Let us consider a system of periodic structure in space and use
a
,
b
, and
c
as the three independent period vectors, forming a right-handed triad, which are also the edge vectors of a cell of the system. Then any vector
r
in Cartesian coordiantes can be written as a linear combination of the period vectors
r
=
u
a
+
v
b
+
w
c
.
Our task is to calculate the scalar coefficients
u
,
v
, and
w
, assuming
r
,
a
,
b
, and
c
are known.
For this purpose, let us calculate the following cell surface area vector
σ
a
=
b
×
c
,
then
b
⋅
σ
a
=
0
,
c
⋅
σ
a
=
0
,
and the volume of the cell is
Ω
=
a
⋅
σ
a
.
If we do a vector inner (dot) product as follows
r
⋅
σ
a
=
u
a
⋅
σ
a
+
v
b
⋅
σ
a
+
w
c
⋅
σ
a
=
u
a
⋅
σ
a
=
u
Ω
,
then we get
u
=
1
Ω
r
⋅
σ
a
.
Similarly,
σ
b
=
c
×
a
,
c
⋅
σ
b
=
0
,
a
⋅
σ
b
=
0
,
b
⋅
σ
b
=
Ω
,
r
⋅
σ
b
=
u
a
⋅
σ
b
+
v
b
⋅
σ
b
+
w
c
⋅
σ
b
=
v
b
⋅
σ
b
=
v
Ω
,
we arrive at
v
=
1
Ω
r
⋅
σ
b
,
and
σ
c
=
a
×
b
,
a
⋅
σ
c
=
0
,
b
⋅
σ
c
=
0
,
c
⋅
σ
c
=
Ω
,
r
⋅
σ
c
=
u
a
⋅
σ
c
+
v
b
⋅
σ
c
+
w
c
⋅
σ
c
=
w
c
⋅
σ
c
=
w
Ω
,
w
=
1
Ω
r
⋅
σ
c
.
If there are many
r
s to be converted with respect to the same period vectors, to speed up, we can have
u
=
r
⋅
σ
a
′
,
v
=
r
⋅
σ
b
′
,
w
=
r
⋅
σ
c
′
,
where
σ
a
′
=
1
Ω
σ
a
,
σ
b
′
=
1
Ω
σ
b
,
σ
c
′
=
1
Ω
σ
c
.
In crystallography, the lengths (
a
,
b
,
c
) of and angles (
α
,
β
,
γ
) between the edge (period) vectors (
a
,
b
,
c
) of the parallelepiped unit cell are known. For simplicty, it is chosen so that edge vector
a
in the positive
x
-axis direction, edge vector
b
in the
x
−
y
plane with positive
y
-axis component, edge vector
c
with positive
z
-axis component in the Cartesian-system, as shown in the figure below.
Then the edge vectors can be written as
a
=
(
a
,
0
,
0
)
,
b
=
(
b
cos
(
γ
)
,
b
sin
(
γ
)
,
0
)
,
c
=
(
c
x
,
c
y
,
c
z
)
,
where all
a
,
b
,
c
,
sin
(
γ
)
,
c
z
are positive. Next, let us express all
c
components with known variables. This can be done with
c
⋅
a
=
a
c
cos
(
β
)
=
c
x
a
,
c
⋅
b
=
b
c
cos
(
α
)
=
c
x
b
cos
(
γ
)
+
c
y
b
sin
(
γ
)
,
c
⋅
c
=
c
2
=
c
x
2
+
c
y
2
+
c
z
2
.
Then
c
x
=
c
cos
(
β
)
,
c
y
=
c
cos
(
α
)
−
cos
(
γ
)
cos
(
β
)
sin
(
γ
)
,
c
z
2
=
c
2
−
c
x
2
−
c
y
2
=
c
2
{
1
−
cos
2
(
β
)
−
[
cos
(
α
)
−
cos
(
γ
)
cos
(
β
)
]
2
sin
2
(
γ
)
}
.
The last one continues
c
z
2
=
c
2
sin
2
(
γ
)
−
sin
2
(
γ
)
cos
2
(
β
)
−
[
cos
(
α
)
−
cos
(
γ
)
cos
(
β
)
]
2
sin
2
(
γ
)
=
c
2
sin
2
(
γ
)
{
sin
2
(
γ
)
−
sin
2
(
γ
)
cos
2
(
β
)
−
[
cos
(
α
)
−
cos
(
γ
)
cos
(
β
)
]
2
}
where
sin
2
(
γ
)
−
sin
2
(
γ
)
cos
2
(
β
)
−
[
cos
(
α
)
−
cos
(
γ
)
cos
(
β
)
]
2
=
sin
2
(
γ
)
−
sin
2
(
γ
)
cos
2
(
β
)
−
cos
2
(
α
)
−
cos
2
(
γ
)
cos
2
(
β
)
+
2
cos
(
α
)
cos
(
γ
)
cos
(
β
)
=
sin
2
(
γ
)
−
cos
2
(
α
)
−
sin
2
(
γ
)
cos
2
(
β
)
−
cos
2
(
γ
)
cos
2
(
β
)
+
2
cos
(
α
)
cos
(
β
)
cos
(
γ
)
=
sin
2
(
γ
)
−
cos
2
(
α
)
−
[
sin
2
(
γ
)
+
cos
2
(
γ
)
]
cos
2
(
β
)
+
2
cos
(
α
)
cos
(
β
)
cos
(
γ
)
=
sin
2
(
γ
)
−
cos
2
(
α
)
−
cos
2
(
β
)
+
2
cos
(
α
)
cos
(
β
)
cos
(
γ
)
=
1
−
cos
2
(
α
)
−
cos
2
(
β
)
−
cos
2
(
γ
)
+
2
cos
(
α
)
cos
(
β
)
cos
(
γ
)
.
Remembering
c
z
,
c
, and
sin
(
γ
)
being positive, one gets
c
z
=
c
sin
(
γ
)
1
−
cos
2
(
α
)
−
cos
2
(
β
)
−
cos
2
(
γ
)
+
2
cos
(
α
)
cos
(
β
)
cos
(
γ
)
.
Since the absolute value of the bottom surface area of the cell is
|
σ
c
|
=
a
b
sin
(
γ
)
,
the volume of the parallelepiped cell can also be expressed as
Ω
=
c
z
|
σ
c
|
=
a
b
c
1
−
cos
2
(
α
)
−
cos
2
(
β
)
−
cos
2
(
γ
)
+
2
cos
(
α
)
cos
(
β
)
cos
(
γ
)
.
Once the volume is calculated as above, one has
c
z
=
Ω
a
b
sin
(
γ
)
.
Now let us summarize the expression of the edge (period) vectors
a
=
(
a
x
,
a
y
,
a
z
)
=
(
a
,
0
,
0
)
,
b
=
(
b
x
,
b
y
,
b
z
)
=
(
b
cos
(
γ
)
,
b
sin
(
γ
)
,
0
)
,
c
=
(
c
x
,
c
y
,
c
z
)
=
(
c
cos
(
β
)
,
c
cos
(
α
)
−
cos
(
β
)
cos
(
γ
)
sin
(
γ
)
,
Ω
a
b
sin
(
γ
)
)
.
Let us calculate the following surface area vector of the cell first
σ
a
=
(
σ
a
,
x
,
σ
a
,
y
,
σ
a
,
z
)
=
b
×
c
,
where
σ
a
,
x
=
b
y
c
z
−
b
z
c
y
=
b
sin
(
γ
)
Ω
a
b
sin
(
γ
)
=
Ω
a
,
σ
a
,
y
=
b
z
c
x
−
b
x
c
z
=
−
b
cos
(
γ
)
Ω
a
b
sin
(
γ
)
=
−
Ω
cos
(
γ
)
a
sin
(
γ
)
,
σ
a
,
z
=
b
x
c
y
−
b
y
c
x
=
b
cos
(
γ
)
c
cos
(
α
)
−
cos
(
β
)
cos
(
γ
)
sin
(
γ
)
−
b
sin
(
γ
)
c
cos
(
β
)
=
b
c
{
cos
(
γ
)
cos
(
α
)
−
cos
(
β
)
cos
(
γ
)
sin
(
γ
)
−
sin
(
γ
)
cos
(
β
)
}
=
b
c
sin
(
γ
)
{
cos
(
γ
)
[
cos
(
α
)
−
cos
(
β
)
cos
(
γ
)
]
−
sin
2
(
γ
)
cos
(
β
)
}
=
b
c
sin
(
γ
)
{
cos
(
γ
)
cos
(
α
)
−
cos
(
β
)
cos
2
(
γ
)
−
sin
2
(
γ
)
cos
(
β
)
}
=
b
c
sin
(
γ
)
{
cos
(
α
)
cos
(
γ
)
−
cos
(
β
)
}
.
Another surface area vector of the cell
σ
b
=
(
σ
b
,
x
,
σ
b
,
y
,
σ
b
,
z
)
=
c
×
a
,
where
σ
b
,
x
=
c
y
a
z
−
c
z
a
y
=
0
,
σ
b
,
y
=
c
z
a
x
−
c
x
a
z
=
a
Ω
a
b
sin
(
γ
)
=
Ω
b
sin
(
γ
)
,
σ
b
,
z
=
c
x
a
y
−
c
y
a
x
=
−
a
c
cos
(
α
)
−
cos
(
β
)
cos
(
γ
)
sin
(
γ
)
=
a
c
sin
(
γ
)
{
cos
(
β
)
cos
(
γ
)
−
cos
(
α
)
}
.
The last surface area vector of the cell
σ
c
=
(
σ
c
,
x
,
σ
c
,
y
,
σ
c
,
z
)
=
a
×
b
,
where
σ
c
,
x
=
a
y
b
z
−
a
z
b
y
=
0
,
σ
c
,
y
=
a
z
b
x
−
a
x
b
z
=
0
,
σ
c
,
z
=
a
x
b
y
−
a
y
b
x
=
a
b
sin
(
γ
)
.
Summarize
σ
a
′
=
1
Ω
σ
a
=
(
1
a
,
−
cos
(
γ
)
a
sin
(
γ
)
,
b
c
cos
(
α
)
cos
(
γ
)
−
cos
(
β
)
Ω
sin
(
γ
)
)
,
σ
b
′
=
1
Ω
σ
b
=
(
0
,
1
b
sin
(
γ
)
,
a
c
cos
(
β
)
cos
(
γ
)
−
cos
(
α
)
Ω
sin
(
γ
)
)
,
σ
c
′
=
1
Ω
σ
c
=
(
0
,
0
,
a
b
sin
(
γ
)
Ω
)
.
As a result
[
u
v
w
]
=
[
1
a
−
cos
(
γ
)
a
sin
(
γ
)
b
c
cos
(
α
)
cos
(
γ
)
−
cos
(
β
)
Ω
sin
(
γ
)
0
1
b
sin
(
γ
)
a
c
cos
(
β
)
cos
(
γ
)
−
cos
(
α
)
Ω
sin
(
γ
)
0
0
a
b
sin
(
γ
)
Ω
]
[
x
y
z
]
where
(
a
,
b
,
c
)
are the components of the arbitrary vector
r
in Cartesian coordiantes.
To return the orthogonal coordinates in ångströms from fractional coordinates, one can employ the first equation on top and the expression of the edge (period) vectors
[
x
y
z
]
=
[
a
b
cos
(
γ
)
c
cos
(
β
)
0
b
sin
(
γ
)
c
cos
(
α
)
−
cos
(
β
)
cos
(
γ
)
sin
(
γ
)
0
0
Ω
a
b
sin
(
γ
)
]
[
u
v
w
]
.
For the special case of a monoclinic cell (a common case) where
α
=
γ
=
90
∘
and
β
>
90
∘
, this gives:
x
=
a
u
+
c
w
cos
(
β
)
,
y
=
b
v
,
z
=
Ω
a
b
w
=
c
w
sin
(
β
)
.
CPMD input
CIF
