[gmx-users] Test Particle Insertion
Javier Cerezo
jcb1 at um.es
Wed May 16 12:21:31 CEST 2012
Hi
Well, according to the link you pointed out, the Widom technique gives
you the excess chemical potential, as we discussed. mu and mu_ideal (in
your link) are not calculated, those are just the reference states
between which the Widom technique calculates the excess chem pot.
As I said, I think U_{n+1} refers to the interaction energy of the
inserte particle with the system, but maybe someone could confirm or
correct.
Javier
El 16/05/12 11:44, Steven Neumann escribió:
>
>
> On Wed, May 16, 2012 at 10:28 AM, Javier Cerezo <jcb1 at um.es
> <mailto: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
>> <mailto: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 <mailto: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 <http://vt.edu> | (540) 231-9080
>>> <tel:%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 <tel:%28%2B34%29868887434>
>>
>> --
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>>
>
> --
> Javier CEREZO BASTIDA
> Ph.D. Student
> Physical Chemistry
> Universidad de Murcia
> 30100, Murcia (SPAIN)
> T: (0034)868887434
>
>
> --
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>
>
--
Javier CEREZO BASTIDA
Ph.D. Student
Physical Chemistry
Universidad de Murcia
30100, Murcia (SPAIN)
T: (0034)868887434
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