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I have a molecular Dynamics simulation and want to calculate the heat capacity with constant volume of my material.

As output parameters I can get temperature, potential energy, kinetic energy, total energy, enthalpy, pressure, volume and density.

I once saw a calculation only with temperature and total energy. But it like a year ago and I cannot find it anymore. (But I guess this would be possible).

I get my data as a huge table with the value for each time step, so I can take the average for the last x timesteps in Matlab or use it in other ways.

My simulation is already on a constant volume, so my output values are under constant volume.

EDIT: I forgot 2 important information. I can calculate the mass of my material per hand and I raised my temperature from 300K to 320K in 0.2 Nanoseconds (I heat up around 1000 atoms. That should explain the fast heating)

Any suggestions?

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Heat capacity is related to fluctuations of energy: $\overline{(\Delta E)^2} = k_B T^2 C_{V,N}$. Dispertion of $E$ can be found from numerical data.

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  • $\begingroup$ Is kb the Boltzmann constant? Therefor it is 1.380e-23 and when I divide through it to get my cv I will end up with a really large number. My cv is now like 4.7e+19 which is definetly wrong. Did I understood something wrong? $\endgroup$ – ChrizZly Dec 12 '17 at 11:09
  • $\begingroup$ I also forgot that I have a solid metal and not a gas, so I cannot even use the Boltzman constant. $\endgroup$ – ChrizZly Dec 12 '17 at 12:27
  • $\begingroup$ Yes, $k_B$ is Boltzmann constant. This constant is needed no matter gas or solid metal. Dispertion of the energy measured in appropriate units is really small, division by $k_B$ must not be a problem. In what units do you measure energy? $\endgroup$ – Gec Dec 12 '17 at 16:01
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If the temperature at any point is changed, the local gradient heat flow is

$\frac{\partial T}{\partial t} = -\frac{1}{\rho C_p} \frac{\partial Q}{\partial x}$

We will use this equation in a moment - the heat energy per unit area is,

$Q = -k \nabla T$

we can create a gradient such that

$\nabla Q = \frac{\partial Q}{\partial x} = -k R T$

Then we can retrieve the definition of the heat flow equation in terms of the Ricci curvature again $R$, keep in mind, $k$ is the thermal conductivity. In the case above, the curvature $R$ has replaced the definition of the gradient $\nabla$. If the temperature at any point changed, the local gradient heat flow is, after we multiply through by $-\frac{1}{\rho C_p}$ we get,

$-\frac{1}{\rho C_p}\nabla Q = -\frac{1}{\rho C_p}\frac{\partial Q}{\partial x} = \frac{k}{\rho C_p} R T = \alpha R T = \frac{\partial T}{\partial t}$

Went off on a bit of a tangent, but if the second equation describes Fouriers relation as heat energy flux through an area, then the volume is found simply as

$\mathbf{Q} = -k \nabla^2 T$

defining the heat-energy density.

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