QuantumATK Forum

QuantumATK => General Questions and Answers => Topic started by: Dipankar Saha on November 23, 2015, 17:29

Title: Force one an atom
Post by: Dipankar Saha on November 23, 2015, 17:29
Hi,
Considering X-Y-Z co-ordinates, when you say force one an atom (say, index 1) as_
 1  [ 0.00023693 -0.0042467  -0.00211885] eV/Ang ......;
what do the negative signs mean?
Best_
Dipankar
Title: Re: Force one an atom
Post by: Anders Blom on November 23, 2015, 23:12
I'm inclined here to answer: what do you think it means?

Of  course it means the force points in a certain direction. The force is a vector, so if it points in +y or -y makes a huge difference.
Title: Re: Force one an atom
Post by: Jess Wellendorff on November 24, 2015, 10:26
So the force convergence criterion is of course a criterion for the maximum length of force vectors.
Title: Re: Force one an atom
Post by: Dipankar Saha on November 25, 2015, 12:35
Dr. Jess Wellendorff,
How do you calculate...? Perhaps, Hellmann–Feynman....!!   The amplitude part is fine...but, for any direction to be positive or, negative... a ref. is needed....
_____
Besides,  how can we verify the zero strain on any particular atom.... ?

Thanks_
Dipankar
Title: Re: Force one an atom
Post by: Jess Wellendorff on November 26, 2015, 10:20
Yes, those are Hellman-Feynman forces. You can find info in any standard DFT textbook. positive or negative direction is simply wrt. to the unit cell coordinate system. There is no such thing as strain on any particular atom, it is a property of the unit cell. Zero strain on the unit cell means that the internal pressure equals the external pressure (which is usually zero).
Title: Re: Force one an atom
Post by: Dipankar Saha on November 26, 2015, 11:52
Ohhh...yesss....it's my mistake.... I should have said 'cell' instead...!! / Thank you Jess for the reply.... :)
Title: Re: Force one an atom
Post by: Dipankar Saha on November 26, 2015, 20:55
Now, can you please tell me your view point about the following :-

The value of the in-plane strain varies significantly with the cell size. For example, if you want to make an interface between 'A' and 'B'.....and you firstly take  a 2x2 supercell of 'A' and  √7x√7 supercell of 'B';  and after that a  4x4 supercell of 'A' and  5x5 supercell of 'B'....; then the strain tensor components (out of the Algorithm that you use) for those two cases will give different  values (sometimes significantly diff.)......

Is it always a good idea to go with the min.  % strain..... and to find a cell size accordingly?? Or, is it..... smaller the individual cells are within an interface...better is the prediction of lattice mismatch ?!!

In case we have information from the Moire-Plot....then it's fine...we may try to go any closer to that.... But, if there are no such exp. evidence...... then, it becomes difficult .....

Thanks in advance..... :)

Regards_
Dipankar
Title: Re: Force one an atom
Post by: Jess Wellendorff on November 27, 2015, 09:14
I think one should in general try to minimize the interface strain at interfaces as much as possible. Real-life interfaces grow such as to minimize the strain, and may therefore not be epitaxial right at the interface. But you still want to use reasonably sized supercells, such that computational time does not explode. So, as always with ab initio simulations, it's a trade-off between modeling reality as closely as possible and the required computational resources.
Title: Re: Force one an atom
Post by: Dipankar Saha on November 27, 2015, 14:56
Thank you very much Jess....for the details.... :)