# Bézier surfaces with prescribed diagonals

A. Arnal111 Email<EMAIL_ADDRESS>ORCID: 0000-0002-3283-3379 J.
Monterde222Email<EMAIL_ADDRESS>
###### Abstract

The affine space of all tensor product Bézier patches of degree $n\times n$
with prescribed main diagonal curves is determined. First, the pair of Bézier
curves which can be diagonals of a Bézier patch is characterized. Besides
prescribing the diagonal curves, other related problems are considered, those
where boundary curves or tangent planes along boundary curves are also
prescribed.

Institut de Matemàtiques i Aplicacions de Castelló (IMAC) and Departament de
Matemàtiques, Universitat Jaume I, E-12071 Castellón, Spain.

Dep. de Matemàtiques, Universitat de València, Burjassot (València), Spain

Keywords: Tensor product Bézier surface; Diagonal curve; Bézier patch;
boundary curve; Geometric Modeling.

##  1 Introduction

One of the paradigmatic problems in the realm of computer design is to build a
surface once its boundary is prescribed. In the case of Bézier surfaces, there
is an extensive literature dealing with the problem of building a Bézier
surface with a prescribed boundary.

These kinds of method are fundamentally based on seeking the minimization of
some functional with a defined geometric meaning in the space of all Bézier
surfaces with that border. Examples of these functionals include the area
functional, which gives rise to a non-linear problem, or quadratic
functionals, such as the Dirichlet functional or similar. All of these methods
can be easily implemented so that the designer is immediately able to
visualize the result of any change to the initial data.

Functional minimization can be performed if all the boundary curves are given,
so the surface shape is quite controlled. Nevertheless, this is not in general
true for PDE surfaces, where the number of boundary curves that can be
prescribed depends on the degree of the PDE (see [7] and related references).
For example, in order to obtain harmonic surfaces only two boundary curves can
be prescribed so the other two get out of control. Hence small changes in the
prescribed data could produce big changes in the shape of the two free
boundary curves.

Therefore, although the natural way of controlling the shape of a surface is
through its boundary, a way of increasing control could be to fix some curves
on the surface. For this purpose, the main diagonal curves of a tensor-product
Bézier surface seems to be the best choice. So, the goal here is to give a
method that allow prescribing not only the boundary curves but the diagonal
curves of the surface.

In fact, diagonal curves play an important role in engineering applications of
surface modeling, such as architectural design or the design and optimization
of quad meshes in architectural geometry [3]. However, very few studies have
been published on constrained modeling of tensor-product surfaces with
diagonal curves, (see [12] and [13]).

Two situations where controlling the diagonal curves of Bézier patches seems
to be fundamental can be found in the literature. The first is related to
conversion from rectangular patches into a triangular scheme.

Based on the concept of diagonal curve, [10] and [4] respectively proposed
S-Patch and BS-Patch. The authors of paper [11] added some requirements for
the main diagonals of a Hermite bicubic patch to be able to adapt to a
triangular scheme. The requirements are related to the degree of the diagonal
curves. Indeed, as stated in [4], conversions between quadrilateral and
triangular meshes need the degree of the diagonal curves to match the degree
of the boundary curves of the quadrilateral meshes. Moreover, the reverse
adaptation, from triangular to rectangular patches, would require building
rectangular patches where not only the boundaries but also the diagonal curves
are prescribed.

The second situation where diagonal curves play a main role is in the design
of surfaces with some geometric property. In papers [8] and [9] the authors
deal with constant mean curvature surfaces. Such kind of surfaces allow
parameterizations with isothermal coordinate lines. That is to say, the two
families of coordinate lines are mutually orthogonal and tangent vectors at an
intersection point are of the same length. In addition to these two families
of coordinate lines, diagonal curves are used not only for better aesthetics,
but also to improve structural stability. The structures that can be built
using these methods could be modified as follows: once the mesh of coordinate
lines and the diagonal lines are obtained, they can be used as prescribed data
to obtain rectangular Bézier patches with additional properties.

A more general situation can be found in papers [5] and [6], where the authors
discuss the problem of adding features to a free form surface by applying one
or several user-defined surface curves, which are seen as editable parameters.
These user-prescribed curves can be diagonal curves, but are not necessarily
so.

Since the surface control net is related to the diagonal curves control
points, (see [2]), the point is that, in order to be prescribed, the boundary
control points and the control points of the diagonal curves in particular
must meet certain conditions that will be shown in this paper. In [12], the
authors considered a similar problem, and provided a method for generating a
surface from free boundaries and diagonal curves given by the user, that is,
without taking into account that boundaries and diagonals are related, so they
have to meet some conditions. They then proposed a correction of this given
information using Lagrange method to minimize the corresponding differences.

Let us recall that when we prescribe $\mathcal{C}^{0}$-boundary conditions,
that is, when the boundary of a $n\times n$ Bézier surface is prescribed, then
$4n$ boundary control points are fixed from a total of $(n+1)^{2}$.
Equivalently, we can say that the affine space of all Bézier surfaces with the
same boundary is parameterized by the $(n-1)^{2}$ interior control points.
Analogously, when we prescribe $\mathcal{C}^{1}$-boundary conditions, that is
to say, when the boundary and the tangent planes along it are prescribed, then
the $4n$ boundary control points and the $4n-8$ control points adjacent to the
boundary are fixed. Equivalently, the affine space of all Bézier surfaces with
the same boundary and tangent planes along it is parameterized by the
$(n-3)^{2}$ interior control points.

In a similar fashion, this paper deals with the following problem: What is the
affine space of all Bézier surfaces with the same diagonals, or with the same
boundaries and diagonals? In other words, is it possible to determine which
control points are fixed if only the diagonals or the diagonals and the
boundaries are prescribed? Which subset of control points parameterizes the
corresponding affine space?

We solve the following problems:

  * a)

Find the conditions, in terms of their control points, that two Bézier curves
must fulfill in order to be the diagonal curves of an indeterminate Bézier
surface. Then, find the conditions, again in terms of these diagonal curves
control points, for them to be the diagonal curves of a Bézier surface with a
fixed border.

  * b)

Once this goal has been achieved, find the dimension of the space of all
Bézier surfaces that have certain boundary and diagonal curves. This dimension
indicates nothing but the number of control points of the Bézier surface that
are free once both the boundary and the diagonals have been prescribed.

Finally we have also considered the above problems for both the two prescribed
diagonals and the prescribed $\mathcal{C}^{1}$-boundary conditions.

Let us remark that free control points of a Bézier surface with a prescribed
boundary and diagonals could be determined by minimizing a functional, such as
the Dirichlet functional or any similar one.

##  2 Admissible pairs of main diagonal curves

Let us consider a tensor product Bézier surface of degree $n\times n$, ${\bf
x}:[0,1]\times[0,1]\to{\mathbb{R}}^{3}$, with control net
${\mathcal{P}}=\\{P_{i,j}\\}_{i,j=0}^{n}$. The pair of main diagonal curves of
this surface are the Bézier curves of degree $2n$:

$\begin{array}[]{rcl}\alpha_{1}(t)&=&{\bf x}(t,t),\qquad\quad
t\in[0,1],\\\\[5.69054pt] \alpha_{2}(t)&=&{\bf x}(t,1-t),\ \quad
t\in[0,1].\end{array}$

Initially, we are interested in the following problem: What are the conditions
that a pair of degree $2n$ Bézier curves must meet in order to be the main
diagonals of a tensor product Bézier surface?

A first condition can be obtained easily on the control polygons, as
illustrated by Figure 1, since the midpoints have to coincide

$\alpha_{1}\left(\frac{1}{2}\right)=\alpha_{2}\left(\frac{1}{2}\right).$ (1)

Figure 1: The two main diagonals of a tensor product Bézier surface. Notice
that both have the same midpoint.

This simple fact has been pointed out in several references, see for instance
[1], page 183, or [12]. The novelty here is that such a condition will be
completely expressed as a condition on the control points of both Bézier
curves.

######  Lemma 1.

The midpoint of two Bézier curves of degree $2n$ with control polygons

${\mathcal{Q}}=\\{Q_{i}\\}_{i=0}^{2n},\qquad{\mathcal{R}}=\\{R_{i}\\}_{i=0}^{2n},$

is the same if and only if

$\sum_{i=0}^{2n}\binom{2n}{i}Q_{i}=\sum_{i=0}^{2n}\binom{2n}{i}R_{i}.$ (2)

###### Proof.

Since

$B^{2n}_{i}\left(\frac{1}{2}\right)=\binom{2n}{i}\left(\frac{1}{2}\right)^{i}\left(1-\frac{1}{2}\right)^{2n-i}=\binom{2n}{i}\frac{1}{2^{2n}},$

then

$\begin{array}[]{rcl}\alpha_{1}\left(\frac{1}{2}\right)&=&\frac{1}{2^{2n}}\sum_{i=0}^{2n}\binom{2n}{i}Q_{i},\\\\[8.53581pt]
\alpha_{2}\left(\frac{1}{2}\right)&=&\frac{1}{2^{2n}}\sum_{i=0}^{2n}\binom{2n}{i}R_{i},\end{array}$

and the result is obtained. ∎

In paper [2] the authors state that the control points of a diagonal curve
verify a particular dependence on the points of the Bézier surface control
net.

######  Lemma 2.

(See [2]) If $\\{Q_{i}\\}_{i=0}^{2n}$ is the control polygon of the first
diagonal curve of a Bézier surface with control net
$\\{P_{i,j}\\}_{i,j=0}^{n}$, the control points, $Q_{k}$, of the diagonal
curve in terms of the control net are

$Q_{k}=\frac{1}{\binom{2n}{k}}\sum_{i=0}^{k}\binom{n}{i}\binom{n}{k-i}P_{i,k-i}\quad\text{for}\quad
k=0,\dots,2n.$

The fact is that $Q_{k}$ depends only on the control points $P_{i,j}$ such
that $i+j=k$. Let us consider only those control points, $Q_{k}$, with $k\leq
n$. The others follow by symmetry. Treat the $P_{i,j}$ with $i+j=k$ as a
degree $k$ Bézier curve and degree-elevate this curve $2n-k$ times, that is,
until it is of degree $2n$. Then, the middle control point on this degree-
elevated curve equals $Q_{k}$, (see [2] and Figure 2).

Figure 2: Control polygon (red) of the diagonal curve of a biquadratic Bézier
surface obtained by repeated degree elevation. The corresponding degree
elevation control points are also in red.

As we said, we are interested in finding the conditions that two Bézier curves
must meet in order to be the main diagonals of some Bézier surface. Obviously,
these curves must have the same midpoint, but, as we will see later in Theorem
1, condition (2) is necessary but not sufficient for two curves to be the
diagonal curves of a Bézier surface. In fact, this condition splits in two
more conditions.

As we will show, this splitting is a consequence of the fact that any diagonal
control point, $Q_{k}$, depends only on the control points in a diagonal line
of the control net, $P_{i,j}$ with $i+j=k$. Thus, condition (2) splits into
two sufficient conditions according to the parity of the indexes of the
control points. We will state this in Theorem 1, but beforehand we will need
the following auxiliary Lemma.

######  Lemma 3.

Let us suppose that $\\{Q_{i}\\}_{i=0}^{2n}$ is the control polygon of the
first diagonal curve, ${\bf x}(t,t)$, and that $\\{R_{i}\\}_{i=0}^{2n}$ of the
second diagonal curve, ${\bf x}(t,1-t)$, of a Bézier surface with control net
$\\{P_{i,j}\\}_{i,j=0}^{n}$, then $\\{(-1)^{i}Q_{i}\\}_{i=0}^{2n}$ and
$\\{(-1)^{n-i}R_{i}\\}_{i=0}^{2n}$ are the control points of the diagonal
curves of a Bézier surface with control net
$\\{(-1)^{i+j}P_{i,j}\\}_{i,j=0}^{n}$.

###### Proof.

As we said before, if $\\{Q_{i}\\}_{i=0}^{2n}$ is the control polygon of the
first diagonal curve of a Bézier surface with control net
$\\{P_{i,j}\\}_{i,j=0}^{n}$ then

$x(t,t)=\sum_{k=0}^{2n}\frac{B^{2n}_{k}(t)}{\binom{2n}{k}}\left(\sum_{i=0}^{k}\binom{n}{i}\binom{n}{k-i}P_{i,k-i}\right)=\sum_{k=0}^{2n}B^{2n}_{k}(t)Q_{k},$

and if we multiply this equation by $(-1)^{k}$ then

$\sum_{k=0}^{2n}\frac{B^{2n}_{k}(t)}{\binom{2n}{k}}\left(\sum_{i=0}^{k}\binom{n}{i}\binom{n}{k-i}(-1)^{i+k-i}P_{i,k-i}\right)=\sum_{k=0}^{2n}B^{2n}_{k}(t)(-1)^{k}Q_{k},$

and the first statement is proved.

The procedure is analogous for the second statement.

∎

Let us recall again the relation between the control net and the control
points of the diagonal curves:

$\displaystyle x(t,t)$
$\displaystyle=\sum_{k=0}^{2n}\frac{B^{2n}_{k}(t)}{\binom{2n}{k}}\left(\sum_{i=0}^{k}\binom{n}{i}\binom{n}{k-i}P_{i,k-i}\right)=\sum_{k=0}^{2n}B^{2n}_{k}(t)Q_{k},$
$\displaystyle x(t,1-t)$
$\displaystyle=\sum_{k=0}^{2n}\frac{B^{2n}_{k}(t)}{\binom{2n}{k}}\left(\sum_{i=0}^{k}\binom{n}{i}\binom{n}{k-i}P_{i,n-k+i}\right)=\sum_{k=0}^{2n}B^{2n}_{k}(t)R_{k}$

and then denote

$\left\\{\begin{array}[]{rcl}D^{n}_{k}({\mathcal{P}})&:=&\sum_{i=0}^{k}\binom{n}{i}\binom{n}{k-i}P_{i,k-i},\\\
E^{n}_{k}({\mathcal{P}})&:=&\sum_{i=0}^{k}\binom{n}{i}\binom{n}{k-i}P_{i,n-k+i},\end{array}\right.$
(3)

where

$Q_{k}=\frac{D^{n}_{k}({\mathcal{P}})}{\binom{2n}{k}}\quad\quad
R_{k}=\frac{E^{n}_{k}({\mathcal{P}})}{\binom{2n}{k}}.$ (4)

Now we will give the necessary and sufficient conditions for two control
polygons to be the control points of the two diagonal curves.

######  Theorem 1.

Two Bézier curves of degree $2n$ with control polygons

${\mathcal{Q}}=\\{Q_{i}\\}_{i=0}^{2n},\qquad{\mathcal{R}}=\\{R_{i}\\}_{i=0}^{2n},$

are the main diagonals of a tensor product Bézier surface of degree $n\times
n$ if and only if, for even $n$

$\begin{array}[]{rcl}\sum_{i=0}^{n}\binom{2n}{2i}Q_{2i}&=&\sum_{i=0}^{n}\binom{2n}{2i}R_{2i},\\\\[8.53581pt]
\sum_{i=0}^{n-1}\binom{2n}{2i+1}Q_{2i+1}&=&\sum_{i=0}^{n-1}\binom{2n}{2i+1}R_{2i+1},\end{array}$
(5)

or, for odd $n$

$\begin{array}[]{rcl}\sum_{i=0}^{n}\binom{2n}{2i}Q_{2i}&=&\sum_{i=0}^{n-1}\binom{2n}{2i+1}R_{2i+1},\\\\[8.53581pt]
\sum_{i=0}^{n-1}\binom{2n}{2i+1}Q_{2i+1}&=&\sum_{i=0}^{n}\binom{2n}{2i}R_{2i}.\end{array}$
(6)

###### Proof.

Let us suppose that $n$ is even. The proof is analogous if $n$ is odd.

First we will prove that if the curves are the main diagonals of a tensor
product Bézier surface with control net $\\{P_{i,j}\\}_{i,j=0}^{n}$ Eq. (5) is
fulfilled. By Lemma 1 we have

$\sum_{i=0}^{2n}\binom{2n}{i}Q_{i}=\sum_{i=0}^{2n}\binom{2n}{i}R_{i},$

which can be rewritten as

$\sum_{i=0}^{n}\binom{2n}{2i}Q_{2i}+\sum_{i=0}^{n-1}\binom{2n}{2i+1}Q_{2i+1}=\sum_{i=0}^{n}\binom{2n}{2i}R_{2i}+\sum_{i=0}^{n-1}\binom{2n}{2i+1}R_{2i+1}.$
(7)

Let us consider the Bézier surface with control net
$\\{(-1)^{i+j}P_{i,j}\\}_{i,j=0}^{n}$. By Lemma 3 the control polygons of the
main diagonals are $\\{(-1)^{i}Q_{i}\\}_{i=0}^{2n}$ and
$\\{(-1)^{n-i}R_{i}\\}_{i=0}^{2n}$, then if we apply Lemma 1 again we have

$\sum_{i=0}^{n}\binom{2n}{2i}Q_{2i}-\sum_{i=0}^{n-1}\binom{2n}{2i+1}Q_{2i+1}=\sum_{i=0}^{n}\binom{2n}{2i}R_{2i}-\sum_{i=0}^{n-1}\binom{2n}{2i+1}R_{2i+1}.$
(8)

Adding and substracting the equations (7) and (8), we get Eq.5.

In order to prove the converse we need to check that if Eq. (5) is fulfilled
then the linear system

$\left\\{\begin{array}[]{rcl}D^{n}_{k}({\mathcal{P}})&=&\binom{2n}{k}Q_{k},\qquad
k=0,\dots,2n\\\\[5.69054pt]
E^{n}_{k}({\mathcal{P}})&=&\binom{2n}{k}R_{k},\qquad
k=0,\dots,2n\end{array},\right.$ (9)

with the surface control points as unknown variables, has a solution. The
conditions in the statement are what we need for the linear system, Eq. (9),
to be compatible. It is easy to check that

$\left\\{\begin{array}[]{lcr}\sum_{i=0}^{n}D^{n}_{2i}({\mathcal{P}})-\sum_{i=0}^{n}E^{n}_{2i}({\mathcal{P}})&=&0\\\\[5.69054pt]
\sum_{i=0}^{n-1}D^{n}_{2i+1}({\mathcal{P}})-\sum_{i=0}^{n-1}E^{n}_{2i+1}({\mathcal{P}})&=&0\end{array}\right..$
(10)

We will show, in addition, that these are the only two vanishing linear
combinations of the left hand side of Eq. (9). Once this is proved then
conditions in Eq. (5), that are same linear combination but of the right hand
side terms of system (9), will imply the compatibility of the system.

So, let us consider an arbitrary vanishing linear combination of the left hand
side in system (9)

$\sum_{i=0}^{n}\lambda_{i}D^{n}_{i}({\mathcal{P}})+\sum_{i=0}^{n}\mu_{i}E^{n}_{i}({\mathcal{P}})=0.$
(11)

Any control point $P_{r,s}$ appears only once in $D^{n}_{r+s}({\mathcal{P}})$
and in $E^{n}_{n+r-s}({\mathcal{P}})$ and with the same coefficient
$\binom{n}{r}\binom{n}{s}$, see Eq. (3). Then, the coefficient of $P_{r,s}$ in
Eq. (11) is

$\binom{n}{r}\binom{n}{s}(\lambda_{r+s}+\mu_{n+r-s}),$

Then, since the linear combination in Eq. (11) must vanish for all $P_{r,s}$
in the control net, $\lambda_{r+s}+\mu_{n+r-s}=0$, the linear combination
coefficients are opposites, as in Eq. (10). This happens for all sums $r+s$,
so we have $\lambda_{r+s}=\lambda_{r-i+s+i}=-\mu_{n+r-s+2i}$, in other words,
all $\lambda_{r+s}$ coefficients with the same parity of subindex are the
opposite of all $\mu_{r+s}$ coefficients with the same parity. Thus, for all
even (odd) subindices $\lambda_{r+s}=\lambda_{even\,(odd)}=-\mu_{r+s}$. Thus

$\begin{array}[]{rcl}\lambda_{even}\,(\sum_{i=0}^{n}D^{n}_{2i}({\mathcal{P}})-\sum_{i=0}^{n}E^{n}_{2i}({\mathcal{P}}))+\lambda_{odd}\,(\sum_{i=0}^{n-1}D^{n}_{2i+1}({\mathcal{P}})-\sum_{i=0}^{n-1}E^{n}_{2i+1}({\mathcal{P}}))=0.\end{array}$

Therefore, this linear combination of equations in system (9) is a linear
combination of the equations in (10). In other words, as mentioned before, Eq.
(10) contains the two only vanishing linear combinations of the left hand side
of Eq. (9).

Since the right hand side verifies the same relation, Eq. (5), we can be sure
that system (9) has a solution, so a Bézier control net that fits the given
diagonals control polygons does exist.

∎

Although we are mainly interested in dealing with both main diagonals
simultaneously, it would be possible to deal with the case where only one
diagonal is prescribed. Notice that although we will not consider it here,
this case is simpler since the fundamental midpoint restriction, in Eq. (1),
does not apply.

######  Remark 1.

The proof of the splitting of Eq. (2) has been obtained thanks to auxiliary
Lemma 2, but there is an alternative way to prove it using rational Bézier
surfaces. As said in [1],(page 183), the two diagonal curves of a rational
Bézier surface, ${\bf x}(u,v)$, not only pass through point ${\bf
x}(\frac{1}{2},\frac{1}{2})$, but also meet one more time:
$\alpha_{1}(\infty)=\alpha_{2}(\infty)$. This second condition can be written
in terms of the (projective) control points of the two diagonals as in Eq. (2)
but after a change of sign of the odd terms. When the rational Bézier surface
reduces to a Bézier surface, in other words, when all the weights are the
same, we obtain Eq. (8).

The relation between the control points of the diagonal curves, given in
Theorem 1, makes it possible in the odd case to solve both central control
points, $Q_{n}$ and $R_{n}$, in terms of the other control points. However, it
is not possible to do the same in the even case since $Q_{n}$ and $R_{n}$
appear in the same equation. We illustrate this relation in the following
corollary.

######  Corollary 1.

${\mathcal{Q}}=\\{Q_{i}\\}_{i=0}^{2n}$ and
${\mathcal{R}}=\\{R_{i}\\}_{i=0}^{2n}$ are the control polygons of the main
diagonals of a Bézier surface if and only if for even $n$

$\begin{array}[]{rcl}Q_{n}&=&\frac{1}{\binom{2n}{n}}\left(\sum_{i=0}^{n}\binom{2n}{2i}R_{2i}-\sum_{i=0,2i\neq
n}^{n}\binom{2n}{2i}Q_{2i}\right),\\\\[8.53581pt]
R_{n+1}&=&\frac{1}{\binom{2n}{n}}\left(\sum_{i=0}^{n-1}\binom{2n}{2i+1}Q_{2i+1}-\sum_{i=0,2i\neq
n}^{n-1}\binom{2n}{2i+1}R_{2i+1}\right),\end{array}$ (12)

and, for odd $n$

$\begin{array}[]{rcl}Q_{n}&=&\frac{1}{\binom{2n}{n}}\left(\sum_{i=0}^{n-1}\binom{2n}{2i+1}R_{2i+1}-\sum_{i=0,2i\neq
n}^{n}\binom{2n}{2i}Q_{2i}\right),\\\\[8.53581pt]
R_{n}&=&\frac{1}{\binom{2n}{n}}\left(\sum_{i=0}^{n-1}\binom{2n}{2i+1}Q_{2i+1}-\sum_{i=0,2i\neq
n}^{n}\binom{2n}{2i}R_{2i}\right).\end{array}$ (13)

At this point, let us remark that user could give two curves the role of
diagonal curves which do not intersect. In that case, the substitution of both
central control points of the given diagonal curves by those computed with Eq.
12 in Corollary 1, for odd degre, would change the given input to a pair of
admissible diagonal curves, see Figure 3. For even degree, the corresponding
control point subtitutions should be done, Eq. (13).

Figure 3: If a designer gave two curves that do not intersect (dashed line),
one could modify them slightly by means of Corollary 1 in order to get two
admissible diagonal curves.

For odd degree, the control point which is modified is the central one, so the
shape of the new diagonal curves is not too different to the initial one. In
the case of even degree, if there was some symmetry in the given diagonals it
could be lost in the modified diagonal curves. Nevertheless, a way of solving
this handicap easily would be to carry out a degree elevation of the even
degree curves given by the user, and then proceed with the degree-elevated odd
case.

An alternative way of adapting the information provided by the user was given
in papers [12] and [13], where the authors minimized the distance between the
given and new diagonal curves. Now, in the same fashion but considering the
necessary conditions in Theorem 1 (or Corollary 1), an analogous minimization
procedure could be performed.

######  Definition 1.

Two tensor product Bézier surfaces of degree $n\times n$ are said to be
diagonals equivalent if they have the same main diagonal curves.

######  Proposition 1.

The class of diagonals equivalent surfaces is an affine subspace that can be
parameterized by $(n-1)^{2}$ control points.

###### Proof.

Let us compute the dimension of the affine subspace of solutions of system
(9). The number of variables, $\\{P_{i,j}\\}_{i,j=0}^{n}$, is $(n+1)^{2}$. The
number of equations of the linear system is $2(2n+1)=4n+2$, but its rank is
two units less, as we have just seen in the proof of Theorem 1. Therefore, the
dimension is $(n+1)^{2}-(4n+2-2)=(n-1)^{2}$. ∎

######  Example 1.

For $n=2$ the two conditions in Eq. (5) are

$\left\\{\begin{array}[]{rcl}Q_{0}+6Q_{2}+Q_{4}&=&R_{0}+6R_{2}+R_{4},\\\
Q_{1}+Q_{3}&=&R_{1}+R_{3}.\end{array}\right.$

Therefore, if we solve it, as stated in Corollary 1,

$\displaystyle Q_{2}=\frac{1}{6}\left(R_{0}+6R_{2}+R_{4}-Q_{0}-Q_{4}\right),$
$\displaystyle R_{3}=Q_{1}+Q_{3}-R_{1},$

then

$\begin{array}[]{rcl}{\mathcal{Q}}&=&\\{Q_{0},Q_{1},\frac{1}{6}\left(R_{0}+6R_{2}+R_{4}-Q_{0}-Q_{4}\right),Q_{3},Q_{4}\\}\\\\[8.53581pt]
{\mathcal{R}}&=&\\{R_{0},R_{1},R_{2},Q_{1}+Q_{3}-R_{1},R_{4}\\}\end{array}$

are the control polygons of two Bézier curves admissible as main diagonals of
a degree $2\times 2$ Bézier surface. Indeed, any Bézier surface with a control
net

$\left(\begin{array}[]{ccc}Q_{0}&\bf{P_{0,1}}&R_{0}\\\
2Q_{1}-\bf{P_{0,1}}&\frac{1}{4}\left(6R_{2}-Q_{0}-Q_{4}\right)&2R_{1}-{\bf
P_{0,1}}\\\ R_{4}&{\bf P_{0,1}}+2Q_{3}-2R_{1}&Q_{4}\\\ \end{array}\right)$

has as its main diagonals the Bézier curves with the control polygons
${\mathcal{Q}}$ and ${\mathcal{R}}$. Notice that once the diagonal curves are
prescribed there is only one degree of freedom, $P_{0,1}$.

######  Example 2.

For $n=3$ from Corollary 1 we have

$\begin{array}[]{rcl}R_{3}&=&\frac{1}{20}\left(Q_{0}+15Q_{2}+15Q_{4}-6R_{1}-6R_{5}\right),\\\\[8.53581pt]
Q_{3}&=&\frac{1}{20}\left(R_{0}+15R_{2}+15R_{4}+R_{6}-6Q_{1}-6Q_{5}\right),\end{array}$

then ${\mathcal{Q}}=\\{Q_{i}\\}_{i=0}^{6}$ and
${\mathcal{R}}=\\{R_{i}\\}_{i=0}^{6},$ are the control polygons of two Bézier
curves that are admissible as main diagonals of a degree $3\times 3$ Bézier
surface with a control net

$\left(\begin{array}[]{cccc}Q_{0}&\bf{P_{0,1}}&\bf{P_{0,2}}&R_{0}\\\
2Q_{1}-\bf{P_{0,1}}&\bf{P_{1,1}}&\bf{P_{1,2}}&2R_{1}-\bf{P_{0,2}}\\\
5Q_{2}-{\bf P_{0,2}}-3{\bf P_{1,1}}&P_{2,1}&P_{2,2}&5R_{2}-{\bf P_{0,1}}-3{\bf
P_{1,2}}\\\ R_{6}&P_{3,1}&P_{3,2}&Q_{6}\\\ \end{array}\right)$

where

$\displaystyle
P_{2,1}=\frac{5}{3}R_{2}+\frac{5}{3}R_{4}-\frac{2}{3}Q_{1}-\frac{2}{3}Q_{5}-{\bf
P_{1,2}}$ $\displaystyle
P_{2,2}=\frac{5}{3}Q_{2}+\frac{5}{3}Q_{4}-\frac{2}{3}R_{1}-\frac{2}{3}R_{5}-{\bf
P_{1,1}}$ $\displaystyle P_{3,1}={\bf P_{0,2}}+3{\bf P_{1,1}}-5Q_{2}+2R_{5}$
$\displaystyle P_{3,2}={\bf P_{0,1}}+3{\bf P_{1,2}}+2Q_{5}-5R_{2}$

Notice that there are four degrees of freedom,
$P_{0,1},P_{0,2},P_{1,1},P_{1,2}$, in other words, the space of all degree $3$
tensor product Bézier surfaces with given diagonals is an affine space of
dimension $2^{2}$. In Figure 4, with the same given diagonal curves, we plot
three different surfaces that belong to this affine space.

Figure 4: $n=3$. Three Bézier surfaces with the same prescribed main
diagonals.

##  3 Generation of a surface with a prescribed boundary and compatible
diagonal curves

Let us define another binary equivalence relation in the set of tensor product
Bézier surfaces.

######  Definition 2.

Two tensor product Bézier surfaces of degree $n\times n$ are said to be
boundary and diagonals equivalent if they have the same boundary and the same
main diagonals curves.

######  Proposition 2.

The class of boundary and diagonals equivalent surfaces is an affine subspace
that can be parameterized by $(n-3)^{2}$ control points.

###### Proof.

Initially, for a degree $n\times n$ Bézier surface with a prescribed boundary
we have $(n-1)^{2}$ free interior control points. However, since the main
diagonal curves must start and end at the corner control points of the net and
moreover, the starting and ending derivatives of the diagonal curves are also
prescribed by the boundary points, that is to say,

$\begin{array}[]{rclcccl}Q_{0}&=&P_{0,0}&&Q_{1}&=&\frac{1}{2}\left(P_{1,0}+P_{0,1}\right)\\\\[2.84526pt]
Q_{2n-1}&=&\frac{1}{2}\left(P_{n-1,n}+P_{n,n-1}\right)&&Q_{2n}&=&P_{n,n}\\\\[2.84526pt]
R_{0}&=&P_{n,0}&&R_{1}&=&\frac{1}{2}\left(P_{n-1,0}+P_{n,1}\right)\\\\[2.84526pt]
R_{2n-1}&=&\frac{1}{2}\left(P_{1,n}+P_{0,n-1}\right)&&R_{2n}&=&P_{0,n}\\\\[2.84526pt]
\end{array}$ (14)

then, the system in Eq. (9), has eight fewer equations,

$\left\\{\begin{array}[]{rcl}D^{n}_{k}({\mathcal{P}})&=&\binom{2n}{k}Q_{k},\qquad
k=2,\dots,2n-2\\\\[5.69054pt]
E^{n}_{k}({\mathcal{P}})&=&\binom{2n}{k}R_{k},\qquad
k=2,\dots,2n-2\end{array}.\right.$ (15)

We show in the proof of Proposition 1, that the rank of the system in Eq. (9)
is $4n$, so now the rank of the reduced system, in Eq. (15), is $4(n-2)$.
Therefore, the number of free control points is $(n-1)^{2}-4(n-2)=(n-3)^{2}$.

∎

According to this result, for degrees $n=2$ and $n=3$, if we prescribe the
boundary and the main diagonal curves, the Bézier surface is totally
determined. The situation changes for $n\geq 4$. For example, for $n=4$ there
is a family of Bézier surfaces, which is parameterized by just one interior
control point, in such a way that any element of this family has the same
prescribed boundary and diagonal curves.

Let us give some low degree examples in order to show the conditions that two
control polygons ${\mathcal{Q}}$ and ${\mathcal{R}}$ must meet in order to be
those of the diagonal curves of a Bézier surface with a prescribed boundary.

######  Example 3.

For $n=3$. When the boundary is prescribed the endpoints of the diagonal
curves and the respective tangent lines at these endpoints are fixed, namely
$Q_{0},Q_{1},Q_{5},Q_{6}$ and $R_{0},R_{1},R_{5},R_{6}$, see Eq. (14).
Therefore, the diagonal curves can be controlled by $Q_{2},Q_{4}$ and
$R_{2},R_{4}$, which can be interpreted as meaning that the curvature at the
endpoints remains free. Hence, we fix the boundary of the Bézier surface as
follows:

$\left(\begin{array}[]{cccc}P_{0,0}&P_{0,1}&P_{0,2}&P_{0,3}\\\\[2.84526pt]
P_{1,0}&\frac{1}{3}\left(5Q_{2}-P_{0,2}-P_{2,0}\right)&\frac{1}{3}\left(5R_{2}-P_{0,1}-P_{2,3}\right)&P_{1,3}\\\\[2.84526pt]
P_{2,0}&\frac{1}{3}\left(5R_{4}-P_{1,0}-P_{3,2}\right)&\frac{1}{3}\left(5Q_{4}-P_{1,3}-P_{3,1}\right)&P_{2,3}\\\\[2.84526pt]
P_{3,0}&P_{3,1}&P_{3,2}&P_{3,3}\\\ \end{array}\right)$

and using Corollary 1 and the conditions at the corners, in Eq. (14), the
control polygons of the main diagonals are

${\mathcal{Q}}=\left\\{P_{0,0},\frac{1}{2}\left(P_{0,1}+P_{1,0}\right),Q_{2},Q_{3},Q_{4},\frac{1}{2}\left(P_{2,3}+P_{3,2}\right),P_{3,3}\right\\}$
${\mathcal{R}}=\left\\{P_{0,3},\frac{1}{2}\left(P_{0,2}+P_{1,3}\right),R_{2},R_{3},R_{4},\frac{1}{2}\left(P_{2,1}+P_{3,1}\right),P_{3,0}\right\\},$

where

$\displaystyle
Q_{3}=\frac{3}{4}\left(R_{2}+R_{4}\right)+\frac{1}{20}\left(P_{0,3}+P_{3,0}\right)-\frac{3}{20}\left(P_{0,1}+P_{1,0}+P_{2,3}+P_{3,2}\right)$
$\displaystyle
R_{3}=\frac{3}{4}(Q_{2}+Q_{4})+\frac{1}{20}\left(P_{0,0}+P_{3,3}\right)-\frac{3}{20}\left(P_{0,2}+P_{1,3}+P_{2,0}+P_{3,1}\right).$

Therefore, if we fix the boundary of a cubic Bézier curve and its (compatible)
diagonals, then the surface is fully determined.

In all the following figures the control points that can be freely prescribed
are represented by large points, while those that are determined (by the large
ones) are represented by small points. The Bézier surface control net is shown
in green and the diagonals control points in red.

For example, in Figure 5, the prescribed boundary control points are, large
green points, and the diagonal curves that can be controlled by $Q_{2},Q_{4}$
and $R_{2},R_{4}$, are large red points. The remaining control points in the
net and in the diagonals polygons are determined and are represented by small
points.

Figure 5: $n=3$. Prescribed diagonal curves and boundary (properly related).

######  Example 4.

For $n=4$. If the boundary of the Bézier surface is prescribed, then the
interior control points are

$\left(\begin{array}[]{ccc}\frac{7Q_{2}}{4}-\frac{3}{8}\left(P_{0,2}+P_{2,0}\right)&\mathbf{P_{1,2}}&\frac{7R_{2}}{4}-\frac{3}{8}\left(P_{0,2}+P_{2,4}\right)\\\\[2.84526pt]
\frac{1}{6}\left(-P_{0,3}-P_{3,0}+14Q_{3}\right)-\mathbf{P_{1,2}}&P_{2,2}&P_{2,3}\\\\[2.84526pt]
\frac{7R_{6}}{4}-\frac{3}{8}\left(P_{2,0}+P_{4,2}\right)&P_{3,2}&\frac{7Q_{6}}{4}-\frac{3}{8}\left(P_{2,4}+P_{4,2}\right)\\\
\end{array}\right)$

where

$\displaystyle
P_{2,2}=\frac{1}{36}\left(-P_{0,0}+6P_{0,2}+6P_{2,0}+6P_{2,4}+6P_{4,2}-P_{4,4}-28Q_{2}-28Q_{6}+70R_{4}\right),$
$\displaystyle
P_{3,2}=\frac{1}{6}\left(P_{0,3}-P_{1,0}+P_{3,0}-P_{4,3}-14Q_{3}+14R_{5}\right)+\mathbf{P_{1,2}},$
$\displaystyle
P_{2,3}=\frac{1}{6}\left(-P_{0,3}+P_{1,0}-P_{1,4}-P_{3,0}-P_{4,1}+P_{4,3}+14Q_{3}+14Q_{5}-14R_{5}\right)-\mathbf{P_{1,2}}.$

Let us remark that, in this case, the diagonals are controlled by the eight
control points $Q_{2},Q_{3},Q_{5},Q_{6},$ $R_{2},R_{4},R_{5},R_{6}$ but, in
addition, $P_{1,2}$ is free. The two diagonals do not depend on $P_{1,2}$.
Indeed, the control polygons are

$\begin{array}[]{lll}{\mathcal{Q}}&=\left\\{P_{0,0},\frac{1}{2}\left(P_{0,1}+P_{1,0}\right),Q_{2},Q_{3},Q_{4},Q_{5},Q_{6},\frac{1}{2}\left(P_{3,4}+P_{4,3}\right),P_{4,4}\right\\},\\\
{\mathcal{R}}&=\left\\{P_{0,4},\frac{1}{2}\left(P_{0,3}+P_{1,4}\right),R_{2},R_{3},R_{4},R_{5},R_{6},\frac{1}{2}\left(P_{3,0}+P_{4,1}\right),P_{4,0}\right\\},\end{array}$

with

$\displaystyle
Q_{4}=R_{4}+\frac{2}{5}\left(R_{2}+R_{6}-Q_{2}-Q_{6}\right)+\frac{1}{70}\left(-P_{0,0}+P_{0,4}+P_{4,0}-P_{4,4}\right),$
$\displaystyle
R_{3}=\frac{1}{14}\left(P_{0,1}-P_{0,3}+P_{1,0}-P_{1,4}-P_{3,0}+P_{3,4}-P_{4,1}+P_{4,3}\right)+Q_{3}+Q_{5}-R_{5}.$

Figure 6: $n=4$. Same boundary curves for all figures but two different pairs
of main diagonals. In each column, with the same diagonal curves, two surfaces
are obtained by changing the free control point $P_{1,2}$, in blue.

##  4 Generation of a surface with a prescribed boundary and tangent planes
along it with compatible diagonal curves

Let us define the following binary equivalence relation in the set of tensor
product Bézier surfaces.

######  Definition 3.

Two tensor product Bézier surfaces of degree $n\times n$ are said to be
$C^{1}$-boundary and diagonals equivalent if they have the same boundary, the
same tangent planes along it and the same main diagonal curves.

######  Proposition 3.

The class of $C^{1}$-boundary and diagonals equivalent surfaces is an affine
subspace that can be parameterized by $(n-5)^{2}$ control points.

###### Proof.

For a degree $n\times n$ Bézier surface with a prescribed $C^{1}$-boundary we
have $(n-3)^{2}$ free interior control points, that is to say, the boundary
control points and their neighboring lines in the net are prescribed. If we
also prescribe the two main diagonals, these $(n-3)^{2}$ control points are
related into a compatible linear system of rank $4(n-4)$, since $16$ equations
from the initial system with rank $4n$, in Eq. 9, only relate prescribed
points. Therefore, the number of free control points is
$(n-3)^{2}-4(n-4)=(n-5)^{2}$. ∎

According to this result and similarly to the previous work, if we prescribe
the boundary, the tangent planes along it and the two diagonal curves, then
the smallest degree surfaces with free control points are degree $6\times 6$
surfaces. For $n=6$ there is a family of Bézier surfaces, parameterized by
just one interior control point such that, any element of this family has the
same prescribed boundary and tangent plane along it and the same prescribed
diagonal curves.

As before, we give some low degree examples.

######  Example 5.

For $n=5$. If we fix the boundary, the tangent planes along it and the main
diagonals we determine the whole surface. Corollary 1 determines the central
control points, $Q_{5}$ and $R_{5}$, of the main diagonals and moreover
$Q_{0},Q_{1},Q_{2},Q_{3},Q_{7},Q_{8},Q_{9},Q_{10}$ and
$R_{0},R_{1},R_{2},R_{3},R_{7},R_{8},R_{9},R_{10}$ depend on prescribed points
of the control net, see Eq. (4). The interior $(n-3)^{2}$ control points of
the net are as follows:

$\left(\begin{array}[]{cc}\frac{1}{20}\left(42Q_{4}-P_{0,4}-10P_{1,3}-10P_{3,1}-P_{4,0}\right)&\frac{1}{20}\left(42R_{4}-P_{0,1}-10P_{1,2}-10P_{3,4}-P_{4,5}\right)\\\
\frac{1}{20}\left(42R_{6}-P_{1,0}-10P_{2,1}-10P_{4,3}-P_{5,4}\right)&\frac{1}{20}\left(42Q_{6}-P_{1,5}-10P_{2,4}-10P_{4,2}-P_{5,1}\right)\\\
\end{array}\right),$

where $Q_{4},Q_{6}$ and $R_{4},R_{6}$ are the way of controlling the shape of
the diagonal curves.

Figure 7: $n=5$. Prescribed boundary and tangent planes along the boundary,
shown as large green points, and the control points $Q_{4},Q_{6}$ and
$R_{4},R_{6}$, shown as large red points, let us control the shape of the
diagonals. Again after prescribing the $C^{1}$-boundary control points and the
diagonal curves the whole surface is determined.

######  Example 6.

For $n=6$. There is a one parameter family of surfaces with the same boundary
and tangent plane along it and with the same diagonal curves. In other words,
the subspace of $C^{1}$-boundary and diagonals equivalent $6\times 6$ Bézier
surfaces can be parameterized by one control point, meaning that the dimension
of the subspace is one. The interior $(n-3)^{2}$ control points are

$\displaystyle
P_{2,2}=\frac{1}{15}\left(33Q_{4}-P_{0,4}-8P_{1,3}-8P_{3,1}-P_{4,0}\right),$
$\displaystyle
P_{2,4}=\frac{1}{15}\left(33R_{4}-P_{0,2}-8P_{1,3}-8P_{3,5}-P_{4,6}\right),$
$\displaystyle
P_{3,2}=\frac{1}{50}\left(132Q_{5}-P_{0,5}-15P_{1,4}-15P_{4,1}-P_{5,0}\right)-{\bf
P_{2,3}},$ $\displaystyle
P_{3,3}=\frac{1}{400}\left(924R_{6}-495Q_{4}-495Q_{8}-P_{0,0}+15P_{0,4}-36P_{1,1}+120P_{1,3}+15P_{2,6}\right.$
$\displaystyle\left.\qquad+120P_{3,1}+120P_{3,5}+15P_{4,0}+120P_{5,3}-36P_{5,5}+15P_{6,2}-P_{6,6}\right),$
$\displaystyle
P_{3,4}=\frac{1}{50}\left(132R_{5}-P_{0,1}-15P_{1,2}-15P_{4,5}-P_{5,6}\right)-{\bf
P_{2,3}},$ $\displaystyle
P_{4,2}=\frac{1}{15}\left(33R_{8}-P_{2,0}-8P_{3,1}-8P_{5,3}-P_{6,4}\right),$
$\displaystyle P_{4,3}={\bf
P_{2,3}}+\frac{1}{50}\left(132Q_{7}-132R_{5}+P_{0,1}+15P_{1,2}-P_{1,6}-15P_{2,5}+15P_{4,5}\right.$
$\displaystyle\left.\qquad-15P_{5,2}+P_{5,6}-P_{6,1}\right),$ $\displaystyle
P_{4,4}=\frac{1}{15}\left(33Q_{8}-P_{2,6}-8P_{3,5}-8P_{5,3}-P_{6,2}\right),$

which depend on the free parameter $P_{2,3}$, $Q_{4},Q_{5},Q_{7},Q_{8}$ and
$R_{4},R_{5},R_{6},R_{8}$ are the diagonal curves control points that can be
prescribed.

Figure 8: $n=6$. Given a boundary, the tangent plane along it and the diagonal
curves of a Bézier surface, $Q_{4},Q_{5},Q_{7},Q_{8}$ and
$R_{4},R_{5},R_{6},R_{8}$ being the diagonal curves control points that can be
prescribed, there is still one free control point, $P_{2,3}$ (black).

After studing all these different methods for generating surfaces from
prescribed information let us make only a comment about multi-patch surfaces.
Two patches could be, of course, continously connected along matching
boundaries, and they could also be $C^{1}$ connected when we prescribe tangent
planes along the boundaries of neighbour patches and they agree. But the thing
is that, since the tangent lines at the diagonal curves at the corner points
depend on the boundary control points, see Eq. (4), if certain continuity
conditions on the diagonal curves are wanted, it implies that conditions on
the prescribed boundary and tangent planes must be imposed. A detailed study
of this issue, but out of our scope here, shows that if we want the diagonals
to be well connected, then the connection of the patches at the common point
to four patches must be $C^{1}$.

In short, if we consider the problem of multi-patches and we want a good
connection of diagonal curves, this entails some conditions on the boundary
and the tangent planes.

##  5 Conclusion

First, in this paper, we study the necessary and sufficient conditions that
two curves must verify in order to constitute the diagonal curves of a tensor
product Bézier surface, which allows us to determine tensor product Bézier
surfaces with prescribed diagonal curves.

The possibility of controlling the diagonals would increase the control of the
resulting surface, so, the goal we achieve here is to provide a method that
makes it possible to prescribe not only the diagonal curves of the surface but
also its boundary, or its boundary and the tangent planes along it.

Whereas for low degrees, it is easy to check that there is only one Bézier
surface with prescribed main diagonals and boundary, this is no longer true
for higher degrees. For fixed diagonals and boundary not a unique Bézier
surface but a family of Bézier surfaces is determined. Such a family has the
structure of an affine subspace whose dimension is also determined here.

In other words, there is an equivalence class of Bézier surfaces that have the
same diagonals, or the same diagonals and boundary as well as having the same
diagonals and meeting the same $C^{1}$-boundary conditions.

In fact, what we are able to prove is that the set of all degree $n\times n$
Bézier surfaces where both the diagonal curves and the boundary are prescribed
is an affine space of dimension $(n-3)^{2}$ whereas the set of surfaces with
prescribed diagonals and $C^{1}$-boundary conditions has the dimension
$(n-5)^{2}$. Therefore, although only for degrees $n\leq 3$ and $n\leq 5$,
respectively, a unique Bézier surface exists as a solution. For higher
degrees, there are free control points, that could also be determined by
functional minimization, considering functionals defined on the surface, such
as the Dirichlet functional.

## Acknowledgements

This work was partially supported by Grant PGC2018-094889-B-100 funded by
MCIN/AEI/ 10.13039/
501100011033 and by “ERDF A way of making Europe”. The first author has also
been funded by Ministerio de Ciencia e Innovación (Spain) through project
PID2019-104927GB-C21, and by Universitat Jaume I (grants UJI-B2019-17 and
GACUJI/2020/05).

We thank the referees for their careful reading of our manuscript and for
taking the time to comment on it and offering constructive suggestions.

## References

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# Dilaton photoproduction in a magnetic dipole field of pulsars and magnetars

M.O. Astashenkov  Physics department Moscow State University, Moscow 119991,
Russia

###### Abstract

According to Einstein-Maxwell-Dilaton theory, the dilaton field $\psi$ can be
produced by electromagnetic fields with non-zero Maxwell invariant. So
electromagnetic wave propagating in an external electromagnetic field is a
typical source of dilaton radiation. For study dilaton photoproduction in
astrophysical conditions it’s interesting to consider plane elliptically
polarized electromagnetic wave propagating in the electromagnetic field of
magnetic dipole ${\bf m}$ of pulsars and magnetars. The dilation field
equation is solved in case $|\psi|\ll 1$. The angular distribution dilaton
radiation is studied in every point of space. It’s shown that spectral
composition of dilatons is similar to spectral composition of plane
electromagnetic wave. Amount of dilaton energy radiated in time and all
directions is greatest in condition
$(B_{1}^{2}-B_{2}^{2})(m_{x}^{2}-m_{y}^{2})\geq 0,$ where $B_{1}$ and $B_{2}$
are electromagnetic wave amplitudes along the axes of polarization ellipse.
This condition is valid for many neutron star systems.

dilaton, neutron star, electromagnetic wave, angular distribution of dilaton
radiation

###### pacs:

23.23.+x, 56.65.Dy

##  I Introduction

Nowadays in scientific literature there are few theories for Goldstone
bosons:arions [1, 2, 3, 4, 5], axions [6, 7, 8] and dilatons [9, 10, 11, 12,
13, 14]. One of the main sources of these particles is electromagnetic fields
and waves. Equations for these bosons in the classic limit have similar form.
In particular, corresponding to string theory system of interacting Maxwell
(U(1) gauge) and scalar (dilaton) fields design, so called Einstein-Maxwell-
Dilaton theory.

According to [15], the action of Einstein-Maxwell-Dilaton theory can be
written as

$S=\int d^{4}x\Big{\\{}a_{0}(\partial\Psi)^{2}+a_{1}e^{-2{\cal
K}\Psi}F^{nm}F_{nm}\Big{\\}},$ (1)

where $\Psi$ is a dilaton field, $a_{0},a_{1}$ and $\cal K$ are gauge
constants and $F_{nm}$ is Maxwell tensor.

In string theory ${\cal K}=1$, the five-dimensional Kaluza-Klein theory
results in the value ${\cal K}=\sqrt{3}$; In this work the constant ${\cal K}$
is arbitrary.

The field equations in Minkowski spacetime obtained from action (1) have the
form:

$\vbox{\hrule\hbox to5.0pt{\vrule width=0.5pt,height=5.0pt\hfill\vrule
width=0.5pt,height=5.0pt} \hrule}\Psi={a_{1}{\cal K}\over a_{0}}e^{-2{\cal
K}\Psi}F_{nm}F^{nm}={2a_{1}{\cal K}\over a_{0}}e^{-2{\cal K}\Psi}\big{[}{\bf
B^{2}}-{\bf E^{2}}\big{]},$ (2) ${\partial\over\partial
x^{n}}\Big{[}e^{-2{\cal K}\Psi}F^{nm}\Big{]}=0\;,$ (3)

where $\bf E$ is the electric field strength and $\bf B$ is magnetic
induction.

As dilaton field hasn’t been discovered, one can assume that dilaton field is
weak in Solar system and $|\psi|\ll 1$. In this case equation (3) can be
expressed as

$\vbox{\hrule\hbox to5.0pt{\vrule width=0.5pt,height=5.0pt\hfill\vrule
width=0.5pt,height=5.0pt} \hrule}\Psi=\frac{2a_{1}{\cal K}}{a_{0}}\big{[}{\bf
B^{2}}-{\bf E^{2}}\big{]}\;.$ (4)

According to equations (3) - (4), the invariant ${\bf B^{2}}-{\bf E^{2}}$ is
the source of the dilaton field. Beside this invariant is equal to zero for
wave zone of every electromagnetic wave, dilaton photoproduction is possible
from near zone or in area where there is a superposition of electromagnetic
fields with non-zero invariant ${\bf B^{2}}-{\bf E^{2}}$.

##  II Basic equation and its solution

Consider a plane elliptically polarized electromagnetic wave with frequency
$\omega$ propagating among axis $z$. Fields E and B of electromagnetic wave
have the following form:

$\displaystyle{\bf B}={\bf e_{x}}B_{1}\cos(kz-\omega t)-{\bf
e_{y}}B_{2}\sin(kz-\omega t),$ (5) $\displaystyle{\bf E}=-{\bf
e_{x}}B_{2}\sin(kz-\omega t)-{\bf e_{y}}B_{1}\cos(kz-\omega t),$

where $k=\omega/c$, $B_{1}$ and $B_{2}$ are the electromagnetic wave
amplitudes among principal axes of the polarization ellipse.

For example, if $B_{1}=B_{2}$ then the wave (5) has circle polarization; if
$B_{1}=0$ or $B_{2}=0$ then the wave (5) is linear polarized. In other cases
wave has elliptical polarization.

Assume that there is a neutron star with radius $R_{S}$ rotating around
magnetic dipole momentum ${\bf m}$ in the origin of the axis. The magnetic
induction ${\bf B}$ of the neuron star for $r>R_{S}$ has the form:

${\bf B}=\frac{3({\bf m}\cdot{\bf r}){\bf r}-{\bf m}r^{2}}{r^{5}}\;,$ (6)

For now a few hundred neutron stars have been discovered [16, 17] whose
rotating axis doesn’t coincide with the axis of magnetic dipole momentum. Such
stars radiate electromagnetic waves in magnetic dipole approximations (pulsars
and magnetars). But there must be neutron stars whose dipole momentum axis
coincides with the rotation axis. In this case there is no electromagnetic
radiation and their magnetic field (6) must be static.

Substituting superposition of electromagnetic fields (5) and (6) to expression
(4) and discarding the time-independent terms, one can obtain

$\displaystyle\vbox{\hrule\hbox to5.0pt{\vrule
width=0.5pt,height=5.0pt\hfill\vrule width=0.5pt,height=5.0pt}
\hrule}\Psi=\frac{4a_{1}{\cal K}}{a_{0}r^{5}}\Big{\\{}3({\bf m}\cdot{\bf
r})\big{[}xB_{1}\cos(kz-\omega t)-yB_{2}\sin(kz-\omega t)\big{]}$ (7)
$\displaystyle-r^{2}\big{[}m_{x}B_{1}\cos(kz-\omega
t)-m_{y}B_{2}\sin(kz-\omega t)\big{]}\Big{\\}}.$

Exact solution of the Eq. (7) is found to be

$\displaystyle\Psi=\frac{2a_{1}{\cal
K}}{ka_{0}}\Big{\\{}k\Big{[}\frac{(m_{x}x+m_{y}y)}{r^{2}(r-z)}-\frac{m_{z}}{r^{2}}\Big{]}\times\Big{[}B_{1}x\cos
k(r-ct)-B_{2}y\sin k(r-ct)\Big{]}$ (8)
$\displaystyle-\Big{[}\frac{(m_{x}x+m_{y}y)(2r-z)}{r^{3}(r-z)^{2}}-\frac{m_{z}}{r^{3}}\Big{]}\times\Big{[}B_{1}x[\sin
k(r-ct)-\sin k(z-ct)]$ $\displaystyle+B_{2}y[\cos k(r-ct)-\cos
k(z-ct)]\;\Big{]}+\frac{1}{r(r-z)}\Big{[}B_{1}m_{x}[\sin k(r-ct)$
$\displaystyle-\sin k(z-ct)]+B_{2}m_{y}[\cos k(r-ct)-\cos
k(z-ct)]\;\Big{]}\Big{\\}}.$

It follows from this expression that spectral composition of dilatons
coincides with the spectral composition of plane electromagnetic wave.

Rewrite expression (8) in spherical coordinates:

$\displaystyle\Psi={2a_{1}{\cal K}\over
ka_{0}}\Big{\\{}k\Big{[}\frac{[m_{x}\cos\phi+m_{y}\sin\phi]\sin\theta}{r(1-\cos\theta)}-{m_{z}\over
r}\Big{]}\Big{[}B_{1}\cos\phi\cos k(r-ct)$ (9) $\displaystyle-
B_{2}\sin\phi\sin
k(r-ct)\Big{]}\sin\theta-\Big{[}\frac{[m_{x}\cos\phi+m_{y}\sin\phi](2-\cos\theta)\sin\theta}{r^{2}(1-\cos\theta)^{2}}-{m_{z}\over
r^{2}}\Big{]}$ $\displaystyle\times\Big{[}B_{1}\cos\phi[\sin k(r-ct)-\sin
k(r\cos\theta-ct)]+B_{2}\sin\phi[\cos k(r-ct)$ $\displaystyle-\cos
k(r\cos\theta-
ct)]\Big{]}\sin\theta+\frac{1}{r^{2}(1-\cos\theta)}\Big{[}B_{1}m_{x}[\sin
k(r-ct)$ $\displaystyle-\sin k(r\cos\theta-ct)]+B_{2}m_{y}[\cos k(r-ct)-\cos
k(r\cos\theta-ct)]\Big{]}\Big{\\}}.$

It should be noted that in limit $\theta\to 0$ dilaton field has no
singularity

$\lim_{\theta\to 0}\Psi={2a_{1}{\cal K}\over a_{0}}\Big{\\{}B_{1}m_{x}\cos
k(r-ct)-B_{2}m_{y}\sin k(r-ct)\Big{\\}}.$ (10)

Thus, dilaton field has finite value everywhere out of neutron star. Using the
expression (9) one can study angular distribution of the dilaton radiation.

##  III Angular distribution of the dilaton radiation

By definition [7, 18], the amount of energy $dI$ emitted by the source per
unit time through the solid angle $d\Omega$ is given by the formula:

$\frac{dI}{d\Omega}=\lim_{r\to\infty}r({\bf W}\cdot{\bf r})\;,$ (11)

where ${\bf W}$ is the energy flux density vector associated with the
components of the stress-energy tensor $T^{ik}$ by the relations:
$W^{\alpha}=T^{0\alpha}$.

For free dilatonic field the stress-energy tensor $T^{ik}$ has the form:

$T^{ik}=2a_{0}g^{in}g^{km}\big{\\{}{\partial\Psi\over\partial
x^{n}}{\partial\Psi\over\partial x^{m}}-{1\over
2}g^{ik}{\partial\Psi\over\partial x^{n}}{\partial\Psi\over\partial
x^{m}}g^{nm}\big{\\}}.$ (12)

It follows that for $a_{0}>0$ dilaton energy density is positive for every
distribution of electromagnetic fields.

$T^{00}=a_{0}\Big{\\{}{1\over c^{2}}\left({\partial\Psi\over\partial
t}\right)^{2}+\left({\partial\Psi\over\partial
x}\right)^{2}+\left({\partial\Psi\over\partial
y}\right)^{2}+\left({\partial\Psi\over\partial z}\right)^{2}\Big{\\}}\geq 0.$
(13)

Angular distribution of dilaton radiation produced by elliptically polarized
electromagnetic wave (5) propagating in magnetic field of neutron star (6) can
be calculated by formula:

${dI\over d\Omega}=r\big{(}{\bf r\ W}\big{)}=-2a_{0}r(\vec{r}\
\vec{\nabla}\Psi){\partial\Psi\over\partial t},$ (14)

In the expression above it is taken into account that the dilaton field $\psi$
has a special point in $\theta=0$.

Substituting (8) to (14) one can obtain angular distribution averaged over the
period of electromagnetic wave $T=2\pi/\omega$:

$\displaystyle\overline{\frac{dI}{d\Omega}}=\frac{4a_{1}^{2}{\cal
K}^{2}c}{a_{0}r^{2}}\Big{\\{}\;4k^{2}r^{2}(B_{1}^{2}\cos^{2}\phi+B_{2}^{2}\sin^{2}\phi)(m_{x}\cos\phi+m_{y}\sin\phi)^{2}+2kr\big{(}$
(15) $\displaystyle B_{1}B_{2}m_{y}\cos\phi-
B_{1}B_{2}m_{x}\sin\phi-2krB_{1}^{2}m_{z}\sin\theta\cos^{2}\phi-2krB_{2}^{2}m_{z}\sin\theta$
$\displaystyle\times\sin^{2}\phi\;\big{)}(m_{x}\cos\phi+m_{y}\sin\phi)+krB_{1}B_{2}m_{z}\sin\theta(m_{x}\sin\phi-
m_{y}\cos\phi)$
$\displaystyle+\;(1-\cos\theta)\times\Big{[}2k^{2}r^{2}m_{z}^{2}(B_{1}^{2}\cos^{2}\phi+B_{2}^{2}\sin^{2}\phi)-4k^{2}r^{2}(B_{1}^{2}\cos^{2}\phi+B_{2}^{2}\sin^{2}\phi)$
$\displaystyle\times(m_{x}\cos\phi+m_{y}\sin\phi)^{2}+kr\big{(}2krB_{1}^{2}m_{z}\sin\theta\cos^{2}\phi+2krB_{2}^{2}m_{z}\sin\theta\sin^{2}\phi$
$\displaystyle+B_{1}B_{2}m_{x}\sin\phi-
B_{1}B_{2}m_{y}\cos\phi\big{)}(m_{x}\cos\phi+m_{y}\sin\phi)\Big{]}$
$\displaystyle+(1-\cos\theta)^{2}\Big{[}k^{2}r^{2}(m_{x}\cos\phi+m_{y}\sin\phi)^{2}-k^{2}r^{2}m_{z}^{2}+\big{(}\;3(m_{x}\cos\phi+m_{y}\sin\phi)^{2}$
$\displaystyle-3m_{z}^{2}-2m_{z}\sin\theta(m_{x}\cos\phi+m_{y}\sin\phi)\;\big{)}[1-\cos
kr(1-\cos\theta)]+\big{(}\;4krm_{z}^{2}+2kr$ $\displaystyle\times
m_{z}\sin\theta(m_{x}\cos\phi+m_{y}\sin\phi)-5kr(m_{x}\cos\phi+m_{y}\sin\phi)^{2}\;\big{)}\sin
kr(1-\cos\theta)\Big{]}$
$\displaystyle\times(B_{1}^{2}\cos^{2}\phi+B_{2}^{2}\sin^{2}\phi)\;+\;(1-\cos\theta)^{3}\times\Big{[}\big{(}m_{z}^{2}-(m_{x}\cos\phi+m_{y}\sin\phi)^{2}\big{)}$
$\displaystyle\times[1-\cos
kr(1-\cos\theta)]+kr(\;(m_{x}\cos\phi+m_{y}\sin\phi)^{2}-m_{z}^{2}\;)\sin
kr(1-\cos\theta)\Big{]}$
$\displaystyle\times\Big{(}B_{1}^{2}\cos^{2}\phi+B_{2}^{2}\sin^{2}\phi\Big{)}+\;\frac{[1-\cos
kr(1-\cos\theta)]}{(1-\cos\theta)^{2}}\times\Big{[}8(B_{1}^{2}\cos^{2}\phi+B_{2}^{2}\sin^{2}\phi)$
$\displaystyle\times(m_{x}\cos\phi+m_{y}\sin\phi)^{2}+2(B_{1}^{2}m_{x}^{2}+B_{2}^{2}m_{y}^{2})-8(B_{1}^{2}m_{x}\cos\phi+B_{2}^{2}m_{y}\sin\phi)$
$\displaystyle\times(m_{x}\cos\phi+m_{y}\sin\phi)\Big{]}+\;\frac{[1-\cos
kr(1-\cos\theta)]}{(1-\cos\theta)}\times\Big{[}2\big{(}B_{1}^{2}m_{x}\cos\phi+B_{2}^{2}m_{y}\sin\phi$
$\displaystyle-3B_{1}^{2}m_{z}\sin\theta\cos^{2}\phi-3B_{2}^{2}m_{z}\sin\theta\sin^{2}\phi+2krB_{1}B_{2}m_{y}\cos\phi-2krB_{1}B_{2}m_{x}\sin\phi\big{)}$
$\displaystyle\times(m_{x}\cos\phi+m_{y}\sin\phi)+3B_{1}^{2}m_{x}m_{z}\sin\theta\cos\phi+3B_{2}^{2}m_{y}m_{z}\sin\phi\sin\theta$
$\displaystyle-2krB_{1}B_{2}m_{y}m_{z}\sin\theta\cos\phi+2krB_{1}B_{2}m_{x}m_{z}\sin\phi\sin\theta-
B_{1}^{2}m_{x}^{2}-B_{2}^{2}m_{y}^{2}\Big{]}$
$\displaystyle+\;\frac{\sin(kr(1-\cos\theta))}{(1-\cos\theta)}\times\Big{[}m_{z}\sin\theta\big{(}B_{1}B_{2}m_{y}\cos\phi-
B_{1}B_{2}m_{x}\sin\phi-2krB_{1}^{2}m_{x}\cos\phi$
$\displaystyle-2krB_{2}^{2}m_{y}\sin\phi\big{)}-8kr(B_{1}^{2}\cos^{2}\phi+B_{2}^{2}\sin^{2}\phi)(m_{x}\cos\phi+m_{y}\sin\phi)^{2}$
$\displaystyle+2\big{(}2krB_{1}^{2}m_{x}\cos\phi+2krB_{2}^{2}m_{y}\sin\phi-
B_{1}B_{2}m_{y}\cos\phi+B_{1}B_{2}m_{x}\sin\phi$
$\displaystyle+2krB_{1}^{2}m_{z}\sin\theta\cos^{2}\phi+2krB_{2}^{2}m_{z}\sin\theta\sin^{2}\phi\big{)}(m_{x}\cos\phi+m_{y}\sin\phi)\Big{]}$
$\displaystyle+\;[1-\cos
kr(1-\cos\theta)]\times\Big{[}\big{(}5B_{1}^{2}m_{x}\cos\phi+5B_{2}^{2}m_{y}\sin\phi-4krB_{1}B_{2}m_{y}\cos\phi$
$\displaystyle+4krB_{1}B_{2}m_{x}\sin\phi+3B_{1}^{2}m_{z}\sin\theta\cos^{2}\phi+3B_{2}^{2}m_{z}\sin\theta\sin^{2}\phi\big{)}(m_{x}\cos\phi$
$\displaystyle+m_{y}\sin\phi)-10(B_{1}^{2}\cos^{2}\phi+B_{2}^{2}\sin^{2}\phi)(m_{x}\cos\phi+m_{y}\sin\phi)^{2}+m_{z}\sin\theta$
$\displaystyle\times\big{(}krB_{1}B_{2}m_{y}\cos\phi-
krB_{1}B_{2}m_{x}\sin\phi-2B_{1}^{2}m_{x}\cos\phi-2B_{2}^{2}m_{y}\sin\phi\big{)}\Big{]}$
$\displaystyle+\;\sin
kr(1-\cos\theta)\times\Big{[}4kr(B_{1}^{2}\cos^{2}\phi+B_{2}^{2}\sin^{2}\phi)(m_{x}\cos\phi+m_{y}\sin\phi)^{2}$
$\displaystyle+\big{(}4krB_{1}^{2}m_{z}\sin\theta\cos^{2}\phi+4krB_{2}^{2}m_{z}\sin\theta\sin^{2}\phi-4krB_{1}^{2}m_{x}\cos\phi$
$\displaystyle-4krB_{2}^{2}m_{y}\sin\phi+B_{1}B_{2}m_{y}\cos\phi-
B_{1}B_{2}m_{x}\sin\phi\big{)}(m_{x}\cos\phi$
$\displaystyle+m_{y}\sin\phi)+krm_{z}\sin\theta(B_{1}^{2}m_{x}\cos\phi+B_{2}^{2}m_{y}\sin\phi)\Big{]}$
$\displaystyle+\;(1-\cos\theta)[1-\cos
kr(1-\cos\theta)]\times\Big{[}2m_{z}^{2}(B_{1}^{2}\cos^{2}\phi+B_{2}^{2}\sin^{2}\phi)$
$\displaystyle+2(B_{1}^{2}\cos^{2}\phi+B_{2}^{2}\sin^{2}\phi)(m_{x}\cos\phi+m_{y}\sin\phi)^{2}+\big{(}4B_{1}^{2}m_{z}\sin\theta\cos^{2}\phi$
$\displaystyle+4B_{2}^{2}m_{z}\sin\theta\sin^{2}\phi-2B_{1}^{2}m_{x}\cos\phi-2B_{2}^{2}m_{y}\sin\phi+krB_{1}B_{2}m_{y}\cos\phi$
$\displaystyle-
krB_{1}B_{2}m_{x}\sin\phi\big{)}(m_{x}\cos\phi+m_{y}\sin\phi)\Big{]}+\;(1-\cos\theta)\sin
kr(1-\cos\theta)$
$\displaystyle\times\Big{[}6kr(B_{1}^{2}\cos^{2}\phi+B_{2}^{2}\sin^{2}\phi)(m_{x}\cos\phi+m_{y}\sin\phi)^{2}-4krm_{z}^{2}(B_{1}^{2}\cos^{2}\phi$
$\displaystyle+B_{2}^{2}\sin^{2}\phi)+kr(B_{1}^{2}m_{x}\cos\phi+B_{2}^{2}m_{y}\sin\phi-7B_{1}^{2}m_{z}\sin\theta\cos^{2}\phi$
$\displaystyle-7B_{2}^{2}m_{z}\sin\theta\sin^{2}\phi)(m_{x}\cos\phi+m_{y}\sin\phi)\Big{]}\Big{\\}}.$

For dilaton radiation forward ($\theta=0$) one can obtain:

$\overline{\frac{dI}{d\Omega}}(\theta=0)={4a_{1}^{2}{\cal
K}^{2}\omega^{2}\over
a_{0}c}\Big{(}B_{1}^{2}m_{x}^{2}+B_{2}^{2}m_{y}^{2}\Big{)}.$ (16)

There is no dilaton radiation backward $dI/d\Omega(\theta=\pi)=0$.

Integrating over angles $\theta$, $\phi$ one can obtain amount of dilaton
energy $\overline{I}$ averaged over the period of electromagnetic wave in main
asymptotic approximation ($r\to\infty$) radiating per unit of time in all
directions:

$\overline{I}=\frac{8\pi a_{1}ck^{2}{\cal
K}}{3a_{0}}\Big{\\{}3(B_{1}^{2}m_{x}^{2}+B_{2}^{2}m_{y}^{2})+2(B_{1}^{2}+B_{2}^{2})m_{z}^{2}+B_{1}^{2}m_{y}^{2}+B_{2}^{2}m_{x}^{2}\Big{\\}}\;.$
(17)

It follows from expression (17) that amount of dilaton energy $\overline{I}$
per unit of time is the greatest in condition
$(B_{1}^{2}-B_{2}^{2})(m_{x}^{2}-m_{y}^{2})>0$.

##  IV conclusion

As it was shown, a plane elliptically polarized electromagnetic wave
propagating in an electromagnetic field of magnetic dipole produces a dilaton
wave, whose amplitude has a special point in $\theta=0$. But dilaton $\psi$
has a finite value at this point. That’s why the dilaton field has finite
value in an area where $r>R_{S}$.

The dilaton radiation forward ($\theta=0$) has the form:

$\overline{\frac{dI}{d\Omega}}(\theta=0)={4a_{1}^{2}{\cal
K}^{2}\omega^{2}\over
a_{0}c}\Big{(}B_{1}^{2}m_{x}^{2}+B_{2}^{2}m_{y}^{2}\Big{)}.$ (18)

There is no dilaton radiation backward $dI/d\Omega(\theta=\pi)=0$.

Simple analysis shows that the amount of dilaton energy $\overline{I}$ per
unit of time averaged over the period of electromagnetic wave is the greatest
in condition $(B_{1}^{2}-B_{2}^{2})(m_{x}^{2}-m_{y}^{2})>0$. This condition is
valid for many superpositions of electromagnetic waves and magnetic dipole
momentums of neuron stars.

##  V Acknowledgements

The research was carried out within the framework of the scientific program of
the National Center for Physics and Mathematics, the project ”Particle Physics
and Cosmology”.

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