reciprocal frames, the vector derivative and curvilinear ... · reciprocal frames, the vector...

Reciprocal Frames, the Vector Derivative andCurvilinear Coordinates.

17th Santalo Summer School 2016, Santander

Joan Lasenby

Signal Processing Group,Engineering Department,

Cambridge, UKand

Trinity CollegeCambridge

jl221@cam.ac.uk, www-sigproc.eng.cam.ac.uk/ ∼ jl

22 August 2016

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Overview

Reciprocal Frames: examples of their use

definition and use of the vector derivative

Curvilinear Coordinates: how reciprocal frames can beused to simplify complicated mathematics.

Summary

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Overview

Summary

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Overview

Summary

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Overview

Summary

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Reciprocal Frames

Many problems in mathematics, physics and engineeringrequire a treatment of non-orthonormal frames.

Take a set of n linearly independent vectors {ek}; these are notnecessarily orthogonal nor of unit length.

Can we find a second set of vectors (in the same space), callthese {ek}, such that

ei·ej = δij

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Reciprocal Frames

ei·ej = δij

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Reciprocal Frames

ei·ej = δij

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Reciprocal Frames

ei·ej = δij

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Reciprocal Frames

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Reciprocal Frames cont....

We call such a frame a reciprocal frame. Note that since anyvector a can be written as a = akek ≡ ∑ akek (ie we are adoptingthe convention that repeated indices are summed over), wehave

ek ·a = ek ·(ajej) = aj(ek ·ej) = ajδkj = ak

Similarly, since we can also write a = akek ≡ ∑ akek

ek ·a = ek ·(ajej) = aj(ek ·ej) = ajδjk = ak

Thus we would be able to recover the components of a givenvector in a similar way to that used for orthonormal frames.

So how do we find a reciprocal frame?

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So how do we find a reciprocal frame?14 / 110

We need, for example, e1 to be orthogonal to the set of vectors{e2, e3, ..., en}. ie e1 must be perpendicular to the hyperplanee2∧e3∧....∧en.

We find this by dualisation, ie multiplication by I [note: I is then-d pseudoscalar for our space]. We form e1 via

e1 = αe2∧e3∧...∧enI

α is a scalar found by dotting with e1:

e1·e1 = 1 = e1·(αe2∧e3∧...∧enI) = α(e1∧e2∧...∧en)I

(this uses a useful GA relation a·(BI) = (a∧B)I).

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e1 = αe2∧e3∧...∧enI

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e1 = αe2∧e3∧...∧enI

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Reciprocal Frames ....

If we letEn = e1∧e2∧...∧en 6= 0

we see that αEnI = 1, so that α = E−1n I−1. Thus giving us

ek = (−1)k+1e1∧e2∧...∧ek∧...∧enE−1n

where the ek notation indicates that ek is missing from the blade.

These reciprocal frames are remarkably useful!

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ek = (−1)k+1e1∧e2∧...∧ek∧...∧enE−1n

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ek = (−1)k+1e1∧e2∧...∧ek∧...∧enE−1n

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ek = (−1)k+1e1∧e2∧...∧ek∧...∧enE−1n

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Exercises 1

1 Show that a·(BI) = (a∧B)I. [Hint: make use of the fact thata·(BrIn) = 〈aBrIn〉n−r−1].

2 For {f1, f2, f3} = {e1, e1 + 2e3, e1 + e2 + e3} show, usingthe given formulae, that the reciprocal frame is given by

{f 1, f 2, f 3} = {e1 −12(e2 + e3),

12(e3 − e2), e2}

[these are the reciprocal frames shown in the earlierpictures]

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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+1e1∧e2∧...∧ek∧...∧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−1en∧en−1∧...∧ek∧...∧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 problemof recovering the rotor which rotates between two 3-dnon-orthonormal frames {ek} and {fk}, ie find R such that

fk = RekR

It is not too hard to show that R can be written as

R = β(1 + fkek)

where the constant β ensures that RR = 1.

A very easy way of recovering rotations.

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fk = RekR

R = β(1 + fkek)

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fk = RekR

R = β(1 + fkek)

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The Vector Derivative

A vector x can be represented in terms of coordinates in twoways:

x = xkek or x = xkek

(Summation implied). Depending on whether we expand interms of a given frame {ek} or its reciprocal {ek}. Thecoefficients in these two frames are therefore given by

xk = ek ·x and xk = ek ·x

Now define the following derivative operator which we call thevector derivative

∇ = ∑k

ek ∂

∂xk ≡ ek ∂

..this is clearly a vector!

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∇ = ∑k

ek ∂

∂xk ≡ ek ∂

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∇ = ∑k

ek ∂

∂xk ≡ ek ∂

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∇ = ∑k

ek ∂

∂xk ≡ ek ∂

..this is clearly a vector!30 / 110

The Vector Derivative, cont...

∇ = ∑k

ek ∂

This is a definition so far, but we will now see how this formarises.

Suppose we have a function acting on vectors, F(x). Usingstandard definitions of rates of change, we can define thedirectional derivative of F in the direction of a vector a as

limε→0

F(x + εa)− F(x)ε

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∇ = ∑k

ek ∂

limε→0

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∇ = ∑k

ek ∂

limε→0

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∇ = ∑k

ek ∂

limε→0

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The Vector Derivative cont....

Now, suppose we want the directional derivative in thedirection of one of our frame vectors, say e1, this is given by

limε→0

F((x1 + ε)e1 + x2e2 + x3e3)− F(x1e1 + x2e2 + x3e3)

which we recognise as∂F(x)

ie the derivative with respect to the first coordinate, keepingthe second and third coordinates constant.

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limε→0

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limε→0

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limε→0

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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 adirection, we clearly need:

ei·∇ =∂

∂xi for i = 1, 2, 3

...which, since ei·ej ∂∂xj =

∂∂xi , gives us the previous form of the

vector derivative:

∇ = ∑k

ek ∂

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ei·∇ =∂

∂xi for i = 1, 2, 3

vector derivative:

∇ = ∑k

ek ∂

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ei·∇ =∂

∂xi for i = 1, 2, 3

vector derivative:

∇ = ∑k

ek ∂

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ei·∇ =∂

∂xi for i = 1, 2, 3

vector derivative:

∇ = ∑k

ek ∂

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It follows now that if we dot ∇ with a, we get the directionalderivative in the a direction:

a·∇ F(x) = limε→0

We will see later that the definition of ∇ is independent of thechoice of frame.

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It follows now that if we dot ∇ with a, we get the directionalderivative in the a direction:

a·∇ F(x) = limε→0

We will see later that the definition of ∇ is independent of thechoice 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.

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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Operating on:

∇J = ∇·J +∇∧J

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Operating on:

∇J = ∇·J +∇∧J

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Operating on:

∇J = ∇·J +∇∧J

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Operating on:

∇J = ∇·J +∇∧J

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Operating on:

∇J = ∇·J +∇∧J

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Curvilinear Coordinates

Curvilinear coordinates are systems where the frame vectorsvary with position – the two most commonly used sets in 3-dare:

r(r, θ, φ) = rer r(ρ, θ, z) = ρeρ + zez

For spherical polars, our position vector is defined in terms of alength r and two angles θ, φ: =⇒ coordinates are (r, θ, φ).For cylindrical polars, our position vector is defined in terms oftwo lengths ρ, z and an angle φ: =⇒ coordinates are (ρ, φ, z).

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For spherical polars, our position vector is defined in terms of alength r and two angles θ, φ: =⇒ coordinates are (r, θ, φ).

For cylindrical polars, our position vector is defined in terms oftwo 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 coordinatesxi, i = 1, ..., n, which are functions of the position vector, r.

Of course, we can also write the position vector, r, as a functionof the coordinates {xi} (as on previous page).

Vary one coordinate while keeping others fixed to create acoordinate curve. We can then create a set of frame vectors, callthem {ei}, by finding the derivatives along these curves:

ei(r) =∂r∂xi ≡ lim

ε→0

r(x1, ..., xi + ε, ..., xn)− r(x1, ..., xi, ..., xn)

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ε→0

r(x1, ..., xi + ε, ..., xn)− r(x1, ..., xi, ..., xn)

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ε→0

r(x1, ..., xi + ε, ..., xn)− r(x1, ..., xi, ..., xn)

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ε→0

r(x1, ..., xi + ε, ..., xn)− r(x1, ..., xi, ..., xn)

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Now recall from earlier that the derivative in the ei direction isei·∇, which is also the partial derivative wrt the xi coordinate:

ei·∇ =∂

It then follows that

(ei·∇)xj ≡ ei·(∇xj) =∂xj

∂xi = δji

Therefore, using the definition of the reciprocal frame(ei·ej = δ

ji), we can deduce that

ej = ∇xj

Thus, we can construct a second, reciprocal, frame from thecoordinates using the vector derivative

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ei·∇ =∂

∂xi = δji

ej = ∇xj

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ei·∇ =∂

∂xi = δji

ej = ∇xj

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ei·∇ =∂

∂xi = δji

ej = ∇xj

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ei·∇ =∂

∂xi = δji

ej = ∇xj

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ei·∇ =∂

∂xi = δji

ej = ∇xj

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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 framevectors as

ei(r) =∂r∂xi

We can then construct a second, reciprocal, frame from thecoordinates via

ej = ∇xj [note : ∇∧ej = ∇∧∇xj = 0]

We see therefore that the vector derivative is crucial in relatingcoordinates to frames – and we will see how this simplifiesmanipulations in curvilinear coordinates.

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ei(r) =∂r∂xi

ej = ∇xj [note : ∇∧ej = ∇∧∇xj = 0]

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ei(r) =∂r∂xi

ej = ∇xj [note : ∇∧ej = ∇∧∇xj = 0]

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ei(r) =∂r∂xi

ej = ∇xj [note : ∇∧ej = ∇∧∇xj = 0]

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ei(r) =∂r∂xi

ej = ∇xj [note : ∇∧ej = ∇∧∇xj = 0]

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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 ∂

∂xi ·(Jjej) = ei·

∂(Jjej)

∂xi [scalar]

Curl of a Vector Function, J

∇∧J = ei ∂

∂xi∧(Jjej) = ei∧

∂(Jjej)

∂xi [bivector]

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∇ψ = ei ∂ψ

∂xi [vector]

∇·J = ei ∂

∂(Jjej)

∂xi [scalar]

∇∧J = ei ∂

∂(Jjej)

∂xi [bivector]

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∇ψ = ei ∂ψ

∂xi [vector]

∇·J = ei ∂

∂(Jjej)

∂xi [scalar]

∇∧J = ei ∂

∂(Jjej)

∂xi [bivector]

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Div, Grad, Curl cont....

Now, we can write the expressions for div and curl in a waywhich makes them easier to relate to the standard expressionsfor derivatives in curvilinear coordinates.

Divergence