In particular, the goal is to obtain a file like an extended xyz one and try training a machine learing potential.<div><br /></div><div>Best regards,</div><div>Riccardo Piombo<br /><br /></div><div class="gmail_quote"><div dir="auto" class="gmail_attr">Il giorno martedì 10 ottobre 2023 alle 10:37:48 UTC+2 riccardo piombo ha scritto:<br/></div><blockquote class="gmail_quote" style="margin: 0 0 0 0.8ex; border-left: 1px solid rgb(204, 204, 204); padding-left: 1ex;">Hi Prof. Hutter,<br>my simulations are also based on the Langevin ensamble. <div>do we have to give up on calculating atomic forces from the xyz trajectory?</div><div><br>Kind regards,<br>Riccardo Piombo<br><br></div><div class="gmail_quote"><div dir="auto" class="gmail_attr">Il giorno martedì 10 ottobre 2023 alle 10:17:17 UTC+2 Simone Ritarossi ha scritto:<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="auto">Hi users,<div dir="auto">I have the same problem and I used Langevin (second-generation Car-Parrinello). In this case?</div><div dir="auto"><br></div><div dir="auto">Simone Ritarossi </div></div><br><div class="gmail_quote"></div><div class="gmail_quote"><div dir="ltr" class="gmail_attr">Il Mar 10 Ott 2023, 08:58 Jürg Hutter <<a rel="nofollow">hut...@chem.uzh.ch</a>> ha scritto:<br></div></div><div class="gmail_quote"><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">Hi<br>
<br>
all integrators in CP2K are based on the Velocity Verlet algorithm.<br>
The different routines can be found in the src/motion directory.<br>
<br>
If your trajectory was generated for a NVE ensemble without any<br>
constraints, you can easily extract the forces from inverting the<br>
Velocity Verlet algorithm.<br>
If you were using any type of thermostat or any geometrical constraints<br>
it will be very difficult or impossible. At least I don't see a straightforward<br>
way.<br>
Depending on you final goal, recalculating the forces at a subset of your<br>
trajectory will be your only option.<br>
<br>
regards<br>
JH<br>
<br>
________________________________________<br>
From: <a rel="noreferrer nofollow">cp...@googlegroups.com</a> <<a rel="noreferrer nofollow">cp...@googlegroups.com</a>> on behalf of riccardo piombo <<a rel="noreferrer nofollow">riccard...@gmail.com</a>><br>
Sent: Monday, October 9, 2023 5:54 PM<br>
To: cp2k<br>
Subject: [CP2K:19326] Extracting Atomic Forces from XYZ Trajectories in CP2K<br>
<br>
Dear CP2K Community,<br>
<br>
We have encountered a challenge while working with MD trajectories in XYZ format. Specifically, we have trajectories containing atomic positions and velocities for each frame, but unfortunately, we do not have the corresponding atomic forces.<br>
<br>
In theory, it is possible to calculate atomic forces based on atomic positions and velocities. However, we are unsure about the specific algorithm and where it can be found within the CP2K codebase.<br>
<br>
We are reaching out to the community to inquire if anyone can provide guidance on:<br>
<br>
1. 1. Locating the algorithm or module responsible for calculating atomic forces within CP2K.<br>
2. 2. Assessing the feasibility of extracting atomic forces from our XYZ trajectories in a reasonably short amount of time.<br>
<br>
Any assistance or insights on this matter would be greatly appreciated. We understand that this may be a non-trivial task, but we are hopeful that the CP2K community's expertise can help us navigate this challenge.<br>
<br>
Thank you in advance for your support and expertise.<br>
<br>
Best regards Riccardo Piombo<br>
<br>
--<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 rel="noreferrer nofollow">cp2k+uns...@googlegroups.com</a><mailto:<a rel="noreferrer nofollow">cp2k+uns...@googlegroups.com</a>>.<br>
To view this discussion on the web visit <a href="https://groups.google.com/d/msgid/cp2k/311dfc7b-f7ec-4d45-a7c4-32e9203b8ee1n%40googlegroups.com" rel="noreferrer noreferrer nofollow" target="_blank" data-saferedirecturl="https://www.google.com/url?hl=it&q=https://groups.google.com/d/msgid/cp2k/311dfc7b-f7ec-4d45-a7c4-32e9203b8ee1n%2540googlegroups.com&source=gmail&ust=1697015848128000&usg=AOvVaw2WpurMpE-MhZxtV41o96-t">https://groups.google.com/d/msgid/cp2k/311dfc7b-f7ec-4d45-a7c4-32e9203b8ee1n%40googlegroups.com</a><<a href="https://groups.google.com/d/msgid/cp2k/311dfc7b-f7ec-4d45-a7c4-32e9203b8ee1n%40googlegroups.com?utm_medium=email&utm_source=footer" rel="noreferrer noreferrer nofollow" target="_blank" data-saferedirecturl="https://www.google.com/url?hl=it&q=https://groups.google.com/d/msgid/cp2k/311dfc7b-f7ec-4d45-a7c4-32e9203b8ee1n%2540googlegroups.com?utm_medium%3Demail%26utm_source%3Dfooter&source=gmail&ust=1697015848128000&usg=AOvVaw3x8FE1ys57-HFKIrITxBTr">https://groups.google.com/d/msgid/cp2k/311dfc7b-f7ec-4d45-a7c4-32e9203b8ee1n%40googlegroups.com?utm_medium=email&utm_source=footer</a>>.<br>
<br>
-- <br>
You received this message because you are subscribed to the Google Groups "cp2k" group.<br></blockquote></div><div class="gmail_quote"><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
To unsubscribe from this group and stop receiving emails from it, send an email to <a rel="noreferrer nofollow">cp2k+uns...@googlegroups.com</a>.<br>
To view this discussion on the web visit <a href="https://groups.google.com/d/msgid/cp2k/ZR0P278MB0759B06368A3D5E374E749D39FCDA%40ZR0P278MB0759.CHEP278.PROD.OUTLOOK.COM" rel="noreferrer noreferrer nofollow" target="_blank" data-saferedirecturl="https://www.google.com/url?hl=it&q=https://groups.google.com/d/msgid/cp2k/ZR0P278MB0759B06368A3D5E374E749D39FCDA%2540ZR0P278MB0759.CHEP278.PROD.OUTLOOK.COM&source=gmail&ust=1697015848128000&usg=AOvVaw3Sg5UqicN-6BewlyZHTcon">https://groups.google.com/d/msgid/cp2k/ZR0P278MB0759B06368A3D5E374E749D39FCDA%40ZR0P278MB0759.CHEP278.PROD.OUTLOOK.COM</a>.<br>
</blockquote></div>
</blockquote></div></blockquote></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+unsubscribe@googlegroups.com">cp2k+unsubscribe@googlegroups.com</a>.<br />
To view this discussion on the web visit <a href="https://groups.google.com/d/msgid/cp2k/c2cef1eb-eb15-4b04-adf2-7b365f2ea301n%40googlegroups.com?utm_medium=email&utm_source=footer">https://groups.google.com/d/msgid/cp2k/c2cef1eb-eb15-4b04-adf2-7b365f2ea301n%40googlegroups.com</a>.<br />