[CP2K-user] Kinetic Energy from Velocities

[Matt W]

Great. The manual is a wiki, you can edit it as you suggest. You should 
edit the development version, though:

https://manual.cp2k.org/trunk/CP2K_INPUT/MOTION/PRINT/VELOCITIES.html

On Wednesday, June 19, 2019 at 8:52:53 PM UTC+1, fei wrote:
>
> 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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