<div dir="ltr"><div>I need to estimate temperature of the non-equlibrium system using kinetic energy <br>which can obtained by generalized equipartition.<br></div><br>Is there any other way so that I can measure it?<br><br> <div><br><br></div></div><div class="gmail_extra"><br><br><div class="gmail_quote">On Thu, Mar 27, 2014 at 7:24 PM, Carlos Campana <span dir="ltr"><<a href="mailto:campa...@gmail.com" target="_blank">campa...@gmail.com</a>></span> wrote:<br> <blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div dir="ltr">Why would you expect equipartition to hold in a non-equilibrium system? If one cannot define the partition function, how are quantity averages to be computed? Might be different when the non-equilibrium system achieves a certain steady-state...<div> These are my 2 cents (haven't dusted my stat mech in a long time...)<span class="HOEnZb"><font color="#888888"><br><div>Carlos</div></font></span></div></div><div class="HOEnZb"><div class="h5"><div class="gmail_extra"> <br><br><div class="gmail_quote">On Thu, Mar 27, 2014 at 3:06 PM, Rasoul Nasiri <span dir="ltr"><<a href="mailto:nasir...@gmail.com" target="_blank">nasir...@gmail.com</a>></span> wrote:<br> <blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div dir="ltr"><div>Hello Matt,</div><div><br></div><div>Many thanks.</div><div><br></div><div>I think even using NVE, one will be dealt with some errors in non-equilibrium situations. Could anybody please comment on it?</div> <div><br></div><div>Probably the reason is related to <mv2> = KBT in vibrational modes.</div><div><br></div><div>Best,</div><div>Rasoul</div></div><div><div><div class="gmail_extra"><br><br> <div class="gmail_quote">On Thu, Mar 27, 2014 at 7:23 PM, Matt W <span dir="ltr"><<a href="mailto:MattWa...@gmail.com" target="_blank">MattWa...@gmail.com</a>></span> wrote:<br> <blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div dir="ltr"><div>Hi, interesting question, and I'd hope of interest to a few.</div><div><br></div>I can't give a proper answer, but I think, in general, if you are not in equilibrium, equipartition is going to break - waving hands:<div> <br></div><div>(i) you can't define your modes properly.<div>(ii) Also, most (all?, maybe some Langevin types recently implemented might escape this) thermostats only work at equilibrium too, so you'd be running NVE or similar where temperature isn't defined properly for a small cluster.<br> </div><div><br></div><div>I guess there are tricks for steady-state thermostatted non-equilibrium processes, but that is well beyond anything I know about.<br><br>Matt<div><div><br><br>On Thursday, March 27, 2014 10:16:07 AM UTC, Rasoul wrote:<blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;padding-left:1ex;border-left-color:rgb(204,204,204);border-left-width:1px;border-left-style:solid"> <div dir="ltr"><div><div><div><div><div>Dear All,<br><br></div><div>Apologize if my question seems to be irrelevant for CP2K users.<br></div><div><br></div>I'm trying to do a non-equilibrium MD simulation using CP2K in which one molecule attacking the nano-droplet.<br> <br></div>My question is whether the equipartition of kinetic energy is satisfied in the code for vibrational, rotational and translational modes at high temperatures and non-equilibrium situations?<br></div>Having said this if <mv2> = KBT works properly for estimation of temperature in the vibrational modes even using small time steps?<br> <br></div><div>Many thanks,<br></div><div><br></div>Best,<br></div>Rasoul<br><div><div><div> <br></div></div></div></div> </blockquote></div></div></div></div></div><span><font color="#888888"> <p></p> -- <br> You received this message because you are subscribed to the Google Groups "cp2k" group.<br> To unsubscribe from this group and stop receiving emails from it, send an email to <a href="mailto:cp2k+uns...@googlegroups.com" target="_blank">cp2k+uns...@googlegroups.com</a>.<br> To post to this group, send email to <a href="mailto:cp...@googlegroups.com" target="_blank">cp...@googlegroups.com</a>.<br> Visit this group at <a href="http://groups.google.com/group/cp2k" target="_blank">http://groups.google.com/group/cp2k</a>.<br> For more options, visit <a href="https://groups.google.com/d/optout" target="_blank">https://groups.google.com/d/optout</a>.<br> </font></span></blockquote></div><br></div> <p></p> -- <br> You received this message because you are subscribed to the Google Groups "cp2k" group.<br> To unsubscribe from this group and stop receiving emails from it, send an email to <a href="mailto:cp2k+uns...@googlegroups.com" target="_blank">cp2k+uns...@googlegroups.com</a>.<br> To post to this group, send email to <a href="mailto:cp...@googlegroups.com" target="_blank">cp...@googlegroups.com</a>.<br> Visit this group at <a href="http://groups.google.com/group/cp2k" target="_blank">http://groups.google.com/group/cp2k</a>.<br> For more options, visit <a href="https://groups.google.com/d/optout" target="_blank">https://groups.google.com/d/optout</a>.<br> </div></div></blockquote></div><br></div> <p></p> -- <br> You received this message because you are subscribed to the Google Groups "cp2k" group.<br> To unsubscribe from this group and stop receiving emails from it, send an email to <a href="mailto:cp2k+uns...@googlegroups.com" target="_blank">cp2k+uns...@googlegroups.com</a>.<br> To post to this group, send email to <a href="mailto:cp...@googlegroups.com" target="_blank">cp...@googlegroups.com</a>.<br> Visit this group at <a href="http://groups.google.com/group/cp2k" target="_blank">http://groups.google.com/group/cp2k</a>.<br> For more options, visit <a href="https://groups.google.com/d/optout" target="_blank">https://groups.google.com/d/optout</a>.<br> </div></div></blockquote></div><br></div>