Orthonormal vector polynomials in a unit circle, Part II : completing the basis set

Chunyu Zhao; James H. Burge

doi:10.1364/OE.16.006586

1. Introduction

In a previous paper,¹ we developed an orthonormal set of vector polynomials over a unit circle, which we call S⃗ polynomials. These polynomials are great for fitting the slope data taken by a Shack-Hartmann sensor. But since they are gradients of linear combinations of Zernike polynomials, they have zero curl, which means they make an incomplete set of vector polynomials such that an arbitrary continuously differentiable vector function defined over a unit circle cannot be represented by linear combinations of these polynomials. Additional vector polynomials must be added to make a complete set. An optical application that needs a complete set of vector polynomials is to fit the mapping distortions of an interferometric null test. See Reference 2 for an example of mapping distortion in a null test and how to correct it. The lowest modes of mapping distortions include translation, scaling and rotation. Polynomial S⃗ ₂ and S⃗ ₃ represent x and y translation, respectively, and S⃗ ₄ represents scaling. But no S⃗ polynomial represents rotation. The reason is that the rotation vector has non-zero curl, while all S⃗ polynomials have zero curl. In this paper, we derived a complementary set of vector polynomials which have zero divergence and non-zero curl. This new set combined with the zero-curl set S⃗ makes a complete set such that it can represent any continuously differentiable vector polynomials defined over a unit circle.

In Section 2, we review the S⃗ polynomials we derived and reported in a previous paper. We then proceed to derive the complementary T⃗ polynomials in Section 3.

2. The S⃗ polynomials

In a previous paper¹, we derived an orthonormal set of vector polynomials over a unit circle. We call this set the S⃗ polynomials. Each S⃗ polynomial is the gradient of a scalar function:

{\vec{S}}_{j} = \nabla ϕ_{j} = \hat{i} \frac{\partial ϕ_{j}}{\partial x} + \hat{j} \frac{\partial ϕ_{j}}{\partial y} .

The scalar functions ϕ_j are linear combination of Zernike polynomials. Following Noll’s notation and numbering scheme,³

ϕ_{j} = \frac{1}{\sqrt{2 n (n + 1)}} Z_{j}, for all j with n = m,

and

ϕ_{j} = \frac{1}{\sqrt{4 n (n + 1)}} (Z_{j} - \sqrt{\frac{n + 1}{n - 1}} Z_{j' (n' = n - 2, m' = m)}), for all j with n \neq m .

Each S⃗ polynomial is then the linear combinations of gradient of Zernike polynomials following (1) and (2). Since gradient of Zernike polynomials can also be represented by Zernike polynomials,³ S⃗ polynomials can be written as linear combinations of Zernike polynomials as well. The first 14 non-trivial S⃗ polynomials are listed in Table 1.

Table 1. List of the first 14 non-trivial S⃗ polynomials as linear combinations of Zernike polynomials.

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If A⃗ and B⃗ are two vector polynomials defined over a unit circle, we define their inner product as

(\vec{A}, \vec{B}) = \frac{1}{π} \iint (\vec{A} • \vec{B}) dxdy,

where integration is over unit circle.

S⃗ polynomials are orthonormal, which means

({\vec{S}}_{i}, {\vec{S}}_{j}) = \frac{1}{π} \iint ((\nabla ϕ_{i}) • (\nabla ϕ_{j})) dxdy = δ_{ij} .

3. Derivation of a complementary set of vector polynomials

Any vector field can be written as⁴

\vec{v} = \nabla ϕ + \nabla \times \vec{P},

where ϕ is a scalar and P⃗ is a vector. The divergence of ν⃗ is then

∇•ν⃗=∇² ϕ+∇•(∇×P⃗)=∇² ϕ,

and the curl of ν⃗ is

∇×ν⃗=∇×(∇×P⃗)=∇(∇•P⃗)-∇² P⃗.

The S⃗ polynomials presented in the previous paper¹ were defined as gradients of scalar functions, so have no curl component and P⃗=0. We complete the basis by adding a second set that has zero divergence, therefore zero ϕ, but non zero P⃗, such that

\vec{T} = \nabla \times \vec{P} = [\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ P_{x} & P_{y} & P_{z} \end{matrix}] .

This set has to be mutually orthogonal as well.

Like the S⃗ polynomials, T⃗ polynomials are vectors defined in x-y plane only. A convenient choice of P⃗ is vectors along z axis only, i.e. P_x=P_y=0. We can use a scalar ψ instead to represent P⃗:

\vec{P} = ψ \hat{k,}

where ψ is a function of x and y: ψ=ψ (x, y). It follows that

{\vec{T}}_{i} = \nabla \times (ψ_{i} \hat{k}) = \hat{i} \frac{\partial ψ_{i}}{\partial y} - \hat{j} \frac{\partial ψ_{i}}{\partial x} .

The inner product of two T⃗ polynomials is then

({\vec{T}}_{i}, {\vec{T}}_{j}) = \iint ((\frac{\partial ψ_{i}}{\partial y}) (\frac{\partial ψ_{j}}{\partial y}) + (- \frac{\partial ψ_{i}}{\partial x}) (- \frac{\partial ψ_{j}}{\partial x})) dxdy

We choose a basis of functions {ψ_i} that we use to generate the T⃗ polynomials to be the same basis as we used to generate the S⃗ polynomials, {ϕ_i}. By Eq. (4), we know that their choice will create T⃗ polynomials that are mutually orthogonal.

From Eqs. (1) and (8), and with ψ_i=ϕ_i, we know that S⃗_i(x,y) and T⃗_i(x,y) have same magnitude and are orthogonal to each other at any point in a unit circle, therefore (S⃗_i,T⃗_i)=0. But the sets S⃗ and T⃗ are not fully independent. For all the j with m=n, we can show that

∇×T⃗_j=-k̂∇² ϕ_j,

and

\nabla^{2} ϕ_{j} \propto \nabla^{2} Z_{j} \propto (\frac{1}{r} \frac{\partial}{\partial r} (r \frac{\partial}{\partial r}) + \frac{1}{r^{2}} \frac{\partial^{2}}{\partial θ^{2}}) [r^{n} (\begin{matrix} \cos n θ \\ \sin n θ \end{matrix})] = 0,

which means T⃗_j has 0 curl and is therefore not linearly independent of S⃗ polynomials. For example, when j=9 or 10, m=n=3: ${\vec{T}}_{9} = \frac{1}{\sqrt{2}} (\hat{i} Z_{6} - \hat{j} Z_{5}) = {\vec{S}}_{10}$ and ${\vec{T}}_{10} = \frac{1}{\sqrt{2}} (\hat{i} Z_{5} - \hat{j} Z_{6}) = {- \vec{S}}_{9}$ .

For any other pair of i and j, (S⃗_i,T⃗_j)=0.

Fig. 1. Relations between the S⃗ and T⃗ polynomials. The Laplacian vector fields are the overlap between S⃗ and T⃗. The dashed circles and associated solid arrows illustrate the local behaviors of the vectors in different sets after subtracting the local constant vector.

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The S⃗ and T⃗ polynomials can be thought of as vector fields in a unit circle. In vector calculus, S⃗ is known as irrotational vector fields which have zero curl everywhere, and T⃗ is known as solenoidal vector fields which have zero divergence everywhere. The two types vector fields have some overlap where both divergence and curls are everywhere zero, which is known as Laplacian vector field. The overall relationship between S⃗ and T⃗ vector fields is illustrated in Figure 1. The overlapped area contains terms derived from corresponding scalar ϕ polynomials whose Laplacian is 0. If ϕ represents wavefront, these terms correspond to a wavefront that has zero net curvature at any point in the pupil.

It is useful to compare the different types of functions defined here. The S⃗ functions are generated from gradients, thus have no curl. Since S⃗ functions are 2-d vectors defined in a plane, mathematically, we can express the curl as line integral along a closed path in the plane:

\oint \vec{S} • d \vec{l} = 0 .

The T⃗ functions have no divergence. Again they are 2-d vectors defined in a plane. Mathematically, we express divergence of a 2-d vector as a line integral over a closed path:

\oint \vec{T} • \hat{n} d l = 0,

where n̂ is the unit normal vector pointing out of the closed path.

The intersection, which includes both S⃗ and T⃗, is of the form that fits both Eqs. (11) and (12), having both zero divergence and zero curl. Graphical depictions of the local behavior of the functions are included in Fig. 1: dashed circles represent infinitesimal region and solid arrows represent local vectors (after a constant vector is subtracted.)

Table 2 lists expressions for the first 15 T⃗ polynomials. The plots of first 12 non-trivial T⃗ polynomials are shown in Table 3. The complete set of orthogonal vector polynomials that fully spans the circular domain can be written as the combined set of S⃗ polynomials and independent T⃗ polynomials since the Laplacian type functions are included in both sets. Care must be taken to ensure that the common functions are not counted twice.

Table 2. Analytical expressions of the first 15 T⃗ polynomials.

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Table 3. Plots of the first 12 non-trivial T⃗ polynomials.

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4. Summary

We derived a set of vector polynomials defined over a unit circle which complements the set we presented in a previous paper. Each set of the vector polynomials is orthonormal over a unit circle. But there are some overlap between the two sets. We can combine the two sets in such a way that the overlapped subset is counted only once, then we obtain a complete set of vector polynomials defined over a unit circle. The combined set is useful for fitting any continuously differentiable vector functions in a circular domain. In particular, it is useful in fitting the mapping distortions often seen in an interferometric null test. We will explore and report its applications in subsequent papers.

If interested, you can request the MATLAB codes for calculating the S⃗ and T⃗ polynomials from Dr. Chunyu Zhao, czhao@optics.arizona.edu.

References and links

1. C. Zhao and J. H. Burge, “Orthonormal vector polynomials in a unit circle, Part I: basis set derived from gradients of Zernike polynomials,” Opt. Express 15, 18014–18024 (2007). [CrossRef] [PubMed]

2. C. Zhao, et al, “Figure measurement of a large optical flat with a Fizeau interferometer and stitching technique,” Proc. SPIE 6293, 62930k, (2006). [CrossRef]

3. R. J. Noll, “Zernike polynomials and atmospheric turbulence”, J. Opt. Soc. Am. 66, 1976, 207–211. [CrossRef]

4. H. F. Davis and A. D. Snider, Introduction to Vector Analysis, (Wm. C. Brown Publisher, 1986).

S⃗ ₂=îZ ₁	${\vec{S}}_{9} = \frac{1}{\sqrt{2}} (\hat{i} Z_{5} + \hat{j} Z_{6})$
S⃗ ₃=ĵZ ₁	${\vec{S}}_{10} = \frac{1}{\sqrt{2}} (\hat{i} Z_{6} - \hat{j} Z_{5})$
${\vec{S}}_{4} = \frac{1}{\sqrt{2}} (\hat{i} Z_{2} + \hat{j} Z_{3})$	${\vec{S}}_{11} = \frac{1}{\sqrt{2}} (\hat{i} Z_{8} + \hat{j} Z_{7})$
${\vec{S}}_{5} = \frac{1}{\sqrt{2}} (\hat{i} Z_{3} + \hat{j} Z_{2})$	${\vec{S}}_{12} = \frac{1}{2} (\hat{i} (Z_{8} + Z_{10}) + \hat{j} (- Z_{7} + Z_{9}))$
${\vec{S}}_{6} = \frac{1}{\sqrt{2}} (\hat{i} Z_{2} - \hat{j} Z_{3})$	${\vec{S}}_{13} = \frac{1}{2} (\hat{i} (Z_{7} + Z_{9}) + \hat{j} (Z_{8} - Z_{10}))$
${\vec{S}}_{7} = \frac{1}{2} (\hat{i} Z_{5} + \hat{j} (\sqrt{2} Z_{4} - Z_{6}))$	${\vec{S}}_{14} = \frac{1}{\sqrt{2}} (\hat{i} Z_{10} - \hat{j} Z_{9})$
${\vec{S}}_{8} = \frac{1}{2} (\hat{i} ({\sqrt{2} Z}_{4} + Z_{6}) + \hat{j} Z_{5})$	${\vec{S}}_{15} = \frac{1}{\sqrt{2}} (\hat{i} Z_{9} + \hat{j} Z_{10})$

T⃗ ₁=0, trivia
T⃗ ₂=-ĵZ ₁=-S⃗ ₃	${\vec{T}}_{9} = \frac{1}{\sqrt{2}} (\hat{i} Z_{6} - \hat{j} Z_{5}) = {\vec{S}}_{10}$
T⃗ ₃=îZ ₁=S⃗ ₂	${\vec{T}}_{10} = \frac{1}{\sqrt{2}} (- \hat{i} Z_{5} - \hat{j} Z_{6}) = - {\vec{S}}_{9}$
${\vec{T}}_{4} = \frac{1}{\sqrt{2}} (\hat{i} Z_{3} - \hat{j} Z_{2})$	${\vec{T}}_{11} = \frac{1}{\sqrt{2}} (\hat{i} Z_{7} - \hat{j} Z_{8})$
${\vec{T}}_{5} = \frac{1}{\sqrt{2}} (\hat{i} Z_{2} - \hat{j} Z_{3}) = {\vec{S}}_{6}$	${\vec{T}}_{12} = \frac{1}{2} (\hat{i} (- Z_{7} + Z_{9}) - \hat{j} (Z_{8} + Z_{10}))$
${\vec{T}}_{6} = \frac{1}{\sqrt{2}} (- \hat{i} Z_{3} - \hat{j} Z_{2}) = - {\vec{S}}_{5}$	${\vec{T}}_{13} = \frac{1}{2} (\hat{i} (Z_{8} - Z_{10}) - \hat{j} (Z_{7} + Z_{9}))$
${\vec{T}}_{7} = \frac{1}{2} (\hat{i} (\sqrt{2} Z_{4} - Z_{6}) - \hat{j} Z_{5})$	${\vec{T}}_{14} = \frac{1}{\sqrt{2}} (- \hat{i} Z_{9} - {\hat{j} Z}_{10}) = - {\vec{S}}_{15}$
${\vec{T}}_{8} = \frac{1}{2} (\hat{i} Z_{5} - \hat{j} ({\sqrt{2} Z}_{4} + Z_{6}))$	${\vec{T}}_{15} = \frac{1}{\sqrt{2}} (\hat{i} Z_{10} - \hat{j} Z_{9}) = {\vec{S}}_{14}$

S⃗ ₂=îZ ₁	${\vec{S}}_{9} = \frac{1}{\sqrt{2}} (\hat{i} Z_{5} + \hat{j} Z_{6})$
S⃗ ₃=ĵZ ₁	${\vec{S}}_{10} = \frac{1}{\sqrt{2}} (\hat{i} Z_{6} - \hat{j} Z_{5})$
${\vec{S}}_{4} = \frac{1}{\sqrt{2}} (\hat{i} Z_{2} + \hat{j} Z_{3})$	${\vec{S}}_{11} = \frac{1}{\sqrt{2}} (\hat{i} Z_{8} + \hat{j} Z_{7})$
${\vec{S}}_{5} = \frac{1}{\sqrt{2}} (\hat{i} Z_{3} + \hat{j} Z_{2})$	${\vec{S}}_{12} = \frac{1}{2} (\hat{i} (Z_{8} + Z_{10}) + \hat{j} (- Z_{7} + Z_{9}))$
${\vec{S}}_{6} = \frac{1}{\sqrt{2}} (\hat{i} Z_{2} - \hat{j} Z_{3})$	${\vec{S}}_{13} = \frac{1}{2} (\hat{i} (Z_{7} + Z_{9}) + \hat{j} (Z_{8} - Z_{10}))$
${\vec{S}}_{7} = \frac{1}{2} (\hat{i} Z_{5} + \hat{j} (\sqrt{2} Z_{4} - Z_{6}))$	${\vec{S}}_{14} = \frac{1}{\sqrt{2}} (\hat{i} Z_{10} - \hat{j} Z_{9})$
${\vec{S}}_{8} = \frac{1}{2} (\hat{i} ({\sqrt{2} Z}_{4} + Z_{6}) + \hat{j} Z_{5})$	${\vec{S}}_{15} = \frac{1}{\sqrt{2}} (\hat{i} Z_{9} + \hat{j} Z_{10})$

T⃗ ₁=0, trivia
T⃗ ₂=-ĵZ ₁=-S⃗ ₃	${\vec{T}}_{9} = \frac{1}{\sqrt{2}} (\hat{i} Z_{6} - \hat{j} Z_{5}) = {\vec{S}}_{10}$
T⃗ ₃=îZ ₁=S⃗ ₂	${\vec{T}}_{10} = \frac{1}{\sqrt{2}} (- \hat{i} Z_{5} - \hat{j} Z_{6}) = - {\vec{S}}_{9}$
${\vec{T}}_{4} = \frac{1}{\sqrt{2}} (\hat{i} Z_{3} - \hat{j} Z_{2})$	${\vec{T}}_{11} = \frac{1}{\sqrt{2}} (\hat{i} Z_{7} - \hat{j} Z_{8})$
${\vec{T}}_{5} = \frac{1}{\sqrt{2}} (\hat{i} Z_{2} - \hat{j} Z_{3}) = {\vec{S}}_{6}$	${\vec{T}}_{12} = \frac{1}{2} (\hat{i} (- Z_{7} + Z_{9}) - \hat{j} (Z_{8} + Z_{10}))$
${\vec{T}}_{6} = \frac{1}{\sqrt{2}} (- \hat{i} Z_{3} - \hat{j} Z_{2}) = - {\vec{S}}_{5}$	${\vec{T}}_{13} = \frac{1}{2} (\hat{i} (Z_{8} - Z_{10}) - \hat{j} (Z_{7} + Z_{9}))$
${\vec{T}}_{7} = \frac{1}{2} (\hat{i} (\sqrt{2} Z_{4} - Z_{6}) - \hat{j} Z_{5})$	${\vec{T}}_{14} = \frac{1}{\sqrt{2}} (- \hat{i} Z_{9} - {\hat{j} Z}_{10}) = - {\vec{S}}_{15}$
${\vec{T}}_{8} = \frac{1}{2} (\hat{i} Z_{5} - \hat{j} ({\sqrt{2} Z}_{4} + Z_{6}))$	${\vec{T}}_{15} = \frac{1}{\sqrt{2}} (\hat{i} Z_{10} - \hat{j} Z_{9}) = {\vec{S}}_{14}$

Orthonormal vector polynomials in a unit circle, Part II : completing the basis set

Abstract

1. Introduction

2. The S⃗ polynomials

3. Derivation of a complementary set of vector polynomials

4. Summary

References and links

Cited By

Figures (1)

Tables (3)

Equations (14)

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