph.jpg|700px|thumb|center|''Figure 1'': Radial Distribution Functions for Solid, Liquid and Gas]]
The three different phases show very different distributions. All three show a very low probability until a distance of r=1. This is due to the Lennard-Jones potential being very high. All three phases also show a peak at 1 which is the equilibrium energy, so it is most favourable for neighbours to sit here.

The solid is in a face centred cubic lattice which means that an atom's neighbours are going to be very particular distances away. This explains why there are sudden peaks and troughs in probability.

The liquid is more diffuse and fits a weak, almost lattice-like, pattern while the Lennard Jones potential is favourable. Further away, the distribution becomes random. It is equally likely to find an atom at any distance at high (r).

The gas is very disordered and is distributed completely randomly, though the probability is higher while there is a favourable Lennard Jones potential. The average probability tends to 1 as the Lennard-Jones potential tends to 0.

[[File:JK2105FCC diagram.jpg|500px|thumb|center|''Figure 2'': Diagram showing the three atoms which the first three peaks on the solid RDF diagram correspond to.]]
[[File:JK2105Integration of solid RDF.jpg|700px|thumb|center|''Figure 3'': Cumulative integration of Solid RDF, used to calculate the coordination number to each lattice position]]

File:JK2105Integration of solid RDF.jpg

2016-11-17T18:32:46Z

Jk2914:

File:JK2105RDF graph.jpg

2016-11-17T18:31:19Z

Jk2914:

Rep:Mod:JK21052

2016-11-17T16:07:29Z

Jk2914:

=Section 3: Equilibration=
==Task 1==
Generating atoms in random positions will place them in situations in which they wouldn't naturally be found. The simulation is therefore deviating from reality. For example, two atoms could be generated very close to each other, a configuration which has very high energy and will give an imperfect simulation.
==Task 2==
Satisfying that the lattice spacing and density match:

<math> Density = \frac{N}{V} </math>

Here the volume is the lattice spacing cubed, while the number of atoms equals one in the unit cell.

<math> Density = \frac{1}{1.07722^3} = 0.800 </math>

Now we are considering a face-centred cubic lattice with a lattice point number density of 1.2. There are four atoms per unit cell, therefore:

<math> 1.2 = \frac{4}{(side length)^3} </math>

So:

<math> Side length = 1.49 </math>

==Task 3==
4000 atoms would be created in the face-centred cubic lattice because each unit cell contains 4 atoms.
==Task 4==
'mass 1 1.0' means set the mass of atom type 1 to 1.0 in reduced units.

'pair_style lj/cut 3.0' sets the interactions between pairs of atoms. lj/cut means set Lennard-Jones interactions for atoms beyond a distance of 3.0 to 0. This function does not compute a Coulombic interaction.

'pair_coeff * * 1.0 1.0' This is setting the coefficients in the Lennard-Jones potential equation. The asterisks mean apply the following coefficients to all atom types. The first 1.0 sets <math> \epsilon = 1.0 </math> and the second 1.0 sets <math> \sigma = 1.0 </math>, both in energy units.
==Task 5==
We will use the Verlet algorithm.
==Task 6==
[[File:JK2105Energyofensemble.jpg|700px|thumb|center|''Figure 1:'' Energy against time for the timestep 0.001]]
[[File:JK2105Temperatureofensemble.jpg|700px|thumb|center|''Figure 2:'' Temperature against time for the timestep 0.001]]
[[File:JK2105Pressureofensemble.jpg|700px|thumb|center|''Figure 3:'' Pressure against time for the timestep 0.001]]
The system reaches equilibrium, within the boundaries <math> t=0.3 </math> and <math> t=0.4 </math>.
[[File:JK2105Energiesalltimesteps.jpg|700px|thumb|center|''Figure 4:'' Energy vs time for all timesteps. Key: 0.0075 = Blue, 0.0025 = Green, 0.015 = Orange, 0.01 = Yellow, 0.001 = Grey]]
The best timesteps appear to be 0.001 and 0.0025 because they converge. The larger of these two is 0.0025 so this is the best timestep to use.

Timestep of 0.015 gave particularly bad results. It bears no resemblance to other data and it does not equilibrate.

Rep:Mod:JK21052

2016-11-17T15:59:42Z

Jk2914:

=Section 3: Equilibration=
==Task 1==
Generating atoms in random positions will place them in situations in which they wouldn't naturally be found. The simulation is therefore deviating from reality. For example, two atoms could be generated very close to each other, a configuration which has very high energy and will give an imperfect simulation.
==Task 2==
Satisfying that the lattice spacing and density match:

<math> Density = \frac{N}{V} </math>

Here the volume is the lattice spacing cubed, while the number of atoms equals one in the unit cell.
<math> Density = \frac{1}{1.07722^3} = 0.800 </math>
==Task 3==
4000 atoms would be created in the face-centred cubic lattice because each unit cell contains 4 atoms.
==Task 4==
'mass 1 1.0' means set the mass of atom type 1 to 1.0 in reduced units.

'pair_style lj/cut 3.0' sets the interactions between pairs of atoms. lj/cut means set Lennard-Jones interactions for atoms beyond a distance of 3.0 to 0. This function does not compute a Coulombic interaction.

'pair_coeff * * 1.0 1.0' This is setting the coefficients in the Lennard-Jones potential equation. The asterisks mean apply the following coefficients to all atom types. The first 1.0 sets <math> \epsilon = 1.0 </math> and the second 1.0 sets <math> \sigma = 1.0 </math>, both in energy units.
==Task 5==
We will use the Verlet algorithm.
==Task 6==
[[File:JK2105Energyofensemble.jpg|700px|thumb|center|''Figure 1:'' Energy against time for the timestep 0.001]]
[[File:JK2105Temperatureofensemble.jpg|700px|thumb|center|''Figure 2:'' Temperature against time for the timestep 0.001]]
[[File:JK2105Pressureofensemble.jpg|700px|thumb|center|''Figure 3:'' Pressure against time for the timestep 0.001]]
The system reaches equilibrium, within the boundaries <math> t=0.3 </math> and <math> t=0.4 </math>.
[[File:JK2105Energiesalltimesteps.jpg|700px|thumb|center|''Figure 4:'' Energy vs time for all timesteps. Key: 0.0075 = Blue, 0.0025 = Green, 0.015 = Orange, 0.01 = Yellow, 0.001 = Grey]]
The best timesteps appear to be 0.001 and 0.0025 because they converge. The larger of these two is 0.0025 so this is the best timestep to use.

Timestep of 0.015 gave particularly bad results. It bears no resemblance to other data and it does not equilibrate.

Rep:Mod:JK21053

2016-11-17T15:48:41Z

Jk2914:

==Part 4: Running Simulations under Specific Conditions==
===Task 1===
Determining <math> \gamma </math>.

We are given:

<math>\frac{1}{2}\sum_i m_i v_i^2 = \frac{3}{2} N k_B T</math>

Substitute <math> \gamma_i v_i </math> for <math> v_i </math> and then substitute <math> \mathfrak{T} </math> for <math> T </math> to give:

<math>\frac{1}{2}\sum_i m_i (\gamma_i v_i)^2 = \frac{3}{2} N k_B \mathfrak{T} </math>

Solve these two equation simultaneously to give:

<math> \gamma^2 = \frac{\mathfrak{T}}{T} </math> and therefore:

<math> \gamma = \sqrt{\frac{\mathfrak{T}}{T}} </math>

===Task 2===
It is easier to explain the numbers in reverse order. The third number dictates how often (in terms of timesteps) to take an average. The second number controls how many input values will be used in the average. The first number controls how far apart the input values will be in the average. In this particular simulation, the average is taken every 100,000 timesteps and 1,000 values contribute to each average, with the input values being found 100 timesteps apart.

This simulation will run for 100,000 timesteps.
===Task 3===
[[File:JK2105Graphofoutputnpt.jpg|700px|thumb|center|''Figure 1'': Density as a function of temperature at <math> p = 2.5 </math> and <math> p = 3.5 </math>.]]
The simulated density is lower. This is because the Ideal Gas Law does not take into account interactions between atoms. Therefore they can get much closer to each other at the same pressure, increasing the density. The simulation considers interactions (as it uses the Lennard-Jones potential), preventing atoms from becoming so close because the potential energy is too high. The discrepancy increases with pressure.

Rep:Mod:JK21054

2016-11-17T15:36:33Z

Jk2914:

==Part 5: Calculating Heat Capacities using Statistical Physics==
===Task 1===
[[File:JK2105Heatcapacitygraph.jpg|700px|thumb|center|''Figure 1'': <math> C_V/V </math> as a function of <math> T </math>.]]
This trend is not expected. Heat capacity normally increases with temperature because the degrees of freedom increase. The simulated trend can be explained by the band gap in the crystal getting smaller as temperature increases. Therefore for a given temperature increase, the amount of energy required to promote to the next translational energy level is smaller, hence heat capacity is smaller.

Here is an input file for the NVT ensemble at p=0.2 and T=2.0:

The heat capacity is larger for a larger density. This is because

[[File:JK2105Nvt0.2,2.0modified.txt|thumb|center]]

Rep:Mod:JK21055

2016-11-17T15:25:28Z

Jk2914: Created page with "==Part 6: Structural properties and the radial distribution function== ===Task 1=== File:JK2105FCC diagram.jpg|500px|thumb|Center|''Figure 1'': Diagram showing the three ato..."

==Part 6: Structural properties and the radial distribution function==
===Task 1===
[[File:JK2105FCC diagram.jpg|500px|thumb|Center|''Figure 1'': Diagram showing the three atoms which the first three peaks on the solid RDF diagram correspond to.]]

File:JK2105FCC diagram.jpg

2016-11-17T15:04:44Z

Jk2914:

Rep:Mod:JK21054

2016-11-17T14:05:40Z

Jk2914:

File:JK2105Nvt0.2,2.0modified.txt

2016-11-17T14:03:39Z

Jk2914:

File:JK2105Heatcapacitygraph.jpg

2016-11-17T13:26:41Z

Jk2914:

Rep:Mod:JK21054

2016-11-16T18:50:53Z

Jk2914: Created page with "==Part 5: Calculating Heat Capacities using Statistical Physics== ===Task 1==="

==Part 5: Calculating Heat Capacities using Statistical Physics==
===Task 1===

Rep:Mod:JK21053

2016-11-16T16:40:53Z

Jk2914:

==Part 4: Running Simulations under Specific Conditions==
===Task 1===
===Task 2===
It is easier to explain the numbers in reverse order. The third number dictates how often (in terms of timesteps) to take an average. The second number controls how many input values will be used in the average. The first number controls how far apart the input values will be in the average. In this particular simulation, the average is taken every 100,000 timesteps and 1,000 values contribute to each average, with the input values being found 100 timesteps apart.

This simulation will run for 100,000 timesteps.
===Task 3===
[[File:JK2105Graphofoutputnpt.jpg|700px|thumb|center|''Figure 1'': Density as a function of temperature at <math> p = 2.5 </math> and <math> p = 3.5 </math>.]]
The simulated density is lower. This is because the Ideal Gas Law does not take into account interactions between atoms. Therefore they can get much closer to each other at the same pressure, increasing the density. The simulation considers interactions (as it uses the Lennard-Jones potential), preventing atoms from becoming so close because the potential energy is too high. The discrepancy increases with pressure.

File:JK2105Graphofoutputnpt.jpg

2016-11-16T16:37:16Z

Jk2914:

Rep:Mod:JK21053

2016-11-16T16:36:52Z

Jk2914:

==Part 4: Running Simulations under Specific Conditions==
===Task 1===
===Task 2===
It is easier to explain the numbers in reverse order. The third number dictates how often (in terms of timesteps) to take an average. The second number controls how many input values will be used in the average. The first number controls how far apart the input values will be in the average. In this particular simulation, the average is taken every 100,000 timesteps and 1,000 values contribute to each average, with the input values being found 100 timesteps apart.

This simulation will run for 100,000 timesteps.
===Task 3===
The simulated density is lower. The Ideal Gas Law does not take into account interactions between atoms. Therefore they can get much closer to each other at the same pressure, increasing the density. The simulation considers interactions (as it uses the Lennard-Jones potential), preventing atoms from becoming so close because the potential energy is too high. The discrepancy increases with pressure.

Rep:Mod:JK21053

2016-11-16T16:15:11Z

Jk2914:

Rep:Mod:JK21053

2016-11-16T14:38:17Z

Jk2914: Created page with "==Part 4: Running Simulations under Specific Conditions== ===Task 1=== ===Task 2=== It is easier to explain the numbers in reverse order. The third number dictates how often (..."

Rep:Mod:JK21052

2016-11-15T17:56:39Z

Jk2914:

=Section 3: Equilibration=
==Task 1==
Generating atoms in random positions will place them in situations in which they wouldn't naturally be found. The simulation is therefore deviating from reality. For example, two atoms could be generated very close to each other, a configuration which has very high energy and will give an imperfect simulation.
==Task 2==
==Task 3==
4000 atoms would be created in the face-centred cubic lattice because each unit cell contains 4 atoms.
==Task 4==
'mass 1 1.0' means set the mass of atom type 1 to 1.0 in reduced units.

'pair_style lj/cut 3.0' sets the interactions between pairs of atoms. lj/cut means set Lennard-Jones interactions for atoms beyond a distance of 3.0 to 0. This function does not compute a Coulombic interaction.

'pair_coeff * * 1.0 1.0' This is setting the coefficients in the Lennard-Jones potential equation. The asterisks mean apply the following coefficients to all atom types. The first 1.0 sets <math> \epsilon = 1.0 </math> and the second 1.0 sets <math> \sigma = 1.0 </math>, both in energy units.
==Task 5==
We will use the Verlet algorithm.
==Task 6==
[[File:JK2105Energyofensemble.jpg|700px|thumb|center|''Figure 1:'' Energy against time for the timestep 0.001]]
[[File:JK2105Temperatureofensemble.jpg|700px|thumb|center|''Figure 2:'' Temperature against time for the timestep 0.001]]
[[File:JK2105Pressureofensemble.jpg|700px|thumb|center|''Figure 3:'' Pressure against time for the timestep 0.001]]
The system reaches equilibrium, within the boundaries <math> t=0.3 </math> and <math> t=0.4 </math>.
[[File:JK2105Energiesalltimesteps.jpg|700px|thumb|center|''Figure 4:'' Energy vs time for all timesteps. Key: 0.0075 = Blue, 0.0025 = Green, 0.015 = Orange, 0.01 = Yellow, 0.001 = Grey]]
The best timesteps appear to be 0.001 and 0.0025 because they converge. The larger of these two is 0.0025 so this is the best timestep to use.

Timestep of 0.015 gave particularly bad results. It bears no resemblance to other data and it does not equilibrate.

File:JK2105Energiesalltimesteps.jpg

2016-11-15T17:38:32Z

Jk2914:

Rep:Mod:JK21052

2016-11-15T17:30:27Z

Jk2914:

File:JK2105Pressureofensemble.jpg

2016-11-15T17:29:48Z

Jk2914:

File:JK2105Temperatureofensemble.jpg

2016-11-15T17:29:01Z

Jk2914:

File:JK2105Energyofensemble.jpg

2016-11-15T17:26:04Z

Jk2914:

Rep:Mod:JK21052

2016-11-15T16:42:11Z

Jk2914:

Rep:Mod:JK21052

2016-11-15T16:26:13Z

Jk2914: Created page with "=Section 3: Equilibration= ==Task 1== Generating atoms in random positions will place them in situations in which they wouldn't naturally be found. The simulation is therefore..."

Rep:Mod:JK2105

2016-11-15T15:20:35Z

Jk2914:

==Section 2: Introduction to Molecular Dynamics Simulation==
===Task 1===
[[File:JK2105Analytical graph.jpg|700px|thumb|center|'''Figure 1''': Analytical is the classical position at time ''t'' ]]
[[File:JK2105Energy graph.jpg|700px|thumb|center|'''Figure 2''': Energy is the Velocity-Verlet position at time ''t'' ]]
[[File:JK2105Error graph.jpg|700px|thumb|center|'''Figure 3''': Error is the absolute difference between the two energy calculations at time ''t'' ]]
===Task 2===
Maximum error is found at Time = 2, 4.9, 8, 11.1, 14.2 for timestep 0.1.
[[File:JK2105Maximum error.jpg|700px|thumb|center|'''Figure 4''': Showing the positions of maximum error, fitted with a trend line and function ]]
Figure 4 shows the function of the maximum error as <math> y = 0.0004x-8 \times 10^{-5} </math> where ''y = '' error and ''x'' = time.
===Task 3===
The timestep must be less than or equal to 0.2 in order to avoid a change in error greater than 0.005, which is a 1% change.
If the energy increases largely, it is an indication that the simulation is incorrect because the total energy of a system experimentally will always want to decrease.
===Task 4===
Finding the separation, <math> r_0 </math>, at which the potential energy is 0:

<math>\phi\left(r\right) = 4\epsilon \left( \frac{\sigma^{12}}{r_0^{12}} - \frac{\sigma^6}{r_0^6} \right) = 0</math>

Assumed that:

<math> ( \frac{\sigma^{12}}{r_0^{12}} - \frac{\sigma^6}{r_0^6}) = 0</math>

If <math>r_0 = \sigma </math> then <math> ( \frac{\sigma^{12}}{r_0^{12}} - \frac{\sigma^6}{r_0^6}) = (1-1) = 0 </math>

Therefore <math>r_0 = \sigma </math>

Next the force was calculated at this separation:

<math> F = \frac{dv}{dr} = 4\epsilon \left( -12 \frac{\sigma^{12}}{r_0^{13}} + 6 \frac{\sigma^6}{r_0^7} \right) </math>. Substitute <math>r_0 = \sigma </math> to obtain <math> F = 4\epsilon - 6\sigma </math>.

Subsequently the equilibrium separation, <math> r_{eq} </math> was calculated:

At equilibrium, <math> F = \frac{dv}{dr} = 4\epsilon \left( -12 \frac{\sigma^{12}}{r_{eq}^{13}} + 6 \frac{\sigma^6}{r_{eq}^7} \right) = 0 </math>.

Assumed that:

<math> ( -12 \frac{\sigma^{12}}{r_{eq}^{13}} + 6 \frac{\sigma^6}{r_{eq}^7}) = 0</math>

And so:

<math> ( 2 \frac{\sigma^{12}}{r_{eq}^{13}} = \frac{\sigma^6}{r_{eq}^7}) </math> so <math> ( 2 \frac{\sigma^{12}}{\sigma^{6}} = \frac{r_{eq}^{13}}{r_{eq}^7}) </math> so <math> 2\sigma^6=r^6 </math>

Substitute back into:

<math>\phi\left(r\right) = 4\epsilon \left( \frac{\sigma^{12}}{r_{eq}^{12}} - \frac{\sigma^6}{r_{eq}^6} \right) </math>

To give:

<math>\phi\left(r\right) = 4\epsilon \left( \frac{\sigma^{12}}{4\sigma^{12}} - \frac{\sigma^6}{2\sigma^6} \right) = 4\epsilon \left( \frac{1}{4} - \frac{1}{2} \right)= -\epsilon </math>

This has shown that <math> \epsilon </math> is the well depth.

Evaluating the integrals:

<math>\int \phi\left(r\right)\mathrm{d}r = 4 \epsilon (\frac{\sigma^6}{7r^7}-\frac{\sigma^{12}}{13r^{13}}) </math>

<math> \sigma=\epsilon=1.0 </math>

Between <math>r=2\sigma</math> and <math> r=\infty </math>:

<math>\int_{2\sigma}^\infty \phi\left(r\right)\mathrm{d}r = -0.0248</math>

Between <math>r=2.5\sigma</math> and <math> r=\infty </math>:

<math>\int_{2.5\sigma}^\infty \phi\left(r\right)\mathrm{d}r = -0.00818</math>

Between <math>r=3\sigma</math> and <math> r=\infty </math>:

<math>\int_{3\sigma}^\infty \phi\left(r\right)\mathrm{d}r = -0.00329</math>

===Task 5===
There is 1 mL of water present, which equals 1 g. First, calculate moles:

<math> Moles = \frac{Mass}{M_r} = \frac{1}{18.0} </math>

Next, multiply by Avogadro's number to calculate the number of molecules present:

<math> Number of molecules = 6.02 \times 10^{23} \times \frac{1}{18.0} = 3.34 \times 10^{22} molecules </math>

The calculation is reversed to find the volume occupied by 10,000 water molecules:

<math> \frac {10000}{6.02 \times 10^{23}} = 1.66 \times 10 ^{-20} moles </math>

Then to calculate mass:

<math> Mass = M_r \times Moles = 18.0 \times 1.66 \times 10^{-20} = 2.99 \times 10 ^{-19} g = 2.99 \times 10^{-19} ml </math>

The simulation clearly models very small volumes.

===Task 6===
The atom moves to <math> (0.5,0.5,0.5)+(0.7,0.6,0.2) = (1.2,1.1,0.7) </math>. Taking into account the ''Periodic Boundary Conditions'', this becomes <math> (0.2,0.1,0.7)
</math>.

===Task 7===
Calculating the LJ cutoff in real units:
<math> r^*=\frac{r}{\sigma} </math>

<math> r^* = 3.2 = \frac{r}{0.34} </math>

<math> r = 1.088 nm </math>

Calculating the well depth, <math> \epsilon </math>:

<math> \frac{\epsilon}{k_B} = 120 K </math>

Using the Boltzmann Constant: <math> 8.31 \times 10^{-3} kJ K^{-1} mol^{-1} </math>

<math> \epsilon = 8.31 \times 10^{-3} \times 120 = 0.997 kJ mol^{-1} </math>

Calculating the temperature in real units, where <math> \frac{\epsilon}{k_B}=120 </math>:

<math> T = \frac{T^* \epsilon}{k_B} = 1.5 \times 120 = 180 K </math>

Rep:Mod:JK2105

2016-11-15T15:06:06Z

Jk2914:

==Section 2: Introduction to Molecular Dynamics Simulation==
===Task 1===
[[File:JK2105Analytical graph.jpg|700px|thumb|left|'''Figure 1''': Analytical is the classical position at time ''t'' ]]
[[File:JK2105Energy graph.jpg|700px|thumb|left|'''Figure 2''': Energy is the Velocity-Verlet position at time ''t'' ]]
[[File:JK2105Error graph.jpg|700px|thumb|left|'''Figure 3''': Error is the absolute difference between the two energy calculations at time ''t'' ]]
===Task 2===
Maximum error is found at Time = 2, 4.9, 8, 11.1, 14.2 for timestep 0.1.
[[File:JK2105Maximum error.jpg|700px|thumb|left|'''Figure 4''': Showing the positions of maximum error, fitted with a trend line and function ]]
Figure 4 shows the function of the maximum error as <math> y = 0.0004x-8 \times 10^{-5} </math> where ''y = '' error and ''x'' = time.
===Task 3===
The timestep must be less than or equal to 0.2 in order to avoid a change in error greater than 0.005, which is a 1% change.
If the energy increases largely, it is an indication that the simulation is incorrect because the total energy of a system experimentally will always want to decrease.
===Task 4===
Finding the separation, <math> r_0 </math>, at which the potential energy is 0:

<math>\phi\left(r\right) = 4\epsilon \left( \frac{\sigma^{12}}{r_0^{12}} - \frac{\sigma^6}{r_0^6} \right) = 0</math>

Assumed that:

<math> ( \frac{\sigma^{12}}{r_0^{12}} - \frac{\sigma^6}{r_0^6}) = 0</math>

If <math>r_0 = \sigma </math> then <math> ( \frac{\sigma^{12}}{r_0^{12}} - \frac{\sigma^6}{r_0^6}) = (1-1) = 0 </math>

Therefore <math>r_0 = \sigma </math>

Next the force was calculated at this separation:

<math> F = \frac{dv}{dr} = 4\epsilon \left( -12 \frac{\sigma^{12}}{r_0^{13}} + 6 \frac{\sigma^6}{r_0^7} \right) </math>. Substitute <math>r_0 = \sigma </math> to obtain <math> F = 4\epsilon - 6\sigma </math>.

Subsequently the equilibrium separation, <math> r_{eq} </math> was calculated:

At equilibrium, <math> F = \frac{dv}{dr} = 4\epsilon \left( -12 \frac{\sigma^{12}}{r_{eq}^{13}} + 6 \frac{\sigma^6}{r_{eq}^7} \right) = 0 </math>.

Assumed that:

<math> ( -12 \frac{\sigma^{12}}{r_{eq}^{13}} + 6 \frac{\sigma^6}{r_{eq}^7}) = 0</math>

And so:

<math> ( 2 \frac{\sigma^{12}}{r_{eq}^{13}} = \frac{\sigma^6}{r_{eq}^7}) </math> so <math> ( 2 \frac{\sigma^{12}}{\sigma^{6}} = \frac{r_{eq}^{13}}{r_{eq}^7}) </math> so <math> 2\sigma^6=r^6 </math>

Substitute back into:

<math>\phi\left(r\right) = 4\epsilon \left( \frac{\sigma^{12}}{r_{eq}^{12}} - \frac{\sigma^6}{r_{eq}^6} \right) </math>

To give:

<math>\phi\left(r\right) = 4\epsilon \left( \frac{\sigma^{12}}{4\sigma^{12}} - \frac{\sigma^6}{2\sigma^6} \right) = 4\epsilon \left( \frac{1}{4} - \frac{1}{2} \right)= -\epsilon </math>

This has shown that <math> -\epsilon </math> is the well depth.

Evaluating the integrals:

<math>\int \phi\left(r\right)\mathrm{d}r = 4 \epsilon (\frac{\sigma^6}{7r^7}-\frac{\sigma^{12}}{13r^{13}}) </math>

<math> \sigma=\epsilon=1.0 </math>

Between <math>r=2\sigma</math> and <math> r=\infty </math>

===Task 5===
There is 1 mL of water present, which equals 1 g. First, calculate moles:

<math> Moles = \frac{Mass}{M_r} = \frac{1}{18.0} </math>

Next, multiply by Avogadro's number to calculate the number of molecules present:

<math> Number of molecules = 6.02 \times 10^{23} \times \frac{1}{18.0} = 3.34 \times 10^{22} molecules </math>

The calculation is reversed to find the volume occupied by 10,000 water molecules:

<math> \frac {10000}{6.02 \times 10^{23}} = 1.66 \times 10 ^{-20} moles </math>

Then to calculate mass:

<math> Mass = M_r \times Moles = 18.0 \times 1.66 \times 10^{-20} = 2.99 \times 10 ^{-19} g = 2.99 \times 10^{-19} ml </math>

The simulation clearly models very small volumes.

===Task 6===
The atom moves to <math> (0.5,0.5,0.5)+(0.7,0.6,0.2) = (1.2,1.1,0.7) </math>. Taking into account the ''Periodic Boundary Conditions'', this becomes <math> (0.2,0.1,0.7)
</math>.

===Task 7===
Calculating the LJ cutoff in real units:
<math> r^*=\frac{r}{\sigma} </math>

<math> r^* = 3.2 = \frac{r}{0.34} </math>

<math> r = 1.088 nm </math>

Calculating the well depth, <math> \epsilon </math>:

<math> \frac{\epsilon}{k_B} = 120 K </math>

Using the Boltzmann Constant: <math> 8.31 \times 10^{-3} kJ K^{-1} mol^{-1} </math>

<math> \epsilon = 8.31 \times 10^{-3} \times 120 = 0.997 kJ mol^{-1} </math>

Calculating the temperature in real units, where <math> \frac{\epsilon}{k_B}=120 </math>:

<math> T = \frac{T^* \epsilon}{k_B} = 1.5 \times 120 = 180 K </math>