Curvilinear coordinates
PH 108: Electricity & Magnetism
Rwitaban Goswami
2020
Cartesian Coordinates: recap
What did we do in Cartesian coordinates?
We just expressed the point location in terms ofx (distance fromyz plane),y (distance from xz plane), z (distance fromxy plane).
How did we get the basis unit vectors ˆı, ˆ, ˆk? We can get them in two ways
ˆıcan be written as the direction in which the point moves if we change x bydx, keepingy &z constant
ˆıcan also be written as the normal vector to the surface defined by keepingx constant, and changingy &z
The first one is called the covariant basis, and is basically found by the tangential direction in the corresponding coordinate path
The second one is called the contravariant basis, and is basically found by the normal to the corresponding surface created by other two coordinates
Cartesian Coordinates: recap
What did we do in Cartesian coordinates?
We just expressed the point location in terms ofx (distance fromyz plane),y (distance from xz plane), z (distance fromxy plane).
How did we get the basis unit vectors ˆı, ˆ, ˆk?
We can get them in two ways
ˆıcan be written as the direction in which the point moves if we change x bydx, keepingy &z constant
ˆıcan also be written as the normal vector to the surface defined by keepingx constant, and changingy &z
The first one is called the covariant basis, and is basically found by the tangential direction in the corresponding coordinate path
The second one is called the contravariant basis, and is basically found by the normal to the corresponding surface created by other two coordinates
Cartesian Coordinates: recap
What did we do in Cartesian coordinates?
We just expressed the point location in terms ofx (distance fromyz plane),y (distance from xz plane), z (distance fromxy plane).
How did we get the basis unit vectors ˆı, ˆ, ˆk?
We can get them in two ways
ˆıcan be written as the direction in which the point moves if we change x bydx, keepingy &z constant
ˆıcan also be written as the normal vector to the surface defined by keepingx constant, and changingy &z
The first one is called the covariant basis, and is basically found by the tangential direction in the corresponding coordinate path
The second one is called the contravariant basis, and is basically found by the normal to the corresponding surface created by other two coordinates
Cartesian Coordinates: recap
What did we do in Cartesian coordinates?
We just expressed the point location in terms ofx (distance fromyz plane),y (distance from xz plane), z (distance fromxy plane).
How did we get the basis unit vectors ˆı, ˆ, ˆk?
We can get them in two ways
ˆıcan be written as the direction in which the point moves if we change x bydx, keepingy & z constant
ˆıcan also be written as the normal vector to the surface defined by keepingx constant, and changingy &z
The first one is called the covariant basis, and is basically found by the tangential direction in the corresponding coordinate path
The second one is called the contravariant basis, and is basically found by the normal to the corresponding surface created by other two coordinates
Cartesian Coordinates: recap
What did we do in Cartesian coordinates?
We just expressed the point location in terms ofx (distance fromyz plane),y (distance from xz plane), z (distance fromxy plane).
How did we get the basis unit vectors ˆı, ˆ, ˆk?
We can get them in two ways
ˆıcan be written as the direction in which the point moves if we change x bydx, keepingy & z constant
ˆıcan also be written as the normal vector to the surface defined by keepingx constant, and changingy &z
The first one is called the covariant basis, and is basically found by the tangential direction in the corresponding coordinate path
The second one is called the contravariant basis, and is basically found by the normal to the corresponding surface created by other two coordinates
Cartesian Coordinates: recap
What did we do in Cartesian coordinates?
We just expressed the point location in terms ofx (distance fromyz plane),y (distance from xz plane), z (distance fromxy plane).
How did we get the basis unit vectors ˆı, ˆ, ˆk?
We can get them in two ways
ˆıcan be written as the direction in which the point moves if we change x bydx, keepingy & z constant
ˆıcan also be written as the normal vector to the surface defined by keepingx constant, and changingy &z
The first one is called the covariant basis, and is basically found by the tangential direction in the corresponding coordinate path
The second one is called the contravariant basis, and is basically found by the normal to the corresponding surface created by other two coordinates
Cartesian Coordinates: recap
What did we do in Cartesian coordinates?
We just expressed the point location in terms ofx (distance fromyz plane),y (distance from xz plane), z (distance fromxy plane).
How did we get the basis unit vectors ˆı, ˆ, ˆk?
We can get them in two ways
ˆıcan be written as the direction in which the point moves if we change x bydx, keepingy & z constant
ˆıcan also be written as the normal vector to the surface defined by keepingx constant, and changingy &z
The first one is called the covariant basis, and is basically found by the tangential direction in the corresponding coordinate path
The second one is called the contravariant basis, and is basically found by the normal to the corresponding surface created by other two coordinates
Curvilinear Coordinates: Spherical Polar
x y
z
(r, θ, φ) φ
θ rcosθ rsinθcosφ rsinθsinφ
In spherical coordinates, the point is expressed in terms of r,θ,φ, which express the quantities which can be seen in the diagram
ˆ r
We are going to define ˆr, ˆθ, ˆφfor such a system. Remember we just said there are 2 ways we can do that. We are going to use the first one, i.e the covariant basis. It turns out that in the case of spherical polar, both yield the same basis vectors. (Note: This is because the basis is orthogonal, which may notbe true in general)
So for ˆr we need how the point location changes whenr is changed bydr
~
r =xˆı+yˆ+zkˆ
∂~r
∂r = ∂x
∂rˆı+∂y
∂r ˆ+∂z
∂r kˆ
∂~r
∂r = sinθcosφˆı+ sinθsinφˆ+ cosθkˆ So ˆr = sinθcosφˆı+ sinθsinφˆ+ cosθkˆ
ˆ r
We are going to define ˆr, ˆθ, ˆφfor such a system. Remember we just said there are 2 ways we can do that. We are going to use the first one, i.e the covariant basis. It turns out that in the case of spherical polar, both yield the same basis vectors. (Note: This is because the basis is orthogonal, which may notbe true in general)
So for ˆr we need how the point location changes whenr is changed bydr
~
r =xˆı+yˆ+zkˆ
∂~r
∂r = ∂x
∂rˆı+∂y
∂r ˆ+∂z
∂r kˆ
∂~r
∂r = sinθcosφˆı+ sinθsinφˆ+ cosθkˆ So ˆr = sinθcosφˆı+ sinθsinφˆ+ cosθkˆ
ˆ r
We are going to define ˆr, ˆθ, ˆφfor such a system. Remember we just said there are 2 ways we can do that. We are going to use the first one, i.e the covariant basis. It turns out that in the case of spherical polar, both yield the same basis vectors. (Note: This is because the basis is orthogonal, which may notbe true in general)
So for ˆr we need how the point location changes whenr is changed bydr
~
r =xˆı+yˆ+zkˆ
∂~r
∂r = ∂x
∂rˆı+∂y
∂r ˆ+∂z
∂r kˆ
∂~r
∂r = sinθcosφˆı+ sinθsinφˆ+ cosθkˆ So ˆr = sinθcosφˆı+ sinθsinφˆ+ cosθkˆ
ˆ r
We are going to define ˆr, ˆθ, ˆφfor such a system. Remember we just said there are 2 ways we can do that. We are going to use the first one, i.e the covariant basis. It turns out that in the case of spherical polar, both yield the same basis vectors. (Note: This is because the basis is orthogonal, which may notbe true in general)
So for ˆr we need how the point location changes whenr is changed bydr
~
r =xˆı+yˆ+zkˆ
∂~r
∂r = ∂x
∂rˆı+∂y
∂r ˆ+∂z
∂r kˆ
∂~r
∂r = sinθcosφˆı+ sinθsinφˆ+ cosθkˆ
So ˆr = sinθcosφˆı+ sinθsinφˆ+ cosθkˆ
ˆ r
We are going to define ˆr, ˆθ, ˆφfor such a system. Remember we just said there are 2 ways we can do that. We are going to use the first one, i.e the covariant basis. It turns out that in the case of spherical polar, both yield the same basis vectors. (Note: This is because the basis is orthogonal, which may notbe true in general)
So for ˆr we need how the point location changes whenr is changed bydr
~
r =xˆı+yˆ+zkˆ
∂~r
∂r = ∂x
∂rˆı+∂y
∂r ˆ+∂z
∂r kˆ
∂~r
∂r = sinθcosφˆı+ sinθsinφˆ+ cosθkˆ So ˆr = sinθcosφˆı+ sinθsinφˆ+ cosθkˆ
θ ˆ
~
r =xˆı+yˆ+zkˆ
∂~r
∂θ = ∂x
∂θˆı+∂y
∂θ ˆ+∂θ
∂θkˆ
∂~r
∂θ =rcosθcosφˆı+rcosθsinφˆ−rsinθkˆ
But
∂~r
∂θ
=r6= 1
So we can say that changing θ bydθmoves the point rdθ in the direction of ˆθ
We call this the scale factorhθ =r. Notice thathr = 1 So ˆθ= h1
θ
∂~r
∂θ = cosθcosφˆı+ cosθsinφˆ−sinθkˆ
θ ˆ
~
r =xˆı+yˆ+zkˆ
∂~r
∂θ = ∂x
∂θˆı+∂y
∂θ ˆ+∂θ
∂θkˆ
∂~r
∂θ =rcosθcosφˆı+rcosθsinφˆ−rsinθkˆ
But
∂~r
∂θ
=r6= 1
So we can say that changing θ bydθmoves the point rdθ in the direction of ˆθ
We call this the scale factorhθ =r. Notice thathr = 1 So ˆθ= h1
θ
∂~r
∂θ = cosθcosφˆı+ cosθsinφˆ−sinθkˆ
θ ˆ
~
r =xˆı+yˆ+zkˆ
∂~r
∂θ = ∂x
∂θˆı+∂y
∂θ ˆ+∂θ
∂θkˆ
∂~r
∂θ =rcosθcosφˆı+rcosθsinφˆ−rsinθkˆ
But
∂~r
∂θ
=r6= 1
So we can say that changing θ bydθmoves the point rdθ in the direction of ˆθ
We call this the scale factorhθ =r. Notice thathr = 1 So ˆθ= h1
θ
∂~r
∂θ = cosθcosφˆı+ cosθsinφˆ−sinθkˆ
θ ˆ
~
r =xˆı+yˆ+zkˆ
∂~r
∂θ = ∂x
∂θˆı+∂y
∂θ ˆ+∂θ
∂θkˆ
∂~r
∂θ =rcosθcosφˆı+rcosθsinφˆ−rsinθkˆ
But
∂~r
∂θ
=r6= 1
So we can say that changing θ bydθmoves the point rdθ in the direction of ˆθ
We call this the scale factorhθ =r. Notice thathr = 1 So ˆθ= h1
θ
∂~r
∂θ = cosθcosφˆı+ cosθsinφˆ−sinθkˆ
θ ˆ
~
r =xˆı+yˆ+zkˆ
∂~r
∂θ = ∂x
∂θˆı+∂y
∂θ ˆ+∂θ
∂θkˆ
∂~r
∂θ =rcosθcosφˆı+rcosθsinφˆ−rsinθkˆ
But
∂~r
∂θ
=r6= 1
So we can say that changing θ bydθmoves the point rdθ in the direction of ˆθ
We call this the scale factorhθ =r. Notice thathr = 1 So ˆθ= h1
θ
∂~r
∂θ = cosθcosφˆı+ cosθsinφˆ−sinθkˆ
θ ˆ
~
r =xˆı+yˆ+zkˆ
∂~r
∂θ = ∂x
∂θˆı+∂y
∂θ ˆ+∂θ
∂θkˆ
∂~r
∂θ =rcosθcosφˆı+rcosθsinφˆ−rsinθkˆ
But
∂~r
∂θ
=r6= 1
So we can say that changing θ bydθmoves the point rdθ in the direction of ˆθ
We call this the scale factorhθ=r. Notice thathr = 1
So ˆθ= h1
θ
∂~r
∂θ = cosθcosφˆı+ cosθsinφˆ−sinθkˆ
θ ˆ
~
r =xˆı+yˆ+zkˆ
∂~r
∂θ = ∂x
∂θˆı+∂y
∂θ ˆ+∂θ
∂θkˆ
∂~r
∂θ =rcosθcosφˆı+rcosθsinφˆ−rsinθkˆ
But
∂~r
∂θ
=r6= 1
So we can say that changing θ bydθmoves the point rdθ in the direction of ˆθ
We call this the scale factorhθ=r. Notice thathr = 1 So ˆθ= h1
θ
∂~r
∂θ = cosθcosφˆı+ cosθsinφˆ−sinθkˆ
φ ˆ
Similarly we can get ˆφ=−sinφˆı+ cosφˆandhφ=rsinθ
A better way to write this
Notice we can write this as
ˆ r θˆ φˆ
=
sinθcosφ sinθsinφ cosθ cosθcosφ cosθsinφ −sinθ
−sinφ cosφ 0
ˆı ˆ
kˆ
ˆ r θˆ φˆ
=M
ˆı ˆ
kˆ
, whereM =
sinθcosφ sinθsinφ cosθ cosθcosφ cosθsinφ −sinθ
−sinφ cosφ 0
We can even find ˆı, ˆ, ˆk in terms of ˆr, ˆθ, ˆφ
ˆı ˆ
kˆ
=M−1
ˆ r θˆ φˆ
, whereM−1 =MT, because M is orthogonal
A better way to write this
Notice we can write this as
ˆ r θˆ φˆ
=
sinθcosφ sinθsinφ cosθ cosθcosφ cosθsinφ −sinθ
−sinφ cosφ 0
ˆı ˆ
kˆ
ˆ r θˆ φˆ
=M
ˆı ˆ
kˆ
, where M =
sinθcosφ sinθsinφ cosθ cosθcosφ cosθsinφ −sinθ
−sinφ cosφ 0
We can even find ˆı, ˆ, ˆk in terms of ˆr, ˆθ, ˆφ
ˆı ˆ
kˆ
=M−1
ˆ r θˆ φˆ
, whereM−1 =MT, because M is orthogonal
A better way to write this
Notice we can write this as
ˆ r θˆ φˆ
=
sinθcosφ sinθsinφ cosθ cosθcosφ cosθsinφ −sinθ
−sinφ cosφ 0
ˆı ˆ
kˆ
ˆ r θˆ φˆ
=M
ˆı ˆ
kˆ
, where M =
sinθcosφ sinθsinφ cosθ cosθcosφ cosθsinφ −sinθ
−sinφ cosφ 0
We can even find ˆı, ˆ, ˆk in terms of ˆr, ˆθ, ˆφ
ˆı ˆ
kˆ
=M−1
ˆ r θˆ φˆ
, where M−1 =MT, because M is orthogonal
Other coordinate systems
The above results are valid only for Spherical Polar system Now do this same thing for
2D Polar
Cylindrical Polar
