Potential kinetic energy and temperature

In summary, the conversation discusses a non-equilibrium molecular dynamics model of MgO and the process of introducing temperature by assigning initial velocities to atoms in random directions. An equation relating atomic velocity and temperature is mentioned, as well as the transfer of kinetic energy into potential energy as the atoms move out of their equilibrium positions. The decrease in velocity is observed to be roughly half the initial temperature, regardless of the specific heat or temperature of the lattice. The potential energy is determined to be "thermal" and related to the equipartition theorem, with half of the initial energy going into the potential energy during equilibration. The equation for kinetic energy is not applicable in this case due to the conversion of energy into potential energy.
  • #1
Hypatio
151
1
I have a working non-equilibrium molecular dynamics model of MgO. I first find the equilibrium positions of the atoms in the lattice and then introduce a temperature by assigning initial velocities in random directions for each atom. An equation relating atomic velocity and temperature is

[itex]mv^2=3k_BT[/itex]
where k_b is the Boltzmann constant.

So for example, oxygen with mass=2.66*10^-26 kg, and with T=2000K each oxygen is assigned a velocity v=1766 m/s.

Running the simulation, however, I observe that as the atoms move out of their T=0K equilibrium positions their velocity decreases until they begin bouncing off zero as the atoms bounce off one another. When the system is in a steady state of vibration the average temperature and atomic velocity in the lattice becomes about 1/2 of the initially assigned temperature, ~880 m/s and ~1000 K in this case.

This seems to be a transfer of kinetic energy into potential energy from moving out of the bottom of the potential energy well, but I'm not sure? Is this associated with a particular concept or phenomenon? Latent heat? The decrease in velocity is roughly 1/2 for all temperatures so I don't think it is related to specific heat. Where is the energy going? Is the actual temperature of the lattice 1000K or 2000K? How do I determine temperature in such a case?
 
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  • #2
Assigning one value for the velocity for all atoms of one type is not reasonable. You'd better create a velocity distribution (Maxwell-Boltzmann).
Regarding what happens later on in your simulation, you need to tell us more about the time step of your simulation. Is it short enough so as not to blow up the simulation. What ensemble are you simulating? If you're simulating a microcanonical ensemble, is the total energy conserved? etc...
 
  • #3
Hi UN,

I would have created a velocity distribution, but it appears that from the initial condition of constant temperature an appropriate distribution quickly occurs in the model.

The timestep is lower than a point of instability, although I think it is blowing up for "high" temperature simulations (>2500 K). I am currently using 1.0 femtoseconds. Significantly decreasing the timestep doesn't result in any clear changes for lower temperature simulations but I will try it for higher temperature.

I believe my simulation is essentially microcanonical. The lattice is isolated and free to expand or contract from vibration, although I will later simulate temperature gradients to evaluate thermal conductivity. It is a 20x20 (400 atom) two-dimensional lattice. I am not certain that energy is conserved although there is no reason to doubt it unless my interpretation of the temperature change is incorrect. The only thermal energy lost should be that related to the momentum of the entire lattice, which is small.

I think that the average velocity of the lattice becomes 1/2 the initial velocity because the potential energy increases when atoms are displaced from their equilibrium positions. It is always 1/2 because the given velocity is the maximum and the minimum is zero at the point of elastic rebound against adjacent atoms. But what does this mean for temperature?

The total energy associated with each atom should be the sum of it's kinetic and potential energy, but is the potential energy thermal? Does it contribute to temperature?
 
  • #4
Yes, the potential energy is "thermal". What you do is starting in a positional configuration corresponding to T=0 and a kinematic configuration corresponding to T=2000 K (or whatever value). When your system equilibrates, some of the kinetic energy goes into potential energy (-> equipartition theorem), resulting in a configuration representative for some temperature 0<T<2000 K. The most obvious solution to your problem seems to be coupling a thermostat to your system. That's a dirty trick from a purist's perspective (but few people a purists nowadays, as dirty tricks get you more publications than formally proper work), but as I understood it you ultimately don't want to work in a microcanonical ensemble, anyways.
 
  • #5
So perhaps, due to equipartition, I am seeing a kinetic 'cooling' on the order of 1/2 because I have 2 degrees of freedom in my two-dimensional model? And the 2k_bT/3 law cannot describe the temperature of my system. Furthermore, this should be related to specific heat. But then why am I getting similar values (1/2 initial temperature) for all temperatures? Is there something in my model that prevents a reasonable approximation of the strong temperature-dependence of specific heat? Maybe I need a larger lattice in which phonons develop?
 
  • #6
My guess (without having seen your simulation): You do not really see cooling. You see equilibration. Assuming the deviations from the ground state positions are not too small, then you can Taylor expand around those. This will give you potential energy terms that are harmonics springs. The number of degrees of freedom (in the sense of the equipartition theorem) for the potential energy therefore equals the number of degrees of freedom in the kinetic energy for this approximated Hamiltonian => half of the energy (that was initially only in the kinetic terms) goes into the potential energy. That's irrespective of temperature (at least for as long as the ansatz with the Taylor expansion holds), the spatial dimension of your system, its size, and the specific heat.
 
  • #7
Well, how is this related to the thermodynamic temperature then? It seems that the equation

[itex]\frac{1}{2}mv^2=\frac{3}{2}k_b T[\itex]

cannot then be true for a particle because half the energy is converted into potential energy.
 

What is potential kinetic energy?

Potential kinetic energy is the energy that an object possesses due to its motion or position. It is the sum of an object's potential energy, which is energy stored in its position, and its kinetic energy, which is energy due to its motion.

What is the relationship between potential kinetic energy and temperature?

The relationship between potential kinetic energy and temperature is indirect. As an object's temperature increases, its particles move faster and have more kinetic energy. This increased kinetic energy can also lead to an increase in potential energy, as the particles may move to higher energy states.

What factors affect potential kinetic energy and temperature?

The factors that affect potential kinetic energy and temperature include the mass and velocity of an object, as well as the type of material and its surroundings. For example, a heavier object will have more potential kinetic energy than a lighter object moving at the same speed.

Can potential kinetic energy be converted into other forms of energy?

Yes, potential kinetic energy can be converted into other forms of energy. For example, when a car brakes, its kinetic energy is converted into heat energy due to friction. Additionally, potential kinetic energy can also be converted into electrical energy through generators.

How is potential kinetic energy and temperature related to the laws of thermodynamics?

Potential kinetic energy and temperature are closely related to the laws of thermodynamics, which govern the behavior of energy in a system. The first law of thermodynamics states that energy cannot be created or destroyed, but can only be converted from one form to another. The second law of thermodynamics states that in any energy conversion, some energy will be lost as heat, resulting in a decrease in potential kinetic energy and an increase in temperature.

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