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Hi Steven.<br>
<br>
<blockquote
cite="mid:CAKZJqQEQg7Hz6039f9AV4KBvq-HcEinOchabSGavcGnuvrzJUg@mail.gmail.com"
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>
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.<br>
<blockquote
cite="mid:CAKZJqQEQg7Hz6039f9AV4KBvq-HcEinOchabSGavcGnuvrzJUg@mail.gmail.com"
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>
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.<br>
<blockquote
cite="mid:CAKZJqQEQg7Hz6039f9AV4KBvq-HcEinOchabSGavcGnuvrzJUg@mail.gmail.com"
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>
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.<br>
<br>
Javier<br>
<br>
<br>
El 15/05/12 09:57, Steven Neumann escribió:
<blockquote
cite="mid:CAKZJqQEQg7Hz6039f9AV4KBvq-HcEinOchabSGavcGnuvrzJUg@mail.gmail.com"
type="cite"><br>
<br>
<div class="gmail_quote">On Mon, May 14, 2012 at 5:05 PM, Justin
A. Lemkul <span dir="ltr"><<a moz-do-not-send="true"
href="mailto:jalemkul@vt.edu" target="_blank">jalemkul@vt.edu</a>></span>
wrote:<br>
<blockquote class="gmail_quote" style="margin:0 0 0
.8ex;border-left:1px #ccc solid;padding-left:1ex">
<div class="HOEnZb">
<div class="h5"><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(<Ve\S-\xb\f{}U\N>/<V>)"<br>
@ s1 legend "f. -kT log<e\S-\xb\f{}U\N>"<br>
@ s2 legend "f. <e\S-\xb\f{}U\N>"<br>
@ s3 legend "f. V"<br>
@ s4 legend "f. <Ue\S-\xb\f{}U\N>"<br>
@ s5 legend "f. <U\sVdW System\Ne\S-\xb\f{}U\N>"<br>
@ s6 legend "f. <U\sdisp c\Ne\S-\xb\f{}U\N>"<br>
@ s7 legend "f. <U\sCoul System\Ne\S-\xb\f{}U\N>"<br>
@ s8 legend "f. <U\sCoul recip\Ne\S-\xb\f{}U\N>"<br>
<br>
@ xaxis label "Time (ps)"<br>
@ yaxis 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. Have you tried
plotting the file to see what it contains? The legends will
be far more obvious if you do.<br>
<br>
-Justin<span class="HOEnZb"><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">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>
<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 class="HOEnZb"><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>
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</div>
<br>
<br>
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<br>
</blockquote>
<br>
<div class="moz-signature">-- <br>
Javier CEREZO BASTIDA<br>
PhD Student<br>
Physical Chemistry<br>
Universidad de Murcia<br>
Murcia (Spain)<br>
Tel: (+34)868887434
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