34
1 Mathematical Physics
1.2 ∇(x y
2
+ xz) =
ˆ
i
∂
∂ x
+ ˆ j
∂
∂ y
+ ˆ
k
∂
∂z
(x y
2
+ xz)
= (y
2
+ z) ˆ
i + (2x y) ˆ j + x ˆ
k
= 2 ˆ
i − 2 ˆ j − ˆ
k, at(−1, 1, 1)
A unit vector normal to the surface is obtained by dividing the above vector
by its magnitude. Hence the unit vector is
(2 ˆ
i − 2 ˆ j − ˆ
k)[(2)
2
+ (−2)
2
+ (−1)
2 ]
−1/2
=
2
3
ˆ
i −
2
3
ˆ j −
1
3
ˆ
k
1.3 F ∝ 1/r
2
∇ . (r
−3 r) = r
−3
∇ . r + r . ∇r
−3
But ∇ . r =
ˆ
i
∂
∂ x
+ ˆ j
∂
∂ y
+ ˆ
k
∂
∂z
.
ˆ
i x + ˆ j y + ˆ
kz
=
∂ x
∂ x
+
∂ y
∂ y
+
∂z
∂z
= 3
r . ∇r
−3
= (x ˆ
i + y ˆ j + z ˆ
k) .
ˆ
i
∂
∂ x
+ ˆ j
∂
∂ y
+ ˆ
k
∂
∂z
(x
2
+ y
2
+ z
2 )
−3/2
= (x ˆ
i + y ˆ j + z ˆ
k) .
−
3
2
. (2x ˆ
i + 2y ˆ j + 2z ˆ
k)(x
2
+ y
2
+ z
2 )
−5/2
= −3(x
2
+ y
2
+ z
2 )(x
2
+ y
2
+ z
2 )
−
5
2 = −3r
−3
Thus ∇ . (r
−3 r) = 3r
−3
− 3r
−3
= 0
1.4 By problem ∇ × A = 0 and ∇ × B = 0, it follows that
B . (∇ × A) = 0
A. (∇ × B) = 0
Subtracting, B . (∇ × A) − A . (∇ × B) = 0
Now ∇ . ( A × B) = B . (∇ × A) − A . (∇ × B)
Therefore ∇ . ( A × B) = 0, so that ( A × B) is solenoidal.
1.5 (a) Curl {r f (r )} = ∇ × {r f (r )} = ∇ × {x f (r ) ˆ
i + y f (r ) ˆ j + z f (r ) ˆ
k}
=
ˆ
i
ˆ j
ˆ
k
∂
∂ x
∂
∂ y
∂
∂z
x f (r ) y f (r ) z f (r )
=
z
∂ f
∂ y
− y
∂ f
∂z
ˆ
i +
x
∂ f
∂z
− z
∂ f
∂ x
ˆ j +
y
∂ f
∂ x
− x
∂ f
∂ y
ˆ
k
But
∂ f
∂ x
=
∂ f
∂r
∂r
∂ x
=
∂ f
∂r
∂(x
2 +y
2 +z
2 )
1/2
∂ x
=
x f
r
Similarly
∂ f
∂ y
=
y f
r
and
∂ f
∂z
=
z f
r
, where prime means differentiation with
respect to r .
Thus,
curl{r f (r )} =
zy f
r
−
yz f
r
ˆ
i +
xz f
r
−
zx f
r
ˆ j +
yx f
r
−
x y f
r
ˆ
k = 0