VECTOR IDENTITIES
4
Notation:
f, g,
are scalars;
A
,
B
, etc., are vectors;
T
is a tensor;
I
is the unit
dyad.
(1)
A
·
B
×
C
=
A
×
B
·
C
=
B
·
C
×
A
=
B
×
C
·
A
=
C
·
A
×
B
=
C
×
A
·
B
(2)
A
×
(
B
×
C
) = (
C
×
B
)
×
A
= (
A
·
C
)
B
−
(
A
·
B
)
C
(3)
A
×
(
B
×
C
) +
B
×
(
C
×
A
) +
C
×
(
A
×
B
) = 0
(4) (
A
×
B
)
·
(
C
×
D
) = (
A
·
C
)(
B
·
D
)
−
(
A
·
D
)(
B
·
C
)
(5) (
A
×
B
)
×
(
C
×
D
) = (
A
×
B
·
D
)
C
−
(
A
×
B
·
C
)
D
(6)
∇
(
f g
) =
∇
(
gf
) =
f
∇
g
+
g
∇
f
(7)
∇ ·
(
f
A
) =
f
∇ ·
A
+
A
· ∇
f
(8)
∇ ×
(
f
A
) =
f
∇ ×
A
+
∇
f
×
A
(9)
∇ ·
(
A
×
B
) =
B
· ∇ ×
A
−
A
· ∇ ×
B
(10)
∇ ×
(
A
×
B
) =
A
(
∇ ·
B
)
−
B
(
∇ ·
A
) + (
B
· ∇
)
A
−
(
A
· ∇
)
B
(11)
A
×
(
∇ ×
B
) = (
∇
B
)
·
A
−
(
A
· ∇
)
B
(12)
∇
(
A
·
B
) =
A
×
(
∇ ×
B
) +
B
×
(
∇ ×
A
) + (
A
· ∇
)
B
+ (
B
· ∇
)
A
(13)
∇
2
f
=
∇ · ∇
f
(14)
∇
2
A
=
∇
(
∇ ·
A
)
− ∇ × ∇ ×
A
(15)
∇ × ∇
f
= 0
(16)
∇ · ∇ ×
A
= 0
If
e
1
, e
2
, e
3
are orthonormal unit vectors, a second-order tensor
T
can be
written in the dyadic form
(17)
T
=
P
i,j
T
ij
e
i
e
j
In cartesian coordinates the divergence of a tensor is a vector with components
(18) (
∇·
T
)
i
=
P
j
(
∂T
ji
/∂x
j
)
[This definition is required for consistency with Eq. (29)]. In general
(19)
∇ ·
(
AB
) = (
∇ ·
A
)
B
+ (
A
· ∇
)
B
(20)
∇ ·
(
f
T
) =
∇
f
·
T
+
f
∇·
T
4
Let
r
=
i
x
+
j
y
+
k
z
be the radius vector of magnitude
r
, from the origin to
the point
x, y, z
. Then
(21)
∇ ·
r
= 3
(22)
∇ ×
r
= 0
(23)
∇
r
=
r
/r
(24)
∇
(1
/r
) =
−
r
/r
3
(25)
∇ ·
(
r
/r
3
) = 4
πδ
(
r
)
(26)
∇
r
=
I
If
V
is a volume enclosed by a surface
S
and
d
S
=
n
dS
, where
n
is the unit
normal outward from
V,
(27)
Z
V
dV
∇
f
=
Z
S
d
S
f
(28)
Z
V
dV
∇ ·
A
=
Z
S
d
S
·
A
(29)
Z
V
dV
∇·
T
=
Z
S
d
S
·
T
(30)
Z
V
dV
∇ ×
A
=
Z
S
d
S
×
A
(31)
Z
V
dV
(
f
∇
2
g
−
g
∇
2
f
) =
Z
S
d
S
·
(
f
∇
g
−
g
∇
f
)
(32)
Z
V
dV
(
A
· ∇ × ∇ ×
B
−
B
· ∇ × ∇ ×
A
)
=
Z
S
d
S
·
(
B
× ∇ ×
A
−
A
× ∇ ×
B
)
If
S
is an open surface bounded by the contour
C
, of which the line element is
d
l
,
(33)
Z
S
d
S
× ∇
f
=
I
C
d
l
f
5
(34)
Z
S
d
S
· ∇ ×
A
=
I
C
d
l
·
A
(35)
Z
S
(
d
S
× ∇
)
×
A
=
I
C
d
l
×
A
(36)
Z
S
d
S
·
(
∇
f
× ∇
g
) =
I
C
f dg
=
−
I
C
gdf
6