Re: [Pw_forum] B_field per Rydberg

[Paolo Giannozzi]

On Mon, Nov 28, 2016 at 8:42 PM, Mems Aghaee  wrote:

why magnetic field is per Rydberg?
>

because everything is in atomic (Rydberg) units and whoever wrote that
quantity was lazy, exactly as you are, since you ask:


> and how can i convert that to Tesla?
>

instead of doing the math

Paolo
-- 
Paolo Giannozzi, Dip. Scienze Matematiche Informatiche e Fisiche,
Univ. Udine, via delle Scienze 208, 33100 Udine, Italy
Phone +39-0432-558216, fax +39-0432-558222
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[Pw_forum] B_field per Rydberg

[Mems Aghaee]

Hiplz look at the end of my scf.out:why magnetic field is per Rydberg?and how 
can i convert that to Tesla?.total energy = -253.04759495 Ry
 Harris-Foulkes estimate = -253.04758781 Ry
 estimated scf accuracy < 0.0452 RyThe total energy is the sum of the 
following terms:one-electron contribution = -127.46773250 Ry
 hartree contribution = 72.38794581 Ry
 xc contribution = -27.02379306 Ry
 ewald contribution = -170.94380974 Ry
 smearing contrib. (-TS) = -0.00020546 Rytotal magnetization = 5.00 Bohr 
mag/cell
 absolute magnetization = 5.00 Bohr mag/cell
 Magnetic field = 0.3285968 
Ryconvergence has been achieved in 6 
iterations.

and my scf.in:
&CONTROL
 calculation = "scf",
 prefix = "Fe",
 pseudo_dir = "./",
 outdir = "./TMP",
/
&SYSTEM
 ibrav = 3,
 celldm(1) = 5.45,
constrained_magnetization = "total",
nspin=2,
fixed_magnetization (3) = 5,
 nat = 1,
 ntyp = 1,
 ecutwfc = 28,
 ecutrho = 300,
 starting_magnetization(1)=0.5
 occupations='smearing'
 smearing='mp'
 degauss=0.015
lambda=1
/
&ELECTRONS
 conv_thr = 1.D-5,
 mixing_beta = 0.3D0,
/
&IONS
/
ATOMIC_SPECIES
Fe 55.845 Fe.pbe-sp-van_ak.UPF
ATOMIC_POSITIONS (angstrom)
Fe 0.0 0.0 0.0
K_POINTS automatic
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