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ACrashCourseinVectorCalculus
SCALARANDVECTORFIELDS
This introductory chapter is a review of mathematical concepts required for the
course. It is assumed that the reader is already familiar with elementary vector
analysis.
Physical quantities that we deal with in electromagnetism can be scalars or vec-
tors.
Ascalar is an entity which only has a magnitude. Examples of scalars are mass,
time, distance, electric charge, electric potential, energy, temperature etc.
Avector is characterized by both magnitude and direction. Examples of vectors
in physics are displacement, velocity, acceleration, force, electric field, magnetic
field etc.
Afield is a quantity which can be specified everywhere in space as a function of
position. The quantity that is specified may be a scalar or a vector. For instance,
we can specify the temperature at every point in a room. The room may, there-
fore, be said to be a region of “temperature field” which is a scalar field because
the temperature T(x;y;z) is a scalar function of the position. An example of a
scalar field in electromagnetism is the electric potential.
In a similar manner, a vector quantity which can be specified at every point in a
region of space is a vector field. For instance, every point on the earth may be
considered to be in the gravitational force field of the earth. we may specify the
field by the magnitude and the direction of acceleration due to gravity (i.e. force
per unit mass ) g(x;y;z) at every point in space. As another example consider
flowofwaterinapipe. Ateach point in the pipe, the water molecule has a veloc-
ity ~v(x;y;z). The water in the pipe may be said to be in a velocity field. There
~
are several examples of vector field in electromagnetism, e.g., the electric field E,
~
the magnetic flux density B etc.
1.1 Coordinate Systems :
Weare familiar with cartesian coordinate system. For systems exhibiting cylin-
drical or spherical symmetry, it is convenient to use respectively the cylindrical
and spherical coordinate systems.
1.1.1 Polar Coordinates :
In two dimensions one defines the polar coordinate (ρ;θ) of a point by defining
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ρ as the radial distance from the origin O and θ as the angle made by the radial
vector with a reference line (usually chosen to coincide with the x-axis of the
cartesian system). The radial unit vector ρˆ and the tangential (or angular) unit
ˆ
vector θ are taken respectively along the direction of increasing distance ρ and
that of increasing angle θ respectively, as shown in the figure.
y
^ ^
θ ρ
θ θ
y ρ
θ
x x
Relationship with the cartesian components are
x = ρcosθ
y = ρsinθ
so that the inverse relationships are
q 2 2
ρ = x +y
−1 y
θ = tan x
Bydefinition,thedistanceρ > 0. wewilltaketherangeofangles θ tobe0 ≤ θ <
2π (It is possible to define the range to be −π ≤ θ < +π). One has to be careful
in using the inverse tangent as the arc-tan function is defined in 0 ≤ θ < π. If y
is negative, one has to add π to the principal value of θ calculated by the arc - tan
function so that the point is in proper quadrant.
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Example1:
~
Avector A has cartesian components A and A . Write the vector in terms of its
x y
radial and tangential components.
Solution :
Let us write
~ ˆ
A=Aρˆ+Aθ
ρ θ
ˆ ˆ ˆ ˆ
Since ρˆand θ are orthonormal basis vectors ρˆ· ρˆ = θ · θ = 1 and ρˆ· θ = 0. Thus
~ ~ ˆ
A =A·ρˆ; A =A·θ
r θ
Notethat (see figure) ρˆmakes an angle θ with the x-axis (ˆı) and π=2 − θ with the
ˆ
y-axis (ˆ). Similarly, the unit vector θ makes π=2 + θ with the x-axis and θ with
the y-axis. Thus
~
A =A·ρˆ = A ˆı·ρˆ+A ˆ·ρˆ
ρ x y
= A cosθ+A cos(π=2−θ)
x y
= A cosθ+A sinθ
x y
~ ˆ ˆ ˆ
A =A·θ = A ˆıθ+A ·θ
θ x y
= A cos(π=2+θ)+A cosθ
x y
= −A sinθ+A cosθ
x y
TheJacobian:
When we transform from one coordinate system to another, the differential ele-
mentalso transform.
For instance, in 2 dimension the ele- ρ
ment of an area is dxdy but in polar dθ
coordinates the element is not dθdρ
but (ρdθ)dρ. This extra factor ρ is dρ
important when we wish to integrate
a function using a different coordi- θ
d ρ
nate system. θ
If f(x;y) is a function of x;y we may express the function in polar coordinates
and write it as g(ρ;θ). However, when we evaluate the integral R f(x;y)dxdy
in polar coordinates, the corresponding integral is R g(r;θ)ρdθdρ. In general, if
x = f(u;v) and y = g(u;v), then, in going from (x;y) to (u;v), the differential
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element dxdy →| J | dudv where J is given by the determinant
∂x ∂x
∂u ∂v
J = ∂y ∂y
∂u ∂v
The differentiations are partial, i.e., while differentiating ∂x=∂u = ∂f(u;v)=∂u,
the variable v is treated as constant. An useful fact is that the Jacobian of the
inverse transformation is 1=J because the detrminant of the inverse of a matrix is
equal to the inverse of the determinant of the original matrix.
Example2: ShowthattheJacobianofthetransformationfromcartesian topolar
coordinates is ρ.
Solution :
Wehave
∂x ∂x
∂ρ ∂θ
J = ∂y ∂y
∂ρ ∂θ
Using x = ρcosθ and y = ρsinθ, we have
cosθ −ρsinθ
J = = ρ
Exercise : sinθ ρcosθ
Show that the Jacobian of the inverse transformation from polar to cartesian is
√ 2 2
1=ρ = 1= x +y .
Example3:
findthearea of a circle of radius R.
Solution :
Taketheorigin to be at the centre of the circle and the plane of the circle to be the
ρ−θplane. Since the area element in the polar coordinates is ρdθdr, the area of
the circle is
Z 2π Z R " 2#R
dθ ρdρ = 2π ρ =πR2
0 0 2 0
- a very well known result !
Example4:
Z −(x2+y2)
Find the integral e dxdy where the region of integration is a unit circle
about the origin.
2
−ρ
Using polar coordinates the integrand becomes e . The range of i integration
for ρ is from 0 to 1 and for θ is from 0 to 2π. The integral is given by
Z 2π Z 1 2 Z 1 2
−ρ −ρ
I = dθ e ρdρ = 2π e ρdρ
0 0 0
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