def sundials_rhs(self, t, y, ydot):
self.t = t
# already synchronized when call this funciton
# self.spin[:]=y[:]
self.compute_effective_field(t)
if self.update_j_fun is not None:
clib.compute_stt_field(self.spin, self.field_stt,
self._jx * self.update_j_fun(t),
self._jy * self.update_j_fun(t),
self.mesh.dx * self.mesh.unit_length,
self.mesh.dy * self.mesh.unit_length,
self.mesh.nx, self.mesh.ny, self.mesh.nz,
self.xperiodic, self.yperiodic)
else:
clib.compute_stt_field(self.spin, self.field_stt, self._jx,
self._jy,
self.mesh.dx * self.mesh.unit_length,
self.mesh.dy * self.mesh.unit_length,
self.mesh.neighbours, self.n)
clib.compute_llg_stt_rhs(ydot, self.spin, self.field, self.field_stt,
self.alpha, self.beta,
self.u0 * self.p / self.Ms_const, self.gamma,
self.n)
def sundials_rhs(self, t, y, ydot):
self.t = t
# already synchronized when call this funciton
# self.spin[:]=y[:]
self.compute_effective_field(t)
if self.update_j_fun is not None:
clib.compute_stt_field(self.spin,
self.field_stt,
self._jx * self.update_j_fun(t),
self._jy * self.update_j_fun(t),
self._jz * self.update_j_fun(t),
self.mesh.dx * self.mesh.unit_length,
self.mesh.dy * self.mesh.unit_length,
self.mesh.dz * self.mesh.unit_length,
self.mesh.neighbours,
self.n
)
else:
clib.compute_stt_field(self.spin,
self.field_stt,
self._jx,
self._jy,
self._jz,
self.mesh.dx * self.mesh.unit_length,
self.mesh.dy * self.mesh.unit_length,
self.mesh.dz * self.mesh.unit_length,
self.mesh.neighbours,
self.n
)
clib.compute_llg_stt_rhs(ydot,
self.spin,
self.field,
self.field_stt,
self._alpha,
self.beta,
self.u0 * self.p / self.Ms_const,
self.gamma,
self.n)
def test_sst_field_1d():
"""
This is a direct test of the STT C library
We create a 1-D 4 spins system along the x direction with the
following components:
s_x s_y s_z spin
[[ 0. 4. 8. ] i = 0
[ 0.8 5. 9. ] i = 1
[ 2. 6. 10. ] i = 2
[ 2.1 7. 11. ]] i = 3
Most of the STT parameters are set to zero, so the field is reduced
to the calculation of:
H_STT = j \cdot \nabla S = ( jx * d S_x / dx + jy * d S_x / dy ,
jx * d S_y / dx + jy * d S_y / dy ,
0)
Therefore, the x component of the field for the 1th spin, for example,
would be:
H_STTx (i=1) = jx * d S_x (i=1) / dx + 0
= [ S_x (i=2) - S_x (i=0) ] / 2 * dx
= ( 2 - 0 ) / 2
= 1
since jx = dx = 1 and the derivatives along y are zero (1D spin chain)
The first test is the spin chain without PBC and the second
test is using PBCs, so the neighbours matrix changes
** Without PBC:
At the extremes, the derivative has a different discretisation
(using only the right or left neighbour and centered at the spin)
which, for the 0th spin, for instance, is
H_STTx (i=0) = jx * d S_x (i=0) / dx + 0
= [ S_x (i=1) - S_x (i=0) ] / dx
= ( 0.8 - 0 )
= 0.8
For further details, see: fidimag/atomistic/lib/stt.c
With PBC: The derivatives at the extremes are performed with the
periodic neighbour, e.g. for the 0th spin, the NN at -x is the 3th spin
"""
# The system is 1d with 4 spins.
nx, ny, nz = 4, 1, 1
n = nx * ny * nz
dx, dy, dz = 1, 1, 1
# NO PBCs -----------------------------------------------------------------
# We can manually construct the neighbours matrix,
# the neighbours order is [-x, +x, -y, +y, -z, +z]
# Since there are no PBCs, for the 0th spin, i.e. ngbs[0],
# the left (-x) NN is -1 and for the 3th spin, the +x NN is -1
# Also, since it is a 1D chain, there are no NNs in y or z
ngbs = [[-1, 1, -1, -1, -1, -1],
[0, 2, -1, -1, -1, -1],
[1, 3, -1, -1, -1, -1],
[2, -1, -1, -1, -1, -1],
]
ngbs = np.array(ngbs, dtype=np.int32)
nxyz = nx * ny * nz * 3
# The spin values to get the matrix:
#
# [[ 0. 4. 8. ]
# [ 1 5. 9. ]
# [ 2. 6. 10. ]
# [ 3 7. 11. ]]
spin = np.array([1.0 * i for i in range(nxyz)]).reshape((-1, 3), order='F').flatten()
field = np.zeros(nxyz)
jx = np.zeros(nxyz)
jy = np.zeros(nxyz)
jz = np.zeros(nxyz)
jx[:] = 1
jy[:] = 2
# Modify the x components of the spins to get the values
# specified in the problem description
spin[3 * 1] = 0.8
spin[3 * 2] = 2
spin[3 * 3] = 2.1
# print spin.reshape(-1, 3)
clib.compute_stt_field(spin,
field,
jx, jy, jz,
dx, dy, dz,
ngbs,
n
)
# print field
assert field[3 * 0] == 0.8
assert field[3 * 1] == 1
assert field[3 * 2] == 0.65
assert abs(field[3 * 3] - 0.1) < 1e-16
# PBCs --------------------------------------------------------------------
# Now we make PBCs in the x direction so the -x NN for
# the 0th spin is the 3rd spin, and the opposite way for the 3rd spin
ngbs = [[3, 1, -1, -1, -1, -1],
[0, 2, -1, -1, -1, -1],
[1, 3, -1, -1, -1, -1],
[2, 0, -1, -1, -1, -1],
]
ngbs = np.array(ngbs, dtype=np.int32)
clib.compute_stt_field(spin,
field,
jx, jy, jz,
dx, dy, dz,
ngbs,
n
)
# print field
assert field[3 * 0] == -0.65
assert field[3 * 3] == -1
def test_sst_field_1d():
"""
This is a direct test of the STT C library
We create a 1-D 4 spins system along the x direction with the
following components:
s_x s_y s_z spin
[[ 0. 4. 8. ] i = 0
[ 0.8 5. 9. ] i = 1
[ 2. 6. 10. ] i = 2
[ 2.1 7. 11. ]] i = 3
Most of the STT parameters are set to zero, so the field is reduced
to the calculation of:
H_STT = j \cdot \nabla S = ( jx * d S_x / dx + jy * d S_x / dy ,
jx * d S_y / dx + jy * d S_y / dy ,
0)
Therefore, the x component of the field for the 1th spin, for example,
would be:
H_STTx (i=1) = jx * d S_x (i=1) / dx + 0
= [ S_x (i=2) - S_x (i=0) ] / 2 * dx
= ( 2 - 0 ) / 2
= 1
since jx = dx = 1 and the derivatives along y are zero (1D spin chain)
The first test is the spin chain without PBC and the second
test is using PBCs, so the neighbours matrix changes
** Without PBC:
At the extremes, the derivative has a different discretisation
(using only the right or left neighbour and centered at the spin)
which, for the 0th spin, for instance, is
H_STTx (i=0) = jx * d S_x (i=0) / dx + 0
= [ S_x (i=1) - S_x (i=0) ] / dx
= ( 0.8 - 0 )
= 0.8
For further details, see: fidimag/atomistic/lib/stt.c
With PBC: The derivatives at the extremes are performed with the
periodic neighbour, e.g. for the 0th spin, the NN at -x is the 3th spin
"""
# The system is 1d with 4 spins.
nx, ny, nz = 4, 1, 1
n = nx * ny * nz
dx, dy = 1, 1
# NO PBCs -----------------------------------------------------------------
# We can manually construct the neighbours matrix,
# the neighbours order is [-x, +x, -y, +y, -z, +z]
# Since there are no PBCs, for the 0th spin, i.e. ngbs[0],
# the left (-x) NN is -1 and for the 3th spin, the +x NN is -1
# Also, since it is a 1D chain, there are no NNs in y or z
ngbs = [[-1, 1, -1, -1, -1, -1],
[0, 2, -1, -1, -1, -1],
[1, 3, -1, -1, -1, -1],
[2, -1, -1, -1, -1, -1],
]
ngbs = np.array(ngbs, dtype=np.int32)
nxyz = nx * ny * nz * 3
# The spin values to get the matrix:
#
# [[ 0. 4. 8. ]
# [ 1 5. 9. ]
# [ 2. 6. 10. ]
# [ 3 7. 11. ]]
spin = np.array([1.0 * i for i in range(nxyz)]).reshape((-1, 3), order='F').flatten()
field = np.zeros(nxyz)
jx = np.zeros(nxyz)
jy = np.zeros(nxyz)
jx[:] = 1
jy[:] = 2
# Modify the x components of the spins to get the values
# specified in the problem description
spin[3 * 1] = 0.8
spin[3 * 2] = 2
spin[3 * 3] = 2.1
# print spin.reshape(-1, 3)
clib.compute_stt_field(spin,
field,
jx, jy,
dx, dy,
ngbs,
n
)
# print field
assert field[3 * 0] == 0.8
assert field[3 * 1] == 1
assert field[3 * 2] == 0.65
assert abs(field[3 * 3] - 0.1) < 1e-16
# PBCs --------------------------------------------------------------------
# Now we make PBCs in the x direction so the -x NN for
# the 0th spin is the 3rd spin, and the opposite way for the 3rd spin
ngbs = [[3, 1, -1, -1, -1, -1],
[0, 2, -1, -1, -1, -1],
[1, 3, -1, -1, -1, -1],
[2, 0, -1, -1, -1, -1],
]
ngbs = np.array(ngbs, dtype=np.int32)
clib.compute_stt_field(spin,
field,
jx, jy,
dx, dy,
ngbs,
n
)
# print field
assert field[3 * 0] == -0.65
assert field[3 * 3] == -1
