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 <j...@um.es <mailto:j...@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 <j...@um.es
    <mailto:j...@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
        <jalem...@vt.edu <mailto:jalem...@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
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        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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