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    Hi<br>
    <br>
    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. <br>
    <br>
    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.<br>
    <br>
    Javier<br>
    <br>
    El 16/05/12 11:44, Steven Neumann escribi&oacute;:
    <blockquote
cite="mid:CAKZJqQGaeYJ=u70Ps4nfrn7MxamyWQvSrwokf2dVL0Kb+e7asQ@mail.gmail.com"
      type="cite"><br>
      <br>
      <div class="gmail_quote">On Wed, May 16, 2012 at 10:28 AM, Javier
        Cerezo <span dir="ltr">&lt;<a moz-do-not-send="true"
            href="mailto:jcb1@um.es" target="_blank">jcb1@um.es</a>&gt;</span>
        wrote:<br>
        <blockquote class="gmail_quote" style="margin:0 0 0
          .8ex;border-left:1px #ccc solid;padding-left:1ex">
          <div bgcolor="#FFFFFF" text="#000000"> 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..)<br>
            <br>
            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&nbsp; 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.<br>
            <br>
            Javier<br>
          </div>
        </blockquote>
        <div><br>
          Thank you.<br>
          In this case I am considering the curve with NPT - with
          volume. <br>
          <br>
          From the equation u=-kTlog(Ve ^ (U*B)/(V))n+1&nbsp; (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:<a moz-do-not-send="true"
href="http://www.sklogwiki.org/SklogWiki/index.php/Widom_test-particle_method">http://www.sklogwiki.org/SklogWiki/index.php/Widom_test-particle_method</a><br>
          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.<br>
          <br>
          Steven<br>
          &nbsp;</div>
        <blockquote class="gmail_quote" style="margin:0pt 0pt 0pt
          0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex">
          <div bgcolor="#FFFFFF" text="#000000"> <br>
            <br>
            El 16/05/12 11:06, Steven Neumann escribi&oacute;:
            <div>
              <div class="h5">
                <blockquote type="cite">Thank you very much! I just saw
                  your response.<br>
                  <br>
                  As I run it in NPT ensemble the plot with volume is
                  important for me. Please, See the plot:<br>
                  <br>
                  <a moz-do-not-send="true"
                    href="http://speedy.sh/CJn5b/tpiN.jpg"
                    target="_blank">http://speedy.sh/CJn5b/tpiN.jpg</a><br>
                  <br>
                  So does the fluctuating red curve make any sesnse then
                  if it does not consider volume?<br>
                  <br>
                  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:<br>
                  u=-kTlog(Ve ^ (-deltaU*B)/(V)) I should calculate it
                  for the system with N molecules and then substract it.
                  <br>
                  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.<br>
                  <br>
                  Steven<br>
                  <br>
                  <br>
                  <div class="gmail_quote">On Tue, May 15, 2012 at 11:59
                    AM, Javier Cerezo <span dir="ltr">&lt;<a
                        moz-do-not-send="true" href="mailto:jcb1@um.es"
                        target="_blank">jcb1@um.es</a>&gt;</span> wrote:<br>
                    <blockquote class="gmail_quote" style="margin:0 0 0
                      .8ex;border-left:1px #ccc solid;padding-left:1ex">
                      <div bgcolor="#FFFFFF" text="#000000"> Hi Steven.
                        <div><br>
                          &nbsp;<br>
                          <blockquote type="cite">
                            <div class="gmail_quote">
                              <div>1. Why this value is divided by nm3?
                                Shall I multiply it by the simulation
                                box?<br>
                              </div>
                            </div>
                          </blockquote>
                        </div>
                        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.
                        <div><br>
                          <blockquote type="cite">
                            <div class="gmail_quote">
                              <div>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?<br>
                              </div>
                            </div>
                          </blockquote>
                        </div>
                        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.
                        <div><br>
                          <blockquote type="cite">
                            <div class="gmail_quote">
                              <div> 3. The value corresponds to the
                                plateau so I should run it for longer
                                time?<br>
                              </div>
                            </div>
                          </blockquote>
                        </div>
                        You are getting a time&amp;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.<br>
                        <br>
                        Javier<br>
                        <br>
                        <br>
                        El 15/05/12 09:57, Steven Neumann escribi&oacute;:
                        <div>
                          <div>
                            <blockquote type="cite"><br>
                              <br>
                              <div class="gmail_quote">On Mon, May 14,
                                2012 at 5:05 PM, Justin A. Lemkul <span
                                  dir="ltr">&lt;<a
                                    moz-do-not-send="true"
                                    href="mailto:jalemkul@vt.edu"
                                    target="_blank">jalemkul@vt.edu</a>&gt;</span>
                                wrote:<br>
                                <blockquote class="gmail_quote"
                                  style="margin:0 0 0
                                  .8ex;border-left:1px #ccc
                                  solid;padding-left:1ex">
                                  <div>
                                    <div><br>
                                      <br>
                                      On 5/14/12 11:53 AM, Steven
                                      Neumann wrote:<br>
                                      <blockquote class="gmail_quote"
                                        style="margin:0 0 0
                                        .8ex;border-left:1px #ccc
                                        solid;padding-left:1ex"> Dear
                                        Gmx Users,<br>
                                        <br>
                                        Did anyone use TPI method for
                                        the calculation of chemical
                                        potential? The tpi.xvg<br>
                                        files consists of:<br>
                                        <br>
                                        @ s0 legend "-kT
                                        log(&lt;Ve\S-\xb\f{}U\N&gt;/&lt;V&gt;)"<br>
                                        @ s1 legend "f. -kT
                                        log&lt;e\S-\xb\f{}U\N&gt;"<br>
                                        @ s2 legend "f.
                                        &lt;e\S-\xb\f{}U\N&gt;"<br>
                                        @ s3 legend "f. V"<br>
                                        @ s4 legend "f.
                                        &lt;Ue\S-\xb\f{}U\N&gt;"<br>
                                        @ s5 legend "f. &lt;U\sVdW
                                        System\Ne\S-\xb\f{}U\N&gt;"<br>
                                        @ s6 legend "f. &lt;U\sdisp
                                        c\Ne\S-\xb\f{}U\N&gt;"<br>
                                        @ s7 legend "f. &lt;U\sCoul
                                        System\Ne\S-\xb\f{}U\N&gt;"<br>
                                        @ s8 legend "f. &lt;U\sCoul
                                        recip\Ne\S-\xb\f{}U\N&gt;"<br>
                                        <br>
                                        @ &nbsp; &nbsp;xaxis &nbsp;label "Time (ps)"<br>
                                        @ &nbsp; &nbsp;yaxis &nbsp;label "(kJ
                                        mol\S-1\N) / (nm\S3\N)"<br>
                                        <br>
                                        Can anyone explain me these
                                        legends? I just want obtain a
                                        value of the excess<br>
                                        chemical potential according to
                                        the equation:<br>
                                        u=-kT log (-deltaV/kT), Which
                                        legend is responsible for this
                                        and what are the<br>
                                        units? kJ/mol? Please, explain
                                        as the above letters does not
                                        mean to me anything?<br>
                                        <br>
                                      </blockquote>
                                      <br>
                                    </div>
                                  </div>
                                  These strings are formatted for
                                  XmGrace. &nbsp;Have you tried plotting the
                                  file to see what it contains? &nbsp;The
                                  legends will be far more obvious if
                                  you do.<br>
                                  <br>
                                  -Justin<span><font color="#888888"><br>
                                    </font></span></blockquote>
                                <div><br>
                                  Thank you Justin. <br>
                                  Can anyone explain me from the plot:<br>
                                  <br>
                                  <a moz-do-not-send="true"
                                    href="http://speedy.sh/Xpnws/tpi.JPG"
                                    target="_blank">http://speedy.sh/Xpnws/tpi.JPG</a><br>
                                  <br>
                                  1. Why this value is divided by nm3?
                                  Shall I multiply it by the simulation
                                  box?<br>
                                  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?<br>
                                  3. The value corresponds to the
                                  plateau so I should run it for longer
                                  time?<br>
                                  <br>
                                  <br>
                                  Thank you,<br>
                                  <br>
                                  Steven<br>
                                  &nbsp;<br>
                                </div>
                                <blockquote class="gmail_quote"
                                  style="margin:0pt 0pt 0pt
                                  0.8ex;border-left:1px solid
                                  rgb(204,204,204);padding-left:1ex"> <span><font
                                      color="#888888"> <br>
                                      -- <br>
========================================<br>
                                      <br>
                                      Justin A. Lemkul, Ph.D.<br>
                                      Department of Biochemistry<br>
                                      Virginia Tech<br>
                                      Blacksburg, VA<br>
                                      jalemkul[at]<a
                                        moz-do-not-send="true"
                                        href="http://vt.edu"
                                        target="_blank">vt.edu</a> | <a
                                        moz-do-not-send="true"
                                        href="tel:%28540%29%20231-9080"
                                        value="+15402319080"
                                        target="_blank">(540) 231-9080</a><br>
                                      <a moz-do-not-send="true"
                                        href="http://www.bevanlab.biochem.vt.edu/Pages/Personal/justin"
                                        target="_blank">http://www.bevanlab.biochem.vt.edu/Pages/Personal/justin</a><br>
                                      <br>
========================================<br>
                                      -- <br>
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                        <span><font color="#888888">
                            <div>-- <br>
                              Javier CEREZO BASTIDA<br>
                              PhD Student<br>
                              Physical Chemistry<br>
                              Universidad de Murcia<br>
                              Murcia (Spain)<br>
                              Tel: <a moz-do-not-send="true"
                                href="tel:%28%2B34%29868887434"
                                value="+34868887434" target="_blank">(+34)868887434</a>
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                          </font></span></div>
                      <br>
                      --<br>
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            <span class="HOEnZb"><font color="#888888">
                <pre cols="72">-- 
Javier CEREZO BASTIDA
Ph.D. Student
Physical Chemistry
Universidad de Murcia
30100, Murcia (SPAIN)
T: (0034)868887434</pre>
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    <br>
    <pre class="moz-signature" cols="72">-- 
Javier CEREZO BASTIDA
Ph.D. Student
Physical Chemistry
Universidad de Murcia
30100, Murcia (SPAIN)
T: (0034)868887434</pre>
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