[CP2K-user] Kinetic Energy from Velocities

[Matt W]

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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