PBC Lecture Notes in Physics – Vector Algebra & Vector Analysis/ A. Kar Gupta
1
Vector Algebra & Vector Analysis
(Combined Lectures, 1st Ed.)
Lecture Notes prepared by-
Dr. Abhijit Kar Gupta
Physics Department, Panskura Banamali College
Panskura R.S., East Midnapore, WB, India, Pin-code: 721152
e-mail: kg.abhi@gmail.com, abhijit_kargupta@rediffmail.com,
Lecture-1
Books to be consulted:
1. Vector Analysis
- Murray R. Spiegel (Schaum Series, McGraw Hill)
2. Mathematical Methods
- Merle C. Potter, Jack Goldberg (Prentice Hall of India)
3. Introduction to Mathematical Physics
- Charlie Harper (PHI)
4. Mathematical Methods for Physicists
- G. Arfken (Academic Pub., Prism Books Pvt. Ltd.)
5. Mathematical Physics
H. K. Dass (S. Chand & Company Ltd.)
We begin from the definition of ‘Vector’.
Vector: A vector is a quantity having both magnitude and direction.
Examples: v (velocity), F (force), m (magnetization) etc.
To express a vector A pictorially:
We have to know the coordinates of the starting point, the coordinates of the end point
with respect to a fixed coordinate system. The arrow (in the picture) indicates the
direction and the length of it (measured in the coordinate system) is the magnitude of the
vector.
Magnitude of a vector A is written as | A | (sometimes only A ).
Any vector along the direction of A but having unit magnitude is called unit vector.
PBC Lecture Notes in Physics – Vector Algebra & Vector Analysis/ A. Kar Gupta
2
A
Unit Vector:
uˆ =
Null Vector:
)
Ο , having magnitude zero.
| A|
Parallelogram law for Vector addition:
(Sometimes called Triangle law of addition)
B
A
C = A+ B
The above law can be extended to add any number of vectors:
Resultant Vector
The resultant vector (sum of all vectors) always starts from the starting point of the 1st
vector and ends at the end of the last vector. This is the end-to-end distance of a series of
‘walks’ (as directed by the vectors).
Applications: Any polygon of vectors, a model Polymer, Random Walk
To construct a Vector Algebra:
•
•
•
Arithmetic is mathematical operations (summation, subtraction, multiplication
and division) with Numbers.
Algebra is mathematical operations with symbols.
Therefore, Vector Algebra must be mathematical operations with Vectors.
Mathematical operations are to be defined. The ‘unit’ and ‘zero’ entities are
already defined. Some basic rules are to be defined to construct the Algebra.
PBC Lecture Notes in Physics – Vector Algebra & Vector Analysis/ A. Kar Gupta
Rules of Vector Algebra:
3
A
1. A + B = B + A (Cumulative law for addition)
B
B
A
2. m A = Am
(Cumulative law for multiplication)
C
3. A + ( B + C ) = ( A + B) + C (Associative law for addition)
4. m(n A) = (mn) A (Associative law for distribution)
A
B
5. (m + n) A = m A + n A (Distributive law for addition)
6. m( A + B ) = m A + m B (Distributive law for multiplication).
Lecture-2
Let us think of a rectangular Cartesian (right handed) coordinate system:
Z
)
The three unit vectors along the x-, y- and z-axes are i ,
)
)
j and k respectively.
)
k
)
j
Y
)
i
X
A vector A has components Ax , Ay and Az along x-, y- and z-axes respectively.
)
)
)
Therefore, one can write A = Ax i + Ay j + Az k .
2
2
2
Magnitude of A is | A | = Ax + Ay + Az .
PBC Lecture Notes in Physics – Vector Algebra & Vector Analysis/ A. Kar Gupta
•
Home Work: Prove the above by Pythagoras theorem.
Definition of Vector Field:
If in a region R(x,y,z) everywhere there is a vector A( x, y, z ) defined then we can say
that a vector field in R(x,y,z) is defined.
Suppose A = xiˆ + yˆj and we have | A |= x 2 + y 2 = a =constant.
Then we have a region of concentric circles for all values of a . The region under each
circular ring is a vector filed where a vector of magnitude of the radius of a vector is
defined.
Lecture-2
Suppose a vector A makes angles α , β and γ with respect to x-, y- and z-axes
respectively.
We can write:
Ay
A
Az
cos α = x , cos β =
, cos γ =
.
| A|
| A|
| A|
Y
A
γ
β
Z
α
X
Therefore, we find
2
cos α + cos β + cos γ =
2
2
2
2
Ax + Ay + Az
| A |2
Problem # 1 Equation of a circle:
2
=
| A |2
| A |2
= 1.
4
PBC Lecture Notes in Physics – Vector Algebra & Vector Analysis/ A. Kar Gupta
Z
r0
P( x0 , y 0 , z 0 )
Q ( x, y , z )
O
r
Y
X
Note: a ‘position vector’ is such that its tip denotes the position of a point and the
starting point is at the centre of the coordinate system.
Let the position vectors are
OP ≡ r0 = x0 iˆ + y 0 ˆj + z 0 kˆ
OQ ≡ r = xiˆ + yˆj + zkˆ
∴ The vector
PQ = r − r0
= ( xiˆ + yˆj + zkˆ) − ( x0 iˆ + y 0 ˆj + z 0 kˆ)
= ( x − x0 )iˆ + ( y − y 0 ) ˆj + ( z − z 0 )kˆ
The magnitude of the vector PQ is
| PQ |= ( x − x0 ) 2 + ( y − y 0 ) 2 + ( z − z 0 ) 2 .
Now if the magnitude of | PQ | is = a (constant) then we can write
( x − x0 ) 2 + ( y − y 0 ) 2 + ( z − z 0 ) 2 = a 2 .
This is an equation of a circle with the centre at ( x0 , y 0 , z 0 ) and radius ‘ a ’.
Problem # 2
Equation of a straight line passing through the points P ( x1 , y1 , z1 ) and
Q ( x 2 , y 2 , z 2 ).
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PBC Lecture Notes in Physics – Vector Algebra & Vector Analysis/ A. Kar Gupta
Z
P ( x1 , y1 , z1 )
R ( x, y , z )
r1
Q( x2 , y 2 , z 2 )
r
r2
O
Y
X
Let R ( x, y, z ) be any point on the straight line which joins P ( x1 , y1 , z1 ) and
Q( x 2 , y 2 , z 2 ).
Now, r1 = x1iˆ + y 1ˆj + z1 kˆ
r2 = x 2 iˆ + y 2 ˆj + z 2 kˆ
r3 = x3iˆ + y 3 ˆj + z 3 kˆ
Let R ( x, y, z ) be such a point on the line PQ that PR = m PQ , where m is a scalar
quantity and a fraction here.
∴ (r − r1 ) = m(r2 − r1 )
⇒ ( xiˆ + yˆj + zkˆ) − ( x iˆ + y ˆj + z kˆ) = m ( x iˆ + y ˆj + z kˆ) − ( x iˆ + y ˆj + z kˆ)
[
⇒ ( x − x )iˆ + ( y − y ) ˆj + ( z − z )kˆ = m[( x
1
1
1
1
1
2
1
2
2
The above three equations yield
x − x1
y − y1
z − z1
.
=
=
x 2 − x1 y 2 − y1 z 2 − z1
1
1
− x1 )iˆ + ( y 2 − y1 ) ˆj + ( z 2
Equating the respective components on both sides,
( x − x1 ) = m( x 2 − x1 )
( y − y1 ) = m( y 2 − y1 )
( z − z1 ) = m( z 2 − z1 ) .
2
]
− z )kˆ ]
1
1
6
PBC Lecture Notes in Physics – Vector Algebra & Vector Analysis/ A. Kar Gupta
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Vector in a Rotated Coordinate system:
Y′
Y
r
X′
X
φ
Let us consider the vector r with respect to the orthogonal Cartesian coordinate system;
the components are x and y .
Now, keeping r fixed we can rotate the coordinate system so that the new system rotates
by an angle φ with respect to the old one.
Resolving the vector r into the components in the new system we get,
x ′ = x cos φ + y sin φ
y ′ = − x sin φ + y cos φ
We can write the above in the following fashion:
 x ′   cos φ
  = 
 y ′   − sin φ
sin φ  x 
 
cos φ  y 
Or in an abstract form: r ′ = M r , where M is a Matrix having 2 rows and 2 columns.
M is a 2 × 2 Matrix. Similarly, the components of the vector form a 2 × 1 Matrix
(column matrix).
In general, any 2 × 2 matrix is defined by
 a11 a12 


a
a
22 
 21
The determinant of the above matrix is = a11 ⋅ a 22 − a 21 ⋅ a12 .
Therefore, the determinant of the matrix M is | M | = cos 2 φ + sin 2 φ = 1 .
PBC Lecture Notes in Physics – Vector Algebra & Vector Analysis/ A. Kar Gupta
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Here in this example, the matrix M is called a Rotational matrix. This is also called an
orthogonal matrix whose determinant is unity.
The magnitude of the vector r and that of r ′ (in the new coordinate system) are same.
This means that the magnitude of a vector is an invariant quantity under rotation.
PRODUCT OF TWO VECTORS
Dot Product (Scalar Product):
B
θ
A
The Scalar or Dot product is defined by A ⋅ B =| A || B | cos θ …………………….(1)
Example:
Work, W = F ⋅ ∆ r (Product of Force and displacement).
Also,
A ⋅ B = ( Ax iˆ + Ay ˆj + Az kˆ) ⋅ ( B x iˆ + B y ˆj + B z kˆ) = Ax B x + Ay B y + Az B z ,…….(2)
where the unit vectors iˆ , ĵ , k̂ satisfy the following relations
iˆ ⋅ iˆ = ˆj ⋅ ˆj = kˆ ⋅ kˆ = 1
iˆ ⋅ ˆj = iˆ ⋅ kˆ = ˆj ⋅ kˆ = 0 .
and
Where the three unit vectors are mutually perpendicular. The above relations are called
orthogonality conditions.
The above orthogonality relations can also be written in a compact form:
uˆ m ⋅ uˆ n = δ mn …………………………………………..(3)
where the indices m and n take three values, m = 1, 2, 3 and n = 1, 2, 3. The unit vectors
û m or û n means that
û1 = î, û 2 = ĵ , û 3 = k̂ .
The symbol in (3) δ mn is called the Kronecker delta defined by
PBC Lecture Notes in Physics – Vector Algebra & Vector Analysis/ A. Kar Gupta
δ mn = 1
=0
for m = n
for m ≠ n
From the definition (2), we could write
Ax = | A |, Ay = 0, Az = 0 and B x = | B | cos θ to arrive at (1).
Also from (2) we can write
2
2
2
A ⋅ A = Ax Ax + Ay Ay + Az Az = Ax + Ay + Az
∴ A ⋅ A = | A | 2 ≡ A 2 , where we write | A | = A , the magnitude of the vector A .
Example:
The kinetic energy, E =
1 2 1
mv = m(v ⋅ v) .
2
2
Application:
Proof of the Law of Cosines
C
φ
B
θ
A
C = A + B
C ⋅C = A + B ⋅ A + B
(
)(
)
= A ⋅ A + B ⋅ B + 2A ⋅ B
∴ C = A 2 + B 2 + 2 AB cos θ = A 2 + B 2 + 2 AB cos(π − φ )
∴ C 2 = A 2 + B 2 − 2 AB cos φ .
2
Cross Product (Vector Product):
C = A× B
The vector C will be perpendicular to the plane of A and B .
The magnitude
| C | = | A | | B | sin θ .
Also the vector C is such that A , B and C form a right-handed system.
With the above choice,
A × B = −B × A .
Relations among the three unit vectors in the right-handed Cartesian
coordinate system:
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PBC Lecture Notes in Physics – Vector Algebra & Vector Analysis/ A. Kar Gupta
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ˆj = kˆ × kˆ = 0
ˆj × kˆ = iˆ , kˆ × iˆ = ˆj
ˆj × iˆ = − kˆ , kˆ × ˆj = −iˆ , iˆ × kˆ = − ˆj .
iˆ × iˆ = ˆj ×
iˆ × ˆj = kˆ ,
Cross product of two vectors is also defined in the following and more useful way:
iˆ
C = C x iˆ + C y ˆj + C z kˆ = A × B = Ax
Bx
ˆj
Ay
By
kˆ
Az , a 3× 3 determinant.
Bz
= iˆ( Ay BZ − B y AZ ) − ˆj ( Ax B z − B x Az ) + kˆ( Ax B y − B x Ay ) .
In the above, the components ( C x , C y , C z ) of the product vector C are easily identified.
Home Work: If A × B = B × A , what is the relation between the two vectors (assume
that they are nonzero vectors)?
Lecture-3
Physical Examples of Vector Product:
Angular Momentum : L = r × p ,
r is position vector, p is linear momentum.
Relation between Linear and Angular velocity:
V =ω×r
Numerical EXAMPLES: product of two vectors
Given two vectors, A = iˆ − kˆ
B = 2iˆ −
iˆ ˆj
kˆ
0
A × B = 1 0 − 1 = iˆ
−1
2 −1 1
ˆj + kˆ
1 −1 ˆ 1 0
−1
− ˆj
+k
2 1
2 −1
1
= iˆ(0 − 1) − ˆj (1 + 2) + kˆ(−1 − 0) = − iˆ − 3 ˆj − kˆ
∴ A × B = (−1) 2 + (−3) 2 + (−1) 2 = 1 + 9 + 1 = 11 ………………………(1)
A ⋅ B = 1 ⋅ 2 + 0 ⋅ (−1) + (−1) ⋅ 1 = 2 − 1 = 1
Therefore, A ⋅ B = | A | . | B | cos θ = 1
PBC Lecture Notes in Physics – Vector Algebra & Vector Analysis/ A. Kar Gupta
We have,
| A | = 12 + 0 2 + (−1) 2 = 2 and
|B| =
∴ cos θ =
2 2 + (−1) 2 + 12 = 6
1
2. 6
=
1
12
⇒ sin θ = 1 − cos 2 θ =
11
12
.
Now we can also write the cross product of the two vectors,
11
∴ A × B = | A | . | B | sin θ = 2 . 6 .
= 11 ………………….(2)
12
The above exercise shows that the definitions of cross products of two vectors in two
different ways in (1) and in (2) yield the same result.
Home-Work Problems
1. Using the following vectors
P = iˆ cos θ + ˆj sin θ ,
Q = iˆ cos φ − ˆj sin φ ,
R = iˆ cos φ + ˆj sin φ
prove the following trigonometric identities:
sin(θ + φ ) = sin θ cos φ + cos θ sin φ
cos(θ + φ ) = cos θ cos φ − sin θ sin φ .
2. Verify that if you have two set of vectors A , B , C and A′ , B ′ , C ′ such that
A′ ⋅ A = B ′ ⋅ B = C ′ ⋅ C = 1 ,
A′ ⋅ B = A′ ⋅ C = B ′ ⋅ A = B ′ ⋅ C = C ′ ⋅ A = C ′ ⋅ B = 0
then
A′ =
B×C
A⋅ B×C
, B′ =
C× A
A⋅ B×C
, C′ =
A× B
A⋅ B×C
.
Note:
A′ , B ′ and C ′ are called Reciprocal set of vectors.
TRIPLE PRODUCT: product of three vectors
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PBC Lecture Notes in Physics – Vector Algebra & Vector Analysis/ A. Kar Gupta
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Product of three vectors can also be either a vector or a scalar. This can be
a combination of dot and cross products or only cross products.
Triple Scalar Product:
A ⋅ ( B × C ) = Ax ( B y C z − B z C y ) + Ay ( B z C x − B x C z ) + Az ( B x C y − B y C x )
= B x (C y Az − C z Ay ) + B y (C z Ax − C x Az ) + B z (C x Ay − C y Ax )
= B ⋅ (C × A)
Thus we can show A ⋅ ( B × C ) = B ⋅ (C × A) = C ⋅ ( A × B ) .
C
B
Cyclically…
A
Also, we can show
A ⋅ ( B × C ) = − A ⋅ (C × B ) = −C ⋅ ( B × A) = − B ⋅ ( A × C )
If C = A × B then
A ⋅ C = A ⋅ ( A × B)
= Ax ( Ay B z − Az B y ) + Ay ( Az B x − Ax B z ) + Az ( Ax B y − Ay B x ) = 0
Similarly, B ⋅ C = B ⋅ ( A × B ) = 0 .
Physical Example:
C
A
B
A ⋅ B × C = Volume of the Parallelepiped by the three vectors A , B and C (taken in
proper order)
PBC Lecture Notes in Physics – Vector Algebra & Vector Analysis/ A. Kar Gupta
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Now if A ⋅ B × C = 0 then we can say that the three vectors are coplanar (lie in the same
plane).
Triple Vector Product:
A × B × C = B( A ⋅ C ) − C ( A ⋅ B)
The above can be proved in a straight forward way, taking A = Ax iˆ + Ay ˆj + Az kˆ etc. and
iˆ
using the rule A × B = Ax
Bx
kˆ
Az and so on.
Bz
ˆj
Ay
By
Do this yourself and check the
formula.
•
Product of Four or more vectors can be obtained by using the rules of
Vector triple products.
Home-Work Problems
1. Prove the following:
(A × B )⋅ (C × D ) = (A ⋅ C )(B ⋅ D )− (A ⋅ D )(B ⋅ C )
and then show that
(A × B )⋅ (A × B ) = ( AB) − (A ⋅ B ) .
2
2
2. Show that
(A × B )× (C × D ) = (A ⋅ B × D )C − (A ⋅ B × C )D
Lecture-4
Vector Operator
GRADIENT:
PBC Lecture Notes in Physics – Vector Algebra & Vector Analysis/ A. Kar Gupta
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Suppose we have a scalar quantity φ (x) , then the differentiation of this with respect to
dφ
the only variable x is
.
dx
Now if φ ( x, y, z ) is scalar function whose value depends on the values of the coordinates
( x, y, z ) then the dependence of φ on each coordinate separately can be expressed
through partial differentiation:
∂φ  ∂φ ( x, y, z ) 
∂φ  ∂φ ( x, y, z ) 
∂φ
 ∂φ ( x, y, z ) 
≡
, 
≡
, 
≡
.




∂z
∂x
∂z
∂x 
∂y 
∂y
 y, z
 x, y
 x,z
We can now construct a new vector A in the following way:
∂φ ˆ ∂φ ∂φ
+ j
+
.
A ≡ ∇φ = iˆ
∂x
∂y ∂z
∂
∂
∂
Here ∇ is an ‘operator’ (a vector differential operator) which is ∇ = iˆ + ˆj + .
∂x
∂y ∂z
∂
∂ ∂
[ Note:
,
,
are also operators; they are ordinary differential operators.]
∂x ∂y ∂z
∇φ is called the GRADIENT of the scalar φ .
Applications:
Let us take the magnitude of the position vector φ = | r | =
x2 + y2 + z2 .
∂φ ˆ ∂φ ∂φ
+ j
+
,
∴ ∇φ = iˆ
∂x
∂y ∂z
∂φ
∂ 2
x
x
where
=
( x + y 2 + z 2 )1 / 2 = 2
= .
2
2 1/ 2
∂x
φ
∂x
(x + y + z )
y
z
∂φ
∂φ
and
Similarly,
=
= .
∂y
φ
∂z
φ
r
r
1
= n̂ .
∴ ∇φ = (iˆx + ˆjy + kˆz ) = =
φ
φ |r|
Here n̂ is a unit vector in the positive direction of the position vector. We have here
φ =| r |= r .
So we can write r = n̂ | r | = r ∇φ = φ ∇φ in this case.
Also note that ∇φ ⋅ r = nˆ ⋅ nˆ r = r . Product of the vector on its own unit vector =
magnitude of the vector itself.
PBC Lecture Notes in Physics – Vector Algebra & Vector Analysis/ A. Kar Gupta
In general, if φ = f (| r |) = f (r ) then
∴ ∇φ =
15
df x
∂φ
∂f (r ) ∂r
=
=
⋅
⋅
∂x
dr r
∂r ∂x
df
dφ
r df
= n̂
= nˆ
.
r dr
dr
dr
Interpretation of ∇φ :
The infinitesimal increment of the position vector d r = iˆdx + ˆjdy + kˆdz
∂φ
∂φ
∂φ
∴ ∇φ ⋅ d r =
dx +
dy +
dz = dφ .
∂x
∂y
∂z
So, the above gives an estimate of the change in the scalar function φ due to the change
(
)
in position r .
Now we can think of a surface where φ ( x, y, z ) = constant.
Example: φ ( x, y, z ) = x 2 + y 2 + z 2 = a 2 , Equation of a Sphere.
Z
φ = constant
∇φ
dr
r
r + dr
Y
X
If r be the position vector of a point on a surface (as shown in the figure) then the
infinitesimal change, d r is on the surface.
We can write
dφ = 0 = ∇φ ⋅ d r .
∴ ∇φ is perpendicular to the vector d r . That means ∇φ is a vector perpendicular to the
surface at the point ( x, y, z ) . If φ is some kind of potential (electrical or something), the
surface is called equipotential surface.
PBC Lecture Notes in Physics – Vector Algebra & Vector Analysis/ A. Kar Gupta
Also, along any direction n̂ , the component of ∇φ is ( ∇φ ⋅ n̂ ). So this is the rate of
change of φ at ( x, y, z ) in that direction.
The DIVERGENCE:
If we have a vector V = iˆV x + ˆjV y + kˆV z differentiable at each point ( x, y, z ) in some
region of space, we can define a scalar quantity called Divergence:
∂V x ∂V y ∂V z
 ∂
∂
∂
+
+
∇ ⋅ V =  iˆ + ˆj + kˆ  ⋅ iˆV x + ˆjV y + kˆV z =
∂x
∂y
∂z
∂y
∂z 
 ∂x
(
)
Examples:
(
)
•
 ∂
∂
∂
∂x ∂y ∂z
∇ ⋅ r =  iˆ + ˆj + kˆ  ⋅ iˆx + ˆjy + kˆz =
+
+
=3
∂y
∂z 
∂x ∂y ∂z
 ∂x
•
∇ ⋅ r f (r ) =
∂
[xf (r )] + ∂ [ yf (r )] + ∂ [zf (r )]
∂z
∂x
∂y
= 3 f (r ) +
x 2 df y 2 df z 2 df
df
+
+
= 3 f (r ) + r
r dr
r dr r dr
dr
[Q r
•
2
= x2 + y2 + z2
]
If we have a combination of a vector ( V ) and a scalar ( φ ) such that A = φV ,
then
∂
∂
∂
∇⋅ A =
(φV x ) + (φV y ) + (φV z )
∂z
∂x
∂y
∂V y ∂φ
∂V
∂V
∂φ
∂φ
Vx + φ x + V y + φ
+
Vz + φ z
∴ ∇ ⋅ (φV ) =
∂x
∂x ∂y
∂y
∂z
∂z
∂V y ∂V z 
 ∂φ
∂φ
∂φ   ∂V

=  V x + V y + V z  + φ  x +
+
∂y
∂z 
∂y
∂z   ∂x
 ∂x
( )
= ∇φ ⋅ V + φ ∇ ⋅ V
( Combination of ‘Divergence’ and ‘Gradient’)
If we would have a vector A = ∇φ then we can write
 ∂
∂
∂   ∂φ ˆ ∂φ ˆ ∂φ 

+ j
+k
∇ ⋅ A = ∇ ⋅ ∇φ =  iˆ + ˆj + kˆ  ⋅  iˆ
∂y
∂z   ∂x
∂y
∂z 
 ∂x
=
∂ 2φ ∂ 2φ ∂ 2φ
+ 2 + 2 = ∇ 2φ
2
∂x
∂y
∂z
(as it is denoted)
16
PBC Lecture Notes in Physics – Vector Algebra & Vector Analysis/ A. Kar Gupta
Interpretation:
The Divergence signifies the flow of ‘something’ out of a volume.
If ∇ ⋅ A = 0 (nothing comes out!), the vector A is called ‘Solenoidal’.
The CURL:
For the vector V we can define Curl as
ˆj
iˆ
kˆ
 ∂

∂
∂
∂
∂
∇ ×V =
= iˆ V z − V y  +
∂z 
∂x ∂y ∂z
 ∂y
Vx V y Vz
ˆj  ∂ V x − ∂ V z  + kˆ ∂ V y − ∂ V x 
∂x   ∂x
∂y 
 ∂z
Example:
If V = r f (r ) then we have
[
]
∇ × V = f (r )∇ × r + ∇ f (r ) × r
Note: ∇ × (φ A) = φ ∇ × A + ∇φ × A
For V = r we can check, ∇ × r = 0.
Also we derived earlier, ∇ f (r ) = nˆ
Therefore, ∇ × r f (r ) =
df (r )
.
dr
df
nˆ × r = 0.
dr
[ n̂ is the unit vector along r ]
Interpretation:
The curl signifies the circulation or rotation of ‘something’ around a loop.
If ∇ × V = 0, then the vector V is called Irrotational.
Home-Work Problem
Use the formula A × B × C = B ( A ⋅ C ) − C ( A ⋅ B ) and find out
∇ × (∇ × A) = − ∇ 2 A + ∇(∇ ⋅ A). This may also be checked from basic definitions.
Try that.
Lecture-5
Applications in Electromagnetic theory: Maxwell’s equations
17
PBC Lecture Notes in Physics – Vector Algebra & Vector Analysis/ A. Kar Gupta
18
∇ ⋅ E = 0 ………………………………(1)
∇ ⋅ H = 0 ………………………………(2)
∇× E = −
∇× H =
∂H
…………………………(3)
∂t
∂E
……………………..…….(4)
∂t
From (3) we can write
 ∂H 
.
∇× ∇× E = ∇×−
 ∂t 


(
)
Now, ∇ × (∇ × E ) = −∇ E + ∇(∇ ⋅ E ) = −∇ E
2
2
Also we can write
2
 ∂H 


 = − ∂ ∇ × H = − ∂  ∂E  = − ∂ E
∇×−
 ∂t 
∂t
∂t  ∂t 
∂t 2


(
∴
∇2 E =
∂2 E
∂t 2
)
[using equation (1)]
[using equation (4)]
……………………….(I)
Similarly, from (4) we get
∂2 H
∇ H=
∂t 2
2
………………………(II)
The relations (I) and (II) are called wave equations. Each component of E ( E x , E y , E z )
and that of H ( H x , H y , H z ) satisfy above equations.
Therefore, we can write for example,
∂ 2 Ex ∂ E y ∂ 2 Ez ∂ 2 Ex
+
+
=
and so on.
∂x 2
∂y 2
∂z 2
∂t 2
That means, in general if φ is any scalar which is any if the components ( E x , E y , E z ) or
2
∂ 2φ ∂ 2φ ∂ 2φ ∂ 2φ
+
+
=
.
∂x 2 ∂y 2 ∂z 2 ∂t 2
This is a wave equation.
( H x , H y , H z ) then
PBC Lecture Notes in Physics – Vector Algebra & Vector Analysis/ A. Kar Gupta
19
Additional problems:
(i) Express iˆ , ĵ , k̂ in terms of a gradient operator.
(ii) Express iˆ , ĵ , k̂ in terms of two gradient operators.
Hints: (i) Put φ = x in the expression of the gradient operator and see.
(ii) Use the relation iˆ × ˆj = kˆ and then use the result of (i).
Vector Integration
We can sum up several vectors and get a new resultant vector. Therefore, we may also
integrate a vector which is defined at every point ( x, y, z ) on a line or on a surface or in a
volume.
Line Integrals:
In a three dimensional Cartesian Coordinate system, the increment of length is
d r = iˆdx + ˆjdy + kˆdz .
Let us have a vector A( x, y, z ) = iˆA + ˆjA + kˆA defined at every point on a continuous
x
y
z
curve C .
P1
C
P2
We have the following type of line integral:
P2
∫ A ⋅ d r = ∫ A ⋅ d r = ∫ (A dx + A dy + A dz ) .
P1
C
C
x
y
z
This represents the integral of the tangential component of A along C from P1 and P2 .
If C is a closed curve (simple curve, no intersection by itself), the integral is denoted by
∫ A ⋅ d r = ∫ (A dx + A dy + A dz ) .
C
x
y
z
PBC Lecture Notes in Physics – Vector Algebra & Vector Analysis/ A. Kar Gupta
20
The above is called circulation of A about C .
Example:
The work done by a force along a path
W =
P2
P2
P1
P1
∫ F ⋅dr =
∫ F dx + F dy + F dz .
x
y
z
Suppose, F = −iˆy + ˆjx , P1 = (0,0) and P1 = (1,1) .
The work done along a path going from point P1 to the point P2 is then
W =
1,1
1
1
0,0
0
0
∫ (− ydx + xdy) = − ∫ ydx + ∫ xdy .
Consider the following path as shown:
Y
(1, 1)
1
1
0
0
∴ W = − ∫ 0.dx + ∫ 1.dy = 1.
(0, 0)
(1, 0)
X
If we choose another path as shown below:
Y
(0, 1)
(1, 1)
1
W = 0 − ∫ 1 / 2.dx +
0
(0, 0)
1
∫ 1.dy = 0.
1/ 2
X
Therefore, we see that this choice of force, the work done depends on the choice of path.
PBC Lecture Notes in Physics – Vector Algebra & Vector Analysis/ A. Kar Gupta
21
On the other hand, if W = ∫ F ⋅ d r is independent of the choice of path C joining any
C
two points P1 and P2 , the force F is called the Conservative force (For any arbitrary
vector A , this is called Conservative vector).
If we have A = ∇φ , where φ is a scalar quantity (called, scalar potential),
∂φ
∂φ
∂φ
A ⋅ d r = Ax dx + Ay dy + Az dz =
dx +
dy +
dz = dφ .
∂z
∂x
∂y
∴
P2
∫ A⋅dr
P2
=
P1
∫ dφ
= φ ( P1 ) − φ ( P2 ).
P1
Thus the value of the above integral depends only on the end points P1 and P2 and not on
the choice of path connecting them.
Now, as A = ∇φ , we have
∇× A = 0
So, this is also a condition for the vector A to be conservative.
•
There can be other two types of line integrals,
∫ φd r and ∫ A × d r .
C
C
Lecture-6
Conservative Force, Potential:
If F be a conservative force field, we may write F = −∇φ , φ is a scalar quantity, we
call it potential (The negative sign we choose deliberately for attractive forces).
d2r
.
dt 2
m d  d r d r  m d 2
dr
d 2r dr
⋅
=
∴ F⋅
=m 2 ⋅
=
v
dt
dt
2 dt  dt dt 
2 dt
dt
m
∴ F ⋅ d r = d (v 2 )
2
Therefore, Work done along any arbitrary path from a point A to the point B is
B
B
m
m 2B 1
1
2
2
W = ∫ F ⋅ d r = ∫ d (v 2 ) =
v = mvB − mv A .
A
2
2
2
2A
A
On the other hand,
We can write F = m a = m
( )
PBC Lecture Notes in Physics – Vector Algebra & Vector Analysis/ A. Kar Gupta
B
B
B
A
A
A
22
∫ F ⋅ d r = − ∫ ∇φ ⋅ d r = − ∫ dφ = φ ( A) − φ ( B) .
1
1
2
2
mvB − mv A .
2
2
Therefore, we can identify φ ( A) as the potential energy at A and φ ( B) as that at B . So,
∴ φ ( A) − φ ( B) =
the negative sign in F = − ∇φ is also explained in the sense of conservation of total
energy:
1
1
φ ( A) + mv A 2 = φ ( B) + mvB 2 = constant.
2
2
A Standard Problem:
If A = iˆAx + ˆjAy + kˆAz is a conservative vector,
(i) Find the scalar potential,
(ii) Find the integration of A along a curve from some point P1 ( x1 , y1 , z1 ) to
P2 ( x 2 , y 2 , z 2 ).
Soln.
(i) We can write A = ∇φ = iˆ
∂φ ˆ ∂φ ∂φ
+ j
+
∂x
∂y ∂z
= iˆAx + ˆjAy + kˆAz .
∂φ
= Ax ( x, y, z ) ……………………(1)
∂x
∂φ
= Ay ( x, y, z ) ……………………(2)
∂y
∂φ
= Az ( x, y, z ) ……………………(3)
∂z
Integrating (1), (2) and (3) respectively,
∴
φ = f1 ( x, y, z ) + c1 ( y, z )
φ = f 2 ( x, y , z ) + c 2 ( x, z )
φ = f 3 ( x , y , z ) + c 3 ( x, y )
Choose the integration constants c1 , c 2 and c3 such that φ turns out to be the same in all
the expressions.
Example:
A = (2 xy + z 3 )iˆ + x 2 ˆj + 3 xz 2 kˆ
Here we have,
∂φ
= 2 xy + z 3 ………………………………………(1)
∂x
∂φ
= x 2 ……………………………………………… (2)
∂y
PBC Lecture Notes in Physics – Vector Algebra & Vector Analysis/ A. Kar Gupta
23
∂φ
= 3xz 2 ……………………………………………..(3)
∂z
Integrating (1), (2) and (3) respectively,
φ = x 2 y + xz 3 + c1 ( y, z )
φ = x 2 y + c 2 ( x, z )
φ = xz 3 + c3 ( x, y )
So, we choose c1 ( y, z ) = 0 , c 2 ( x, z ) = xz 3 and c3 ( x, y ) = x 2 y such that
φ ( x, y, z ) = x 2 y + xz 3 .
One may also add some arbitrary constant (independent of x, y, z) with it.
Surface Integrals:
Consider the area element dS . This area element can be treated as a vector d S ≡ nˆ dS ,
where n̂ is a unit normal vector to indicate the positive direction.
Z
dS
Y
X
There are two conventions:
(i) If the surface is CLOSED, we take the outward normal as positive
(ii) If the surface is OPEN surface, the positive normal depends on the direction in which
the perimeter of the open surface is traversed (which is dependent on the handedness of
the coordinate system).
PBC Lecture Notes in Physics – Vector Algebra & Vector Analysis/ A. Kar Gupta
24
Z
Y
X
If a vector A is defined on the surface, the surface integral
∫ A⋅dS
can be interpreted as
a flow or flux through the given surface.
How to solve a surface integral:
Suppose, we have an arbitrary surface S . If we can project the surface S over any plane,
and the area of the projected surface is R , then the integration of a vector A over S is
equal to the integration of A over R .
Z
A
dS
D
Y
B
R
C
X
∫ A⋅dS
is the flow of “something” through the surface S . So, if we consider ABCD be
an imaginary pipe, then the flow through S must be equal to the flow through R or any
other cross-section in that.
Surface area elements in the
XY -plane is dxdy ,
YZ -plane is dydz ,
ZX -plane is dzdx .
PBC Lecture Notes in Physics – Vector Algebra & Vector Analysis/ A. Kar Gupta
25
In this example, to calculate the flow of A through R , we need to estimate the projected
area of dS on R , which is dxdy .
We can write,
k̂
dxdy = dS (nˆ ⋅ kˆ)
dS
n̂
dxdy
Therefore,
dxdy
∫∫ A ⋅ nˆdS = ∫∫ A ⋅ nˆ (nˆ ⋅ kˆ)
S
R
[ Note: Double integrals have been used as we have to integrate over two variables ]
Lecture-7
Example: Surface Integral
Evaluate
∫∫ A ⋅ nˆdS , where
S
A =
(
)
1 ˆ ˆ
xi + yj and S is the surface of the cylinder
4
x 2 + y 2 = 16 included in the first octant bounded between z = 0 and z = 5 .
Z
z=5
n̂
ĵ
R
xz-plane
Y
x=0
x=4
X
z=0
PBC Lecture Notes in Physics – Vector Algebra & Vector Analysis/ A. Kar Gupta
26
Area element on the xz -plane is dxdz . The entire area of the 1st quadrant of the cylinder
can be projected onto the xz -plane, where x ranges from x = 0 to x = 4 and z ranges
from z = 0 to z = 5 .
So, the area element dS of the given surface is related to corresponding area element
dxdz on xz -plane by
dxdz
dS =
.
nˆ ⋅ ˆj
Therefore,
dxdz
∫∫S A ⋅ nˆdS = ∫∫R A ⋅ nˆ nˆ ⋅ ˆj
( )
( )
The normal vector to the surface x 2 + y 2 = 16 at any point ( x, y ) is
∇( x 2 + y 2 ) = 2 xiˆ + 2 yˆj . [Consider this is a equipotential surface.]
The unit normal is
1
2 xiˆ + 2 yˆj
nˆ =
= xiˆ + yˆj .
2
2
4
(2 x) + (2 y )
(
)
1 2
(x + y 2 ) = 1
16
y
xiˆ + yˆj ˆ
nˆ ⋅ ˆj =
⋅j =
4
4
∴ A ⋅ nˆ =
Hence the surface integral is
4
5
4
1
∫∫R y dxdz = 4 x∫=0 16 − x 2 dx z∫=0dz
4
= 20 ∫
0
1
16 − x 2
dx
= 20 π
= 10 π
Note:
Substitute x = 4 cos θ , the integral reduces to
0
− ∫ dθ =
π
π
2
2
∴ The Integral of the vector A over the entire surface of the cylinder is
∫∫ A ⋅ nˆdS
= 4 × 10π = 40π .
S
However, the area of the surface of the cylinder is = 2πrl = 2π × 4 × 5 = 40π .
Therefore, we can say that the flow or flux of a unit normal vector (note here, A ≡ nˆ )
through the entire surface is just the actual area of the surface.
There are other different kinds of surface integrals which we encounter very
commonly: ∫ φd S , ∫ A × d S .
PBC Lecture Notes in Physics – Vector Algebra & Vector Analysis/ A. Kar Gupta
27
Homework Problems:
1. Evaluate
∫∫ A ⋅ nˆdS
over the entire surface S of the region bounded by the cylinder
S
x + z = 9 , x = 0 , y = 0 , z = 0 and z = 8 , if A = 6 ziˆ + (2 x + y ) ˆj − xkˆ .
2
2
2. Evaluate
∫∫
x 2 + y 2 dxdy over the region R in the xy -plane bounded by
R
x + y = 36 .
2
2
Volume Integrals:
We encounter the following Volume integrals:
∫ ∇ ⋅ Adτ , ∫ ∇φdτ , ∫ ∇ × Adτ ,
where dτ is the differential volume.
Let us consider an Integral of first kind.
∇ ⋅ A is the divergence of vector A (or flow out of some volume).
∴
∫ ∇ ⋅ Adτ
is the divergence or flow out of some volume ∫ dτ .
If d S is the vector area element of this volume then the flow through the entire area of
that volume is also ∫ A ⋅ d S , If we think it physically, the flow of something out of some
volume has to come out through the entire surface enclosing that volume.
Therefore, ∫ ∇ ⋅ Adτ = ∫ A ⋅ d S .
Thus we correspond between volume and surface integrals. This is Gauss’ Divergence
formula.
This is not proved here (see any text book).
Calculation of ordinary Surface and Volume Integrals
Area of a Circle: x 2 + y 2 = a 2
a
 a2 − x2 
π


A = ∫∫ dxdy = ∫
dy dx = 2 ∫ a 2 − x 2 dx = 2. a 2 . = πa 2 .
∫
2
x =− a  y =− a 2 − x 2 
−a


a
Volume of a Sphere: x 2 + y 2 + z 2 = a 2
 a2 − x2  a2 − x2 − y2  

V = ∫  ∫
dz dy dx =
∫
 


x =− a y =− a 2 − x 2 z =− a 2 − x 2 − y 2
 


a
[ Substitute y =
 a2 − x2

2
2
2

dx
−
−
2
a
x
y
dy
∫ ∫2 2

−a 
 − a −x
a
a 2 − x 2 sin θ , the integral in the third bracket becomes
π /2
PBC Lecture Notes in Physics – Vector Algebra & Vector Analysis/ A. Kar Gupta
28
Home Work Problem:
# Let us take a vector A =
1 ˆ ˆ
( xi + yj + zkˆ) to evaluate
a
∫∫ A ⋅ nˆdS
over a sphere
S
x 2 + y 2 + z 2 = a 2 above the xy-plane.
Hints:
• Solve the problem either directly or by applying Gauss’ Divergence theorem.
• To solve the integral directly one has to calculate the unit normal vector to the surface
S (the hemisphere above the XY-plane).
Unit normal, n̂ =
∇φ
| ∇φ |
.
Here,
∇φ = ∇( x 2 + y 2 + z 2 ) = 2 xiˆ + 2 yˆj + 2 zkˆ .
1 ˆ ˆ
2 xiˆ + 2 yˆj + 2 zkˆ
∴ n̂ =
xi + yj + zkˆ
=
2
2
2
a
(2 x) + (2 y ) + (2 z )
z
∴ A ⋅ nˆ = 1 , nˆ ⋅ kˆ =
a
The projection of the infinitesimal area dS onto the xy-plane is dxdy . k̂ is the unit normal
(
vector to the xy-plane.
)
PBC Lecture Notes in Physics – Vector Algebra & Vector Analysis/ A. Kar Gupta
Therefore,
∫∫ A ⋅ nˆdS = ∫∫ (A ⋅ nˆ ) (nˆ ⋅ kˆ )
dxdy
[ as shown earlier in Lecture-6]
R
S
1
= a ∫∫ dxdy
z
a
=a
∫
a2 − x2
∫
1
x =− a y =− a 2 − x 2
a2 − x2 − y2
dydx .
Lecture-8
GREEN’S FORMULA:
Let C be a simple closed curve in the XY-plane. R is the region bounded by the closed
curve. Suppose, M ( x, y ) and N ( x, y ) are two differentiable functions in the region R.
We can then evaluate the following integral,
∂M ( x, y )
I 1 = ∫∫
dxdy =
∂y
R
 y2 ( x ) ∂M 
∫  y = y∫ ( x ) ∂y dy dx
x=a 
 1

b
Here y = y1 ( x) represents the lower portion APB of the closed curve C and y = y 2 ( x)
represents the upper portion AQB of the curve.
Y
y 2 ( x)
q
A
Q
C
B
R
p
P y1 ( x)
X
a
b
∴ I1 =
∫ [M ( x, y)]
x=a
y2 ( x )
b
dx =
y = y1 ( x )
b
∫ [M ( x, y
a
2
) − M ( x, y1 )]dx
29
PBC Lecture Notes in Physics – Vector Algebra & Vector Analysis/ A. Kar Gupta
b
a
a
b
= − ∫ M ( x, y1 )dx − ∫ M ( x, y 2 )dx = − ∫ M ( x, y )dx
∴
∫ M ( x, y)dx
C
C
= − ∫∫
R
∂M ( x, y )
dxdy …………………………….(1)
∂y
Similarly,
∂N ( x, y )
I 2 = ∫∫
dxdy =
∂x
R
 x2 ( y ) ∂N 
∫  x= x∫( y ) ∂x dxdy
y= p 
 1

q
Here x = x1 ( y ) represents the left portion PAQ of the closed curve C and x = x 2 ( y )
represents the right portion PBQ of the curve.
x2 ( y )
q
∴ I2 =
∫ [N ( x, y)]
y= p
∫ N (x
p
∴
dy =
p
2
∫ [N ( x
2
, y )dy + ∫ N ( x1 , y )dy =
q
∫ N ( x, y)dx = ∫∫
C
R
, y ) − N ( x1 , y )]dy
p
x = x1 ( y )
q
=
q
∫ N ( x, y)dy
C
∂N ( x, y )
dxdy …………………………….(2)
∂x
Adding (1) and (2),
 ∂N ∂M 
∫C Mdx + Ndy = ∫∫R  ∂x − ∂y dxdy …………………………..(3)
This is Green’s Formula in the xy-plane.
Green’s Formula (3) can be written in the vector notation:
We can write
M ( x, y )dx + N ( x, y )dy = M ( x, y )iˆ + N ( x, y ) ˆj ⋅ (dxiˆ + dyˆj ) = A ⋅ d r
(
)
Here A = M ( x, y )iˆ + N ( x, y ) ˆj
d r = iˆdx + ˆjdy
iˆ
∂
Then ∇ × A =
∂x
M
ˆj
∂
∂y
N
kˆ
∂
∂z
0
on a plane
30
PBC Lecture Notes in Physics – Vector Algebra & Vector Analysis/ A. Kar Gupta
∂N ˆ ∂M ˆ ∂N ∂M
+ j
+ k 
−
∂z
∂z
∂y
 ∂x
 ∂N ∂M 
.
∴ ∇ × A ⋅ kˆ = 
−
∂y 
 ∂x
Therefore, formula (3) can be written as
= − iˆ
(
)
31



∫ A ⋅ d r = ∫∫ (∇ × A)⋅ kˆdR ,……………………………………(4)
C
R
where dR = dxdy .
The above formula (4) when generalized for an arbitrary surface S bounded by a simple
curve C (both sides of the surface are open) can be written as
∫ A ⋅ d r = ∫∫ (∇ × A)⋅ nˆdS
C
……….This is Stoke’s Formula.
S
-------------------------------------------------------------------------------------------------------Problems:
(To solve by using Green’s Formula and Stokes Formula)
#1. Use Green’s Theorem to evaluate ∫ ( x 2 + xy )dx + ( x 2 + y 2 )dy where C is the Square
C
formed by the lines y = ±1 , x = ±1 .
Solution
∫ (Mdx + Ndy)
C
=
 ∂N
∫∫  ∂x
−
∂M
∂y
∂
∂ 2
= ∫ ∫  x2 + y2 −
x + xy
∂x
∂y
−1 −1 
1 1
(
)
(

dxdy


dxdy =

)
#2. Using Stoke’s Theorem to evaluate
1 1
∫ ∫ xdxdy
−1 −1
1
1
−1
−1
= ∫ xdx ∫ dy = 0.
∫ [(2 x − y)dx − yz
2
]
dy − y 2 zdz , where C is a
C
circle x 2 + y 2 = 1 corresponding to a sphere of unit radius.
Solution
Here we identify the vector, A = (2 x − y )iˆ − yz 2 ˆj − y 2 zkˆ and ∇ × A = kˆ .
dxdy
∴ ∫ A ⋅ d r = ∫∫ kˆ ⋅ nˆ dS = ∫∫ dS = ∫∫
= ∫∫ dxdy = π .
(nˆ ⋅ kˆ)
C
PBC Lecture Notes in Physics – Vector Algebra & Vector Analysis/ A. Kar Gupta
#3. Use Stoke’s Theorem to evaluate
∫ [( x + 2 y)dx + ( x − z )dy + ( y − z )dz ] where C is the
C
boundary of the triangle with vertices (2,0,0), (0,3,0) and (0,0,6) oriented in the anticlockwise direction.
Solution
Z
C
(0,0,6)
B
O
A
Y
(0,3,0)
(2,0,0)
X
∫ A ⋅ d r = ∫ [( x + 2 y)dx + ( x − z )dy + ( y − z )dz ]
C
C
[
][
= ∫ ( x + 2 y )iˆ + ( x − z ) ˆj + ( y − z )kˆ ⋅ iˆdx + ˆjdy + kˆdz
]
C
Here, A = ( x + 2 y )iˆ + 9 x − z ) ˆj + ( y − z )kˆ
ˆj
iˆ
kˆ
∂
∂
∂
∴∇× A =
= 2iˆ − kˆ .
∂x
∂y
∂z
x + 2y x − z y − z
x y z
+ + = 1.
2 3 6
∇φ
x y z
, where φ = + + .
The normal vector n̂ =
2 3 6
∇φ
The equation of the plane ABC (a triangle) is
1 ˆ
(3i + 2 ˆj + kˆ) , ∴ n̂ =
6
5
∴ (∇ × A) ⋅ nˆ =
14
∇φ =
1
14
32
(3iˆ + 2 ˆj + kˆ)
PBC Lecture Notes in Physics – Vector Algebra & Vector Analysis/ A. Kar Gupta
Now applying Stoke’s theorem:
∫ A ⋅ d r = ∫∫ (∇ × A)⋅ nˆdS
C
S
=
R
k⋅
5
∫∫ dS
14
S
=
dxdy
5
∫∫
14 (nˆ ⋅ kˆ)
R
dxdy
5
∫∫ ˆ
14
=
1
(3iˆ + 2 ˆj + kˆ)
14
= 5∫∫ dxdy = 5 × projection of the area ABC on the XY-plane
R
1
= 5 × Area of triangle OAB = 5 × × (2 × 3) =15.
2
33