Well if you restart every step you'd have the result you want (but probably a bit slow ... maybe with a python interface?).<div>If you run multiple steps and then adjust the temp by 10K etc. Depends whether you just want to get a structure or if you are trying to do something quantitative.</div><div>Matt<br /><br /></div><div class="gmail_quote"><div dir="auto" class="gmail_attr">On Friday 12 April 2024 at 06:50:05 UTC+1 Vahiya Mitanshu wrote:<br/></div><blockquote class="gmail_quote" style="margin: 0 0 0 0.8ex; border-left: 1px solid rgb(204, 204, 204); padding-left: 1ex;">Is it jusifiable to do so? i mean rather than performing quenching is it valid to use thermostat at different temperature to bring down the temperature to the room temperature?<div><br><br></div><div class="gmail_quote"><div dir="auto" class="gmail_attr">On Thursday 11 April 2024 at 3:24:32 pm UTC+5:30 Matt Watkins wrote:<br></div><blockquote class="gmail_quote" style="margin:0 0 0 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex">Currently there is no way to get a linear decrease only exponential. You can approximately do it by restarting runs with different thermostat temperatures.<div>Matt<br><br></div><div class="gmail_quote"><div dir="auto" class="gmail_attr">On Wednesday 10 April 2024 at 09:47:56 UTC+1 Vahiya Mitanshu wrote:<br></div><blockquote class="gmail_quote" style="margin:0 0 0 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex"><div dir="ltr">Hello cp2k users,<br>I want to perform a fast quench on my atomic system, so that it can be in an amorphous state. but i don't know how exactly the annealing parameter works. I have started the run for quenching, The temperature is decreasing but I want to fix that rate. Right now I don't know which equation I should use so that I can take my temperature to my targeted temperature in a particular timestep. So if anyone has any ideas please let me know. I have attached my input file for melt and quench so if something is wrong in that, I can make modifications.<div><br>Thanking in advance,</div><div>Mitanshu Vahiya</div></div>
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