EFOR - Revision history

Wolloch at 14:25, 17 January 2025

2025-01-17T14:25:17Z

← Older revision		Revision as of 14:25, 17 January 2025
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	Description: {{TAG\|EFOR}} sets external forces in eV/<span>Å</span> on each atom in the cartesian x-, y-, and z-directions. The order of the ions is equivalent to the order in the {{FILE\|POSCAR}} file, and for each ion, x-, y-, and z-components have to be given.		Description: {{TAG\|EFOR}} sets external forces in eV/<span>Å</span> on each atom in the cartesian x-, y-, and z-directions. The order of the ions is equivalent to the order in the {{FILE\|POSCAR}} file, and for each ion, x-, y-, and z-components have to be given.
	{{NB\|~~mind\|This tag is only available as of VASP.~~6.5.0.}}		{{Available\|6.5.0}}
	----		----
	The array of external forces will be added to the [[forces]] computed internally for each ionic step ([[ionic minimization]] or [[molecular dynamics]]).		The array of external forces will be added to the [[forces]] computed internally for each ionic step ([[ionic minimization]] or [[molecular dynamics]]).

Wolloch: Wolloch moved page Construction:EFOR to EFOR

2024-12-18T09:52:16Z

Wolloch moved page Construction:EFOR to EFOR

← Older revision	Revision as of 09:52, 18 December 2024
(No difference)

Wolloch: /* Related Tags and Sections */

2024-12-18T09:51:09Z

Related Tags and Sections

← Older revision		Revision as of 09:51, 18 December 2024
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	----		----

	~~<!--~~[[Category:INCAR tag]][[Category:Forces]][[Category:Symmetry]]~~-->~~		[[Category:INCAR tag]][[Category:Forces]][[Category:Symmetry]]

Wolloch: /* Example */

2024-10-21T09:43:13Z

Example

← Older revision		Revision as of 09:43, 21 October 2024
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	==Example==		==Example==

	As an example, ~~we compress~~ an Al(001) slab under ~~a pressure of~~ 4 GPa. At ~39500 times the atmospheric pressure at sea level, this is considerable, but pressures in [https://en.wikipedia.org/wiki/Diamond_anvil_cell diamond anvil cells] can be two orders of magnitude larger. The four-layer thick slab corresponding to the following {{FILE\|POSCAR}} is already relaxed without external forces:		As an example, the compression of an Al(001) slab under 4 GPa of pressure is chosen. At ~39500 times the atmospheric pressure at sea level, this is considerable, but pressures in [https://en.wikipedia.org/wiki/Diamond_anvil_cell diamond anvil cells] can be two orders of magnitude larger. The four-layer thick slab corresponding to the following {{FILE\|POSCAR}} is already relaxed without external forces:
	Al 001 slab		Al 001 slab
	1.00000000000000		1.00000000000000
Line 106:		Line 106:
	0.4999999999382325 0.0000000000000000 0.3864211449750742 T T T		0.4999999999382325 0.0000000000000000 0.3864211449750742 T T T

	~~We have~~ to ~~convert the pressure~~ into the correct units. Forces in VASP are given in eV/<span>Å</span>, and an area in <span>Å<sup>2</sup></span> thus ~~we need~~ to ~~convert GPa~~ to (eV/<span>Å</span>)/<span>Å<sup>2</sup></span>, or eV <span>Å<sup>-3</sup></span>:		The pressure has to be converted into the correct units. Forces in VASP are given in eV/<span>Å</span>, and an area in <span>Å<sup>2</sup></span>, thus GPa needs to be converted to (eV/<span>Å</span>)/<span>Å<sup>2</sup></span>, or eV <span>Å<sup>-3</sup></span>:
	::<math>		::<math>
	\mathrm{GPa} = 10^9 \mathrm{Pa} = 10^9 \frac{\mathrm{N}}{\mathrm{m}^2} = 10^9 \frac{\mathrm{Nm}}{\mathrm{m}^3} = 10^9 \frac{6.2415 \times 10^{18}\mathrm{eV}}{10^{30} \AA^3} = 6.2415 \times 10^{-3} \frac{\mathrm{eV}/\AA}{\AA^2}		\mathrm{GPa} = 10^9 \mathrm{Pa} = 10^9 \frac{\mathrm{N}}{\mathrm{m}^2} = 10^9 \frac{\mathrm{Nm}}{\mathrm{m}^3} = 10^9 \frac{6.2415 \times 10^{18}\mathrm{eV}}{10^{30} \AA^3} = 6.2415 \times 10^{-3} \frac{\mathrm{eV}/\AA}{\AA^2}
	</math>		</math>
	~~with~~ a lattice parameter of 4.047<span>Å</span>, the surface area A of ~~our~~ slab is about 16.381 <span>Å<sup>2</sup></span>. Thus the total force in eV/<span>Å</span> we have to put on ~~our~~ top layer to compress the slab with 4 GPa of pressure is:		With a lattice parameter of 4.047<span>Å</span>, the surface area A of the slab is about 16.381 <span>Å<sup>2</sup></span>. Thus the total force in eV/<span>Å</span> we have to put on the top layer to compress the slab with 4 GPa of pressure is:
	::<math>		::<math>
	4 \times A \times 6.2415 \times 10^{-3} \simeq 0.409		4 \times A \times 6.2415 \times 10^{-3} \simeq 0.409 \quad ,
	</math>		</math>
	or 0.20448 eV/<span>Å</span> per surface atom. ~~We can now write our~~ {{FILE\|INCAR}} file:		or 0.20448 eV/<span>Å</span> per surface atom. The {{FILE\|INCAR}} file can now be written:

	ENCUT = 250		ENCUT = 250
Line 127:		Line 127:
	EDIFF = 1E-6		EDIFF = 1E-6
	EFOR = 20 0.40896 170 -0.20448 2*0 -0.20448		EFOR = 20 0.40896 170 -0.20448 2*0 -0.20448

			Note that the forces on the top atoms are balanced by putting the opposing force on one atom only. This is feasible since both atoms in the bottom layer are fixed by selective dynamics flags.

	Using a 5x5x1 Γ-centered mesh in the {{FILE\|KPOINTS}} file, and the standard PBE Al {{FILE\|POTCAR}}, the calculation converges in less than 10 ionic steps.		Using a 5x5x1 Γ-centered mesh in the {{FILE\|KPOINTS}} file, and the standard PBE Al {{FILE\|POTCAR}}, the calculation converges in less than 10 ionic steps.

	~~4 GPa translate in forces of ~0.2 eV/<span>Å</span>, which is only 20 times the convergence criterion of 0.01 eV/<span>Å</span> for the forces in our calculation. So the~~ slab is compressed by less than 2%, from 6.21 <span>Å</span> to 6.10 <span>Å</span>. This relatively small compression at relatively high pressure is due to ~~two~~ factors:		The slab is compressed by less than 2%, from 6.21 <span>Å</span> to 6.10 <span>Å</span>. This relatively small compression at relatively high pressure is due to three factors:
	* We do not allow the material to [https://en.wikipedia.org/wiki/Poisson%27s_ratio expand in the direction normal to the pressure] by keeping the cell shape fixed.		* 4 GPa corresponds to forces of ~0.2 eV/<span>Å</span>, which is only 20 times the negligible force limit of 0.01 eV/<span>Å</span> in this calculation.
	* Intra-atomic forces in close-packed metallic ~~perfect~~ crystals are powerful, and lattice imperfections usually mediate deformations in real materials.		* We do not allow the material to [https://en.wikipedia.org/wiki/Poisson%27s_ratio expand in the direction normal to the pressure] by keeping the cell shape fixed with [[ISIF]] = 2.
			* Intra-atomic forces in perfect close-packed metallic crystals are powerful, and lattice imperfections usually mediate deformations in real materials.

	For layered materials, molecules on surfaces, and other, less strongly bound systems, similar forces will have larger effects, however.		For layered materials, molecules on surfaces, and other, less strongly-bound systems, similar forces will have larger effects, however.

	== Related Tags and Sections ==		== Related Tags and Sections ==

Wolloch: /* Example */

2024-10-21T09:28:01Z

Example

← Older revision		Revision as of 09:28, 21 October 2024
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	0.4999999999382325 0.0000000000000000 0.3864211449750742 T T T		0.4999999999382325 0.0000000000000000 0.3864211449750742 T T T

	~~Next, we~~ have to convert the pressure into the correct units. Forces in VASP are given in eV/<span>Å</span>, and an area in <span>Å<sup>2</sup></span> thus we need to convert GPa to (eV/<span>Å</span>)/<span>Å<sup>2</sup></span>, or eV <span>Å<sup>-3</sup></span>:		We have to convert the pressure into the correct units. Forces in VASP are given in eV/<span>Å</span>, and an area in <span>Å<sup>2</sup></span> thus we need to convert GPa to (eV/<span>Å</span>)/<span>Å<sup>2</sup></span>, or eV <span>Å<sup>-3</sup></span>:
	::<math>		::<math>
	\mathrm{GPa} = 10^9 \mathrm{Pa} = 10^9 \frac{\mathrm{N}}{\mathrm{m}^2} = 10^9 \frac{\mathrm{Nm}}{\mathrm{m}^3} = 10^9 \frac{6.2415 \times 10^{18}\mathrm{eV}}{10^{30} \AA^3} = 6.2415 \times 10^{-3} \frac{\mathrm{eV}/\AA}{\AA^2}		\mathrm{GPa} = 10^9 \mathrm{Pa} = 10^9 \frac{\mathrm{N}}{\mathrm{m}^2} = 10^9 \frac{\mathrm{Nm}}{\mathrm{m}^3} = 10^9 \frac{6.2415 \times 10^{18}\mathrm{eV}}{10^{30} \AA^3} = 6.2415 \times 10^{-3} \frac{\mathrm{eV}/\AA}{\AA^2}

Wolloch: /* Example */

2024-10-21T09:27:25Z

Example

← Older revision		Revision as of 09:27, 21 October 2024
Line 87:		Line 87:
	==Example==		==Example==

	As an example, we compress an Al(001) slab ~~with~~ a pressure of 4 GPa. At ~39500 times the atmospheric pressure at sea level, this is considerable, but pressures in [https://en.wikipedia.org/wiki/Diamond_anvil_cell diamond anvil cells] can be two orders of magnitude larger. The four-layer thick slab corresponding to the following {{FILE\|POSCAR}} is already relaxed without external forces:		As an example, we compress an Al(001) slab under a pressure of 4 GPa. At ~39500 times the atmospheric pressure at sea level, this is considerable, but pressures in [https://en.wikipedia.org/wiki/Diamond_anvil_cell diamond anvil cells] can be two orders of magnitude larger. The four-layer thick slab corresponding to the following {{FILE\|POSCAR}} is already relaxed without external forces:
	Al 001 slab		Al 001 slab
	1.00000000000000		1.00000000000000

Wolloch: /* Example */

2024-10-21T09:27:12Z

Example

← Older revision		Revision as of 09:27, 21 October 2024
Line 87:		Line 87:
	==Example==		==Example==

	As an example, we ~~want to~~ compress an Al(001) slab with a pressure of 4 GPa. At ~39500 times the atmospheric pressure at sea level, this is considerable, but pressures in [https://en.wikipedia.org/wiki/Diamond_anvil_cell diamond anvil cells] can be two orders of magnitude larger. The four-layer thick slab corresponding to the following {{FILE\|POSCAR}} is already relaxed without external forces:		As an example, we compress an Al(001) slab with a pressure of 4 GPa. At ~39500 times the atmospheric pressure at sea level, this is considerable, but pressures in [https://en.wikipedia.org/wiki/Diamond_anvil_cell diamond anvil cells] can be two orders of magnitude larger. The four-layer thick slab corresponding to the following {{FILE\|POSCAR}} is already relaxed without external forces:
	Al 001 slab		Al 001 slab
	1.00000000000000		1.00000000000000

Wolloch at 09:26, 21 October 2024

2024-10-21T09:26:53Z

@@ Line 87: / Line 87: @@
 ==Example==
-In this example, we want to compress an Al(001) slab with a pressure of 2 GPa. First, we have to convert the pressure into the correct units. Forces in VASP are given in <math>\mathrm{eV}/\AA</math>, and length in <math>\AA</math>, so a pressure would be of unit <math>(\mathrm{eV}/\AA)/\AA^2</math>. The full conversion from GPa is:
+As an example, we want to compress an Al(001) slab with a pressure of 4 GPa. At ~39500 times the atmospheric pressure at sea level, this is considerable, but pressures in [https://en.wikipedia.org/wiki/Diamond_anvil_cell diamond anvil cells] can be two orders of magnitude larger. The four-layer thick slab corresponding to the following {{FILE|POSCAR}} is already relaxed without external forces:
 ::<math>
-\mathrm{GPa} = 10^9 \mathrm{Pa} = 10^9 \frac{\mathrm{N}}{\mathrm{m}^2} = 10^9 \frac{\mathrm{Nm}}{\mathrm{m}^3} = 10^9 \frac{6.2415 \times 10^{18}\mathrm{eV}}{10^{30} \AA^3} = 6.2415 10^{-3} \frac{\mathrm{eV}/\AA}{\AA^2}
+\mathrm{GPa} = 10^9 \mathrm{Pa} = 10^9 \frac{\mathrm{N}}{\mathrm{m}^2} = 10^9 \frac{\mathrm{Nm}}{\mathrm{m}^3} = 10^9 \frac{6.2415 \times 10^{18}\mathrm{eV}}{10^{30} \AA^3} = 6.2415 \times 10^{-3} \frac{\mathrm{eV}/\AA}{\AA^2}
 </math>
 == Related Tags and Sections ==

Wolloch: /* Example */

2024-10-18T16:16:47Z

Example

← Older revision		Revision as of 16:16, 18 October 2024
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	total drift: 0.000000 -0.000000 -0.000000		total drift: 0.000000 -0.000000 -0.000000
	==Example==		==Example==
	~~bla~~
			In this example, we want to compress an Al(001) slab with a pressure of 2 GPa. First, we have to convert the pressure into the correct units. Forces in VASP are given in <math>\mathrm{eV}/\AA</math>, and length in <math>\AA</math>, so a pressure would be of unit <math>(\mathrm{eV}/\AA)/\AA^2</math>. The full conversion from GPa is:
			::<math>
			\mathrm{GPa} = 10^9 \mathrm{Pa} = 10^9 \frac{\mathrm{N}}{\mathrm{m}^2} = 10^9 \frac{\mathrm{Nm}}{\mathrm{m}^3} = 10^9 \frac{6.2415 \times 10^{18}\mathrm{eV}}{10^{30} \AA^3} = 6.2415 10^{-3} \frac{\mathrm{eV}/\AA}{\AA^2}
			</math>

	== Related Tags and Sections ==		== Related Tags and Sections ==
	{{FILE\|ICONST}}		{{FILE\|ICONST}}

Wolloch: /* Related Tags and Sections */

2024-10-17T10:07:44Z

Related Tags and Sections

← Older revision		Revision as of 10:07, 17 October 2024
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	----		----

	[[Category:INCAR tag]][[Category:Forces]][[Category:Symmetry]]		<!--[[Category:INCAR tag]][[Category:Forces]][[Category:Symmetry]]-->