<br><br><div class="gmail_quote">On Wed, May 16, 2012 at 10:28 AM, Javier Cerezo <span dir="ltr"><<a href="mailto:jcb1@um.es" target="_blank">jcb1@um.es</a>></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 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 (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 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> </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 escribió:
<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 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"><<a href="mailto:jcb1@um.es" target="_blank">jcb1@um.es</a>></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>
<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&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:
<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"><<a 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>
<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(<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><font color="#888888"><br>
</font></span></blockquote>
<div><br>
Thank you Justin. <br>
Can anyone explain me from the plot:<br>
<br>
<a 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>
<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 href="http://vt.edu" target="_blank">vt.edu</a>
| <a href="tel:%28540%29%20231-9080" value="+15402319080" target="_blank">(540)
231-9080</a><br>
<a href="http://www.bevanlab.biochem.vt.edu/Pages/Personal/justin" target="_blank">http://www.bevanlab.biochem.vt.edu/Pages/Personal/justin</a><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 href="tel:%28%2B34%29868887434" value="+34868887434" target="_blank">(+34)868887434</a> </div>
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</div></div><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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