Physics Asked by ChrizZly on March 29, 2021
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?
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.
Answered by Gec on March 29, 2021
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.
Answered by Gareth Meredith on March 29, 2021
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