<?xml version="1.0"?>
<feed xmlns="http://www.w3.org/2005/Atom" xml:lang="en">
	<id>https://chemwiki.ch.ic.ac.uk/api.php?action=feedcontributions&amp;feedformat=atom&amp;user=Hy2915</id>
	<title>ChemWiki - User contributions [en]</title>
	<link rel="self" type="application/atom+xml" href="https://chemwiki.ch.ic.ac.uk/api.php?action=feedcontributions&amp;feedformat=atom&amp;user=Hy2915"/>
	<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/wiki/Special:Contributions/Hy2915"/>
	<updated>2026-05-17T11:05:52Z</updated>
	<subtitle>User contributions</subtitle>
	<generator>MediaWiki 1.43.0</generator>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01095534&amp;diff=638483</id>
		<title>01095534</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01095534&amp;diff=638483"/>
		<updated>2017-11-08T09:58:58Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Open the file HO.xls. In it, the velocity-Verlet algorithm is used to model the behaviour of a classical harmonic oscillator. Complete the three columns &amp;quot;ANALYTICAL&amp;quot;, &amp;quot;ERROR&amp;quot;, and &amp;quot;ENERGY&amp;quot;: &amp;quot;ANALYTICAL&amp;quot; should contain the value of the classical solution for the position at time &amp;lt;math&amp;gt;t&amp;lt;/math&amp;gt;, &amp;quot;ERROR&amp;quot; should contain the &#039;&#039;absolute&#039;&#039; difference between &amp;quot;ANALYTICAL&amp;quot; and the velocity-Verlet solution (i.e. ERROR should always be positive -- make sure you leave the half step rows blank!), and &amp;quot;ENERGY&amp;quot; should contain the total energy of the oscillator for the velocity-Verlet solution. Remember that the position of a classical harmonic oscillator is given by &amp;lt;math&amp;gt; x\left(t\right) = A\cos\left(\omega t + \phi\right)&amp;lt;/math&amp;gt; (the values of &amp;lt;math&amp;gt;A&amp;lt;/math&amp;gt;, &amp;lt;math&amp;gt;\omega&amp;lt;/math&amp;gt;, and &amp;lt;math&amp;gt;\phi&amp;lt;/math&amp;gt; are worked out for you in the sheet).&#039;&#039;&#039;&lt;br /&gt;
[[File:1.png|thumb|centre|Energy vs Time Diagram.]]&lt;br /&gt;
[[File:2.png|thumb|centre|Position vs Time Diagram.]]&lt;br /&gt;
[[File:3.png|thumb|centre|Error vs Time Diagram.]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: For the default timestep value, 0.1, estimate the positions of the maxima in the ERROR column as a function of time. Make a plot showing these values as a function of time, and fit an appropriate function to the data.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
the maximum error occur on the maximum distance from original position &lt;br /&gt;
[[File:4.png|thumb|the black area show the position with maximum error.]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Experiment with different values of the timestep. What sort of a timestep do you need to use to ensure that the total energy does not change by more than 1% over the course of your &amp;quot;simulation&amp;quot;? Why do you think it is important to monitor the total energy of a physical system when modelling its behaviour numerically?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
more discrete time step the change of total energy will be lower, and its better to monitor the instant change of the model.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK:&amp;lt;/big&amp;gt; For a single Lennard-Jones interaction, &amp;lt;math&amp;gt;\phi\left(r\right) = 4\epsilon \left( \frac{\sigma^{12}}{r^{12}} - \frac{\sigma^6}{r^6} \right)&amp;lt;/math&amp;gt;, find the separation, &amp;lt;math&amp;gt;r_0&amp;lt;/math&amp;gt;, at which the potential energy is zero. What is the force at this separation? Find the equilibrium separation, &amp;lt;math&amp;gt;r_{eq}&amp;lt;/math&amp;gt;, and work out the well depth (&amp;lt;math&amp;gt;\phi\left(r_{eq}\right)&amp;lt;/math&amp;gt;). Evaluate the integrals &amp;lt;math&amp;gt;\int_{2\sigma}^\infty \phi\left(r\right)\mathrm{d}r&amp;lt;/math&amp;gt;, &amp;lt;math&amp;gt;\int_{2.5\sigma}^\infty \phi\left(r\right)\mathrm{d}r&amp;lt;/math&amp;gt;, and &amp;lt;math&amp;gt;\int_{3\sigma}^\infty \phi\left(r\right)\mathrm{d}r&amp;lt;/math&amp;gt; when &amp;lt;math&amp;gt;\sigma = \epsilon = 1.0&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
at r0 the particle are close together the repulsion force is going to be extremely large,&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Estimate the number of water molecules in 1ml of water under standard conditions. Estimate the volume of  water molecules under standard conditions. &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3.346*10^22  molecules   and   10000molecules of water around 2.99*10^-19 ml&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: The Lennard-Jones parameters for argon are . If the LJ cutoff is , what is it in real units? What is the well depth in ? What is the reduced temperature  in real units?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
r=1.088nm well depth=-6.16*10^-24 reduced temperature T= 180K in real unit&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Why do you think giving atoms random starting coordinates causes problems in simulations? Hint: what happens if two atoms happen to be generated close together?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Because when two atoms are simulated closed together they may counted as a larger molecule since no interaction when start the model.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Satisfy yourself that this lattice spacing corresponds to a number density of lattice points of . Consider instead a face-centred cubic lattice with a lattice point number density of 1.2. What is the side length of the cubic unit cell?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
1.144&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Using the LAMMPS manual, find the purpose of the following commands in the input script:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
mass 1 1.0&lt;br /&gt;
pair_style lj/cut 3.0&lt;br /&gt;
pair_coeff * * 1.0 1.0&lt;br /&gt;
&lt;br /&gt;
means only one atom type and mass is 1.0&lt;br /&gt;
&lt;br /&gt;
the pair style compute the standard 12/6 Lennard-Jones potential  and 3 stands for coefficients for 3 pairs of atom types&lt;br /&gt;
&lt;br /&gt;
the atoms set the coefficients for multiple pairs of atom types and the coefficient of the pairing is 1&lt;br /&gt;
&lt;br /&gt;
=== Running the simulation ===&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Look at the lines below.&#039;&#039;&#039; &lt;br /&gt;
 ### SPECIFY TIMESTEP ###&lt;br /&gt;
 variable timestep equal 0.001&lt;br /&gt;
 variable n_steps equal floor(100/${timestep})&lt;br /&gt;
 timestep ${timestep}&lt;br /&gt;
 &lt;br /&gt;
 &amp;lt;nowiki&amp;gt;### RUN SIMULATION ###&lt;br /&gt;
 run ${n_steps}&lt;br /&gt;
 run 100000&amp;lt;/nowiki&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;The second line (starting &amp;quot;variable timestep...&amp;quot;) tells LAMMPS that if it encounters the text ${timestep} on a subsequent line, it should replace it by the value given. In this case, the value ${timestep} is always replaced by 0.001. In light of this, what do you think the purpose of these lines is? Why not just write:&#039;&#039;&#039; &lt;br /&gt;
 timestep 0.001&lt;br /&gt;
 run 100000&lt;br /&gt;
&#039;&#039;&#039;Ask the demonstrator if you need help.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;nowiki&amp;gt; &amp;lt;/nowiki&amp;gt;this command is to let the time step continuously going with time step 0.001&lt;br /&gt;
&lt;br /&gt;
timestep 0.001&lt;br /&gt;
run 100000&lt;br /&gt;
&lt;br /&gt;
this will let result a repeatation  100000times with 0.001 time step&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: make plots of the energy, temperature, and pressure, against time for the 0.001 timestep experiment (attach a picture to your report). Does the simulation reach equilibrium? How long does this take? When you have done this, make a single plot which shows the energy versus time for all of the timesteps (again, attach a picture to your report). Choosing a timestep is a balancing act: the shorter the timestep, the more accurately the results of your simulation will reflect the physical reality; short timesteps, however, mean that the same number of simulation steps cover a shorter amount of actual time, and this is very unhelpful if the process you want to study requires observation over a long time. Of the five timesteps that you used, which is the largest to give acceptable results? Which one of the five is a &#039;&#039;particularly&#039;&#039; bad choice? Why?&#039;&#039;&#039;&lt;br /&gt;
[[File:Hy1.png|thumb|centre|pressure as function of time at0.001 timestep.]]&lt;br /&gt;
[[File:Hy2.png|thumb|centre|temperature as function of time at0.001 timestep.]]&lt;br /&gt;
[[File:Hy3.png|thumb|centre|total energy as function of time at0.001 timestep.]]&lt;br /&gt;
it reach the equilibrium around 0.17s after the simulation .&lt;br /&gt;
&lt;br /&gt;
ILL CHOOSE  the 0.0025 time step .   fewer than 0.001 time step and still keep the accuracy ,&lt;br /&gt;
[[File:Hy4.png|thumb|centre|total energy as function of time at different timestep.]]&lt;br /&gt;
0.015 is the bad choice which cant monitor the equilibrium of the model&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Choose 5 temperatures (above the critical temperature &amp;lt;math&amp;gt;T^* = 1.5&amp;lt;/math&amp;gt;), and two pressures (you can get a good idea of what a reasonable pressure is in Lennard-Jones units by looking at the average pressure of your simulations from the last section). This gives ten phase points &amp;amp;mdash; five temperatures at each pressure. Create 10 copies of npt.in, and modify each to run a simulation at one of your chosen &amp;lt;math&amp;gt;\left(p, T\right)&amp;lt;/math&amp;gt; points. You should be able to use the results of the previous section to choose a timestep. Submit these ten jobs to the HPC portal. While you wait for them to finish, you should read the next section.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
temperature T = 1.6, 1.8, 2.0, 2.2, 2.4&lt;br /&gt;
&lt;br /&gt;
Pressure P=2.6, 2.8&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: We need to choose &amp;lt;math&amp;gt;\gamma&amp;lt;/math&amp;gt; so that the temperature is correct &amp;lt;math&amp;gt;T = \mathfrak{T}&amp;lt;/math&amp;gt; if we multiply every velocity &amp;lt;math&amp;gt;\gamma&amp;lt;/math&amp;gt;. We can write two equations:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\frac{1}{2}\sum_i m_i v_i^2 = \frac{3}{2} N k_B T&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\frac{1}{2}\sum_i m_i \left(\gamma v_i\right)^2 = \frac{3}{2} N k_B \mathfrak{T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solve these to determine &amp;lt;math&amp;gt;\gamma&amp;lt;/math&amp;gt;.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
gamma is the difference between the real temperature and target temperature. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Use the manual page to find out the importance of the three numbers &#039;&#039;100 1000 100000&#039;&#039;. How often will values of the temperature, etc., be sampled for the average? How many measurements contribute to the average? Looking to the following line, how much time will you simulate?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
use input values every 100 timesteps&lt;br /&gt;
&lt;br /&gt;
1000 of times to use input values for calculating averages&lt;br /&gt;
&lt;br /&gt;
calculate averages every 100000 timesteps&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: When your simulations have finished, download the log files as before. At the end of the log file, LAMMPS will output the values and errors for the pressure, temperature, and density &amp;lt;math&amp;gt;\left(\frac{N}{V}\right)&amp;lt;/math&amp;gt;. Use software of your choice to plot the density as a function of temperature for both of the pressures that you simulated.  Your graph(s) should include error bars in both the x and y directions. You should also include a line corresponding to the density predicted by the ideal gas law at that pressure. Is your simulated density lower or higher? Justify this. Does the discrepancy increase or decrease with pressure?&#039;&#039;&#039;&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01095534&amp;diff=638464</id>
		<title>01095534</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01095534&amp;diff=638464"/>
		<updated>2017-11-08T09:48:56Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Open the file HO.xls. In it, the velocity-Verlet algorithm is used to model the behaviour of a classical harmonic oscillator. Complete the three columns &amp;quot;ANALYTICAL&amp;quot;, &amp;quot;ERROR&amp;quot;, and &amp;quot;ENERGY&amp;quot;: &amp;quot;ANALYTICAL&amp;quot; should contain the value of the classical solution for the position at time , &amp;quot;ERROR&amp;quot; should contain the &#039;&#039;absolute&#039;&#039; difference between &amp;quot;ANALYTICAL&amp;quot; and the velocity-Verlet solution (i.e. ERROR should always be positive -- make sure you leave the half step rows blank!), and &amp;quot;ENERGY&amp;quot; should contain the total energy of the oscillator for the velocity-Verlet solution. Remember that the position of a classical harmonic oscillator is given by  (the values of , , and  are worked out for you in the sheet).&lt;br /&gt;
[[File:1.png|thumb|centre|Energy vs Time Diagram.]]&lt;br /&gt;
[[File:2.png|thumb|centre|Position vs Time Diagram.]]&lt;br /&gt;
[[File:3.png|thumb|centre|Error vs Time Diagram.]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: For the default timestep value, 0.1, estimate the positions of the maxima in the ERROR column as a function of time. Make a plot showing these values as a function of time, and fit an appropriate function to the data.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
the maximum error occur on the maximum distance from original position &lt;br /&gt;
[[File:4.png|thumb|the black area show the position with maximum error.]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Experiment with different values of the timestep. What sort of a timestep do you need to use to ensure that the total energy does not change by more than 1% over the course of your &amp;quot;simulation&amp;quot;? Why do you think it is important to monitor the total energy of a physical system when modelling its behaviour numerically?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
more discrete time step the change of total energy will be lower, and its better to monitor the instant change of the model.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;For a single Lennard-Jones interaction, , find the separation, , at which the potential energy is zero. What is the force at this separation? Find the equilibrium separation, , and work out the well depth (). Evaluate the integrals , , and  when .&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
at r0 the particle are close together the repulsion force is going to be extremely large,&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Estimate the number of water molecules in 1ml of water under standard conditions. Estimate the volume of  water molecules under standard conditions. &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3.346*10^22  molecules   and   10000molecules of water around 2.99*10^-19 ml&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: The Lennard-Jones parameters for argon are . If the LJ cutoff is , what is it in real units? What is the well depth in ? What is the reduced temperature  in real units?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
r=1.088nm well depth=-6.16*10^-24 reduced temperature T= 180K in real unit&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Why do you think giving atoms random starting coordinates causes problems in simulations? Hint: what happens if two atoms happen to be generated close together?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Because when two atoms are simulated closed together they may counted as a larger molecule since no interaction when start the model.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Satisfy yourself that this lattice spacing corresponds to a number density of lattice points of . Consider instead a face-centred cubic lattice with a lattice point number density of 1.2. What is the side length of the cubic unit cell?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
1.144&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Using the LAMMPS manual, find the purpose of the following commands in the input script:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
mass 1 1.0&lt;br /&gt;
pair_style lj/cut 3.0&lt;br /&gt;
pair_coeff * * 1.0 1.0&lt;br /&gt;
&lt;br /&gt;
means only one atom type and mass is 1.0&lt;br /&gt;
&lt;br /&gt;
the pair style compute the standard 12/6 Lennard-Jones potential  and 3 stands for coefficients for 3 pairs of atom types&lt;br /&gt;
&lt;br /&gt;
the atoms set the coefficients for multiple pairs of atom types and the coefficient of the pairing is 1&lt;br /&gt;
&lt;br /&gt;
=== Running the simulation ===&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Look at the lines below.&#039;&#039;&#039; &lt;br /&gt;
 ### SPECIFY TIMESTEP ###&lt;br /&gt;
 variable timestep equal 0.001&lt;br /&gt;
 variable n_steps equal floor(100/${timestep})&lt;br /&gt;
 timestep ${timestep}&lt;br /&gt;
 &lt;br /&gt;
 &amp;lt;nowiki&amp;gt;### RUN SIMULATION ###&lt;br /&gt;
 run ${n_steps}&lt;br /&gt;
 run 100000&amp;lt;/nowiki&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;The second line (starting &amp;quot;variable timestep...&amp;quot;) tells LAMMPS that if it encounters the text ${timestep} on a subsequent line, it should replace it by the value given. In this case, the value ${timestep} is always replaced by 0.001. In light of this, what do you think the purpose of these lines is? Why not just write:&#039;&#039;&#039; &lt;br /&gt;
 timestep 0.001&lt;br /&gt;
 run 100000&lt;br /&gt;
&#039;&#039;&#039;Ask the demonstrator if you need help.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;nowiki&amp;gt; &amp;lt;/nowiki&amp;gt;this command is to let the time step continuously going with time step 0.001&lt;br /&gt;
&lt;br /&gt;
timestep 0.001&lt;br /&gt;
run 100000&lt;br /&gt;
&lt;br /&gt;
this will let result a repeatation  100000times with 0.001 time step&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: make plots of the energy, temperature, and pressure, against time for the 0.001 timestep experiment (attach a picture to your report). Does the simulation reach equilibrium? How long does this take? When you have done this, make a single plot which shows the energy versus time for all of the timesteps (again, attach a picture to your report). Choosing a timestep is a balancing act: the shorter the timestep, the more accurately the results of your simulation will reflect the physical reality; short timesteps, however, mean that the same number of simulation steps cover a shorter amount of actual time, and this is very unhelpful if the process you want to study requires observation over a long time. Of the five timesteps that you used, which is the largest to give acceptable results? Which one of the five is a &#039;&#039;particularly&#039;&#039; bad choice? Why?&#039;&#039;&#039;&lt;br /&gt;
[[File:Hy1.png|thumb|centre|pressure as function of time at0.001 timestep.]]&lt;br /&gt;
[[File:Hy2.png|thumb|centre|temperature as function of time at0.001 timestep.]]&lt;br /&gt;
[[File:Hy3.png|thumb|centre|total energy as function of time at0.001 timestep.]]&lt;br /&gt;
it reach the equilibrium around 0.17s after the simulation .&lt;br /&gt;
&lt;br /&gt;
ILL CHOOSE  the 0.0025 time step .   fewer than 0.001 time step and still keep the accuracy ,&lt;br /&gt;
[[File:Hy4.png|thumb|centre|total energy as function of time at different timestep.]]&lt;br /&gt;
0.015 is the bad choice which cant monitor the equilibrium of the model&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Choose 5 temperatures (above the critical temperature &amp;lt;math&amp;gt;T^* = 1.5&amp;lt;/math&amp;gt;), and two pressures (you can get a good idea of what a reasonable pressure is in Lennard-Jones units by looking at the average pressure of your simulations from the last section). This gives ten phase points &amp;amp;mdash; five temperatures at each pressure. Create 10 copies of npt.in, and modify each to run a simulation at one of your chosen &amp;lt;math&amp;gt;\left(p, T\right)&amp;lt;/math&amp;gt; points. You should be able to use the results of the previous section to choose a timestep. Submit these ten jobs to the HPC portal. While you wait for them to finish, you should read the next section.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
temperature T = 1.6, 1.8, 2.0, 2.2, 2.4&lt;br /&gt;
&lt;br /&gt;
Pressure P=2.6, 2.8&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: We need to choose &amp;lt;math&amp;gt;\gamma&amp;lt;/math&amp;gt; so that the temperature is correct &amp;lt;math&amp;gt;T = \mathfrak{T}&amp;lt;/math&amp;gt; if we multiply every velocity &amp;lt;math&amp;gt;\gamma&amp;lt;/math&amp;gt;. We can write two equations:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\frac{1}{2}\sum_i m_i v_i^2 = \frac{3}{2} N k_B T&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\frac{1}{2}\sum_i m_i \left(\gamma v_i\right)^2 = \frac{3}{2} N k_B \mathfrak{T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solve these to determine &amp;lt;math&amp;gt;\gamma&amp;lt;/math&amp;gt;.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
gamma is the difference between the real temperature and target temperature. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Use the manual page to find out the importance of the three numbers &#039;&#039;100 1000 100000&#039;&#039;. How often will values of the temperature, etc., be sampled for the average? How many measurements contribute to the average? Looking to the following line, how much time will you simulate?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
use input values every 100 timesteps&lt;br /&gt;
&lt;br /&gt;
1000 of times to use input values for calculating averages&lt;br /&gt;
&lt;br /&gt;
calculate averages every 100000 timesteps&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: When your simulations have finished, download the log files as before. At the end of the log file, LAMMPS will output the values and errors for the pressure, temperature, and density &amp;lt;math&amp;gt;\left(\frac{N}{V}\right)&amp;lt;/math&amp;gt;. Use software of your choice to plot the density as a function of temperature for both of the pressures that you simulated.  Your graph(s) should include error bars in both the x and y directions. You should also include a line corresponding to the density predicted by the ideal gas law at that pressure. Is your simulated density lower or higher? Justify this. Does the discrepancy increase or decrease with pressure?&#039;&#039;&#039;&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:Hy4.png&amp;diff=638462</id>
		<title>File:Hy4.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:Hy4.png&amp;diff=638462"/>
		<updated>2017-11-08T09:48:26Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:Hy3.png&amp;diff=638459</id>
		<title>File:Hy3.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:Hy3.png&amp;diff=638459"/>
		<updated>2017-11-08T09:47:24Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:Hy2.png&amp;diff=638458</id>
		<title>File:Hy2.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:Hy2.png&amp;diff=638458"/>
		<updated>2017-11-08T09:46:39Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:Hy1.png&amp;diff=638456</id>
		<title>File:Hy1.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:Hy1.png&amp;diff=638456"/>
		<updated>2017-11-08T09:42:50Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:4.png&amp;diff=638450</id>
		<title>File:4.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:4.png&amp;diff=638450"/>
		<updated>2017-11-08T09:37:40Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: Hy2915 uploaded a new version of File:4.png&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:3.png&amp;diff=638449</id>
		<title>File:3.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:3.png&amp;diff=638449"/>
		<updated>2017-11-08T09:35:23Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: Hy2915 uploaded a new version of File:3.png&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:3.png&amp;diff=638448</id>
		<title>File:3.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:3.png&amp;diff=638448"/>
		<updated>2017-11-08T09:35:20Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: Hy2915 uploaded a new version of File:3.png&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:2.png&amp;diff=638446</id>
		<title>File:2.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:2.png&amp;diff=638446"/>
		<updated>2017-11-08T09:33:45Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: Hy2915 uploaded a new version of File:2.png&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:1.png&amp;diff=638441</id>
		<title>File:1.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:1.png&amp;diff=638441"/>
		<updated>2017-11-08T09:24:54Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: Hy2915 uploaded a new version of File:1.png&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01095534&amp;diff=638422</id>
		<title>01095534</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01095534&amp;diff=638422"/>
		<updated>2017-11-08T08:37:25Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Open the file HO.xls. In it, the velocity-Verlet algorithm is used to model the behaviour of a classical harmonic oscillator. Complete the three columns &amp;quot;ANALYTICAL&amp;quot;, &amp;quot;ERROR&amp;quot;, and &amp;quot;ENERGY&amp;quot;: &amp;quot;ANALYTICAL&amp;quot; should contain the value of the classical solution for the position at time , &amp;quot;ERROR&amp;quot; should contain the &#039;&#039;absolute&#039;&#039; difference between &amp;quot;ANALYTICAL&amp;quot; and the velocity-Verlet solution (i.e. ERROR should always be positive -- make sure you leave the half step rows blank!), and &amp;quot;ENERGY&amp;quot; should contain the total energy of the oscillator for the velocity-Verlet solution. Remember that the position of a classical harmonic oscillator is given by  (the values of , , and  are worked out for you in the sheet).&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: For the default timestep value, 0.1, estimate the positions of the maxima in the ERROR column as a function of time. Make a plot showing these values as a function of time, and fit an appropriate function to the data.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
the maximum error occur on the maximum distance from original position &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Experiment with different values of the timestep. What sort of a timestep do you need to use to ensure that the total energy does not change by more than 1% over the course of your &amp;quot;simulation&amp;quot;? Why do you think it is important to monitor the total energy of a physical system when modelling its behaviour numerically?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
more discrete time step the change of total energy will be lower, and its better to monitor the instant change of the model.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;For a single Lennard-Jones interaction, , find the separation, , at which the potential energy is zero. What is the force at this separation? Find the equilibrium separation, , and work out the well depth (). Evaluate the integrals , , and  when .&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
at r0 the particle are close together the repulsion force is going to be extremely large,&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Estimate the number of water molecules in 1ml of water under standard conditions. Estimate the volume of  water molecules under standard conditions. &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3.346*10^22  molecules   and   10000molecules of water around 2.99*10^-19 ml&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: The Lennard-Jones parameters for argon are . If the LJ cutoff is , what is it in real units? What is the well depth in ? What is the reduced temperature  in real units?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
r=1.088nm well depth=-6.16*10^-24 reduced temperature T= 180K in real unit&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Why do you think giving atoms random starting coordinates causes problems in simulations? Hint: what happens if two atoms happen to be generated close together?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Because when two atoms are simulated closed together they may counted as a larger molecule since no interaction when start the model.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Satisfy yourself that this lattice spacing corresponds to a number density of lattice points of . Consider instead a face-centred cubic lattice with a lattice point number density of 1.2. What is the side length of the cubic unit cell?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
1.144&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Using the LAMMPS manual, find the purpose of the following commands in the input script:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
mass 1 1.0&lt;br /&gt;
pair_style lj/cut 3.0&lt;br /&gt;
pair_coeff * * 1.0 1.0&lt;br /&gt;
&lt;br /&gt;
means only one atom type and mass is 1.0&lt;br /&gt;
&lt;br /&gt;
the pair style compute the standard 12/6 Lennard-Jones potential  and 3 stands for coefficients for 3 pairs of atom types&lt;br /&gt;
&lt;br /&gt;
the atoms set the coefficients for multiple pairs of atom types and the coefficient of the pairing is 1&lt;br /&gt;
&lt;br /&gt;
=== Running the simulation ===&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Look at the lines below.&#039;&#039;&#039; &lt;br /&gt;
 ### SPECIFY TIMESTEP ###&lt;br /&gt;
 variable timestep equal 0.001&lt;br /&gt;
 variable n_steps equal floor(100/${timestep})&lt;br /&gt;
 timestep ${timestep}&lt;br /&gt;
 &lt;br /&gt;
 &amp;lt;nowiki&amp;gt;### RUN SIMULATION ###&lt;br /&gt;
 run ${n_steps}&lt;br /&gt;
 run 100000&amp;lt;/nowiki&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;The second line (starting &amp;quot;variable timestep...&amp;quot;) tells LAMMPS that if it encounters the text ${timestep} on a subsequent line, it should replace it by the value given. In this case, the value ${timestep} is always replaced by 0.001. In light of this, what do you think the purpose of these lines is? Why not just write:&#039;&#039;&#039; &lt;br /&gt;
 timestep 0.001&lt;br /&gt;
 run 100000&lt;br /&gt;
&#039;&#039;&#039;Ask the demonstrator if you need help.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;nowiki&amp;gt; &amp;lt;/nowiki&amp;gt;this command is to let the time step continuously going with time step 0.001&lt;br /&gt;
&lt;br /&gt;
timestep 0.001&lt;br /&gt;
run 100000&lt;br /&gt;
&lt;br /&gt;
this will let result a repeatation  100000times with 0.001 time step&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: make plots of the energy, temperature, and pressure, against time for the 0.001 timestep experiment (attach a picture to your report). Does the simulation reach equilibrium? How long does this take? When you have done this, make a single plot which shows the energy versus time for all of the timesteps (again, attach a picture to your report). Choosing a timestep is a balancing act: the shorter the timestep, the more accurately the results of your simulation will reflect the physical reality; short timesteps, however, mean that the same number of simulation steps cover a shorter amount of actual time, and this is very unhelpful if the process you want to study requires observation over a long time. Of the five timesteps that you used, which is the largest to give acceptable results? Which one of the five is a &#039;&#039;particularly&#039;&#039; bad choice? Why?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
it reach the equilibrium around 0.17s after the simulation .&lt;br /&gt;
&lt;br /&gt;
ILL CHOOSE  the 0.0025 time step .   fewer than 0.001 time step and still keep the accuracy ,&lt;br /&gt;
&lt;br /&gt;
0.015 is the bad choice which cant monitor the equilibrium of the model&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Choose 5 temperatures (above the critical temperature &amp;lt;math&amp;gt;T^* = 1.5&amp;lt;/math&amp;gt;), and two pressures (you can get a good idea of what a reasonable pressure is in Lennard-Jones units by looking at the average pressure of your simulations from the last section). This gives ten phase points &amp;amp;mdash; five temperatures at each pressure. Create 10 copies of npt.in, and modify each to run a simulation at one of your chosen &amp;lt;math&amp;gt;\left(p, T\right)&amp;lt;/math&amp;gt; points. You should be able to use the results of the previous section to choose a timestep. Submit these ten jobs to the HPC portal. While you wait for them to finish, you should read the next section.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
temperature T = 1.6, 1.8, 2.0, 2.2, 2.4&lt;br /&gt;
&lt;br /&gt;
Pressure P=2.6, 2.8&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: We need to choose &amp;lt;math&amp;gt;\gamma&amp;lt;/math&amp;gt; so that the temperature is correct &amp;lt;math&amp;gt;T = \mathfrak{T}&amp;lt;/math&amp;gt; if we multiply every velocity &amp;lt;math&amp;gt;\gamma&amp;lt;/math&amp;gt;. We can write two equations:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\frac{1}{2}\sum_i m_i v_i^2 = \frac{3}{2} N k_B T&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\frac{1}{2}\sum_i m_i \left(\gamma v_i\right)^2 = \frac{3}{2} N k_B \mathfrak{T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solve these to determine &amp;lt;math&amp;gt;\gamma&amp;lt;/math&amp;gt;.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
gamma is the difference between the real temperature and target temperature. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Use the manual page to find out the importance of the three numbers &#039;&#039;100 1000 100000&#039;&#039;. How often will values of the temperature, etc., be sampled for the average? How many measurements contribute to the average? Looking to the following line, how much time will you simulate?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
use input values every 100 timesteps&lt;br /&gt;
&lt;br /&gt;
1000 of times to use input values for calculating averages&lt;br /&gt;
&lt;br /&gt;
calculate averages every 100000 timesteps&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: When your simulations have finished, download the log files as before. At the end of the log file, LAMMPS will output the values and errors for the pressure, temperature, and density &amp;lt;math&amp;gt;\left(\frac{N}{V}\right)&amp;lt;/math&amp;gt;. Use software of your choice to plot the density as a function of temperature for both of the pressures that you simulated.  Your graph(s) should include error bars in both the x and y directions. You should also include a line corresponding to the density predicted by the ideal gas law at that pressure. Is your simulated density lower or higher? Justify this. Does the discrepancy increase or decrease with pressure?&#039;&#039;&#039;&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01095534&amp;diff=638421</id>
		<title>01095534</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01095534&amp;diff=638421"/>
		<updated>2017-11-08T08:35:19Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Open the file HO.xls. In it, the velocity-Verlet algorithm is used to model the behaviour of a classical harmonic oscillator. Complete the three columns &amp;quot;ANALYTICAL&amp;quot;, &amp;quot;ERROR&amp;quot;, and &amp;quot;ENERGY&amp;quot;: &amp;quot;ANALYTICAL&amp;quot; should contain the value of the classical solution for the position at time , &amp;quot;ERROR&amp;quot; should contain the &#039;&#039;absolute&#039;&#039; difference between &amp;quot;ANALYTICAL&amp;quot; and the velocity-Verlet solution (i.e. ERROR should always be positive -- make sure you leave the half step rows blank!), and &amp;quot;ENERGY&amp;quot; should contain the total energy of the oscillator for the velocity-Verlet solution. Remember that the position of a classical harmonic oscillator is given by  (the values of , , and  are worked out for you in the sheet).&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: For the default timestep value, 0.1, estimate the positions of the maxima in the ERROR column as a function of time. Make a plot showing these values as a function of time, and fit an appropriate function to the data.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
the maximum error occur on the maximum distance from original position &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Experiment with different values of the timestep. What sort of a timestep do you need to use to ensure that the total energy does not change by more than 1% over the course of your &amp;quot;simulation&amp;quot;? Why do you think it is important to monitor the total energy of a physical system when modelling its behaviour numerically?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
more discrete time step the change of total energy will be lower, and its better to monitor the instant change of the model.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;For a single Lennard-Jones interaction, , find the separation, , at which the potential energy is zero. What is the force at this separation? Find the equilibrium separation, , and work out the well depth (). Evaluate the integrals , , and  when .&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
at r0 the particle are close together the repulsion force is going to be extremely large,&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Estimate the number of water molecules in 1ml of water under standard conditions. Estimate the volume of  water molecules under standard conditions. &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3.346*10^22  molecules   and   10000molecules of water around 2.99*10^-19 ml&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: The Lennard-Jones parameters for argon are . If the LJ cutoff is , what is it in real units? What is the well depth in ? What is the reduced temperature  in real units?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
r=1.088nm well depth=-6.16*10^-24 reduced temperature T= 180K in real unit&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Why do you think giving atoms random starting coordinates causes problems in simulations? Hint: what happens if two atoms happen to be generated close together?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Because when two atoms are simulated closed together they may counted as a larger molecule since no interaction when start the model.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Satisfy yourself that this lattice spacing corresponds to a number density of lattice points of . Consider instead a face-centred cubic lattice with a lattice point number density of 1.2. What is the side length of the cubic unit cell?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
1.144&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Using the LAMMPS manual, find the purpose of the following commands in the input script:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
mass 1 1.0&lt;br /&gt;
pair_style lj/cut 3.0&lt;br /&gt;
pair_coeff * * 1.0 1.0&lt;br /&gt;
&lt;br /&gt;
means only one atom type and mass is 1.0&lt;br /&gt;
&lt;br /&gt;
the pair style compute the standard 12/6 Lennard-Jones potential  and 3 stands for coefficients for 3 pairs of atom types&lt;br /&gt;
&lt;br /&gt;
the atoms set the coefficients for multiple pairs of atom types and the coefficient of the pairing is 1&lt;br /&gt;
&lt;br /&gt;
=== Running the simulation ===&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Look at the lines below.&#039;&#039;&#039; &lt;br /&gt;
 ### SPECIFY TIMESTEP ###&lt;br /&gt;
 variable timestep equal 0.001&lt;br /&gt;
 variable n_steps equal floor(100/${timestep})&lt;br /&gt;
 timestep ${timestep}&lt;br /&gt;
 &lt;br /&gt;
 &amp;lt;nowiki&amp;gt;### RUN SIMULATION ###&lt;br /&gt;
 run ${n_steps}&lt;br /&gt;
 run 100000&amp;lt;/nowiki&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;The second line (starting &amp;quot;variable timestep...&amp;quot;) tells LAMMPS that if it encounters the text ${timestep} on a subsequent line, it should replace it by the value given. In this case, the value ${timestep} is always replaced by 0.001. In light of this, what do you think the purpose of these lines is? Why not just write:&#039;&#039;&#039; &lt;br /&gt;
 timestep 0.001&lt;br /&gt;
 run 100000&lt;br /&gt;
&#039;&#039;&#039;Ask the demonstrator if you need help.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;nowiki&amp;gt; &amp;lt;/nowiki&amp;gt;this command is to let the time step continuously going with time step 0.001&lt;br /&gt;
&lt;br /&gt;
timestep 0.001&lt;br /&gt;
run 100000&lt;br /&gt;
&lt;br /&gt;
this will let result a repeatation  100000times with 0.001 time step&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: make plots of the energy, temperature, and pressure, against time for the 0.001 timestep experiment (attach a picture to your report). Does the simulation reach equilibrium? How long does this take? When you have done this, make a single plot which shows the energy versus time for all of the timesteps (again, attach a picture to your report). Choosing a timestep is a balancing act: the shorter the timestep, the more accurately the results of your simulation will reflect the physical reality; short timesteps, however, mean that the same number of simulation steps cover a shorter amount of actual time, and this is very unhelpful if the process you want to study requires observation over a long time. Of the five timesteps that you used, which is the largest to give acceptable results? Which one of the five is a &#039;&#039;particularly&#039;&#039; bad choice? Why?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
it reach the equilibrium around 0.17s after the simulation .&lt;br /&gt;
&lt;br /&gt;
ILL CHOOSE  the 0.0025 time step .   fewer than 0.001 time step and still keep the accuracy ,&lt;br /&gt;
&lt;br /&gt;
0.015 is the bad choice which cant monitor the equilibrium of the model&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Choose 5 temperatures (above the critical temperature &amp;lt;math&amp;gt;T^* = 1.5&amp;lt;/math&amp;gt;), and two pressures (you can get a good idea of what a reasonable pressure is in Lennard-Jones units by looking at the average pressure of your simulations from the last section). This gives ten phase points &amp;amp;mdash; five temperatures at each pressure. Create 10 copies of npt.in, and modify each to run a simulation at one of your chosen &amp;lt;math&amp;gt;\left(p, T\right)&amp;lt;/math&amp;gt; points. You should be able to use the results of the previous section to choose a timestep. Submit these ten jobs to the HPC portal. While you wait for them to finish, you should read the next section.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
temperature T = 1.6, 1.8, 2.0, 2.2, 2.4&lt;br /&gt;
&lt;br /&gt;
Pressure P=2.6, 2.8&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: We need to choose &amp;lt;math&amp;gt;\gamma&amp;lt;/math&amp;gt; so that the temperature is correct &amp;lt;math&amp;gt;T = \mathfrak{T}&amp;lt;/math&amp;gt; if we multiply every velocity &amp;lt;math&amp;gt;\gamma&amp;lt;/math&amp;gt;. We can write two equations:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\frac{1}{2}\sum_i m_i v_i^2 = \frac{3}{2} N k_B T&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\frac{1}{2}\sum_i m_i \left(\gamma v_i\right)^2 = \frac{3}{2} N k_B \mathfrak{T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solve these to determine &amp;lt;math&amp;gt;\gamma&amp;lt;/math&amp;gt;.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
gamma is the difference between the real temperature and target temperature. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Use the manual page to find out the importance of the three numbers &#039;&#039;100 1000 100000&#039;&#039;. How often will values of the temperature, etc., be sampled for the average? How many measurements contribute to the average? Looking to the following line, how much time will you simulate?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
use input values every 100 timesteps&lt;br /&gt;
&lt;br /&gt;
1000 of times to use input values for calculating averages&lt;br /&gt;
&lt;br /&gt;
calculate averages every 100000 timesteps&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01095534&amp;diff=638418</id>
		<title>01095534</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01095534&amp;diff=638418"/>
		<updated>2017-11-08T08:29:38Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Open the file HO.xls. In it, the velocity-Verlet algorithm is used to model the behaviour of a classical harmonic oscillator. Complete the three columns &amp;quot;ANALYTICAL&amp;quot;, &amp;quot;ERROR&amp;quot;, and &amp;quot;ENERGY&amp;quot;: &amp;quot;ANALYTICAL&amp;quot; should contain the value of the classical solution for the position at time , &amp;quot;ERROR&amp;quot; should contain the &#039;&#039;absolute&#039;&#039; difference between &amp;quot;ANALYTICAL&amp;quot; and the velocity-Verlet solution (i.e. ERROR should always be positive -- make sure you leave the half step rows blank!), and &amp;quot;ENERGY&amp;quot; should contain the total energy of the oscillator for the velocity-Verlet solution. Remember that the position of a classical harmonic oscillator is given by  (the values of , , and  are worked out for you in the sheet).&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: For the default timestep value, 0.1, estimate the positions of the maxima in the ERROR column as a function of time. Make a plot showing these values as a function of time, and fit an appropriate function to the data.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
the maximum error occur on the maximum distance from original position &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Experiment with different values of the timestep. What sort of a timestep do you need to use to ensure that the total energy does not change by more than 1% over the course of your &amp;quot;simulation&amp;quot;? Why do you think it is important to monitor the total energy of a physical system when modelling its behaviour numerically?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
more discrete time step the change of total energy will be lower, and its better to monitor the instant change of the model.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;For a single Lennard-Jones interaction, , find the separation, , at which the potential energy is zero. What is the force at this separation? Find the equilibrium separation, , and work out the well depth (). Evaluate the integrals , , and  when .&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
at r0 the particle are close together the repulsion force is going to be extremely large,&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Estimate the number of water molecules in 1ml of water under standard conditions. Estimate the volume of  water molecules under standard conditions. &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3.346*10^22  molecules   and   10000molecules of water around 2.99*10^-19 ml&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: The Lennard-Jones parameters for argon are . If the LJ cutoff is , what is it in real units? What is the well depth in ? What is the reduced temperature  in real units?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
r=1.088nm well depth=-6.16*10^-24 reduced temperature T= 180K in real unit&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Why do you think giving atoms random starting coordinates causes problems in simulations? Hint: what happens if two atoms happen to be generated close together?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Because when two atoms are simulated closed together they may counted as a larger molecule since no interaction when start the model.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Satisfy yourself that this lattice spacing corresponds to a number density of lattice points of . Consider instead a face-centred cubic lattice with a lattice point number density of 1.2. What is the side length of the cubic unit cell?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
1.144&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Using the LAMMPS manual, find the purpose of the following commands in the input script:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
mass 1 1.0&lt;br /&gt;
pair_style lj/cut 3.0&lt;br /&gt;
pair_coeff * * 1.0 1.0&lt;br /&gt;
&lt;br /&gt;
means only one atom type and mass is 1.0&lt;br /&gt;
&lt;br /&gt;
the pair style compute the standard 12/6 Lennard-Jones potential  and 3 stands for coefficients for 3 pairs of atom types&lt;br /&gt;
&lt;br /&gt;
the atoms set the coefficients for multiple pairs of atom types and the coefficient of the pairing is 1&lt;br /&gt;
&lt;br /&gt;
=== Running the simulation ===&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Look at the lines below.&#039;&#039;&#039; &lt;br /&gt;
 ### SPECIFY TIMESTEP ###&lt;br /&gt;
 variable timestep equal 0.001&lt;br /&gt;
 variable n_steps equal floor(100/${timestep})&lt;br /&gt;
 timestep ${timestep}&lt;br /&gt;
 &lt;br /&gt;
 &amp;lt;nowiki&amp;gt;### RUN SIMULATION ###&lt;br /&gt;
 run ${n_steps}&lt;br /&gt;
 run 100000&amp;lt;/nowiki&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;The second line (starting &amp;quot;variable timestep...&amp;quot;) tells LAMMPS that if it encounters the text ${timestep} on a subsequent line, it should replace it by the value given. In this case, the value ${timestep} is always replaced by 0.001. In light of this, what do you think the purpose of these lines is? Why not just write:&#039;&#039;&#039; &lt;br /&gt;
 timestep 0.001&lt;br /&gt;
 run 100000&lt;br /&gt;
&#039;&#039;&#039;Ask the demonstrator if you need help.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;nowiki&amp;gt; &amp;lt;/nowiki&amp;gt;this command is to let the time step continuously going with time step 0.001&lt;br /&gt;
&lt;br /&gt;
timestep 0.001&lt;br /&gt;
run 100000&lt;br /&gt;
&lt;br /&gt;
this will let result a repeatation  100000times with 0.001 time step&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: make plots of the energy, temperature, and pressure, against time for the 0.001 timestep experiment (attach a picture to your report). Does the simulation reach equilibrium? How long does this take? When you have done this, make a single plot which shows the energy versus time for all of the timesteps (again, attach a picture to your report). Choosing a timestep is a balancing act: the shorter the timestep, the more accurately the results of your simulation will reflect the physical reality; short timesteps, however, mean that the same number of simulation steps cover a shorter amount of actual time, and this is very unhelpful if the process you want to study requires observation over a long time. Of the five timesteps that you used, which is the largest to give acceptable results? Which one of the five is a &#039;&#039;particularly&#039;&#039; bad choice? Why?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
it reach the equilibrium around 0.17s after the simulation .&lt;br /&gt;
&lt;br /&gt;
ILL CHOOSE  the 0.0025 time step .   fewer than 0.001 time step and still keep the accuracy ,&lt;br /&gt;
&lt;br /&gt;
0.015 is the bad choice which cant monitor the equilibrium of the model&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Choose 5 temperatures (above the critical temperature &amp;lt;math&amp;gt;T^* = 1.5&amp;lt;/math&amp;gt;), and two pressures (you can get a good idea of what a reasonable pressure is in Lennard-Jones units by looking at the average pressure of your simulations from the last section). This gives ten phase points &amp;amp;mdash; five temperatures at each pressure. Create 10 copies of npt.in, and modify each to run a simulation at one of your chosen &amp;lt;math&amp;gt;\left(p, T\right)&amp;lt;/math&amp;gt; points. You should be able to use the results of the previous section to choose a timestep. Submit these ten jobs to the HPC portal. While you wait for them to finish, you should read the next section.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: We need to choose &amp;lt;math&amp;gt;\gamma&amp;lt;/math&amp;gt; so that the temperature is correct &amp;lt;math&amp;gt;T = \mathfrak{T}&amp;lt;/math&amp;gt; if we multiply every velocity &amp;lt;math&amp;gt;\gamma&amp;lt;/math&amp;gt;. We can write two equations:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\frac{1}{2}\sum_i m_i v_i^2 = \frac{3}{2} N k_B T&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\frac{1}{2}\sum_i m_i \left(\gamma v_i\right)^2 = \frac{3}{2} N k_B \mathfrak{T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solve these to determine &amp;lt;math&amp;gt;\gamma&amp;lt;/math&amp;gt;.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
gamma is the difference between the real temperature and target temperature. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Use the manual page to find out the importance of the three numbers &#039;&#039;100 1000 100000&#039;&#039;. How often will values of the temperature, etc., be sampled for the average? How many measurements contribute to the average? Looking to the following line, how much time will you simulate?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
use input values every 100 timesteps&lt;br /&gt;
&lt;br /&gt;
1000 of times to use input values for calculating averages&lt;br /&gt;
&lt;br /&gt;
calculate averages every 100000 timesteps&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01095534&amp;diff=638417</id>
		<title>01095534</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01095534&amp;diff=638417"/>
		<updated>2017-11-08T08:20:32Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Open the file HO.xls. In it, the velocity-Verlet algorithm is used to model the behaviour of a classical harmonic oscillator. Complete the three columns &amp;quot;ANALYTICAL&amp;quot;, &amp;quot;ERROR&amp;quot;, and &amp;quot;ENERGY&amp;quot;: &amp;quot;ANALYTICAL&amp;quot; should contain the value of the classical solution for the position at time , &amp;quot;ERROR&amp;quot; should contain the &#039;&#039;absolute&#039;&#039; difference between &amp;quot;ANALYTICAL&amp;quot; and the velocity-Verlet solution (i.e. ERROR should always be positive -- make sure you leave the half step rows blank!), and &amp;quot;ENERGY&amp;quot; should contain the total energy of the oscillator for the velocity-Verlet solution. Remember that the position of a classical harmonic oscillator is given by  (the values of , , and  are worked out for you in the sheet).&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: For the default timestep value, 0.1, estimate the positions of the maxima in the ERROR column as a function of time. Make a plot showing these values as a function of time, and fit an appropriate function to the data.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
the maximum error occur on the maximum distance from original position &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Experiment with different values of the timestep. What sort of a timestep do you need to use to ensure that the total energy does not change by more than 1% over the course of your &amp;quot;simulation&amp;quot;? Why do you think it is important to monitor the total energy of a physical system when modelling its behaviour numerically?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
more discrete time step the change of total energy will be lower, and its better to monitor the instant change of the model.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;For a single Lennard-Jones interaction, , find the separation, , at which the potential energy is zero. What is the force at this separation? Find the equilibrium separation, , and work out the well depth (). Evaluate the integrals , , and  when .&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
at r0 the particle are close together the repulsion force is going to be extremely large,&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Estimate the number of water molecules in 1ml of water under standard conditions. Estimate the volume of  water molecules under standard conditions. &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3.346*10^22  molecules   and   10000molecules of water around 2.99*10^-19 ml&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: The Lennard-Jones parameters for argon are . If the LJ cutoff is , what is it in real units? What is the well depth in ? What is the reduced temperature  in real units?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
r=1.088nm well depth=-6.16*10^-24 reduced temperature T= 180K in real unit&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Why do you think giving atoms random starting coordinates causes problems in simulations? Hint: what happens if two atoms happen to be generated close together?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Because when two atoms are simulated closed together they may counted as a larger molecule since no interaction when start the model.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Satisfy yourself that this lattice spacing corresponds to a number density of lattice points of . Consider instead a face-centred cubic lattice with a lattice point number density of 1.2. What is the side length of the cubic unit cell?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
1.144&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Using the LAMMPS manual, find the purpose of the following commands in the input script:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
mass 1 1.0&lt;br /&gt;
pair_style lj/cut 3.0&lt;br /&gt;
pair_coeff * * 1.0 1.0&lt;br /&gt;
&lt;br /&gt;
means only one atom type and mass is 1.0&lt;br /&gt;
&lt;br /&gt;
the pair style compute the standard 12/6 Lennard-Jones potential  and 3 stands for coefficients for 3 pairs of atom types&lt;br /&gt;
&lt;br /&gt;
the atoms set the coefficients for multiple pairs of atom types and the coefficient of the pairing is 1&lt;br /&gt;
&lt;br /&gt;
=== Running the simulation ===&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Look at the lines below.&#039;&#039;&#039; &lt;br /&gt;
 ### SPECIFY TIMESTEP ###&lt;br /&gt;
 variable timestep equal 0.001&lt;br /&gt;
 variable n_steps equal floor(100/${timestep})&lt;br /&gt;
 timestep ${timestep}&lt;br /&gt;
 &lt;br /&gt;
 &amp;lt;nowiki&amp;gt;### RUN SIMULATION ###&lt;br /&gt;
 run ${n_steps}&lt;br /&gt;
 run 100000&amp;lt;/nowiki&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;The second line (starting &amp;quot;variable timestep...&amp;quot;) tells LAMMPS that if it encounters the text ${timestep} on a subsequent line, it should replace it by the value given. In this case, the value ${timestep} is always replaced by 0.001. In light of this, what do you think the purpose of these lines is? Why not just write:&#039;&#039;&#039; &lt;br /&gt;
 timestep 0.001&lt;br /&gt;
 run 100000&lt;br /&gt;
&#039;&#039;&#039;Ask the demonstrator if you need help.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;nowiki&amp;gt; &amp;lt;/nowiki&amp;gt;this command is to let the time step continuously going with time step 0.001&lt;br /&gt;
&lt;br /&gt;
timestep 0.001&lt;br /&gt;
run 100000&lt;br /&gt;
&lt;br /&gt;
this will let result a repeatation  100000times with 0.001 time step&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: make plots of the energy, temperature, and pressure, against time for the 0.001 timestep experiment (attach a picture to your report). Does the simulation reach equilibrium? How long does this take? When you have done this, make a single plot which shows the energy versus time for all of the timesteps (again, attach a picture to your report). Choosing a timestep is a balancing act: the shorter the timestep, the more accurately the results of your simulation will reflect the physical reality; short timesteps, however, mean that the same number of simulation steps cover a shorter amount of actual time, and this is very unhelpful if the process you want to study requires observation over a long time. Of the five timesteps that you used, which is the largest to give acceptable results? Which one of the five is a &#039;&#039;particularly&#039;&#039; bad choice? Why?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
it reach the equilibrium around 0.17s after the simulation .&lt;br /&gt;
&lt;br /&gt;
ILL CHOOSE  the 0.0025 time step .   fewer than 0.001 time step and still keep the accuracy ,&lt;br /&gt;
&lt;br /&gt;
0.015 is the bad choice which cant monitor the equilibrium of the model&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: We need to choose &amp;lt;math&amp;gt;\gamma&amp;lt;/math&amp;gt; so that the temperature is correct &amp;lt;math&amp;gt;T = \mathfrak{T}&amp;lt;/math&amp;gt; if we multiply every velocity &amp;lt;math&amp;gt;\gamma&amp;lt;/math&amp;gt;. We can write two equations:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\frac{1}{2}\sum_i m_i v_i^2 = \frac{3}{2} N k_B T&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\frac{1}{2}\sum_i m_i \left(\gamma v_i\right)^2 = \frac{3}{2} N k_B \mathfrak{T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solve these to determine &amp;lt;math&amp;gt;\gamma&amp;lt;/math&amp;gt;.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Use the manual page to find out the importance of the three numbers &#039;&#039;100 1000 100000&#039;&#039;. How often will values of the temperature, etc., be sampled for the average? How many measurements contribute to the average? Looking to the following line, how much time will you simulate?&#039;&#039;&#039;&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01095534&amp;diff=638379</id>
		<title>01095534</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01095534&amp;diff=638379"/>
		<updated>2017-11-08T04:11:07Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Open the file HO.xls. In it, the velocity-Verlet algorithm is used to model the behaviour of a classical harmonic oscillator. Complete the three columns &amp;quot;ANALYTICAL&amp;quot;, &amp;quot;ERROR&amp;quot;, and &amp;quot;ENERGY&amp;quot;: &amp;quot;ANALYTICAL&amp;quot; should contain the value of the classical solution for the position at time , &amp;quot;ERROR&amp;quot; should contain the &#039;&#039;absolute&#039;&#039; difference between &amp;quot;ANALYTICAL&amp;quot; and the velocity-Verlet solution (i.e. ERROR should always be positive -- make sure you leave the half step rows blank!), and &amp;quot;ENERGY&amp;quot; should contain the total energy of the oscillator for the velocity-Verlet solution. Remember that the position of a classical harmonic oscillator is given by  (the values of , , and  are worked out for you in the sheet).&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: For the default timestep value, 0.1, estimate the positions of the maxima in the ERROR column as a function of time. Make a plot showing these values as a function of time, and fit an appropriate function to the data.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
the maximum error occur on the maximum distance from original position &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Experiment with different values of the timestep. What sort of a timestep do you need to use to ensure that the total energy does not change by more than 1% over the course of your &amp;quot;simulation&amp;quot;? Why do you think it is important to monitor the total energy of a physical system when modelling its behaviour numerically?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
more discrete time step the change of total energy will be lower, and its better to monitor the instant change of the model.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;For a single Lennard-Jones interaction, , find the separation, , at which the potential energy is zero. What is the force at this separation? Find the equilibrium separation, , and work out the well depth (). Evaluate the integrals , , and  when .&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
at r0 the particle are close together the repulsion force is going to be extremely large,&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Estimate the number of water molecules in 1ml of water under standard conditions. Estimate the volume of  water molecules under standard conditions. &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3.346*10^22  molecules   and   10000molecules of water around 2.99*10^-19 ml&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: The Lennard-Jones parameters for argon are . If the LJ cutoff is , what is it in real units? What is the well depth in ? What is the reduced temperature  in real units?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
r=1.088nm well depth=-6.16*10^-24 reduced temperature T= 180K in real unit&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Why do you think giving atoms random starting coordinates causes problems in simulations? Hint: what happens if two atoms happen to be generated close together?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Because when two atoms are simulated closed together they may counted as a larger molecule since no interaction when start the model.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Satisfy yourself that this lattice spacing corresponds to a number density of lattice points of . Consider instead a face-centred cubic lattice with a lattice point number density of 1.2. What is the side length of the cubic unit cell?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
1.144&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Using the LAMMPS manual, find the purpose of the following commands in the input script:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
mass 1 1.0&lt;br /&gt;
pair_style lj/cut 3.0&lt;br /&gt;
pair_coeff * * 1.0 1.0&lt;br /&gt;
&lt;br /&gt;
means only one atom type and mass is 1.0&lt;br /&gt;
&lt;br /&gt;
the pair style compute the standard 12/6 Lennard-Jones potential  and 3 stands for coefficients for 3 pairs of atom types&lt;br /&gt;
&lt;br /&gt;
the atoms set the coefficients for multiple pairs of atom types and the coefficient of the pairing is 1&lt;br /&gt;
&lt;br /&gt;
=== Running the simulation ===&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Look at the lines below.&#039;&#039;&#039; &lt;br /&gt;
 ### SPECIFY TIMESTEP ###&lt;br /&gt;
 variable timestep equal 0.001&lt;br /&gt;
 variable n_steps equal floor(100/${timestep})&lt;br /&gt;
 timestep ${timestep}&lt;br /&gt;
 &lt;br /&gt;
 &amp;lt;nowiki&amp;gt;### RUN SIMULATION ###&lt;br /&gt;
 run ${n_steps}&lt;br /&gt;
 run 100000&amp;lt;/nowiki&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;The second line (starting &amp;quot;variable timestep...&amp;quot;) tells LAMMPS that if it encounters the text ${timestep} on a subsequent line, it should replace it by the value given. In this case, the value ${timestep} is always replaced by 0.001. In light of this, what do you think the purpose of these lines is? Why not just write:&#039;&#039;&#039; &lt;br /&gt;
 timestep 0.001&lt;br /&gt;
 run 100000&lt;br /&gt;
&#039;&#039;&#039;Ask the demonstrator if you need help.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;nowiki&amp;gt; &amp;lt;/nowiki&amp;gt;this command is to let the time step continuously going with time step 0.001&lt;br /&gt;
&lt;br /&gt;
timestep 0.001&lt;br /&gt;
run 100000&lt;br /&gt;
&lt;br /&gt;
this will let result a repeatation  100000times with 0.001 time step&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: make plots of the energy, temperature, and pressure, against time for the 0.001 timestep experiment (attach a picture to your report). Does the simulation reach equilibrium? How long does this take? When you have done this, make a single plot which shows the energy versus time for all of the timesteps (again, attach a picture to your report). Choosing a timestep is a balancing act: the shorter the timestep, the more accurately the results of your simulation will reflect the physical reality; short timesteps, however, mean that the same number of simulation steps cover a shorter amount of actual time, and this is very unhelpful if the process you want to study requires observation over a long time. Of the five timesteps that you used, which is the largest to give acceptable results? Which one of the five is a &#039;&#039;particularly&#039;&#039; bad choice? Why?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
it reach the equilibrium around 0.17s after the simulation .&lt;br /&gt;
&lt;br /&gt;
ILL CHOOSE  the 0.0025 time step .   fewer than 0.001 time step and still keep the accuracy ,&lt;br /&gt;
&lt;br /&gt;
0.015 is the bad choice which cant monitor the equilibrium of the model&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01095534&amp;diff=638367</id>
		<title>01095534</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01095534&amp;diff=638367"/>
		<updated>2017-11-08T03:33:05Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Open the file HO.xls. In it, the velocity-Verlet algorithm is used to model the behaviour of a classical harmonic oscillator. Complete the three columns &amp;quot;ANALYTICAL&amp;quot;, &amp;quot;ERROR&amp;quot;, and &amp;quot;ENERGY&amp;quot;: &amp;quot;ANALYTICAL&amp;quot; should contain the value of the classical solution for the position at time , &amp;quot;ERROR&amp;quot; should contain the &#039;&#039;absolute&#039;&#039; difference between &amp;quot;ANALYTICAL&amp;quot; and the velocity-Verlet solution (i.e. ERROR should always be positive -- make sure you leave the half step rows blank!), and &amp;quot;ENERGY&amp;quot; should contain the total energy of the oscillator for the velocity-Verlet solution. Remember that the position of a classical harmonic oscillator is given by  (the values of , , and  are worked out for you in the sheet).&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: For the default timestep value, 0.1, estimate the positions of the maxima in the ERROR column as a function of time. Make a plot showing these values as a function of time, and fit an appropriate function to the data.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
the maximum error occur on the maximum distance from original position &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Experiment with different values of the timestep. What sort of a timestep do you need to use to ensure that the total energy does not change by more than 1% over the course of your &amp;quot;simulation&amp;quot;? Why do you think it is important to monitor the total energy of a physical system when modelling its behaviour numerically?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
more discrete time step the change of total energy will be lower, and its better to monitor the instant change of the model.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;For a single Lennard-Jones interaction, , find the separation, , at which the potential energy is zero. What is the force at this separation? Find the equilibrium separation, , and work out the well depth (). Evaluate the integrals , , and  when .&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
at r0 the particle are close together the repulsion force is going to be extremely large,&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Estimate the number of water molecules in 1ml of water under standard conditions. Estimate the volume of  water molecules under standard conditions. &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3.346*10^22  molecules   and   10000molecules of water around 2.99*10^-19 ml&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: The Lennard-Jones parameters for argon are . If the LJ cutoff is , what is it in real units? What is the well depth in ? What is the reduced temperature  in real units?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
r=1.088nm well depth=-6.16*10^-24 reduced temperature T= 180K in real unit&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Why do you think giving atoms random starting coordinates causes problems in simulations? Hint: what happens if two atoms happen to be generated close together?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Because when two atoms are simulated closed together they may counted as a larger molecule since no interaction when start the model.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Satisfy yourself that this lattice spacing corresponds to a number density of lattice points of . Consider instead a face-centred cubic lattice with a lattice point number density of 1.2. What is the side length of the cubic unit cell?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
1.144&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Using the LAMMPS manual, find the purpose of the following commands in the input script:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
mass 1 1.0&lt;br /&gt;
pair_style lj/cut 3.0&lt;br /&gt;
pair_coeff * * 1.0 1.0&lt;br /&gt;
&lt;br /&gt;
means only one atom type and mass is 1.0&lt;br /&gt;
&lt;br /&gt;
the pair style compute the standard 12/6 Lennard-Jones potential  and 3 stands for coefficients for 3 pairs of atom types&lt;br /&gt;
&lt;br /&gt;
the atoms set the coefficients for multiple pairs of atom types and the coefficient of the pairing is 1&lt;br /&gt;
&lt;br /&gt;
=== Running the simulation ===&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Look at the lines below.&#039;&#039;&#039; &lt;br /&gt;
 ### SPECIFY TIMESTEP ###&lt;br /&gt;
 variable timestep equal 0.001&lt;br /&gt;
 variable n_steps equal floor(100/${timestep})&lt;br /&gt;
 timestep ${timestep}&lt;br /&gt;
 &lt;br /&gt;
 &amp;lt;nowiki&amp;gt;### RUN SIMULATION ###&lt;br /&gt;
 run ${n_steps}&lt;br /&gt;
 run 100000&amp;lt;/nowiki&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;The second line (starting &amp;quot;variable timestep...&amp;quot;) tells LAMMPS that if it encounters the text ${timestep} on a subsequent line, it should replace it by the value given. In this case, the value ${timestep} is always replaced by 0.001. In light of this, what do you think the purpose of these lines is? Why not just write:&#039;&#039;&#039; &lt;br /&gt;
 timestep 0.001&lt;br /&gt;
 run 100000&lt;br /&gt;
&#039;&#039;&#039;Ask the demonstrator if you need help.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;nowiki&amp;gt; &amp;lt;/nowiki&amp;gt;this command is to let the time step continuously going with time step 0.001&lt;br /&gt;
&lt;br /&gt;
timestep 0.001&lt;br /&gt;
run 100000&lt;br /&gt;
&lt;br /&gt;
this will let result a repeatation  100000times with 0.001 time step&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01095534&amp;diff=638169</id>
		<title>01095534</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01095534&amp;diff=638169"/>
		<updated>2017-11-07T23:23:23Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Open the file HO.xls. In it, the velocity-Verlet algorithm is used to model the behaviour of a classical harmonic oscillator. Complete the three columns &amp;quot;ANALYTICAL&amp;quot;, &amp;quot;ERROR&amp;quot;, and &amp;quot;ENERGY&amp;quot;: &amp;quot;ANALYTICAL&amp;quot; should contain the value of the classical solution for the position at time , &amp;quot;ERROR&amp;quot; should contain the &#039;&#039;absolute&#039;&#039; difference between &amp;quot;ANALYTICAL&amp;quot; and the velocity-Verlet solution (i.e. ERROR should always be positive -- make sure you leave the half step rows blank!), and &amp;quot;ENERGY&amp;quot; should contain the total energy of the oscillator for the velocity-Verlet solution. Remember that the position of a classical harmonic oscillator is given by  (the values of , , and  are worked out for you in the sheet).&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: For the default timestep value, 0.1, estimate the positions of the maxima in the ERROR column as a function of time. Make a plot showing these values as a function of time, and fit an appropriate function to the data.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
the maximum error occur on the maximum distance from original position &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Experiment with different values of the timestep. What sort of a timestep do you need to use to ensure that the total energy does not change by more than 1% over the course of your &amp;quot;simulation&amp;quot;? Why do you think it is important to monitor the total energy of a physical system when modelling its behaviour numerically?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
more discrete time step the change of total energy will be lower, and its better to monitor the instant change of the model.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;For a single Lennard-Jones interaction, , find the separation, , at which the potential energy is zero. What is the force at this separation? Find the equilibrium separation, , and work out the well depth (). Evaluate the integrals , , and  when .&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
at r0 the particle are close together the repulsion force is going to be extremely large,&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Estimate the number of water molecules in 1ml of water under standard conditions. Estimate the volume of  water molecules under standard conditions. &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3.346*10^22  molecules   and   10000molecules of water around 2.99*10^-19 ml&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: The Lennard-Jones parameters for argon are . If the LJ cutoff is , what is it in real units? What is the well depth in ? What is the reduced temperature  in real units?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
r=1.088nm well depth=-6.16*10^-24 reduced temperature T= 180K in real unit&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Why do you think giving atoms random starting coordinates causes problems in simulations? Hint: what happens if two atoms happen to be generated close together?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Because when two atoms are simulated closed together they may counted as a larger molecule since no interaction when start the model.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Satisfy yourself that this lattice spacing corresponds to a number density of lattice points of . Consider instead a face-centred cubic lattice with a lattice point number density of 1.2. What is the side length of the cubic unit cell?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
1.144&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Using the LAMMPS manual, find the purpose of the following commands in the input script:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
mass 1 1.0&lt;br /&gt;
pair_style lj/cut 3.0&lt;br /&gt;
pair_coeff * * 1.0 1.0&lt;br /&gt;
&lt;br /&gt;
means only one atom type and mass is 1.0&lt;br /&gt;
&lt;br /&gt;
the pair style compute the standard 12/6 Lennard-Jones potential  and 3 stands for coefficients for 3 pairs of atom types&lt;br /&gt;
&lt;br /&gt;
the atoms set the coefficients for multiple pairs of atom types and the coefficient of the pairing is 1&lt;br /&gt;
&lt;br /&gt;
=== Running the simulation ===&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Look at the lines below.&#039;&#039;&#039; &lt;br /&gt;
 ### SPECIFY TIMESTEP ###&lt;br /&gt;
 variable timestep equal 0.001&lt;br /&gt;
 variable n_steps equal floor(100/${timestep})&lt;br /&gt;
 timestep ${timestep}&lt;br /&gt;
 &lt;br /&gt;
 &amp;lt;nowiki&amp;gt;### RUN SIMULATION ###&lt;br /&gt;
 run ${n_steps}&lt;br /&gt;
 run 100000&amp;lt;/nowiki&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;The second line (starting &amp;quot;variable timestep...&amp;quot;) tells LAMMPS that if it encounters the text ${timestep} on a subsequent line, it should replace it by the value given. In this case, the value ${timestep} is always replaced by 0.001. In light of this, what do you think the purpose of these lines is? Why not just write:&#039;&#039;&#039; &lt;br /&gt;
 timestep 0.001&lt;br /&gt;
 run 100000&lt;br /&gt;
&#039;&#039;&#039;Ask the demonstrator if you need help.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;nowiki&amp;gt; &amp;lt;/nowiki&amp;gt;this command is to let the time step continuously going with time step 0.001&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01095534&amp;diff=638133</id>
		<title>01095534</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01095534&amp;diff=638133"/>
		<updated>2017-11-07T22:48:06Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Open the file HO.xls. In it, the velocity-Verlet algorithm is used to model the behaviour of a classical harmonic oscillator. Complete the three columns &amp;quot;ANALYTICAL&amp;quot;, &amp;quot;ERROR&amp;quot;, and &amp;quot;ENERGY&amp;quot;: &amp;quot;ANALYTICAL&amp;quot; should contain the value of the classical solution for the position at time , &amp;quot;ERROR&amp;quot; should contain the &#039;&#039;absolute&#039;&#039; difference between &amp;quot;ANALYTICAL&amp;quot; and the velocity-Verlet solution (i.e. ERROR should always be positive -- make sure you leave the half step rows blank!), and &amp;quot;ENERGY&amp;quot; should contain the total energy of the oscillator for the velocity-Verlet solution. Remember that the position of a classical harmonic oscillator is given by  (the values of , , and  are worked out for you in the sheet).&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: For the default timestep value, 0.1, estimate the positions of the maxima in the ERROR column as a function of time. Make a plot showing these values as a function of time, and fit an appropriate function to the data.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
the maximum error occur on the maximum distance from original position &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Experiment with different values of the timestep. What sort of a timestep do you need to use to ensure that the total energy does not change by more than 1% over the course of your &amp;quot;simulation&amp;quot;? Why do you think it is important to monitor the total energy of a physical system when modelling its behaviour numerically?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
more discrete time step the change of total energy will be lower, and its better to monitor the instant change of the model.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;For a single Lennard-Jones interaction, , find the separation, , at which the potential energy is zero. What is the force at this separation? Find the equilibrium separation, , and work out the well depth (). Evaluate the integrals , , and  when .&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
at r0 the particle are close together the repulsion force is going to be extremely large,&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Estimate the number of water molecules in 1ml of water under standard conditions. Estimate the volume of  water molecules under standard conditions. &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3.346*10^22  molecules   and   10000molecules of water around 2.99*10^-19 ml&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: The Lennard-Jones parameters for argon are . If the LJ cutoff is , what is it in real units? What is the well depth in ? What is the reduced temperature  in real units?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
r=1.088nm well depth=-6.16*10^-24 reduced temperature T= 180K in real unit&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Why do you think giving atoms random starting coordinates causes problems in simulations? Hint: what happens if two atoms happen to be generated close together?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Because when two atoms are simulated closed together they may counted as a larger molecule since no interaction when start the model.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Satisfy yourself that this lattice spacing corresponds to a number density of lattice points of . Consider instead a face-centred cubic lattice with a lattice point number density of 1.2. What is the side length of the cubic unit cell?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
1.144&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Using the LAMMPS manual, find the purpose of the following commands in the input script:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
mass 1 1.0&lt;br /&gt;
pair_style lj/cut 3.0&lt;br /&gt;
pair_coeff * * 1.0 1.0&lt;br /&gt;
&lt;br /&gt;
means only one atom type and mass is 1.0&lt;br /&gt;
&lt;br /&gt;
the pair style compute the standard 12/6 Lennard-Jones potential  and 3 stands for coefficients for 3 pairs of atom types&lt;br /&gt;
&lt;br /&gt;
the atoms set the coefficients for multiple pairs of atom types and the coefficient of the pairing is 1&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01095534&amp;diff=637725</id>
		<title>01095534</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01095534&amp;diff=637725"/>
		<updated>2017-11-07T14:52:06Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Open the file HO.xls. In it, the velocity-Verlet algorithm is used to model the behaviour of a classical harmonic oscillator. Complete the three columns &amp;quot;ANALYTICAL&amp;quot;, &amp;quot;ERROR&amp;quot;, and &amp;quot;ENERGY&amp;quot;: &amp;quot;ANALYTICAL&amp;quot; should contain the value of the classical solution for the position at time , &amp;quot;ERROR&amp;quot; should contain the &#039;&#039;absolute&#039;&#039; difference between &amp;quot;ANALYTICAL&amp;quot; and the velocity-Verlet solution (i.e. ERROR should always be positive -- make sure you leave the half step rows blank!), and &amp;quot;ENERGY&amp;quot; should contain the total energy of the oscillator for the velocity-Verlet solution. Remember that the position of a classical harmonic oscillator is given by  (the values of , , and  are worked out for you in the sheet).&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: For the default timestep value, 0.1, estimate the positions of the maxima in the ERROR column as a function of time. Make a plot showing these values as a function of time, and fit an appropriate function to the data.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
the maximum error occur on the maximum distance from original position &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Experiment with different values of the timestep. What sort of a timestep do you need to use to ensure that the total energy does not change by more than 1% over the course of your &amp;quot;simulation&amp;quot;? Why do you think it is important to monitor the total energy of a physical system when modelling its behaviour numerically?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
more discrete time step the change of total energy will be lower, and its better to monitor the instant change of the model.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;For a single Lennard-Jones interaction, , find the separation, , at which the potential energy is zero. What is the force at this separation? Find the equilibrium separation, , and work out the well depth (). Evaluate the integrals , , and  when .&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
at r0 the particle are close together the repulsion force is going to be extremely large,&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Estimate the number of water molecules in 1ml of water under standard conditions. Estimate the volume of  water molecules under standard conditions. &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3.346*10^22  molecules   and   10000molecules of water around 2.99*10^-19 ml&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: The Lennard-Jones parameters for argon are . If the LJ cutoff is , what is it in real units? What is the well depth in ? What is the reduced temperature  in real units?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
r=1.088nm well depth=-6.16*10^-24 reduced temperature T= 180K in real unit&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Why do you think giving atoms random starting coordinates causes problems in simulations? Hint: what happens if two atoms happen to be generated close together?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Because when two atoms are simulated closed together they may counted as a larger molecule since no interaction when start the model.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Satisfy yourself that this lattice spacing corresponds to a number density of lattice points of . Consider instead a face-centred cubic lattice with a lattice point number density of 1.2. What is the side length of the cubic unit cell?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
1.144&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Using the LAMMPS manual, find the purpose of the following commands in the input script:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
mass 1 1.0&lt;br /&gt;
pair_style lj/cut 3.0&lt;br /&gt;
pair_coeff * * 1.0 1.0&lt;br /&gt;
&lt;br /&gt;
means only one atom type and mass is 1.0&lt;br /&gt;
&lt;br /&gt;
the pair style compute the standard 12/6 Lennard-Jones potential  and 3 stands for coefficients for 3 pairs of atom types&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01095534&amp;diff=637717</id>
		<title>01095534</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01095534&amp;diff=637717"/>
		<updated>2017-11-07T14:43:19Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Open the file HO.xls. In it, the velocity-Verlet algorithm is used to model the behaviour of a classical harmonic oscillator. Complete the three columns &amp;quot;ANALYTICAL&amp;quot;, &amp;quot;ERROR&amp;quot;, and &amp;quot;ENERGY&amp;quot;: &amp;quot;ANALYTICAL&amp;quot; should contain the value of the classical solution for the position at time , &amp;quot;ERROR&amp;quot; should contain the &#039;&#039;absolute&#039;&#039; difference between &amp;quot;ANALYTICAL&amp;quot; and the velocity-Verlet solution (i.e. ERROR should always be positive -- make sure you leave the half step rows blank!), and &amp;quot;ENERGY&amp;quot; should contain the total energy of the oscillator for the velocity-Verlet solution. Remember that the position of a classical harmonic oscillator is given by  (the values of , , and  are worked out for you in the sheet).&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: For the default timestep value, 0.1, estimate the positions of the maxima in the ERROR column as a function of time. Make a plot showing these values as a function of time, and fit an appropriate function to the data.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
the maximum error occur on the maximum distance from original position &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Experiment with different values of the timestep. What sort of a timestep do you need to use to ensure that the total energy does not change by more than 1% over the course of your &amp;quot;simulation&amp;quot;? Why do you think it is important to monitor the total energy of a physical system when modelling its behaviour numerically?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
more discrete time step the change of total energy will be lower, and its better to monitor the instant change of the model.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;For a single Lennard-Jones interaction, , find the separation, , at which the potential energy is zero. What is the force at this separation? Find the equilibrium separation, , and work out the well depth (). Evaluate the integrals , , and  when .&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
at r0 the particle are close together the repulsion force is going to be extremely large,&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Estimate the number of water molecules in 1ml of water under standard conditions. Estimate the volume of  water molecules under standard conditions. &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3.346*10^22  molecules   and   10000molecules of water around 2.99*10^-19 ml&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: The Lennard-Jones parameters for argon are . If the LJ cutoff is , what is it in real units? What is the well depth in ? What is the reduced temperature  in real units?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
r=1.088nm well depth=-6.16*10^-24 reduced temperature T= 180K in real unit&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Why do you think giving atoms random starting coordinates causes problems in simulations? Hint: what happens if two atoms happen to be generated close together?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Because when two atoms are simulated closed together they may counted as a larger molecule since no interaction when start the model.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Satisfy yourself that this lattice spacing corresponds to a number density of lattice points of . Consider instead a face-centred cubic lattice with a lattice point number density of 1.2. What is the side length of the cubic unit cell?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
1.144&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Using the LAMMPS manual, find the purpose of the following commands in the input script:&#039;&#039;&#039;&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01095534&amp;diff=637542</id>
		<title>01095534</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01095534&amp;diff=637542"/>
		<updated>2017-11-07T09:44:54Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Open the file HO.xls. In it, the velocity-Verlet algorithm is used to model the behaviour of a classical harmonic oscillator. Complete the three columns &amp;quot;ANALYTICAL&amp;quot;, &amp;quot;ERROR&amp;quot;, and &amp;quot;ENERGY&amp;quot;: &amp;quot;ANALYTICAL&amp;quot; should contain the value of the classical solution for the position at time , &amp;quot;ERROR&amp;quot; should contain the &#039;&#039;absolute&#039;&#039; difference between &amp;quot;ANALYTICAL&amp;quot; and the velocity-Verlet solution (i.e. ERROR should always be positive -- make sure you leave the half step rows blank!), and &amp;quot;ENERGY&amp;quot; should contain the total energy of the oscillator for the velocity-Verlet solution. Remember that the position of a classical harmonic oscillator is given by  (the values of , , and  are worked out for you in the sheet).&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: For the default timestep value, 0.1, estimate the positions of the maxima in the ERROR column as a function of time. Make a plot showing these values as a function of time, and fit an appropriate function to the data.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
the maximum error occur on the maximum distance from original position &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Experiment with different values of the timestep. What sort of a timestep do you need to use to ensure that the total energy does not change by more than 1% over the course of your &amp;quot;simulation&amp;quot;? Why do you think it is important to monitor the total energy of a physical system when modelling its behaviour numerically?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
more discrete time step the change of total energy will be lower, and its better to monitor the instant change of the model.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;For a single Lennard-Jones interaction, , find the separation, , at which the potential energy is zero. What is the force at this separation? Find the equilibrium separation, , and work out the well depth (). Evaluate the integrals , , and  when .&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
at r0 the particle are close together the repulsion force is going to be extremely large,&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;TASK: Estimate the number of water molecules in 1ml of water under standard conditions. Estimate the volume of  water molecules under standard conditions. &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3.346*10^22  molecules   and   10000molecules of water around 2.99*10^-19 ml&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01095534&amp;diff=637522</id>
		<title>01095534</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01095534&amp;diff=637522"/>
		<updated>2017-11-07T09:15:07Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Open the file HO.xls. In it, the velocity-Verlet algorithm is used to model the behaviour of a classical harmonic oscillator. Complete the three columns &amp;quot;ANALYTICAL&amp;quot;, &amp;quot;ERROR&amp;quot;, and &amp;quot;ENERGY&amp;quot;: &amp;quot;ANALYTICAL&amp;quot; should contain the value of the classical solution for the position at time , &amp;quot;ERROR&amp;quot; should contain the &#039;&#039;absolute&#039;&#039; difference between &amp;quot;ANALYTICAL&amp;quot; and the velocity-Verlet solution (i.e. ERROR should always be positive -- make sure you leave the half step rows blank!), and &amp;quot;ENERGY&amp;quot; should contain the total energy of the oscillator for the velocity-Verlet solution. Remember that the position of a classical harmonic oscillator is given by  (the values of , , and  are worked out for you in the sheet).&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: For the default timestep value, 0.1, estimate the positions of the maxima in the ERROR column as a function of time. Make a plot showing these values as a function of time, and fit an appropriate function to the data.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
the maximum error occur on the maximum distance from original position &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Experiment with different values of the timestep. What sort of a timestep do you need to use to ensure that the total energy does not change by more than 1% over the course of your &amp;quot;simulation&amp;quot;? Why do you think it is important to monitor the total energy of a physical system when modelling its behaviour numerically?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
more discrete time step the change of total energy will be lower, and its better to monitor the instant change of the model.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;For a single Lennard-Jones interaction, , find the separation, , at which the potential energy is zero. What is the force at this separation? Find the equilibrium separation, , and work out the well depth (). Evaluate the integrals , , and  when .&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
at r0 the particle are close together the repulsion force is going to be extremely large,&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01095534&amp;diff=637456</id>
		<title>01095534</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01095534&amp;diff=637456"/>
		<updated>2017-11-06T23:28:01Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: Created page with &amp;quot;&amp;#039;&amp;#039;&amp;#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Open the file HO.xls. In it, the velocity-Verlet algorithm is used to model the behaviour of a classical harmonic oscillator. Complete the three columns &amp;quot;A...&amp;quot;&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Open the file HO.xls. In it, the velocity-Verlet algorithm is used to model the behaviour of a classical harmonic oscillator. Complete the three columns &amp;quot;ANALYTICAL&amp;quot;, &amp;quot;ERROR&amp;quot;, and &amp;quot;ENERGY&amp;quot;: &amp;quot;ANALYTICAL&amp;quot; should contain the value of the classical solution for the position at time , &amp;quot;ERROR&amp;quot; should contain the &#039;&#039;absolute&#039;&#039; difference between &amp;quot;ANALYTICAL&amp;quot; and the velocity-Verlet solution (i.e. ERROR should always be positive -- make sure you leave the half step rows blank!), and &amp;quot;ENERGY&amp;quot; should contain the total energy of the oscillator for the velocity-Verlet solution. Remember that the position of a classical harmonic oscillator is given by  (the values of , , and  are worked out for you in the sheet).&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: For the default timestep value, 0.1, estimate the positions of the maxima in the ERROR column as a function of time. Make a plot showing these values as a function of time, and fit an appropriate function to the data.&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;big&amp;gt;TASK&amp;lt;/big&amp;gt;: Experiment with different values of the timestep. What sort of a timestep do you need to use to ensure that the total energy does not change by more than 1% over the course of your &amp;quot;simulation&amp;quot;? Why do you think it is important to monitor the total energy of a physical system when modelling its behaviour numerically?&#039;&#039;&#039;&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:everything01095534&amp;diff=615499</id>
		<title>MRD:everything01095534</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:everything01095534&amp;diff=615499"/>
		<updated>2017-05-05T15:48:15Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: /* Activation Energy */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== Molecular Reaction Dynamics of Triatomic systems ==&lt;br /&gt;
this wiki page is to explain triatomic interaction dynamic and illustrate it on specific example&lt;br /&gt;
&lt;br /&gt;
=== background Theory ===&lt;br /&gt;
The nucleus, is extremely heavy and large compare to electron in the atom. So when an atom moving toward or backward to another atom, the relationship can be shown by the equation below:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; {dp_i \over dt} = - { \partial V(r_1,r_2,...)\over \partial r_i}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The force (variation of momentum &#039;&#039;&#039;p&#039;&#039;&#039;) acting on a given intervatomic coordinate &#039;&#039;&#039;r&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt;&#039;&#039;&#039; will depend on the derivative of the potential energy surface with respect to that coordinate.&lt;br /&gt;
&lt;br /&gt;
In this concept only three atoms on a straight line will be considered.  &lt;br /&gt;
&lt;br /&gt;
== H+H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; System ==&lt;br /&gt;
In this context, three hydrogen atoms are aligned on a straight line to use Matlab programming the potential energy change of the atoms under different condition&lt;br /&gt;
[[File:Plot1.png|500px|thumb|center|contour plot, r1=0.74A, r2=2.30A, p1=0 p2=-2.70]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
contour graph when the distance between A and B is equal to 0.74 which is just like the normal bond length of a normal H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule. In the H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; model, the initial conditions are set as &#039;&#039;&#039;r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;&#039;&#039;&#039; = 0.74 ,&#039;&#039;&#039;r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;&#039;&#039;&#039;&amp;lt;sub&amp;gt; &amp;lt;/sub&amp;gt;= 2.30 and the momenta &#039;&#039;&#039;p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;&#039;&#039;&#039;=0 &#039;&#039;&#039; p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;&#039;&#039;&#039; = -2.7.&lt;br /&gt;
&lt;br /&gt;
[[File:haiyang2.png|500px|thumb|center|intermolecular distance vs time, r1=0.74A, r2=2.30A, p1=0 p2=-2.70]]&lt;br /&gt;
&lt;br /&gt;
This is the figure which shows when the time goes to 0.38 time units, a repetitive oscillation behavior is happened, showing that the transition state was also suppose to be found at 0.38 time units. So the r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is 0.9148A then. But based on the result of animation, the true r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is 0.90775A but not 0.9148A.&lt;br /&gt;
[[File:haiyang3.png|500px|thumb|left|Kinetic Energy vs Time, r1=0.74A, r2=2.30A, p1=0 p2=-2.70]]&lt;br /&gt;
[[File:haiyang4.png|520px|thumb|center|potential Energy vs Time, r1=0.74A, r2=2.30A, p1=0 p2=-2.70]]&lt;br /&gt;
&lt;br /&gt;
The &amp;quot;Potential Energy vs Time&amp;quot; and &amp;quot;Kinetic Energy vs Time&amp;quot; proved that at transition state the potential energy gives rise to the highest value while kinetic energy give rise to the lowest value. As well, an oscillatory behavior rising after the transition state.&lt;br /&gt;
&lt;br /&gt;
The total gradient of the minimum (at the start and the end point in reactant) on the potential energy surface is equal to &#039;&#039;&#039;0 as well as &#039;&#039;&#039;at the transition state point. But the difference between the minimum and the transition state point is that transition state point has a negative second derivative towards product and a positive second derivative towards reactant; while at the minimum point in the whole system (at the beginning point in the reactant) have a &#039;&#039;&#039;zero&#039;&#039;&#039; second derivative value.Hence the potential energy difference between minimum energy and transition state is the &#039;&#039;&#039;activation eneregy&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
=== Trajectories from r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; locating the transition state ===&lt;br /&gt;
Due to H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; potential surface is symmetric(left and right of the transition state), assuming the transition state must have &#039;&#039;&#039;r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;&#039;&#039;&#039; &#039;&#039;&#039;=&#039;&#039;&#039; &#039;&#039;&#039;0.90775A&#039;&#039;&#039; (r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; at the transition state position), the relative stationary state of motion is observed. If the approaching proton atom is placed at the ridge of potential surface, there is no gradient in the direction at right angle to the ridge; therefore the proton sphere would oscillate about that region without falling down. As shown in the figures below, one figure shows that the reactant proton is oscillating at the transition state position; while another shows that there is not any relative change in terms of the distance for the three protons against time.&lt;br /&gt;
&lt;br /&gt;
[[File:haiyang5.png|300px|thumb|center|the potential energy surface is symmetric through transition state]]&lt;br /&gt;
&lt;br /&gt;
=== Calculating the reaction path by using Trajectories from r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+ δ ===&lt;br /&gt;
[[File:haiyang6.png|400px|thumb|right|Surface Plot, r1=r2=0.90775A p1=p2=0]]&lt;br /&gt;
[[File:haiyang7.png|400px|thumb|left|Internuclear Distance vs Time, r1 = 0.91775, r2 = 0.90775 p1 = p2 = 0]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
This model will favour H&amp;lt;sub&amp;gt;A&amp;lt;/sub&amp;gt;-H&amp;lt;sub&amp;gt;B&amp;lt;/sub&amp;gt; + H&amp;lt;sub&amp;gt;C&amp;lt;/sub&amp;gt; as reactant proton C is very close to the central proton. In this section, a initial system of r1 = 0.91775, r2 = 0.90775. p1 = 0 p2 = 0 is considered.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;MEP calculation type&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
[[File:haiyang8.png|400px|thumb|center|MEP Simulating Contour Plot, r1 = 0.91775, r2 = 0.90775 p1 = p2 = 0]]&lt;br /&gt;
&lt;br /&gt;
In this case, there is a little derivation from the transition state position, say 0.01A derivation for a first try, for one of the protons. The MEP calculation type is set so as to simulate the motion for this circumstance. The MEP contour plot is given .&lt;br /&gt;
&lt;br /&gt;
===unreactive trajectories===&lt;br /&gt;
From the previous calculations we can conclude that trajectories with initial conditions in the range r1 = 0.74, r2 = 2.0, with -1.5 &amp;lt; p1 &amp;lt; -0.8 and p2 = -2.5 are reactive&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;For the initial positions r1 = 0.74 and r2 = 2.0, run trajectories with the following momenta combination:&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=&amp;quot;1&amp;quot;&lt;br /&gt;
|+ caption&lt;br /&gt;
! p1!! p2!! surface plot!! internucler plot!! conclusion&lt;br /&gt;
|-&lt;br /&gt;
| -1.25 || -2.5||[[File:haiyang8.png|300px|center]]||[[File:haiyang10.png|300px|center]]|| r1 and r2 never be equal so unreactive&lt;br /&gt;
|-&lt;br /&gt;
| -1.5 || -2.0||[[File:haiyang11.png|300px|center]]||[[File:haiyang12.png|300px|center]]|| r1 and r2 never be equal so unreactive&lt;br /&gt;
|-&lt;br /&gt;
| -1.5 || -2.5||[[File:haiyang13.png|300px|center]]||[[File:haiyang14.png|300px|center]]|| r1 and r2 has a equal value at a time so transition state formed ,reaction reactive&lt;br /&gt;
|-&lt;br /&gt;
| -2.5 || -5.0||[[File:Haiyang15.png|300px|center]]||[[File:Haiyang16.png|300px|center]]||r1 and r2 has a equal value at a time so transition state formed ,reaction reactive,however the momenta is too large that the particle collide back again. the reaction is reversed &lt;br /&gt;
|-&lt;br /&gt;
| -2.5 || -5.2||[[File:Haiyang17.png|300px|center]]||[[File:Haiyang18.png|300px|center]]||r1 and r2 has a equal value at a time so transition state formed ,reaction reactive,however the momenta is too large that the particle collide within the transition state area several times.&lt;br /&gt;
|-&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
== H-H-F System ==&lt;br /&gt;
the reaction is exothermic by broken a H-H bond and form a H-F bond. It can be proof by the potential energy surface, when hydrogen bonded to fluorine the minimum energy is significantly lower than the minimum energy when hydrogens are bonded &lt;br /&gt;
[[File:haiyang19.png|thumb|center|minimum energy along potential energy surface of F-H-H system]]&lt;br /&gt;
===Transition State===&lt;br /&gt;
As we talk above the transition state is the maximum value of the minimum band, after many postulating the transition state is test as when fluorine-hydrogen distance is 1.810Å from the middle hydrogen(H-H bond existing)&lt;br /&gt;
[[File:haiyang20.png|thumb|center|atom will not move at transition state, so only a spot shown at TS H-F distance 1.810Å]]&lt;br /&gt;
===Activation Energy===&lt;br /&gt;
According to Hammond&#039;s postulate, the transition state is close to the reactant so, by slightly change the distance between fluorine molecule and hydrogen atom in the middle the atom will back to the position.&lt;br /&gt;
[[File:haiyang21.png|thumb|center|energy change when move fluorine far from transition state distance.]]&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang21.png&amp;diff=615493</id>
		<title>File:Haiyang21.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang21.png&amp;diff=615493"/>
		<updated>2017-05-05T15:46:21Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:everything01095534&amp;diff=615488</id>
		<title>MRD:everything01095534</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:everything01095534&amp;diff=615488"/>
		<updated>2017-05-05T15:45:27Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: /* H-H-F System */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== Molecular Reaction Dynamics of Triatomic systems ==&lt;br /&gt;
this wiki page is to explain triatomic interaction dynamic and illustrate it on specific example&lt;br /&gt;
&lt;br /&gt;
=== background Theory ===&lt;br /&gt;
The nucleus, is extremely heavy and large compare to electron in the atom. So when an atom moving toward or backward to another atom, the relationship can be shown by the equation below:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; {dp_i \over dt} = - { \partial V(r_1,r_2,...)\over \partial r_i}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The force (variation of momentum &#039;&#039;&#039;p&#039;&#039;&#039;) acting on a given intervatomic coordinate &#039;&#039;&#039;r&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt;&#039;&#039;&#039; will depend on the derivative of the potential energy surface with respect to that coordinate.&lt;br /&gt;
&lt;br /&gt;
In this concept only three atoms on a straight line will be considered.  &lt;br /&gt;
&lt;br /&gt;
== H+H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; System ==&lt;br /&gt;
In this context, three hydrogen atoms are aligned on a straight line to use Matlab programming the potential energy change of the atoms under different condition&lt;br /&gt;
[[File:Plot1.png|500px|thumb|center|contour plot, r1=0.74A, r2=2.30A, p1=0 p2=-2.70]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
contour graph when the distance between A and B is equal to 0.74 which is just like the normal bond length of a normal H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule. In the H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; model, the initial conditions are set as &#039;&#039;&#039;r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;&#039;&#039;&#039; = 0.74 ,&#039;&#039;&#039;r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;&#039;&#039;&#039;&amp;lt;sub&amp;gt; &amp;lt;/sub&amp;gt;= 2.30 and the momenta &#039;&#039;&#039;p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;&#039;&#039;&#039;=0 &#039;&#039;&#039; p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;&#039;&#039;&#039; = -2.7.&lt;br /&gt;
&lt;br /&gt;
[[File:haiyang2.png|500px|thumb|center|intermolecular distance vs time, r1=0.74A, r2=2.30A, p1=0 p2=-2.70]]&lt;br /&gt;
&lt;br /&gt;
This is the figure which shows when the time goes to 0.38 time units, a repetitive oscillation behavior is happened, showing that the transition state was also suppose to be found at 0.38 time units. So the r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is 0.9148A then. But based on the result of animation, the true r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is 0.90775A but not 0.9148A.&lt;br /&gt;
[[File:haiyang3.png|500px|thumb|left|Kinetic Energy vs Time, r1=0.74A, r2=2.30A, p1=0 p2=-2.70]]&lt;br /&gt;
[[File:haiyang4.png|520px|thumb|center|potential Energy vs Time, r1=0.74A, r2=2.30A, p1=0 p2=-2.70]]&lt;br /&gt;
&lt;br /&gt;
The &amp;quot;Potential Energy vs Time&amp;quot; and &amp;quot;Kinetic Energy vs Time&amp;quot; proved that at transition state the potential energy gives rise to the highest value while kinetic energy give rise to the lowest value. As well, an oscillatory behavior rising after the transition state.&lt;br /&gt;
&lt;br /&gt;
The total gradient of the minimum (at the start and the end point in reactant) on the potential energy surface is equal to &#039;&#039;&#039;0 as well as &#039;&#039;&#039;at the transition state point. But the difference between the minimum and the transition state point is that transition state point has a negative second derivative towards product and a positive second derivative towards reactant; while at the minimum point in the whole system (at the beginning point in the reactant) have a &#039;&#039;&#039;zero&#039;&#039;&#039; second derivative value.Hence the potential energy difference between minimum energy and transition state is the &#039;&#039;&#039;activation eneregy&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
=== Trajectories from r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; locating the transition state ===&lt;br /&gt;
Due to H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; potential surface is symmetric(left and right of the transition state), assuming the transition state must have &#039;&#039;&#039;r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;&#039;&#039;&#039; &#039;&#039;&#039;=&#039;&#039;&#039; &#039;&#039;&#039;0.90775A&#039;&#039;&#039; (r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; at the transition state position), the relative stationary state of motion is observed. If the approaching proton atom is placed at the ridge of potential surface, there is no gradient in the direction at right angle to the ridge; therefore the proton sphere would oscillate about that region without falling down. As shown in the figures below, one figure shows that the reactant proton is oscillating at the transition state position; while another shows that there is not any relative change in terms of the distance for the three protons against time.&lt;br /&gt;
&lt;br /&gt;
[[File:haiyang5.png|300px|thumb|center|the potential energy surface is symmetric through transition state]]&lt;br /&gt;
&lt;br /&gt;
=== Calculating the reaction path by using Trajectories from r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+ δ ===&lt;br /&gt;
[[File:haiyang6.png|400px|thumb|right|Surface Plot, r1=r2=0.90775A p1=p2=0]]&lt;br /&gt;
[[File:haiyang7.png|400px|thumb|left|Internuclear Distance vs Time, r1 = 0.91775, r2 = 0.90775 p1 = p2 = 0]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
This model will favour H&amp;lt;sub&amp;gt;A&amp;lt;/sub&amp;gt;-H&amp;lt;sub&amp;gt;B&amp;lt;/sub&amp;gt; + H&amp;lt;sub&amp;gt;C&amp;lt;/sub&amp;gt; as reactant proton C is very close to the central proton. In this section, a initial system of r1 = 0.91775, r2 = 0.90775. p1 = 0 p2 = 0 is considered.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;MEP calculation type&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
[[File:haiyang8.png|400px|thumb|center|MEP Simulating Contour Plot, r1 = 0.91775, r2 = 0.90775 p1 = p2 = 0]]&lt;br /&gt;
&lt;br /&gt;
In this case, there is a little derivation from the transition state position, say 0.01A derivation for a first try, for one of the protons. The MEP calculation type is set so as to simulate the motion for this circumstance. The MEP contour plot is given .&lt;br /&gt;
&lt;br /&gt;
===unreactive trajectories===&lt;br /&gt;
From the previous calculations we can conclude that trajectories with initial conditions in the range r1 = 0.74, r2 = 2.0, with -1.5 &amp;lt; p1 &amp;lt; -0.8 and p2 = -2.5 are reactive&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;For the initial positions r1 = 0.74 and r2 = 2.0, run trajectories with the following momenta combination:&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=&amp;quot;1&amp;quot;&lt;br /&gt;
|+ caption&lt;br /&gt;
! p1!! p2!! surface plot!! internucler plot!! conclusion&lt;br /&gt;
|-&lt;br /&gt;
| -1.25 || -2.5||[[File:haiyang8.png|300px|center]]||[[File:haiyang10.png|300px|center]]|| r1 and r2 never be equal so unreactive&lt;br /&gt;
|-&lt;br /&gt;
| -1.5 || -2.0||[[File:haiyang11.png|300px|center]]||[[File:haiyang12.png|300px|center]]|| r1 and r2 never be equal so unreactive&lt;br /&gt;
|-&lt;br /&gt;
| -1.5 || -2.5||[[File:haiyang13.png|300px|center]]||[[File:haiyang14.png|300px|center]]|| r1 and r2 has a equal value at a time so transition state formed ,reaction reactive&lt;br /&gt;
|-&lt;br /&gt;
| -2.5 || -5.0||[[File:Haiyang15.png|300px|center]]||[[File:Haiyang16.png|300px|center]]||r1 and r2 has a equal value at a time so transition state formed ,reaction reactive,however the momenta is too large that the particle collide back again. the reaction is reversed &lt;br /&gt;
|-&lt;br /&gt;
| -2.5 || -5.2||[[File:Haiyang17.png|300px|center]]||[[File:Haiyang18.png|300px|center]]||r1 and r2 has a equal value at a time so transition state formed ,reaction reactive,however the momenta is too large that the particle collide within the transition state area several times.&lt;br /&gt;
|-&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
== H-H-F System ==&lt;br /&gt;
the reaction is exothermic by broken a H-H bond and form a H-F bond. It can be proof by the potential energy surface, when hydrogen bonded to fluorine the minimum energy is significantly lower than the minimum energy when hydrogens are bonded &lt;br /&gt;
[[File:haiyang19.png|thumb|center|minimum energy along potential energy surface of F-H-H system]]&lt;br /&gt;
===Transition State===&lt;br /&gt;
As we talk above the transition state is the maximum value of the minimum band, after many postulating the transition state is test as when fluorine-hydrogen distance is 1.810Å from the middle hydrogen(H-H bond existing)&lt;br /&gt;
[[File:haiyang20.png|thumb|center|atom will not move at transition state, so only a spot shown at TS H-F distance 1.810Å]]&lt;br /&gt;
===Activation Energy===&lt;br /&gt;
According to Hammond&#039;s postulate, the transition state is close to the reactant so, by slightly change the distance between fluorine molecule and hydrogen atom in the middle the atom will back to the position.&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang20.png&amp;diff=615386</id>
		<title>File:Haiyang20.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang20.png&amp;diff=615386"/>
		<updated>2017-05-05T15:19:51Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang19.png&amp;diff=615356</id>
		<title>File:Haiyang19.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang19.png&amp;diff=615356"/>
		<updated>2017-05-05T15:11:43Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:everything01095534&amp;diff=615342</id>
		<title>MRD:everything01095534</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:everything01095534&amp;diff=615342"/>
		<updated>2017-05-05T15:06:21Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: /* H+H2 System */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== Molecular Reaction Dynamics of Triatomic systems ==&lt;br /&gt;
this wiki page is to explain triatomic interaction dynamic and illustrate it on specific example&lt;br /&gt;
&lt;br /&gt;
=== background Theory ===&lt;br /&gt;
The nucleus, is extremely heavy and large compare to electron in the atom. So when an atom moving toward or backward to another atom, the relationship can be shown by the equation below:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; {dp_i \over dt} = - { \partial V(r_1,r_2,...)\over \partial r_i}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The force (variation of momentum &#039;&#039;&#039;p&#039;&#039;&#039;) acting on a given intervatomic coordinate &#039;&#039;&#039;r&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt;&#039;&#039;&#039; will depend on the derivative of the potential energy surface with respect to that coordinate.&lt;br /&gt;
&lt;br /&gt;
In this concept only three atoms on a straight line will be considered.  &lt;br /&gt;
&lt;br /&gt;
== H+H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; System ==&lt;br /&gt;
In this context, three hydrogen atoms are aligned on a straight line to use Matlab programming the potential energy change of the atoms under different condition&lt;br /&gt;
[[File:Plot1.png|500px|thumb|center|contour plot, r1=0.74A, r2=2.30A, p1=0 p2=-2.70]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
contour graph when the distance between A and B is equal to 0.74 which is just like the normal bond length of a normal H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule. In the H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; model, the initial conditions are set as &#039;&#039;&#039;r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;&#039;&#039;&#039; = 0.74 ,&#039;&#039;&#039;r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;&#039;&#039;&#039;&amp;lt;sub&amp;gt; &amp;lt;/sub&amp;gt;= 2.30 and the momenta &#039;&#039;&#039;p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;&#039;&#039;&#039;=0 &#039;&#039;&#039; p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;&#039;&#039;&#039; = -2.7.&lt;br /&gt;
&lt;br /&gt;
[[File:haiyang2.png|500px|thumb|center|intermolecular distance vs time, r1=0.74A, r2=2.30A, p1=0 p2=-2.70]]&lt;br /&gt;
&lt;br /&gt;
This is the figure which shows when the time goes to 0.38 time units, a repetitive oscillation behavior is happened, showing that the transition state was also suppose to be found at 0.38 time units. So the r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is 0.9148A then. But based on the result of animation, the true r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is 0.90775A but not 0.9148A.&lt;br /&gt;
[[File:haiyang3.png|500px|thumb|left|Kinetic Energy vs Time, r1=0.74A, r2=2.30A, p1=0 p2=-2.70]]&lt;br /&gt;
[[File:haiyang4.png|520px|thumb|center|potential Energy vs Time, r1=0.74A, r2=2.30A, p1=0 p2=-2.70]]&lt;br /&gt;
&lt;br /&gt;
The &amp;quot;Potential Energy vs Time&amp;quot; and &amp;quot;Kinetic Energy vs Time&amp;quot; proved that at transition state the potential energy gives rise to the highest value while kinetic energy give rise to the lowest value. As well, an oscillatory behavior rising after the transition state.&lt;br /&gt;
&lt;br /&gt;
The total gradient of the minimum (at the start and the end point in reactant) on the potential energy surface is equal to &#039;&#039;&#039;0 as well as &#039;&#039;&#039;at the transition state point. But the difference between the minimum and the transition state point is that transition state point has a negative second derivative towards product and a positive second derivative towards reactant; while at the minimum point in the whole system (at the beginning point in the reactant) have a &#039;&#039;&#039;zero&#039;&#039;&#039; second derivative value.Hence the potential energy difference between minimum energy and transition state is the &#039;&#039;&#039;activation eneregy&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
=== Trajectories from r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; locating the transition state ===&lt;br /&gt;
Due to H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; potential surface is symmetric(left and right of the transition state), assuming the transition state must have &#039;&#039;&#039;r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;&#039;&#039;&#039; &#039;&#039;&#039;=&#039;&#039;&#039; &#039;&#039;&#039;0.90775A&#039;&#039;&#039; (r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; at the transition state position), the relative stationary state of motion is observed. If the approaching proton atom is placed at the ridge of potential surface, there is no gradient in the direction at right angle to the ridge; therefore the proton sphere would oscillate about that region without falling down. As shown in the figures below, one figure shows that the reactant proton is oscillating at the transition state position; while another shows that there is not any relative change in terms of the distance for the three protons against time.&lt;br /&gt;
&lt;br /&gt;
[[File:haiyang5.png|300px|thumb|center|the potential energy surface is symmetric through transition state]]&lt;br /&gt;
&lt;br /&gt;
=== Calculating the reaction path by using Trajectories from r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+ δ ===&lt;br /&gt;
[[File:haiyang6.png|400px|thumb|right|Surface Plot, r1=r2=0.90775A p1=p2=0]]&lt;br /&gt;
[[File:haiyang7.png|400px|thumb|left|Internuclear Distance vs Time, r1 = 0.91775, r2 = 0.90775 p1 = p2 = 0]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
This model will favour H&amp;lt;sub&amp;gt;A&amp;lt;/sub&amp;gt;-H&amp;lt;sub&amp;gt;B&amp;lt;/sub&amp;gt; + H&amp;lt;sub&amp;gt;C&amp;lt;/sub&amp;gt; as reactant proton C is very close to the central proton. In this section, a initial system of r1 = 0.91775, r2 = 0.90775. p1 = 0 p2 = 0 is considered.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;MEP calculation type&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
[[File:haiyang8.png|400px|thumb|center|MEP Simulating Contour Plot, r1 = 0.91775, r2 = 0.90775 p1 = p2 = 0]]&lt;br /&gt;
&lt;br /&gt;
In this case, there is a little derivation from the transition state position, say 0.01A derivation for a first try, for one of the protons. The MEP calculation type is set so as to simulate the motion for this circumstance. The MEP contour plot is given .&lt;br /&gt;
&lt;br /&gt;
===unreactive trajectories===&lt;br /&gt;
From the previous calculations we can conclude that trajectories with initial conditions in the range r1 = 0.74, r2 = 2.0, with -1.5 &amp;lt; p1 &amp;lt; -0.8 and p2 = -2.5 are reactive&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;For the initial positions r1 = 0.74 and r2 = 2.0, run trajectories with the following momenta combination:&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=&amp;quot;1&amp;quot;&lt;br /&gt;
|+ caption&lt;br /&gt;
! p1!! p2!! surface plot!! internucler plot!! conclusion&lt;br /&gt;
|-&lt;br /&gt;
| -1.25 || -2.5||[[File:haiyang8.png|300px|center]]||[[File:haiyang10.png|300px|center]]|| r1 and r2 never be equal so unreactive&lt;br /&gt;
|-&lt;br /&gt;
| -1.5 || -2.0||[[File:haiyang11.png|300px|center]]||[[File:haiyang12.png|300px|center]]|| r1 and r2 never be equal so unreactive&lt;br /&gt;
|-&lt;br /&gt;
| -1.5 || -2.5||[[File:haiyang13.png|300px|center]]||[[File:haiyang14.png|300px|center]]|| r1 and r2 has a equal value at a time so transition state formed ,reaction reactive&lt;br /&gt;
|-&lt;br /&gt;
| -2.5 || -5.0||[[File:Haiyang15.png|300px|center]]||[[File:Haiyang16.png|300px|center]]||r1 and r2 has a equal value at a time so transition state formed ,reaction reactive,however the momenta is too large that the particle collide back again. the reaction is reversed &lt;br /&gt;
|-&lt;br /&gt;
| -2.5 || -5.2||[[File:Haiyang17.png|300px|center]]||[[File:Haiyang18.png|300px|center]]||r1 and r2 has a equal value at a time so transition state formed ,reaction reactive,however the momenta is too large that the particle collide within the transition state area several times.&lt;br /&gt;
|-&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
== H-H-F System ==&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:everything01095534&amp;diff=615333</id>
		<title>MRD:everything01095534</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:everything01095534&amp;diff=615333"/>
		<updated>2017-05-05T14:59:42Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== Molecular Reaction Dynamics of Triatomic systems ==&lt;br /&gt;
this wiki page is to explain triatomic interaction dynamic and illustrate it on specific example&lt;br /&gt;
&lt;br /&gt;
=== background Theory ===&lt;br /&gt;
The nucleus, is extremely heavy and large compare to electron in the atom. So when an atom moving toward or backward to another atom, the relationship can be shown by the equation below:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; {dp_i \over dt} = - { \partial V(r_1,r_2,...)\over \partial r_i}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The force (variation of momentum &#039;&#039;&#039;p&#039;&#039;&#039;) acting on a given intervatomic coordinate &#039;&#039;&#039;r&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt;&#039;&#039;&#039; will depend on the derivative of the potential energy surface with respect to that coordinate.&lt;br /&gt;
&lt;br /&gt;
In this concept only three atoms on a straight line will be considered.  &lt;br /&gt;
&lt;br /&gt;
=== H+H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; System ===&lt;br /&gt;
In this context, three hydrogen atoms are aligned on a straight line to use Matlab programming the potential energy change of the atoms under different condition&lt;br /&gt;
[[File:Plot1.png|500px|thumb|center|contour plot, r1=0.74A, r2=2.30A, p1=0 p2=-2.70]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
contour graph when the distance between A and B is equal to 0.74 which is just like the normal bond length of a normal H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule. In the H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; model, the initial conditions are set as &#039;&#039;&#039;r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;&#039;&#039;&#039; = 0.74 ,&#039;&#039;&#039;r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;&#039;&#039;&#039;&amp;lt;sub&amp;gt; &amp;lt;/sub&amp;gt;= 2.30 and the momenta &#039;&#039;&#039;p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;&#039;&#039;&#039;=0 &#039;&#039;&#039; p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;&#039;&#039;&#039; = -2.7.&lt;br /&gt;
&lt;br /&gt;
[[File:haiyang2.png|500px|thumb|center|intermolecular distance vs time, r1=0.74A, r2=2.30A, p1=0 p2=-2.70]]&lt;br /&gt;
&lt;br /&gt;
This is the figure which shows when the time goes to 0.38 time units, a repetitive oscillation behavior is happened, showing that the transition state was also suppose to be found at 0.38 time units. So the r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is 0.9148A then. But based on the result of animation, the true r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is 0.90775A but not 0.9148A.&lt;br /&gt;
[[File:haiyang3.png|500px|thumb|left|Kinetic Energy vs Time, r1=0.74A, r2=2.30A, p1=0 p2=-2.70]]&lt;br /&gt;
[[File:haiyang4.png|520px|thumb|center|potential Energy vs Time, r1=0.74A, r2=2.30A, p1=0 p2=-2.70]]&lt;br /&gt;
&lt;br /&gt;
The &amp;quot;Potential Energy vs Time&amp;quot; and &amp;quot;Kinetic Energy vs Time&amp;quot; proved that at transition state the potential energy gives rise to the highest value while kinetic energy give rise to the lowest value. As well, an oscillatory behavior rising after the transition state.&lt;br /&gt;
&lt;br /&gt;
The total gradient of the minimum (at the start and the end point in reactant) on the potential energy surface is equal to &#039;&#039;&#039;0 as well as &#039;&#039;&#039;at the transition state point. But the difference between the minimum and the transition state point is that transition state point has a negative second derivative towards product and a positive second derivative towards reactant; while at the minimum point in the whole system (at the beginning point in the reactant) have a &#039;&#039;&#039;zero&#039;&#039;&#039; second derivative value.Hence the potential energy difference between minimum energy and transition state is the &#039;&#039;&#039;activation eneregy&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
=== Trajectories from r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; locating the transition state ===&lt;br /&gt;
Due to H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; potential surface is symmetric(left and right of the transition state), assuming the transition state must have &#039;&#039;&#039;r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;&#039;&#039;&#039; &#039;&#039;&#039;=&#039;&#039;&#039; &#039;&#039;&#039;0.90775A&#039;&#039;&#039; (r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; at the transition state position), the relative stationary state of motion is observed. If the approaching proton atom is placed at the ridge of potential surface, there is no gradient in the direction at right angle to the ridge; therefore the proton sphere would oscillate about that region without falling down. As shown in the figures below, one figure shows that the reactant proton is oscillating at the transition state position; while another shows that there is not any relative change in terms of the distance for the three protons against time.&lt;br /&gt;
&lt;br /&gt;
[[File:haiyang5.png|300px|thumb|center|the potential energy surface is symmetric through transition state]]&lt;br /&gt;
&lt;br /&gt;
=== Calculating the reaction path by using Trajectories from r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+ δ ===&lt;br /&gt;
[[File:haiyang6.png|400px|thumb|right|Surface Plot, r1=r2=0.90775A p1=p2=0]]&lt;br /&gt;
[[File:haiyang7.png|400px|thumb|left|Internuclear Distance vs Time, r1 = 0.91775, r2 = 0.90775 p1 = p2 = 0]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
This model will favour H&amp;lt;sub&amp;gt;A&amp;lt;/sub&amp;gt;-H&amp;lt;sub&amp;gt;B&amp;lt;/sub&amp;gt; + H&amp;lt;sub&amp;gt;C&amp;lt;/sub&amp;gt; as reactant proton C is very close to the central proton. In this section, a initial system of r1 = 0.91775, r2 = 0.90775. p1 = 0 p2 = 0 is considered.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;MEP calculation type&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
[[File:haiyang8.png|400px|thumb|center|MEP Simulating Contour Plot, r1 = 0.91775, r2 = 0.90775 p1 = p2 = 0]]&lt;br /&gt;
&lt;br /&gt;
In this case, there is a little derivation from the transition state position, say 0.01A derivation for a first try, for one of the protons. The MEP calculation type is set so as to simulate the motion for this circumstance. The MEP contour plot is given .&lt;br /&gt;
&lt;br /&gt;
===unreactive trajectories===&lt;br /&gt;
From the previous calculations we can conclude that trajectories with initial conditions in the range r1 = 0.74, r2 = 2.0, with -1.5 &amp;lt; p1 &amp;lt; -0.8 and p2 = -2.5 are reactive&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;For the initial positions r1 = 0.74 and r2 = 2.0, run trajectories with the following momenta combination:&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=&amp;quot;1&amp;quot;&lt;br /&gt;
|+ caption&lt;br /&gt;
! p1!! p2!! surface plot!! internucler plot!! conclusion&lt;br /&gt;
|-&lt;br /&gt;
| -1.25 || -2.5||[[File:haiyang8.png|300px|center]]||[[File:haiyang10.png|300px|center]]|| r1 and r2 never be equal so unreactive&lt;br /&gt;
|-&lt;br /&gt;
| -1.5 || -2.0||[[File:haiyang11.png|300px|center]]||[[File:haiyang12.png|300px|center]]|| r1 and r2 never be equal so unreactive&lt;br /&gt;
|-&lt;br /&gt;
| -1.5 || -2.5||[[File:haiyang13.png|300px|center]]||[[File:haiyang14.png|300px|center]]|| r1 and r2 has a equal value at a time so transition state formed ,reaction reactive&lt;br /&gt;
|-&lt;br /&gt;
| -2.5 || -5.0||[[File:Haiyang15.png|300px|center]]||[[File:Haiyang16.png|300px|center]]||r1 and r2 has a equal value at a time so transition state formed ,reaction reactive,however the momenta is too large that the particle collide back again. the reaction is reversed &lt;br /&gt;
|-&lt;br /&gt;
| -2.5 || -5.2||[[File:Haiyang17.png|300px|center]]||[[File:Haiyang18.png|300px|center]]||r1 and r2 has a equal value at a time so transition state formed ,reaction reactive,however the momenta is too large that the particle collide within the transition state area several times.&lt;br /&gt;
|-&lt;br /&gt;
|}&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang18.png&amp;diff=615331</id>
		<title>File:Haiyang18.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang18.png&amp;diff=615331"/>
		<updated>2017-05-05T14:58:27Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: Hy2915 uploaded a new version of File:Haiyang18.png&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang18.png&amp;diff=615328</id>
		<title>File:Haiyang18.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang18.png&amp;diff=615328"/>
		<updated>2017-05-05T14:57:33Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang17.png&amp;diff=615327</id>
		<title>File:Haiyang17.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang17.png&amp;diff=615327"/>
		<updated>2017-05-05T14:57:19Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang16.png&amp;diff=615307</id>
		<title>File:Haiyang16.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang16.png&amp;diff=615307"/>
		<updated>2017-05-05T14:52:34Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang15.png&amp;diff=615306</id>
		<title>File:Haiyang15.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang15.png&amp;diff=615306"/>
		<updated>2017-05-05T14:52:25Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang14.png&amp;diff=615294</id>
		<title>File:Haiyang14.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang14.png&amp;diff=615294"/>
		<updated>2017-05-05T14:49:03Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: Hy2915 uploaded a new version of File:Haiyang14.png&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang14.png&amp;diff=615287</id>
		<title>File:Haiyang14.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang14.png&amp;diff=615287"/>
		<updated>2017-05-05T14:47:17Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang13.png&amp;diff=615278</id>
		<title>File:Haiyang13.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang13.png&amp;diff=615278"/>
		<updated>2017-05-05T14:45:13Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang12.png&amp;diff=615261</id>
		<title>File:Haiyang12.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang12.png&amp;diff=615261"/>
		<updated>2017-05-05T14:42:58Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang11.png&amp;diff=615257</id>
		<title>File:Haiyang11.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang11.png&amp;diff=615257"/>
		<updated>2017-05-05T14:41:56Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang10.png&amp;diff=615237</id>
		<title>File:Haiyang10.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang10.png&amp;diff=615237"/>
		<updated>2017-05-05T14:38:06Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang8.png&amp;diff=615223</id>
		<title>File:Haiyang8.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang8.png&amp;diff=615223"/>
		<updated>2017-05-05T14:35:37Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: Hy2915 uploaded a new version of File:Haiyang8.png&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang8.png&amp;diff=615170</id>
		<title>File:Haiyang8.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang8.png&amp;diff=615170"/>
		<updated>2017-05-05T14:22:18Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang7.png&amp;diff=615145</id>
		<title>File:Haiyang7.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang7.png&amp;diff=615145"/>
		<updated>2017-05-05T14:16:20Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang6.png&amp;diff=615122</id>
		<title>File:Haiyang6.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang6.png&amp;diff=615122"/>
		<updated>2017-05-05T14:10:24Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang5.png&amp;diff=615105</id>
		<title>File:Haiyang5.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang5.png&amp;diff=615105"/>
		<updated>2017-05-05T14:05:24Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:everything01095534&amp;diff=615099</id>
		<title>MRD:everything01095534</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:everything01095534&amp;diff=615099"/>
		<updated>2017-05-05T14:03:47Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== Molecular Reaction Dynamics of Triatomic systems ==&lt;br /&gt;
this wiki page is to explain triatomic interaction dynamic and illustrate it on specific example&lt;br /&gt;
&lt;br /&gt;
=== background Theory ===&lt;br /&gt;
The nucleus, is extremely heavy and large compare to electron in the atom. So when an atom moving toward or backward to another atom, the relationship can be shown by the equation below:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; {dp_i \over dt} = - { \partial V(r_1,r_2,...)\over \partial r_i}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The force (variation of momentum &#039;&#039;&#039;p&#039;&#039;&#039;) acting on a given intervatomic coordinate &#039;&#039;&#039;r&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt;&#039;&#039;&#039; will depend on the derivative of the potential energy surface with respect to that coordinate.&lt;br /&gt;
&lt;br /&gt;
In this concept only three atoms on a straight line will be considered.  &lt;br /&gt;
&lt;br /&gt;
=== H+H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; System ===&lt;br /&gt;
In this context, three hydrogen atoms are aligned on a straight line to use Matlab programming the potential energy change of the atoms under different condition&lt;br /&gt;
[[File:Plot1.png|500px|thumb|center|contour plot, r1=0.74A, r2=2.30A, p1=0 p2=-2.70]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
contour graph when the distance between A and B is equal to 0.74 which is just like the normal bond length of a normal H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule. In the H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; model, the initial conditions are set as &#039;&#039;&#039;r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;&#039;&#039;&#039; = 0.74 ,&#039;&#039;&#039;r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;&#039;&#039;&#039;&amp;lt;sub&amp;gt; &amp;lt;/sub&amp;gt;= 2.30 and the momenta &#039;&#039;&#039;p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;&#039;&#039;&#039;=0 &#039;&#039;&#039; p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;&#039;&#039;&#039; = -2.7.&lt;br /&gt;
&lt;br /&gt;
[[File:haiyang2.png|500px|thumb|center|intermolecular distance vs time, r1=0.74A, r2=2.30A, p1=0 p2=-2.70]]&lt;br /&gt;
&lt;br /&gt;
This is the figure which shows when the time goes to 0.38 time units, a repetitive oscillation behavior is happened, showing that the transition state was also suppose to be found at 0.38 time units. So the r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is 0.9148A then. But based on the result of animation, the true r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is 0.90775A but not 0.9148A.&lt;br /&gt;
[[File:haiyang3.png|500px|thumb|left|Kinetic Energy vs Time, r1=0.74A, r2=2.30A, p1=0 p2=-2.70]]&lt;br /&gt;
[[File:haiyang4.png|520px|thumb|center|Potential Energy vs Time, r1=0.74A, r2=2.30A, p1=0 p2=-2.70]]&lt;br /&gt;
&lt;br /&gt;
The &amp;quot;Potential Energy vs Time&amp;quot; and &amp;quot;Kinetic Energy vs Time&amp;quot; proved that at transition state the potential energy gives rise to the highest value while kinetic energy give rise to the lowest value. As well, an oscillatory behavior rising after the transition state.&lt;br /&gt;
&lt;br /&gt;
The total gradient of the minimum (at the start and the end point in reactant) on the potential energy surface is equal to &#039;&#039;&#039;0 as well as &#039;&#039;&#039;at the transition state point. But the difference between the minimum and the transition state point is that transition state point has a negative second derivative towards product and a positive second derivative towards reactant; while at the minimum point in the whole system (at the beginning point in the reactant) have a &#039;&#039;&#039;zero&#039;&#039;&#039; second derivative value.Hence the potential energy difference between minimum energy and transition state is the &#039;&#039;&#039;activation eneregy&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
=== Trajectories from r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; locating the transition state ===&lt;br /&gt;
Due to H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; potential surface is symmetric(left and right of the transition state), assuming the transition state must have &#039;&#039;&#039;r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;&#039;&#039;&#039; &#039;&#039;&#039;=&#039;&#039;&#039; &#039;&#039;&#039;0.90775A&#039;&#039;&#039; (r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; at the transition state position), the relative stationary state of motion is observed. If the approaching proton atom is placed at the ridge of potential surface, there is no gradient in the direction at right angle to the ridge; therefore the proton sphere would oscillate about that region without falling down. As shown in the figures below, one figure shows that the reactant proton is oscillating at the transition state position; while another shows that there is not any relative change in terms of the distance for the three protons against time.&lt;br /&gt;
&lt;br /&gt;
=== Calculating the reaction path by using Trajectories from r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+ δ ===&lt;br /&gt;
This model will favour H&amp;lt;sub&amp;gt;A&amp;lt;/sub&amp;gt;-H&amp;lt;sub&amp;gt;B&amp;lt;/sub&amp;gt; + H&amp;lt;sub&amp;gt;C&amp;lt;/sub&amp;gt; as reactant proton C is very close to the central proton. In this section, a initial system of r1 = 0.91775, r2 = 0.90775. p1 = 0 p2 = 0 is considered.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;MEP calculation type&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
In this case, there is a little derivation from the transition state position, say 0.01A derivation for a first try, for one of the protons. The MEP calculation type is set so as to simulate the motion for this circumstance. The MEP contour plot is given below.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Dynamics calculation type&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
When dynamics calculation type is applied again, the trajectories gained is quite different from MEP calculation type. See the figures below, which show the averaged moment and the final positions.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;&#039;&#039;&#039; = 5.281A&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;&#039;&#039;&#039; = 0.7455A&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;&#039;&#039;&#039; = 2.481 (Calculated as an average of the maxima 1.564 and minima 0.9109 at a large t)&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;&#039;&#039;&#039; = 1.237&lt;br /&gt;
&lt;br /&gt;
Because &#039;&#039;&#039;MEP&#039;&#039;&#039; is considered that there is &#039;&#039;&#039;a step-energy down&#039;&#039;&#039; where the main difference is, comparing to the &#039;&#039;&#039;dynamics calculation type&#039;&#039;&#039;. The &#039;&#039;&#039;MEP&#039;&#039;&#039; just calculate all the minimum energy following the trajectories while &#039;&#039;&#039;dynamics calculation type&#039;&#039;&#039; includes all the vibrating situation with the trajectories.&lt;br /&gt;
&lt;br /&gt;
=== Calculating the reaction path by using Trajectories from r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; + δ r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; ===&lt;br /&gt;
Here, we are going to swap the parameters from what we just did in &#039;&#039;&#039;section 1.3&#039;&#039;&#039;,&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;&#039;&#039;&#039; = 0.7455A&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;&#039;&#039;&#039; = 5.281A&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;&#039;&#039;&#039; =1.237 (Calculated as an average of the maxima 1.564 and minima 0.9109 at a large t)&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;&#039;&#039;&#039; = 2.481&lt;br /&gt;
&lt;br /&gt;
1.&#039;&#039;&#039;What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.&#039;&#039;&#039;&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang4.png&amp;diff=615061</id>
		<title>File:Haiyang4.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang4.png&amp;diff=615061"/>
		<updated>2017-05-05T13:54:50Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang3.png&amp;diff=615048</id>
		<title>File:Haiyang3.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:Haiyang3.png&amp;diff=615048"/>
		<updated>2017-05-05T13:52:19Z</updated>

		<summary type="html">&lt;p&gt;Hy2915: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Hy2915</name></author>
	</entry>
</feed>