Reciprocal Frames, the Vector Derivative and
Curvilinear Coordinates.
17th Santaló Summer School 2016, Santander
Joan Lasenby
Signal Processing Group,
Engineering Department,
Cambridge, UK
and
Trinity College
Cambridge
[email protected],
www-sigproc.eng.cam.ac.uk/ ∼ jl
22 August 2016
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Overview
Reciprocal Frames: examples of their use
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Overview
Reciprocal Frames: examples of their use
definition and use of the vector derivative
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Overview
Reciprocal Frames: examples of their use
definition and use of the vector derivative
Curvilinear Coordinates: how reciprocal frames can be
used to simplify complicated mathematics.
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Overview
Reciprocal Frames: examples of their use
definition and use of the vector derivative
Curvilinear Coordinates: how reciprocal frames can be
used to simplify complicated mathematics.
Summary
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Reciprocal Frames
Many problems in mathematics, physics and engineering
require a treatment of non-orthonormal frames.
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Reciprocal Frames
Many problems in mathematics, physics and engineering
require a treatment of non-orthonormal frames.
Take a set of n linearly independent vectors {ek }; these are not
necessarily orthogonal nor of unit length.
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Reciprocal Frames
Many problems in mathematics, physics and engineering
require a treatment of non-orthonormal frames.
Take a set of n linearly independent vectors {ek }; these are not
necessarily orthogonal nor of unit length.
Can we find a second set of vectors (in the same space), call
these {ek }, such that
ei · ej = δji
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Reciprocal Frames
Many problems in mathematics, physics and engineering
require a treatment of non-orthonormal frames.
Take a set of n linearly independent vectors {ek }; these are not
necessarily orthogonal nor of unit length.
Can we find a second set of vectors (in the same space), call
these {ek }, such that
ei · ej = δji
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Reciprocal Frames
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Reciprocal Frames cont....
We call such a frame a reciprocal frame. Note that since any
vector a can be written as a = ak ek ≡ ∑ ak ek (ie we are adopting
the convention that repeated indices are summed over), we
have
ek · a = ek ·(aj ej ) = aj (ek · ej ) = aj δjk = ak
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Reciprocal Frames cont....
We call such a frame a reciprocal frame. Note that since any
vector a can be written as a = ak ek ≡ ∑ ak ek (ie we are adopting
the convention that repeated indices are summed over), we
have
ek · a = ek ·(aj ej ) = aj (ek · ej ) = aj δjk = ak
Similarly, since we can also write a = ak ek ≡ ∑ ak ek
j
ek · a = ek ·(aj ej ) = aj (ek · ej ) = aj δk = ak
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Reciprocal Frames cont....
We call such a frame a reciprocal frame. Note that since any
vector a can be written as a = ak ek ≡ ∑ ak ek (ie we are adopting
the convention that repeated indices are summed over), we
have
ek · a = ek ·(aj ej ) = aj (ek · ej ) = aj δjk = ak
Similarly, since we can also write a = ak ek ≡ ∑ ak ek
j
ek · a = ek ·(aj ej ) = aj (ek · ej ) = aj δk = ak
Thus we would be able to recover the components of a given
vector in a similar way to that used for orthonormal frames.
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Reciprocal Frames cont....
We call such a frame a reciprocal frame. Note that since any
vector a can be written as a = ak ek ≡ ∑ ak ek (ie we are adopting
the convention that repeated indices are summed over), we
have
ek · a = ek ·(aj ej ) = aj (ek · ej ) = aj δjk = ak
Similarly, since we can also write a = ak ek ≡ ∑ ak ek
j
ek · a = ek ·(aj ej ) = aj (ek · ej ) = aj δk = ak
Thus we would be able to recover the components of a given
vector in a similar way to that used for orthonormal frames.
So how do we find a reciprocal frame?
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Reciprocal Frames cont....
We need, for example, e1 to be orthogonal to the set of vectors
{e2 , e3 , ..., en }. ie e1 must be perpendicular to the hyperplane
e2 ∧ e3 ∧ .... ∧ en .
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Reciprocal Frames cont....
We need, for example, e1 to be orthogonal to the set of vectors
{e2 , e3 , ..., en }. ie e1 must be perpendicular to the hyperplane
e2 ∧ e3 ∧ .... ∧ en .
We find this by dualisation, ie multiplication by I [note: I is the
n-d pseudoscalar for our space]. We form e1 via
e1 = αe2 ∧ e3 ∧ ... ∧ en I
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Reciprocal Frames cont....
We need, for example, e1 to be orthogonal to the set of vectors
{e2 , e3 , ..., en }. ie e1 must be perpendicular to the hyperplane
e2 ∧ e3 ∧ .... ∧ en .
We find this by dualisation, ie multiplication by I [note: I is the
n-d pseudoscalar for our space]. We form e1 via
e1 = αe2 ∧ e3 ∧ ... ∧ en I
α is a scalar found by dotting with e1 :
e1 · e1 = 1 = e1 ·(αe2 ∧ e3 ∧ ... ∧ en I ) = α(e1 ∧ e2 ∧ ... ∧ en )I
(this uses a useful GA relation a ·(BI ) = (a ∧ B)I).
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Reciprocal Frames ....
If we let
En = e1 ∧ e2 ∧ ... ∧ en 6= 0
we see that αEn I = 1, so that α = En−1 I −1 . Thus giving us
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Reciprocal Frames ....
If we let
En = e1 ∧ e2 ∧ ... ∧ en 6= 0
we see that αEn I = 1, so that α = En−1 I −1 . Thus giving us
ek = (−1)k+1 e1 ∧ e2 ∧ ... ∧ ěk ∧ ... ∧ en En−1
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Reciprocal Frames ....
If we let
En = e1 ∧ e2 ∧ ... ∧ en 6= 0
we see that αEn I = 1, so that α = En−1 I −1 . Thus giving us
ek = (−1)k+1 e1 ∧ e2 ∧ ... ∧ ěk ∧ ... ∧ en En−1
where the ěk notation indicates that ek is missing from the blade.
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Reciprocal Frames ....
If we let
En = e1 ∧ e2 ∧ ... ∧ en 6= 0
we see that αEn I = 1, so that α = En−1 I −1 . Thus giving us
ek = (−1)k+1 e1 ∧ e2 ∧ ... ∧ ěk ∧ ... ∧ en En−1
where the ěk notation indicates that ek is missing from the blade.
These reciprocal frames are remarkably useful!
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Exercises 1
1
Show that a ·(BI ) = (a ∧ B)I. [Hint: make use of the fact that
a ·(Br In ) = haBr In in−r−1 ].
2
For {f1 , f2 , f3 } = {e1 , e1 + 2e3 , e1 + e2 + e3 } show, using
the given formulae, that the reciprocal frame is given by
1
1
{f 1 , f 2 , f 3 } = {e1 − (e2 + e3 ), (e3 − e2 ), e2 }
2
2
[these are the reciprocal frames shown in the earlier
pictures]
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Exercises 2
1
Interchanging the role of frame and reciprocal frame,
verify that we can write the frame vectors as
ek = (−1)k+1 e1 ∧ e2 ∧ ... ∧ ěk ∧ ... ∧ en {En }−1
where En = e1 ∧ e2 ∧ ... ∧ en 6= 0.
2
Now show that we can move vectors through each other
(changing sign) to give
ek = (−1)k−1 en ∧ en−1 ∧ ... ∧ ěk ∧ ... ∧ e1 {IV }
where {En }−1 = IV, and V is therefore a volume factor.
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Example: Recovering a Rotor in 3-d
As an example of using reciprocal frames, consider the problem
of recovering the rotor which rotates between two 3-d
non-orthonormal frames {ek } and {fk }, ie find R such that
fk = Rek R̃
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Example: Recovering a Rotor in 3-d
As an example of using reciprocal frames, consider the problem
of recovering the rotor which rotates between two 3-d
non-orthonormal frames {ek } and {fk }, ie find R such that
fk = Rek R̃
It is not too hard to show that R can be written as
R = β(1 + fk ek )
where the constant β ensures that RR̃ = 1.
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Example: Recovering a Rotor in 3-d
As an example of using reciprocal frames, consider the problem
of recovering the rotor which rotates between two 3-d
non-orthonormal frames {ek } and {fk }, ie find R such that
fk = Rek R̃
It is not too hard to show that R can be written as
R = β(1 + fk ek )
where the constant β ensures that RR̃ = 1.
A very easy way of recovering rotations.
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The Vector Derivative
A vector x can be represented in terms of coordinates in two
ways:
x = xk ek or x = xk ek
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The Vector Derivative
A vector x can be represented in terms of coordinates in two
ways:
x = xk ek or x = xk ek
(Summation implied). Depending on whether we expand in
terms of a given frame {ek } or its reciprocal {ek }. The
coefficients in these two frames are therefore given by
xk = ek · x and xk = ek · x
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The Vector Derivative
A vector x can be represented in terms of coordinates in two
ways:
x = xk ek or x = xk ek
(Summation implied). Depending on whether we expand in
terms of a given frame {ek } or its reciprocal {ek }. The
coefficients in these two frames are therefore given by
xk = ek · x and xk = ek · x
Now define the following derivative operator which we call the
vector derivative
∇ = ∑ ek
k
∂
∂
≡ ek k
∂xk
∂x
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The Vector Derivative
A vector x can be represented in terms of coordinates in two
ways:
x = xk ek or x = xk ek
(Summation implied). Depending on whether we expand in
terms of a given frame {ek } or its reciprocal {ek }. The
coefficients in these two frames are therefore given by
xk = ek · x and xk = ek · x
Now define the following derivative operator which we call the
vector derivative
∇ = ∑ ek
k
∂
∂
≡ ek k
∂xk
∂x
..this is clearly a vector!
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The Vector Derivative, cont...
∇ = ∑ ek
k
∂
∂xk
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The Vector Derivative, cont...
∇ = ∑ ek
k
∂
∂xk
This is a definition so far, but we will now see how this form
arises.
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The Vector Derivative, cont...
∇ = ∑ ek
k
∂
∂xk
This is a definition so far, but we will now see how this form
arises.
Suppose we have a function acting on vectors, F(x). Using
standard definitions of rates of change, we can define the
directional derivative of F in the direction of a vector a as
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The Vector Derivative, cont...
∇ = ∑ ek
k
∂
∂xk
This is a definition so far, but we will now see how this form
arises.
Suppose we have a function acting on vectors, F(x). Using
standard definitions of rates of change, we can define the
directional derivative of F in the direction of a vector a as
F(x + ea) − F(x)
e →0
e
lim
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The Vector Derivative cont....
Now, suppose we want the directional derivative in the
direction of one of our frame vectors, say e1 , this is given by
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The Vector Derivative cont....
Now, suppose we want the directional derivative in the
direction of one of our frame vectors, say e1 , this is given by
F((x1 + e)e1 + x2 e2 + x3 e3 ) − F(x1 e1 + x2 e2 + x3 e3 )
e →0
e
lim
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The Vector Derivative cont....
Now, suppose we want the directional derivative in the
direction of one of our frame vectors, say e1 , this is given by
F((x1 + e)e1 + x2 e2 + x3 e3 ) − F(x1 e1 + x2 e2 + x3 e3 )
e →0
e
lim
which we recognise as
∂F(x)
∂x1
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The Vector Derivative cont....
Now, suppose we want the directional derivative in the
direction of one of our frame vectors, say e1 , this is given by
F((x1 + e)e1 + x2 e2 + x3 e3 ) − F(x1 e1 + x2 e2 + x3 e3 )
e →0
e
lim
which we recognise as
∂F(x)
∂x1
ie the derivative with respect to the first coordinate, keeping
the second and third coordinates constant.
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Vector Derivative cont.....
So, if we wish to define a gradient operator, ∇, such that
(a ·∇)F(x) gives the directional derivative of F in the a
direction, we clearly need:
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Vector Derivative cont.....
So, if we wish to define a gradient operator, ∇, such that
(a ·∇)F(x) gives the directional derivative of F in the a
direction, we clearly need:
ei ·∇ =
∂
for i = 1, 2, 3
∂xi
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Vector Derivative cont.....
So, if we wish to define a gradient operator, ∇, such that
(a ·∇)F(x) gives the directional derivative of F in the a
direction, we clearly need:
ei ·∇ =
...which, since ei · ej ∂x∂ j =
vector derivative:
∂
for i = 1, 2, 3
∂xi
∂
,
∂xi
gives us the previous form of the
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Vector Derivative cont.....
So, if we wish to define a gradient operator, ∇, such that
(a ·∇)F(x) gives the directional derivative of F in the a
direction, we clearly need:
ei ·∇ =
...which, since ei · ej ∂x∂ j =
vector derivative:
∂
for i = 1, 2, 3
∂xi
∂
,
∂xi
gives us the previous form of the
∇ = ∑ ek
k
∂
∂xk
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The Vector Derivative cont....
It follows now that if we dot ∇ with a, we get the directional
derivative in the a direction:
F(x + ea) − F(x)
e →0
e
a ·∇ F(x) = lim
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The Vector Derivative cont....
It follows now that if we dot ∇ with a, we get the directional
derivative in the a direction:
F(x + ea) − F(x)
e →0
e
a ·∇ F(x) = lim
We will see later that the definition of ∇ is independent of the
choice of frame.
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Operating on Scalar and Vector Fields
Operating on:
A Scalar Field φ(x): it gives ∇φ which is the gradient.
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Operating on Scalar and Vector Fields
Operating on:
A Scalar Field φ(x): it gives ∇φ which is the gradient.
A Vector Field J (x): it gives ∇J. This is a geometric product
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Operating on Scalar and Vector Fields
Operating on:
A Scalar Field φ(x): it gives ∇φ which is the gradient.
A Vector Field J (x): it gives ∇J. This is a geometric product
Scalar part gives divergence
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Operating on Scalar and Vector Fields
Operating on:
A Scalar Field φ(x): it gives ∇φ which is the gradient.
A Vector Field J (x): it gives ∇J. This is a geometric product
Scalar part gives divergence
Bivector part gives curl
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Operating on Scalar and Vector Fields
Operating on:
A Scalar Field φ(x): it gives ∇φ which is the gradient.
A Vector Field J (x): it gives ∇J. This is a geometric product
Scalar part gives divergence
Bivector part gives curl
∇J = ∇· J + ∇∧ J
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Operating on Scalar and Vector Fields
Operating on:
A Scalar Field φ(x): it gives ∇φ which is the gradient.
A Vector Field J (x): it gives ∇J. This is a geometric product
Scalar part gives divergence
Bivector part gives curl
∇J = ∇· J + ∇∧ J
See later discussions of electromagnetism.
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Curvilinear Coordinates
Curvilinear coordinates are systems where the frame vectors
vary with position – the two most commonly used sets in 3-d
are:
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Curvilinear Coordinates
Curvilinear coordinates are systems where the frame vectors
vary with position – the two most commonly used sets in 3-d
are:
r(r, θ, φ) = rer
r(ρ, θ, z) = ρeρ + zez
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Curvilinear Coordinates
Curvilinear coordinates are systems where the frame vectors
vary with position – the two most commonly used sets in 3-d
are:
r(r, θ, φ) = rer
r(ρ, θ, z) = ρeρ + zez
For spherical polars, our position vector is defined in terms of a
length r and two angles θ, φ: =⇒ coordinates are (r, θ, φ).
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Curvilinear Coordinates
Curvilinear coordinates are systems where the frame vectors
vary with position – the two most commonly used sets in 3-d
are:
r(r, θ, φ) = rer
r(ρ, θ, z) = ρeρ + zez
For spherical polars, our position vector is defined in terms of a
length r and two angles θ, φ: =⇒ coordinates are (r, θ, φ).
For cylindrical polars, our position vector is defined in terms of
two lengths ρ, z and an angle φ: =⇒ coordinates are (ρ, φ, z).
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Curvilinear Coordinates cont....
In a general curvilinear setup we will have coordinates
xi , i = 1, ..., n, which are functions of the position vector, r.
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Curvilinear Coordinates cont....
In a general curvilinear setup we will have coordinates
xi , i = 1, ..., n, which are functions of the position vector, r.
Of course, we can also write the position vector, r, as a function
of the coordinates {xi } (as on previous page).
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Curvilinear Coordinates cont....
In a general curvilinear setup we will have coordinates
xi , i = 1, ..., n, which are functions of the position vector, r.
Of course, we can also write the position vector, r, as a function
of the coordinates {xi } (as on previous page).
Vary one coordinate while keeping others fixed to create a
coordinate curve. We can then create a set of frame vectors, call
them {ei }, by finding the derivatives along these curves:
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Curvilinear Coordinates cont....
In a general curvilinear setup we will have coordinates
xi , i = 1, ..., n, which are functions of the position vector, r.
Of course, we can also write the position vector, r, as a function
of the coordinates {xi } (as on previous page).
Vary one coordinate while keeping others fixed to create a
coordinate curve. We can then create a set of frame vectors, call
them {ei }, by finding the derivatives along these curves:
ei ( r ) =
∂r
r(x1 , ..., xi + e, ..., xn ) − r(x1 , ..., xi , ..., xn )
≡
lim
e →0
e
∂xi
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Curvilinear Coordinates cont....
Now recall from earlier that the derivative in the ei direction is
ei ·∇, which is also the partial derivative wrt the xi coordinate:
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Curvilinear Coordinates cont....
Now recall from earlier that the derivative in the ei direction is
ei ·∇, which is also the partial derivative wrt the xi coordinate:
ei ·∇ =
∂
∂xi
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Curvilinear Coordinates cont....
Now recall from earlier that the derivative in the ei direction is
ei ·∇, which is also the partial derivative wrt the xi coordinate:
ei ·∇ =
∂
∂xi
It then follows that
(ei ·∇)xj ≡ ei ·(∇xj ) =
∂xj
j
= δi
i
∂x
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Curvilinear Coordinates cont....
Now recall from earlier that the derivative in the ei direction is
ei ·∇, which is also the partial derivative wrt the xi coordinate:
ei ·∇ =
∂
∂xi
It then follows that
(ei ·∇)xj ≡ ei ·(∇xj ) =
∂xj
j
= δi
i
∂x
Therefore, using the definition of the reciprocal frame
j
(ei · ej = δi ), we can deduce that
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Curvilinear Coordinates cont....
Now recall from earlier that the derivative in the ei direction is
ei ·∇, which is also the partial derivative wrt the xi coordinate:
ei ·∇ =
∂
∂xi
It then follows that
(ei ·∇)xj ≡ ei ·(∇xj ) =
∂xj
j
= δi
i
∂x
Therefore, using the definition of the reciprocal frame
j
(ei · ej = δi ), we can deduce that
ej = ∇ xj
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Curvilinear Coordinates cont....
Now recall from earlier that the derivative in the ei direction is
ei ·∇, which is also the partial derivative wrt the xi coordinate:
ei ·∇ =
∂
∂xi
It then follows that
(ei ·∇)xj ≡ ei ·(∇xj ) =
∂xj
j
= δi
i
∂x
Therefore, using the definition of the reciprocal frame
j
(ei · ej = δi ), we can deduce that
ej = ∇ xj
Thus, we can construct a second, reciprocal, frame from the
coordinates using the vector derivative
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Curvilinear Coordinates : Summary
Given coordinates {xi , i = 1, ..., n}, which any position vector, r
[note, use boldface to distinguish from distance from origin],
can be expressed in terms of, we can define a set of frame
vectors as
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Curvilinear Coordinates : Summary
Given coordinates {xi , i = 1, ..., n}, which any position vector, r
[note, use boldface to distinguish from distance from origin],
can be expressed in terms of, we can define a set of frame
vectors as
∂r
ei (r) = i
∂x
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Curvilinear Coordinates : Summary
Given coordinates {xi , i = 1, ..., n}, which any position vector, r
[note, use boldface to distinguish from distance from origin],
can be expressed in terms of, we can define a set of frame
vectors as
∂r
ei (r) = i
∂x
We can then construct a second, reciprocal, frame from the
coordinates via
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Curvilinear Coordinates : Summary
Given coordinates {xi , i = 1, ..., n}, which any position vector, r
[note, use boldface to distinguish from distance from origin],
can be expressed in terms of, we can define a set of frame
vectors as
∂r
ei (r) = i
∂x
We can then construct a second, reciprocal, frame from the
coordinates via
ej = ∇xj
[note : ∇∧ ej = ∇∧∇xj = 0]
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Curvilinear Coordinates : Summary
Given coordinates {xi , i = 1, ..., n}, which any position vector, r
[note, use boldface to distinguish from distance from origin],
can be expressed in terms of, we can define a set of frame
vectors as
∂r
ei (r) = i
∂x
We can then construct a second, reciprocal, frame from the
coordinates via
ej = ∇xj
[note : ∇∧ ej = ∇∧∇xj = 0]
We see therefore that the vector derivative is crucial in relating
coordinates to frames – and we will see how this simplifies
manipulations in curvilinear coordinates.
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Div, Grad, Curl in Curvilinear Coordinates
Gradient of a Scalar Function, ψ
∇ψ = ei
∂ψ
∂xi
[vector]
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Div, Grad, Curl in Curvilinear Coordinates
Gradient of a Scalar Function, ψ
∇ψ = ei
∂ψ
∂xi
[vector]
Divergence of a Vector Function, J
∇· J = ei
j
∂
j
i ∂ ( J ej )
·(
J
e
)
=
e
·
j
∂xi
∂xi
[scalar]
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Div, Grad, Curl in Curvilinear Coordinates
Gradient of a Scalar Function, ψ
∇ψ = ei
∂ψ
∂xi
[vector]
Divergence of a Vector Function, J
∇· J = ei
j
∂
j
i ∂ ( J ej )
·(
J
e
)
=
e
·
j
∂xi
∂xi
[scalar]
Curl of a Vector Function, J
∇∧ J = ei
j
∂
j
i ∂ (J ej )
∧(
J
e
)
=
e
∧
j
∂xi
∂xi
[bivector]
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Div, Grad, Curl cont....
Now, we can write the expressions for div and curl in a way
which makes them easier to relate to the standard expressions
for derivatives in curvilinear coordinates.
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Div, Grad, Curl cont....
Now, we can write the expressions for div and curl in a way
which makes them easier to relate to the standard expressions
for derivatives in curvilinear coordinates.
Divergence
∇· J = ∇·(Ji ei ) = ei ·(∇Ji ) + Ji (∇· ei )
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Div, Grad, Curl cont....
Now, we can write the expressions for div and curl in a way
which makes them easier to relate to the standard expressions
for derivatives in curvilinear coordinates.
Divergence
∇· J = ∇·(Ji ei ) = ei ·(∇Ji ) + Ji (∇· ei )
(this is a simple application of the chain rule)
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Div, Grad, Curl cont....
Now, we can write the expressions for div and curl in a way
which makes them easier to relate to the standard expressions
for derivatives in curvilinear coordinates.
Divergence
∇· J = ∇·(Ji ei ) = ei ·(∇Ji ) + Ji (∇· ei )
(this is a simple application of the chain rule)
Now, take the pseudovector (n − 1-blade)
P = (−1)k−1 en ∧ en−1 ∧ ... ∧ ěk ∧ ... ∧ e1 , and recall that ei = PIV [See
Exercises 2]. So that (where hXi denotes the scalar part of X)
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Div, Grad, Curl cont....
Now, we can write the expressions for div and curl in a way
which makes them easier to relate to the standard expressions
for derivatives in curvilinear coordinates.
Divergence
∇· J = ∇·(Ji ei ) = ei ·(∇Ji ) + Ji (∇· ei )
(this is a simple application of the chain rule)
Now, take the pseudovector (n − 1-blade)
P = (−1)k−1 en ∧ en−1 ∧ ... ∧ ěk ∧ ... ∧ e1 , and recall that ei = PIV [See
Exercises 2]. So that (where hXi denotes the scalar part of X)
∇· ei = h∇(PIV )i = h(∇P)IV i + hPI (∇V )i
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Div, Grad, Curl cont....
After some manipulation (which will be outlined in the
following exercises) we are able to write
∇· J = ei ·(∇Ji ) + Ji (ei ·∇(ln V ))
= ei ·(∇Ji ) + J ·(∇(ln V )) =
1 ∂
(VJi )
V ∂xi
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Div, Grad, Curl cont....
After some manipulation (which will be outlined in the
following exercises) we are able to write
∇· J = ei ·(∇Ji ) + Ji (ei ·∇(ln V ))
= ei ·(∇Ji ) + J ·(∇(ln V )) =
∇· J =
1 ∂
(VJi )
V ∂xi
1 ∂
(VJi )
V ∂xi
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Div, Grad, Curl cont....
We therefore have the following expressions: Gradient of a
Scalar Function, ψ
∇ψ = ei
∂ψ
∂xi
[vector]
80 / 110
Div, Grad, Curl cont....
We therefore have the following expressions: Gradient of a
Scalar Function, ψ
∇ψ = ei
∂ψ
∂xi
[vector]
Divergence of a Vector Function, J
∇· J =
1 ∂
(VJi )
V ∂xi
[scalar]
81 / 110
Div, Grad, Curl cont....
We therefore have the following expressions: Gradient of a
Scalar Function, ψ
∇ψ = ei
∂ψ
∂xi
[vector]
Divergence of a Vector Function, J
∇· J =
1 ∂
(VJi )
V ∂xi
[scalar]
Curl of a Vector Function, J
∇∧ J = (∇Ji )∧ ei
[bivector]
82 / 110
Exercises 3
1
Since IV is a pseudoscalar, show that
h(∇P)IV i = h(∇∧ P)IV )i
2
Using the fact that ei = PIV, show that
PI (∇V ) = ei ∇(ln V )
3
Verify that ∇∧ a ∧ b = (∇∧ a)∧ b − a ∧(∇∧ b) ,
and then, using our previous result of ∇∧ ei = 0, show that
∇∧ P = 0
and therefore that
∇· ei = ei ·∇(ln V )
83 / 110
Exercises 4
1
By expanding ∇∧ J as
˙
˙
∇∧ J = ∇∧(Ji ei ) = ∇∧(
J̇i ei ) + ∇∧(
Ji e˙i )
explain how we obtain the result ∇∧ J = (∇Ji )∧ ei
84 / 110
An Example: Spherical Polars in 3d
Recall our coordinates are (r, θ, φ), and we also have an
orthogonal set of unit vectors (êr , êθ , êφ ) as shown in the
∂r
diagram. Thus, we can define a frame via ei = ∂x
i to be
85 / 110
An Example: Spherical Polars in 3d
Recall our coordinates are (r, θ, φ), and we also have an
orthogonal set of unit vectors (êr , êθ , êφ ) as shown in the
∂r
diagram. Thus, we can define a frame via ei = ∂x
i to be
er =
∂r
∂(rêr )
=
= êr
∂r
∂r
86 / 110
An Example: Spherical Polars in 3d
Recall our coordinates are (r, θ, φ), and we also have an
orthogonal set of unit vectors (êr , êθ , êφ ) as shown in the
∂r
diagram. Thus, we can define a frame via ei = ∂x
i to be
er =
eθ =
∂r
∂(rêr )
=
= êr
∂r
∂r
∂r
∂(rêr )
∂êr
=
=r
= rêθ
∂θ
∂θ
∂θ
87 / 110
An Example: Spherical Polars in 3d
Recall our coordinates are (r, θ, φ), and we also have an
orthogonal set of unit vectors (êr , êθ , êφ ) as shown in the
∂r
diagram. Thus, we can define a frame via ei = ∂x
i to be
er =
eθ =
eφ =
∂r
∂(rêr )
=
= êr
∂r
∂r
∂r
∂(rêr )
∂êr
=
=r
= rêθ
∂θ
∂θ
∂θ
∂(cos θêz + sin θêρ )
∂r
∂(rêr )
∂êr
=
=r
=r
= r sin θêφ
∂φ
∂φ
∂φ
∂φ
88 / 110
An Example: Spherical Polars in 3d cont...
From the definition of reciprocal frame we therefore see that
the reciprocal vectors are given by
89 / 110
An Example: Spherical Polars in 3d cont...
From the definition of reciprocal frame we therefore see that
the reciprocal vectors are given by
er = êr
eθ =
1
êθ
r
eφ =
1
êφ
r sin θ
(check that ei · ej = δji ).
90 / 110
An Example: Spherical Polars in 3d cont...
From the definition of reciprocal frame we therefore see that
the reciprocal vectors are given by
er = êr
eθ =
1
êθ
r
eφ =
1
êφ
r sin θ
(check that ei · ej = δji ).
Now we can use our previous formulae to give us grad, div
and curl in spherical polars.
91 / 110
An Example: Spherical Polars in 3d cont...
From the definition of reciprocal frame we therefore see that
the reciprocal vectors are given by
er = êr
eθ =
1
êθ
r
eφ =
1
êφ
r sin θ
(check that ei · ej = δji ).
Now we can use our previous formulae to give us grad, div
and curl in spherical polars.
Gradient
∇ ψ = ei
∂ψ
1 ∂ψ
1 ∂ψ
∂ψ
=
êr +
êθ +
êφ
∂r
r ∂θ
r sin θ ∂φ
∂xi
92 / 110
An Example: Spherical Polars in 3d cont...
From the definition of reciprocal frame we therefore see that
the reciprocal vectors are given by
er = êr
eθ =
1
êθ
r
eφ =
1
êφ
r sin θ
(check that ei · ej = δji ).
Now we can use our previous formulae to give us grad, div
and curl in spherical polars.
Gradient
∇ ψ = ei
∂ψ
1 ∂ψ
1 ∂ψ
∂ψ
=
êr +
êθ +
êφ
∂r
r ∂θ
r sin θ ∂φ
∂xi
Which agrees with the formula given in tables etc.
93 / 110
An Example: Spherical Polars in 3d cont...
Divergence
∇·J =
1
1 ∂(VJi )
∂(r2 sin θJr )
1
∂(r2 sin θJ θ )
1
∂(r2 sin θJ φ )
= 2
+ 2
+ 2
i
V ∂x
∂r
∂θ
∂φ
r sin θ
r sin θ
r sin θ
=
1 ∂(r2 Jr )
1 ∂(sin θJ θ ) ∂J φ
+
+
2
r
∂r
sin θ
∂θ
∂φ
94 / 110
An Example: Spherical Polars in 3d cont...
Divergence
∇·J =
1
1 ∂(VJi )
∂(r2 sin θJr )
1
∂(r2 sin θJ θ )
1
∂(r2 sin θJ φ )
= 2
+ 2
+ 2
i
V ∂x
∂r
∂θ
∂φ
r sin θ
r sin θ
r sin θ
=
1 ∂(r2 Jr )
1 ∂(sin θJ θ ) ∂J φ
+
+
2
r
∂r
sin θ
∂θ
∂φ
Since V = −r2 sin θ (see exercises).
95 / 110
An Example: Spherical Polars in 3d cont...
Divergence
∇·J =
1
1 ∂(VJi )
∂(r2 sin θJr )
1
∂(r2 sin θJ θ )
1
∂(r2 sin θJ φ )
= 2
+ 2
+ 2
i
V ∂x
∂r
∂θ
∂φ
r sin θ
r sin θ
r sin θ
=
1 ∂(r2 Jr )
1 ∂(sin θJ θ ) ∂J φ
+
+
2
r
∂r
sin θ
∂θ
∂φ
Since V = −r2 sin θ (see exercises).
Now, note that
J = Jr er + J θ eθ + J φ eφ = Jr êr + J θ (rêθ ) + J φ (r sin θ )êφ = Ĵr êr + Ĵθ êθ + Ĵφ êφ
96 / 110
An Example: Spherical Polars in 3d cont...
Divergence
∇·J =
1
1 ∂(VJi )
∂(r2 sin θJr )
1
∂(r2 sin θJ θ )
1
∂(r2 sin θJ φ )
= 2
+ 2
+ 2
i
V ∂x
∂r
∂θ
∂φ
r sin θ
r sin θ
r sin θ
=
1 ∂(r2 Jr )
1 ∂(sin θJ θ ) ∂J φ
+
+
2
r
∂r
sin θ
∂θ
∂φ
Since V = −r2 sin θ (see exercises).
Now, note that
J = Jr er + J θ eθ + J φ eφ = Jr êr + J θ (rêθ ) + J φ (r sin θ )êφ = Ĵr êr + Ĵθ êθ + Ĵφ êφ
Which agrees with the formula given in tables etc.
∇· J =
1 ∂(r2 Ĵr )
1 ∂(sin θ Ĵθ )
1 ∂Ĵφ
+
+
2
r
∂r
r sin θ
∂θ
r sin θ ∂φ
97 / 110
An Example: Spherical Polars in 3d cont...
Curl
∇∧ J = (∇Ji )∧ ei =
∂Jφ
∂Jφ
∂Jr
∂Jθ
∂Jθ
∂Jr
r
θ
θ
φ
−
(e ∧e ) +
−
(e ∧e ) +
−
(eφ ∧er )
∂θ
∂r
∂θ
∂φ
∂φ
∂r
98 / 110
An Example: Spherical Polars in 3d cont...
Curl
∇∧ J = (∇Ji )∧ ei =
∂Jφ
∂Jφ
∂Jr
∂Jθ
∂Jθ
∂Jr
r
θ
θ
φ
−
(e ∧e ) +
−
(e ∧e ) +
−
(eφ ∧er )
∂θ
∂r
∂θ
∂φ
∂φ
∂r
Now look at the second component, noting that
1
eθ ∧ eφ = 1r êθ ∧ r sin
θ êφ =
1
ê I,
r2 sin θ r
99 / 110
An Example: Spherical Polars in 3d cont...
Curl
∇∧ J = (∇Ji )∧ ei =
∂Jφ
∂Jφ
∂Jr
∂Jθ
∂Jθ
∂Jr
r
θ
θ
φ
−
(e ∧e ) +
−
(e ∧e ) +
−
(eφ ∧er )
∂θ
∂r
∂θ
∂φ
∂φ
∂r
Now look at the second component, noting that
1
1
eθ ∧ eφ = 1r êθ ∧ r sin
θ êφ = r2 sin θ êr I,
"
#
∂Jφ
∂
(
r
sin
θ
Ĵ
)
∂Jθ
∂
(
r
Ĵ
)
1
φ
θ
êr I
−
(eθ ∧ eφ ) =
−
2
∂θ
∂φ
∂θ
∂φ
r sin θ
100 / 110
An Example: Spherical Polars in 3d cont...
Curl
∇∧ J = (∇Ji )∧ ei =
∂Jφ
∂Jφ
∂Jr
∂Jθ
∂Jθ
∂Jr
r
θ
θ
φ
−
(e ∧e ) +
−
(e ∧e ) +
−
(eφ ∧er )
∂θ
∂r
∂θ
∂φ
∂φ
∂r
Now look at the second component, noting that
1
1
eθ ∧ eφ = 1r êθ ∧ r sin
θ êφ = r2 sin θ êr I,
"
#
∂Jφ
∂
(
r
sin
θ
Ĵ
)
∂Jθ
∂
(
r
Ĵ
)
1
φ
θ
êr I
−
(eθ ∧ eφ ) =
−
2
∂θ
∂φ
∂θ
∂φ
r sin θ
which can be written to agree with conventional tabulated
form (though we have a bivector and not a vector):
101 / 110
An Example: Spherical Polars in 3d cont...
Curl
∇∧ J = (∇Ji )∧ ei =
∂Jφ
∂Jφ
∂Jr
∂Jθ
∂Jθ
∂Jr
r
θ
θ
φ
−
(e ∧e ) +
−
(e ∧e ) +
−
(eφ ∧er )
∂θ
∂r
∂θ
∂φ
∂φ
∂r
Now look at the second component, noting that
1
1
eθ ∧ eφ = 1r êθ ∧ r sin
θ êφ = r2 sin θ êr I,
"
#
∂Jφ
∂
(
r
sin
θ
Ĵ
)
∂Jθ
∂
(
r
Ĵ
)
1
φ
θ
êr I
−
(eθ ∧ eφ ) =
−
2
∂θ
∂φ
∂θ
∂φ
r sin θ
which can be written to agree with conventional tabulated
form (though we have a bivector and not a vector):
"
#
∂(sin θ Ĵφ ) ∂(Ĵθ )
1
−
êr I
r sin θ
∂θ
∂φ
102 / 110
Exercises 5
1
2
3
For 3d spherical polars, show that V = −r2 sin θ, where
VI = (En )−1 and En = er ∧ eθ ∧ eφ .
Show that the eφ ∧ er component of ∇∧ J can be written as:
"
#
1
1 ∂(Ĵr ) ∂(rĴφ )
−
êθ I
r sin θ ∂φ
∂r
Show that the er ∧ eθ component on ∇∧ J can be written as:
1 1 ∂(rĴθ ) ∂(Ĵr )
−
êφ I
r r ∂r
∂θ
Check these against standard tabulated formulae.
103 / 110
Connection with conventional Lamé Coefficients
Conventionally, sets of Lamé Coefficients are defined to be
∂r hi = i ∂x
104 / 110
Connection with conventional Lamé Coefficients
Conventionally, sets of Lamé Coefficients are defined to be
∂r hi = i ∂x
In our language we therefore have hi = |ei |, ie simply the
moduli of the frame vectors defined by the coordinates.
105 / 110
Connection with conventional Lamé Coefficients
Conventionally, sets of Lamé Coefficients are defined to be
∂r hi = i ∂x
In our language we therefore have hi = |ei |, ie simply the
moduli of the frame vectors defined by the coordinates.
All expressions of div, grad, curl etc in terms of the hi s, can then
be directly related to the expressions we derive in GA.
106 / 110
Summary
We have
Defined the concept of a Reciprocal Frame
107 / 110
Summary
We have
Defined the concept of a Reciprocal Frame
Shown how to construct Reciprocal Frames
108 / 110
Summary
We have
Defined the concept of a Reciprocal Frame
Shown how to construct Reciprocal Frames
Using reciprocal frames, defined and motivated the form
of the Vector Derivative
109 / 110
Summary
We have
Defined the concept of a Reciprocal Frame
Shown how to construct Reciprocal Frames
Using reciprocal frames, defined and motivated the form
of the Vector Derivative
Shown how to relate coordinates, frame vectors and
reciprocal frame vectors in curvilinear coordinate systems.
110 / 110