[CP2K-user] Kinetic Energy from Velocities

[fei]

This is in the &MOTION &PRINT &VELOCITIES section of the input file.


See:

https://manual.cp2k.org/cp2k-6_1-branch/CP2K_INPUT/MOTION/PRINT/VELOCITIES.html#UNIT


“Section VELOCITIES

Controls the output of the velocities. The default unit for velocities is 
bohr/au_time. The au_time is derived from the hbar value (1.054e-34 J*sec) 
and the value of the hartree unit of energy (27.21 eV or 4.359e-18 J) as 
hbar/Ehartree = 2.42e-17 sec = 0.0242 fs. Having an atom with a mass m in 
AMU the kinetic energy 1/2mv^2 will be obtained in Hartree (i.e. au) 
multiplying by 911.447 .

Section path: CP2K_INPUT / MOTION / PRINT / VELOCITIES”

 

To be clear, we think that the velocities and kinetic energy are correct, 
but the manual is wrong or at least misleading in the conversion between 
these. 



This is the argument:

Assume that you have an atom with a mass of 1 AMU and velocity of 1 in 
bohr/au_time then the kinetic energy is:

 

1/2 *1 AMU *(1 bohr/au_time)^2 = 911.447 Hartree

 

Instead, an atom with mass “M” in AMU and velocity “v” in bohr/au_time will 
have a kinetic energy of


911.447 * M *v^2 in Hartree


Hence, the conversion factor between Ke=1/2 M v^2 with M in AMU and v in 
 bohr/au_time is not 911.447 as we read from the manual but instead 
(911.447 * 2).



*We believe the manual should say: “Having an atom with a mass m in AMU the 
kinetic energy 1/2mv^2 will be obtained in Hartree (i.e. au) multiplying by 
1822.894”*




On Wednesday, June 19, 2019 at 12:34:29 PM UTC-5, Matt W wrote:
>
> Hello,
>
> where in the manual are you referring to?
>
> Matt
>
> On Wednesday, June 19, 2019 at 6:17:06 PM UTC+1, fei wrote:
>>
>> Hi, to make things easier in demonstrating what Claudio is describing I 
>> put a Li atom in a box and requested that its initial temperature be 1200K. 
>>
>>  
>>
>> CP2k writes:
>>
>>  
>>
>> #     Step Nr.          Time[fs]        Kin.[a.u.]          
>> Temp[K]            Pot.[a.u.]        Cons Qty[a.u.]        UsedTime[s]
>>
>>          0            0.000000         *0.005700268 *     
>> 1200.000000000         0.000000000         0.011400535         0.000000000
>>
>>  
>>
>> The velocity reported for that single atoms is:
>>
>>        1
>>
>>  i =        0, time =        0.000, E =         0.0000000000
>>
>>  Li         *0.0000000000        0.0000000000        0.0009492302*
>>
>>  
>>
>> For that single atom we have:
>>
>>
>> 0.5*m*v^2=0.5*6.914*0.0009492302^2=*0.00000312705*  This is in CP2K's 
>> atomic units.
>>
>>
>>
>> Now if I follow the manual’s instruction “Having an atom with a mass m in 
>> AMU the kinetic energy 1/2mv^2 will be obtained in Hartree (i.e. au) 
>> multiplying by 911.447”
>>
>> I get:
>>
>>
>> 0.00000312705*911.447=*0.00285014242* 
>>
>>  
>>
>> which is exactly half of the kinetic energy reported by CP2K.
>>
>>
>> [0.005700268 /  2]  see beginning of this same post.
>>
>> Could someone kindly clarify how this is possible/correct?
>>
>> Also, the reported temperature matches the reported kinetic energy but 
>> obviously not what one would derive from the velocity.
>>
>>
>> Thank you,
>>
>> Fei
>>
>>
>>
>> On Tuesday, June 18, 2019 at 7:50:04 PM UTC-5, Claudio wrote:
>>
>>> Dear all, 
>>>
>>> first thanks in advance for your help.
>>>
>>> I have a very simple question (perhaps a units misunderstanding on my 
>>> part). I am simulating classically in the NVT ensemble a simple AB type 
>>> molten salt with a polarizable potential. When I write out the 
>>> velocities along my simulation and I try to compute the kinetic energy 
>>> using these via the expression:
>>>
>>> 1/2 Sum_i { m_i v_i dot v_i} and multiply by the conversion 
>>> factor  911.447 --as per the manual-- 
>>>
>>> "Having an atom with a mass m in AMU the kinetic energy 1/2mv^2 will be 
>>> obtained in Hartree (i.e. au) multiplying by 911.447"
>>>
>>> I get a number in Hartrees that is exactly 1/2 of that printed out by 
>>> CP2K each frame. I can't seem to understand how to reconcile these two 
>>> numbers. Is there something I am overlooking?
>>>
>>> Thanks once again for your kind help and consideration. I can provide 
>>> the velocities of a snapshot and a simple code to compute the kinetic 
>>> energy if needed, but this is so simple that I doubt the problem is in the 
>>> code.
>>>
>>> Best regards
>>>
>>> Claudio
>>>
>>
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