Hi Bill,
simply because all the contributions to the energy of the molecule which
are
quadratic, contributes as 1/2*k*T
so each contribution to the total energy in the shape like k/2(r-r0)^2
contributes as 1/2*k*T
see here :
https://en.wikipedia.org/wiki/Equipartition_theorem#Potential_energy_and_harmonic_oscillators
But as I wrote, the easiest way to calculate instantaneous temperature of
the system composed of complex molecules should be probably to apply
equipartition theorem just on translational
part of the kinetic energy. Then we do not need to care about the internal
structure of molecules. Am I right ?
Best wishes,
Marek
Dne Mon, 01 Apr 2019 02:57:02 +0200 Bill Ross <ross.cgl.ucsf.edu>
napsal/-a:
> Why would bond (potential) energy be part of temperature? Asking for a
> friend. :-)
>
> Bill
>
> On 3/31/19 5:35 PM, Marek Maly wrote:
>> Hello,
>>
>> I would like to know how exactly the instantaneous temperature is
>> calculated in Amber.
>>
>> I assume that the Equipartition theorem is used but which degrees of
>> freedom are taken in account in case of more complicated molecules
>> (flexible models) ?
>>
>> Could be possible to describe it more in detail on relatively simple
>> molecular system composed just of water molecules (flexible molecular
>> model of course with bond and bond angle harmonic potentials) or
>> eventually to provide the relavant reference ?
>>
>> My guess is, that the averages of kinetic energy <E_kin> or bond energy
>> (if harmonic approximation is used) <E_bond> or the average of both
>> energies <E_kin+E_bond> of such molecule could be connected with the
>> instantaneous temperature using Equipartition theorem this way.
>>
>> <E_kin> = 9*0.5*k*T
>> <E_bond> = 3*0.5*k*T
>> <E_kin+E_bond> = 12*0.5*k*T
>>
>> but I am not sure.
>>
>> Thank you in advance,
>>
>> Best wishes,
>>
>> Marek
>>
>>
>>
>>
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Received on Sun Mar 31 2019 - 20:30:02 PDT
