[gmx-users] Test Particle Insertion

Wed May 16 11:44:14 CEST 2012

On Wed, May 16, 2012 at 10:28 AM, Javier Cerezo <jcb1 at um.es> wrote:

>  About the red curve, I guess fluctuations might be directly related to
> volume fluctuations, you can extract the volume over time from g_energy
> (boxXX*boxYY**boxZZ) and compare. (just another comment, now I am not very
> sure about the "f." that precedes the red line legend..)
>
> About the interpretation of the quantities, the Widom technique does not
> provide you with an absolute value of the chemical potential but directly
> with the excess chemical potential. So, mu=-kTlog(Ve ^ (U*B)/(V))n+1 is the
> excess chemical potential, where (if I recall correctly) U_{n+1} is the the
> interaction energy  between the inserted particle and the rest of the
> system. You don't need (and should not do) such post-processing operations
> that you proposed to get the excess chemical potential.
>
> Javier
>

Thank you.
In this case I am considering the curve with NPT - with volume.

>From the equation u=-kTlog(Ve ^ (U*B)/(V))n+1  (the one on the plot - if it
is correct! Or it should be with delta?) we will obtain the chemical
potential of the system with N+1 molecules. To obtain the excess we need to
have chemical potential of the system wit N particles and the substract it
according to the equation:
http://www.sklogwiki.org/SklogWiki/index.php/Widom_test-particle_method
If it is a mistake and there is deltaU this is the exceess, if not this is
only for N+1. Please, correct me if I am wrong.

Steven

>
>
> El 16/05/12 11:06, Steven Neumann escribió:
>
> Thank you very much! I just saw your response.
>
> As I run it in NPT ensemble the plot with volume is important for me.
> Please, See the plot:
>
> http://speedy.sh/CJn5b/tpiN.jpg
>
> So does the fluctuating red curve make any sesnse then if it does not
> consider volume?
>
> Another thing: this is chemical potential of the system with extra water
> molecule (N+1), right (u=-kTlog(Ve ^ (U*B)/(V))n+1? So if I want to obtain
> the excess chemical potential:
> u=-kTlog(Ve ^ (-deltaU*B)/(V)) I should calculate it for the system with N
> molecules and then substract it.
> Is it calculated somewhere or I should use g_energy of my previosu system
> and calculate the total potential energy then -kTlog... of this values and
> then substract it? Please correct me if I am wrong.
>
> Steven
>
>
> On Tue, May 15, 2012 at 11:59 AM, Javier Cerezo <jcb1 at um.es> wrote:
>
>>  Hi Steven.
>>
>>
>>
>>  1. Why this value is divided by nm3? Shall I multiply it by the
>> simulation box?
>>
>>  It is not not divided by nm3. The legend for "y" axis is not appropriate
>> for your plot. Keep in mind that the same graph is used to represent lots
>> of quantities (you can plot all of them with xmgrace -nxy tpi.xvg). The "y"
>> axis is not the same for all, but only one label is possible, so developers
>> have to chose which label to place on the axis. But this is just a label,
>> don't give much importance to it and analyse you results (including units)
>> according to the equations and the standard units in gromacs.
>>
>>  2. Why e^(-BU) is multiplied by V? I just want to have the excess
>> chemical potential: u=-kTlog(e ^ (-deltaU*B) - so how can I get deltaU?
>>
>>  The volume appears in the expression of the excess chemical potential if
>> you are running a NpT ensemble. The second plot (if you use xmgrace -nxy
>> tpi.xvg) does not contain the volume.
>>
>>   3. The value corresponds to the plateau so I should run it for longer
>> time?
>>
>>  You are getting a time&ensemble average and for large sampling (and
>> large simulation times), this average should converge. So, the final value
>> you will get is the last point of the graph, it up to you to say if it is
>> converged. So you can try to enlarge the number of points sampled, if the
>> shape does not change you are sampling correctly every snapshot, then take
>> longer simulation times if you want to converge your results.
>>
>> Javier
>>
>>
>> El 15/05/12 09:57, Steven Neumann escribió:
>>
>>
>>
>> On Mon, May 14, 2012 at 5:05 PM, Justin A. Lemkul <jalemkul at vt.edu>wrote:
>>
>>>
>>>
>>> On 5/14/12 11:53 AM, Steven Neumann wrote:
>>>
>>>> Dear Gmx Users,
>>>>
>>>> Did anyone use TPI method for the calculation of chemical potential?
>>>> The tpi.xvg
>>>> files consists of:
>>>>
>>>> @ s0 legend "-kT log(<Ve\S-\xb\f{}U\N>/<V>)"
>>>> @ s1 legend "f. -kT log<e\S-\xb\f{}U\N>"
>>>> @ s2 legend "f. <e\S-\xb\f{}U\N>"
>>>> @ s3 legend "f. V"
>>>> @ s4 legend "f. <Ue\S-\xb\f{}U\N>"
>>>> @ s5 legend "f. <U\sVdW System\Ne\S-\xb\f{}U\N>"
>>>> @ s6 legend "f. <U\sdisp c\Ne\S-\xb\f{}U\N>"
>>>> @ s7 legend "f. <U\sCoul System\Ne\S-\xb\f{}U\N>"
>>>> @ s8 legend "f. <U\sCoul recip\Ne\S-\xb\f{}U\N>"
>>>>
>>>> @    xaxis  label "Time (ps)"
>>>> @    yaxis  label "(kJ mol\S-1\N) / (nm\S3\N)"
>>>>
>>>> Can anyone explain me these legends? I just want obtain a value of the
>>>> excess
>>>> chemical potential according to the equation:
>>>> u=-kT log (-deltaV/kT), Which legend is responsible for this and what
>>>> are the
>>>> units? kJ/mol? Please, explain as the above letters does not mean to me
>>>> anything?
>>>>
>>>>
>>>  These strings are formatted for XmGrace.  Have you tried plotting the
>>> file to see what it contains?  The legends will be far more obvious if you
>>> do.
>>>
>>> -Justin
>>>
>>
>> Thank you Justin.
>> Can anyone explain me from the plot:
>>
>> http://speedy.sh/Xpnws/tpi.JPG
>>
>> 1. Why this value is divided by nm3? Shall I multiply it by the
>> simulation box?
>> 2. Why e^(-BU) is multiplied by V? I just want to have the excess
>> chemical potential: u=-kTlog(e ^ (-deltaU*B) - so how can I get deltaU?
>> 3. The value corresponds to the plateau so I should run it for longer
>> time?
>>
>>
>> Thank you,
>>
>> Steven
>>
>>
>>>
>>> --
>>> ========================================
>>>
>>> Justin A. Lemkul, Ph.D.
>>> Department of Biochemistry
>>> Virginia Tech
>>> Blacksburg, VA
>>> jalemkul[at]vt.edu | (540) 231-9080 <%28540%29%20231-9080>
>>> http://www.bevanlab.biochem.vt.edu/Pages/Personal/justin
>>>
>>> ========================================
>>> --
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>>
>>
>>
>>
>>   --
>> Javier CEREZO BASTIDA
>> PhD Student
>> Physical Chemistry
>> Universidad de Murcia
>> Murcia (Spain)
>> Tel: (+34)868887434
>>
>> --
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>
>
>
>
> --
> Javier CEREZO BASTIDA
> Ph.D. Student
> Physical Chemistry
> Universidad de Murcia
> 30100, Murcia (SPAIN)
> T: (0034)868887434
>
>
> --
> gmx-users mailing list    gmx-users at gromacs.org
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