Descartes’ Perfect Lens
Mark B. Villarino
Depto. de Matemática, Universidad de Costa Rica,
2060 San José, Costa Rica
February 12, 2013
Abstract
We give a new, elementary, purely analytical development of Descartes’ theorem
that a smooth connected surface is a perfect focusing lens if and only if it is a connected
subset of the ovoid obtained by revolving a cartesian oval around its axis of symmetry.
Contents
1 Introduction 1
2 The Analytical Problem (in two dimensions) 3
2.1 Both Fixed Points Are Finite . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2.1.1 The Differential Equation . . . . . . . . . . . . . . . . . . . . . . . . 4
2.1.2 The Solution of the Differential Equation . . . . . . . . . . . . . . . . 6
2.2 One Fixed Point At Infinity; One Fixed Point Finite . . . . . . . . . . . . . 7
2.2.1 The Differential Equation . . . . . . . . . . . . . . . . . . . . . . . . 7
2.2.2 The Solution of the Differential Equation . . . . . . . . . . . . . . . . 8
2.3 Both Fixed Points are at Infinity . . . . . . . . . . . . . . . . . . . . . . . . 9
3 Descartes’ Theorem 9
3.1 Drucker’s Characterization of a Surface of Revolution . . . . . . . . . . . . 9
3.2 Proof of Descartes’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
1 Introduction
Almost two thousand three hundred years ago, the hellenistic mathematician Diokles [4]
gave the first proof that a mirror in the shape of a paraboloid of revolution reflects all incident
light rays, which are parallel to its axis of symmetry, to a single point, which Kepler [8],
in 1604, called the focus.
http://arxiv.org/abs/0704.1059v4
After the advent of the calculus it was possible to prove that the only such reflecting
surface is generated by revolving a (proper or degenerate) parabola around its axis of sym-
metry. This is a very famous and well-known result, and is treated in many easily accessible
sources. See, for example, Spiegel [16].
All these proofs are based on Heron’s Law of Reflection, θI = θRf , where θRf is
the angle the reflected ray makes with the normal to the reflecting surface at the point of
incidence of the incoming ray and θI is the angle the incident ray makes with the normal.
One transforms Heron’s equation into the ordinary differential equation (ODE) of
the cross-section curve of the mirror. Drucker’s paper [5] would seem to be the final word
on the subject.
Unfortunately and surprisingly, the corresponding result for a lens, instead of a mirror,
is less well-known, at least among mathematicians (although [11] is pleasant attempt to alter
that). Yet the case of a lens, too, is quite fascinating and is treatable by elementary means.
The purpose of this paper is to remedy the situation and fill this gap.
Indeed, it all started in 1637, when Descartes [3] asked for the refractive analogue of
the parabolic mirror:
Which shape of lens will focus all rays from one radiant point source to one
single image point?
We will call such a lens a perfect lens .
Descartes discovered that the cross-section curve of the perfect lens, assumed to be
a surface of revolution, is a fourth degree curve known today as the cartesian oval . It
can be defined as the locus of points the ”weighted” sum of whose distances from two fixed
points is a constant:
d1 + nd2 = c (1.0.1)
where d1 and d2 are the distances from any point on the curve to the two fixed points, called
the foci, and n is a constant. If one focus is at the origin and the other is at the point (b, 0)
where b > 0 , the equation can be written:
(1 − n2)(x2 + y2) + 2n2bx + c2 − n2b2
= 4c2(x2 + y2) (1.0.2)
If n = ±1, the oval is the conic section :
x − b
c2−b2
) = 1 (1.0.3)
More information on cartesian ovals can be found in [14] and [15] and [18].
Descartes’ own treatment, which is not altogether easy to read (see [1]), shows that
the oval is a solution, but does not show that it is the only solution.
The only treatments of Descartes’ result that we have seen in the literature do not
appear in books on mathematics (!), but rather on optics (see Hecht [7] and Klein [9])
and use Fermat’s Principle: A light ray traverses the path between two points which takes
the least time.
A non-trivial computation, based on the calculus of variations, shows that the time the
light ray takes to go from the radiant point to the image point is constant for every point
of the cross-section curve of the perfect lens (since if the time were different in two points of
the curve, it would not be minimal), and therefore its equation is that of the cartesian oval.
Moreover such treatments make physical assumptions about the velocity of light in differ-
ent media, while, as our treatment will show, the problem is really one in pure mathematics.
We have not seen any treatment of the subject which is founded purely on Snell’s law
of refraction, which describes the relationship between the angle of incidence and the angle
of refraction when light passes the boundary between two isotropic media (media in which
the path of a light ray is a straight line). The law states:
If θI is the angle the incident ray makes with the normal to the boundary at the
point of refraction, and if θR is the angle the refracted ray makes with the normal,
then at all points of the boundary the ratio
sin θI
sin θR
where n, called the index of refraction, is constant .
Such a treatment of Descartes’ theorem would seem desireable, since it is the immediate
generalization of the corresponding treatment of the perfect reflective mirror .
In this paper, we will present a new, self-contained, elementary, purely analytical proof,
based on Snell’s Law and Drucker’s paper [5], of the following complete form of Descartes’
theorem:
Theorem 1. (Descartes’ Theorem) A smooth connected surface is a perfect lens if and
only if it is a connected subset of the ovoid obtained by revolving a cartesian oval around its
axis of symmetry.
2 The Analytical Problem (in two dimensions)
We begin by solving the following two-dimensional purely analytical problem:
It is required to find the equation, f(x, y) = 0, of a smooth connected curve, C,
for which the straight lines from two fixed points cut the normal in two angles
whose sines are in constant ratio.
Please note the absence of physical modeling. The problem is purely mathematical, as its
its solution.
We will find and solve an ordinary differential equation (ODE) for which the equa-
tion of the curve is the general solution. The ODE, in fact, will be a restatement of Snell’s
Definition 1. We call any curve C that solves the problem a perfect two-dimensional
lens with respect to the points F and F ′.
2.1 Both Fixed Points Are Finite
2.1.1 The Differential Equation
We assume a cartesian coordinate system in the xy-plane.
Let the two fixed points be O(0, 0) and B(b, 0) with b > 0 (Here is we use the assumption
that F and F ′ are finite and distinct). Let P(x, y) be a variable point on the curve f(x, y) = 0.
We assume P is in the first quadrant and we assume that the curve is concave downwards
at P (that is, if y(x) is the function defined implicitly by the equation f(x, y) = 0, then
y′′(x0) < 0) . Let l1 be the length of the line segment OP and let l2 be the length of PB.
Let MN be the normal to the curve where N is on the concave side of the curve and P is
between M and N. Let θ1 := MP̂O, the angle that OP forms with the normal MN , and let
θ2 := BP̂N , the angle that BP forms with the normal MN . Let PT be the tangent line
to the curve at P where T is the point on the x-axis where the tangent line crosses it. Let
φ := P T̂B, the angle, measured counter clockwise, the tangent line forms with the x-axis.
We will use the geometry of the figure to obtain formulas for sin θ1 and sin θ2 in terms
of x, y, and the derivative, y′. When we substitute these expressions into Snell’s Law, we
obtain the desired differential equation for the curve f(x, y) = 0.
By the law of cosines applied twice to the △OPB
cos PÔB =
b2 + l2
cos PB̂O :=
b2 + l2
cos PÔB = cos(θ1 + φ − 90) cosPB̂O = cos(90 − θ2 − φ)
= sin(θ1 + φ) = sin(θ2 + φ)
So, we obtain our fundamental formulas:
sin(θ1 + φ) =
b2 + l2
sin(θ2 + φ) :=
b2 + l2
(2.1.1)
Moreover, it is evident that
< θ1 + φ < π 0 6 θ2 + φ <
(2.1.2)
b2 + l2
b2 + l2
2b2 − 2bx
b − x
Therefore, equations (2.1.1) become
sin(θ1 + φ) =
sin(θ2 + φ) =
b − x
By (2.1.1) and (2.1.2) and the definition of the arcsin function, we obtain
π − (θ1 + φ) = arcsin
θ2 + φ = arcsin
b − x
and therefore
θ1 = (π − φ) − arcsin
θ2 = arcsin
b − x
whence,
sin θ1 = sin(π − φ) cos
arcsin
− cos(π − φ) sin
arcsin
= sin φ
+ cos φ
1 + y′2
1 + y′2
sin θ2 = cos φ sin
arcsin
b − x
− sin φ cos
arcsin
b − x
1 + y′2
b − x
1 + y′2
b − x
1 + y′2
b − x
(1 + y′2)
Now, our assumption is that Snell’s Law holds, i.e., that
sin θ1
sin θ2
where n is a constant, holds for every point P (x, y) of the curve. Substituting our two
formulas for sin θ1 and sin θ2 into this equation gives us the equation:
1 + y′2
1 + y′2
1 + y′2
b − x
(1 + y′2)
= n (2.1.3)
Solving this equation (2.1.3) for y′ we obtain the differential equation of the curve:
b − x
) (2.1.4)
2.1.2 The Solution of the Differential Equation
We use the “arrow” notation. “P ⇒ Q” means “the proposition P (logically) implies the
proposition Q.”
(2.1.4) ⇒
b − x
yy′ + x
nyy′ − n(b − x)
yy′ + x
x2 + y2
+ n ·
−(b − x) + yy′
(b − x)2 + y2
2yy′ + 2x
x2 + y2
+ n ·
−2(b − x) + 2yy′
(b − x)2 + y2
x2 + y2 + n
(b − x)2 + y2
x2 + y2 + n
(b − x)2 + y2 = c
for some (arbitrary) constant c. We have therefore proved:
Theorem 2. The general solution for the differential equation (2.1.4) of the perfect two-
dimensional lens, C, with respect to the points F = (0, 0) and F ′ = (b, 0), where b > 0, is
given by the equation:
x2 + y2 + n
(b − x)2 + y2 = c. (2.1.5)
As we saw, this is the equation (1.0.1) of a cartesian oval with foci at the points (0, 0)
and (b, 0).
We have assumed that b is finite in this analysis, i.e., that the two foci are a finite distance
apart.
Now we consider the limiting cases where one or both foci are “at infinity.” We will see
that we obtain proper or degenerate conic sections for these cases.
2.2 One Fixed Point At Infinity; One Fixed Point Finite
2.2.1 The Differential Equation
We will slightly alter the treatment for the case of two finite foci. To do so, we begin with
the following:
Definition 2. A point at infinity is specified by means of a line through the origin. The
line joining P to a point at infinity is the line through P parallel to the given line.
Points at infinity are not considered to be on C.
We assume that the fixed point F is at −∞ along the x-axis and that the fixed point F ′
is at the point (b, 0) of the x-axis, where b > 0.
Intuitively, this means that a beam of light from −∞, parallel to the x-axis, is brought to
a point focus at (b, 0) by a single refracting curve, f(x, y) = 0, of index n.
The line joining P to the point at infinity is the line parallel to the x-axis through P . θ1
is the angle the horizontal line through P (x, y) makes with the normal while θ2 is the angle
PF makes with the normal. Finally, l be the length of PF .
Then, the earlier derivation of the ODE is applicable. We need only observe that
θ1 + φ =
So, substituting our two new formulas for sin θ1 and sin θ2 into Snell’s Law gives us, instead
of (2.1.3), the new equation:
1 + y′2
1 + y′2
b − x
(1 + y′2)
After some rearrangement, we obtain the differential equation of the curve:
(b − x) − yy′
(b − x)2 + y2
· n = 0. (2.2.1)
2.2.2 The Solution of the Differential Equation
The ODE (2.2.1) can be solved by the same computations as we did for the ODE (2.1.4)
which lead us to the
Theorem 3. The general solution for the differential equation (2.2.1) of the perfect two-
dimensional lens, C, with the radiant point at −∞ is given by the equation:
x + n
(b − x)2 + y2 = c. (2.2.2)
where c is an arbitrary constant.
We observe that the equation (2.2.2) has the following interesting interpretation. The
equation (2.2.2) says that the ratio of the distance of the point P from the line x =
c − bn
1 − n
its distance from the point (b, 0) is the constant ±n, and thererefore, by the focus-directrix
definition, is a conic section.
This theorem takes a more elegant form if we assume that the curve C passes through the
origin. Then, the constant c = nb and, after rationalizing (2.2.2), we obtain ([10], problem
B-10, Chapter 20):
Theorem 4. The general solution for the differential equation (2.2.1) of the perfect two-
dimensional lens, C, is a conic section whose focus is the point where the light is focused
and whose excentricity is the reciprocal of the index of refraction.
1. If n2 6= 1, C given by the equation:
x − nb
) = 1 (2.2.3)
Therefore C is an ellipse if n2 > 1 or an hiperbola if n2 < 1, either one of which is
centered at
n + 1
2. If n = 1, then C is the segment of the x-axis given by 0 6 x 6 b.
3. If n = −1, then C is the parabola
y2 = 4bx (2.2.4)
The reader should compare this result with that of the form of the perfect reflecting
mirror already cited in [5]. If n < 0, then we get reflection instead of refraction.
Maesumi [11] used Fermat’s Principle to treat this case in a very elegant paper, al-
though his definition of the index of refraction is the reciprocal of our (standard) one.
2.3 Both Fixed Points are at Infinity
Keeping the notation of the case of the radiant point at −∞, we assume that the refracted
rays form a parallel beam in the direction such that
θ2 + φ = Constant,
but, this means that
sin(θ2 + φ) =
b − x
(b − x)2 + y2
where C is some constant. But the condition that C goes through the origin means that
C = 1,
and rationalizing the resulting equation we obtain:
Theorem 5. If both fixed points are at infinity, then the perfect lens C has the equation:
x = 0 (2.3.1)
That is, it is the vertical y-axis.
3 Descartes’ Theorem
3.1 Drucker’s Characterization of a Surface of Revolution
In 1992 [5] Drucker published a very interesting paper in which he treated the problem
of finding all perfect mirrors, i.e., mirrors which reflect all rays issuing from one radiant
point to one image point.
After showing that the two dimensional curve with the perfect reflecting property is a
proper or degenerate conic section, he (implicitly) proved the following characterization of a
surface of revolution. Drucker, himself, did not state it explicitly.
Theorem 6. Let F and F ′ be two fixed points. If, for each point P of the smooth connected
surface S the normal ~N at P lies in the subspace spanned by the vectors
FP and
F ′P , then
S is a surface of revolution whose axis of revolution is the line through F and F ′.
Proof. We offer a new proof of Drucker’s theorem. It is based on an idea in Salmon
[15] which goes back to Monge [12].
Since, by definition, the normal MN is in the subspace spanned by FP and F ′P , it is in
the plane of △FPF ′.
If MN is always parallel to FF ′, then S is a plane which is perpendicular to FF ′. We
exclude this degenerate case for the rest of the argument. (See (2.3.1)).
Therefore MN is not always parallel to FF ′. Thus, the infinite line MN intersects FF ′
at some point. This is the characteristic property of the surface S.
Let (α, β, γ) be a point on FF ′ and let (l, m, n) be the line’s direction numbers where we
assume l · m · n 6= 0. The corollaries deal with the case where one or more coefficients are
equal to zero. Then, the equation of the line FF ′ is
x − α
y − β
z − γ
= t (3.1.1)
where t is the common value of the three fractions.
F (x, y, z) = 0 (3.1.2)
be the equation of S, where F is a continuously differentiable function of x, y, and z in some
open set R, and let (x0, y0, z0) be the point P on S..
Since MN is normal to S at P , its equation is:
x − x0
Fx(x0, y0, z0)
y − y0
Fy(x0, y0, z0)
z − z0
Fz(x0, y0, z0)
= T (3.1.3)
where T is the common value of the three fractions, and where Fx(x0, y0, z0) ≡
evaluated
in (x0, y0, z0), and where the other denominators have a similar interpretation. We assume
that all three denominators are different from zero. The corollaries deal with the cases where
the denominators are equal to zero.
Solving equations (3.1.1) and (3.1.3) for x, y, and z, and then equating the values ob-
tained, we get the following homogeneous linear system for the unknowns t, T , and 1:
lt − Fx(x0, y0, z0)T + (α − x0) · 1 = 0
mt − Fy(x0, y0, z0)T + (β − y0) · 1 = 0
nt − Fz(x0, y0, z0)T + (γ − z0) · 1 = 0
The analytical condition that this system have a nontrivial solution, which it does by as-
sumption, is that the determinant of their coefficients vanish:
∣∣∣∣∣∣
Fx(x0, y0, z0) Fy(x0, y0, z0) Fz(x0, y0, z0)
l m n
x0 − α y0 − β z0 − γ
∣∣∣∣∣∣
= 0 (3.1.4)
The determinant on the left-hand side of (3.1.4) is (one half of ) the Jacobian of the three
functions
Ω := F (x, y, z), u := lx + my + nz, v := (x − α)2 + (y − β)2 + (z − γ)2, (3.1.5)
evaluated at the point P of S.
But the point P is totally arbitrary, which means the Jacobian (3.1.4) vanishes in a full
neighborhood of P , since F is a continuously differentiable function of x, y, and z in some
open set R. According to a classical theorem (see Buck [2], Goursat [6], Osgood [13],
and Taylor [17], if the Jacobian of the three functions vanishes identically, then three
functions are functionally dependent.
That means that there is a function, Ω(u, v), of the two variables u and v, defined and
continuously differentiable in a neighborhood of the point (u0, v0), where
u0 := lx0 + my0 + nz0, v0 := (x0 − α)2 + (y0 − β)2 + (z0 − γ)2
for which the equation
F (x, y, z) = Ω
lx + my + nz, (x − α)2 + (y − β)2 + (z − γ)2
(3.1.6)
holds identically in a neighborhood of (x0, y0, z0).
Now, the equation
lx + my + nz = u (3.1.7)
represents a plane which cuts the line FF ′ (represented by (3.1.1)) perpendicularly, while
the equation
(x − α)2 + (y − β)2 + (z − γ)2 = v (3.1.8)
represents a sphere of radius
v and with center (α, β, γ) on the line FF ′.
The points (x, y, z) which are on the plane, (3.1.7), and on the sphere, (3.1.8), simulta-
neously, are on their circle of intersection and this circle has its center on the line FF ′.
Therefore, the equation (3.1.2) of S, i.e., Ω(u, v) = 0, represents a surface generated
by a circle of variable radius whose center moves along the line FF ′ and whose plane is
perpendicular to that line.
Thus, every planar transverse section of S, perpendicular to FF ′, consists of one or more
circles whose centers are on the line FF ′.
That is, S is a surface of revolution with axis FF ′.
This completes the proof of Drucker’s theorem.
Corollary 1. If the z-axis is the axis of revolution, we may take the origin as the point
(α, β, γ), and the equation (3.1.2) becomes
F (x, y, z) = Ω
z, x2 + y2 + z2
(3.1.9)
There are similar simplifications in (3.1.6) if we take the other coordinate axes as the
axis of revolution.
Corollary 2. If Fx(x0, y0, z0) ≡ 0, then the normal is everywhere perpendicular to the x −
axis and the equation (3.1.2) becomes the cylinder of revolution:
F (x, y, z) = Ω
my + nz, (y − β)2 + (z − γ)2
(3.1.10)
There are similar simplifications if the other components of the normal are zero.
Remark 1. In order to apply the theorem on functional dependence which we used in
the above proof, we have to make sure that we comply with all the hypotheses. The only
one, which we did not explicitly state in the body of the proof is, using the notations of
(3.1.7) and (3.1.8), is that at least one of the three jacobians
∂(u, v)
∂(x, y)
∂(u, v)
∂(y, z)
∂(u, v)
∂(x, z)
, (3.1.11)
is different from zero at (x0, y0, z0).
We claim that even more is true in our case. We will prove that at least two of the
jacobians (3.1.11) are different from zero.
Suppose, to the contrary, that at least two of them are equal to zero, say
∂(u, v)
∂(x, y)
∂(u, v)
∂(y, z)
= 0 (3.1.12)
This leads to
x − α
y − β
y − β
z − γ
, (3.1.13)
respectively. By (3.1.13)
x − α
y − β
z − γ
(3.1.14)
which is the equation of the axis FF ′. But, this means that S is just the straight line axis,
which is excluded by the hypothesis that S is a smooth surface. Therefore, at least two of
the jacobians (3.1.11) are different from zero and the theorem on functional dependence is
applicable.
Remark 2. The proof shows that the characteristic property of a surface S of revolution is
that the normal to any point of S intersects the axis of revolution.
3.2 Proof of Descartes’ Theorem
We adapt Drucker’s definition
Definition 3. Let S be a smooth connected surface and let F and F ′ be points not in S. We
say that S is a perfect lens relative to F and F ′ if, for each point P in S:
1. the normal ~N at P lies in the subspace spanned by the vectors
FP and
F ′P , and
2. the sines of the angles which
FP and
F ′P form with that normal are in constant
ratio for every point P in S.
By condition 2 of the definition, the cross-section of S sliced out by the xy-plane is a
plane curve C which is a perfect two-dimensional lens relative to F and F ′.
That means that C is either (part of) a cartesian oval, or (part of) a conic section,
or a degenerate case of either one.
Therefore, by condition 1 and Drucker’s Theorem, a three dimensional perfect lens S
is (part of) a surface of revolution with axis FF ′ obtained by rotating a two-dimensional
perfect lens S around it.
This completes the proof of Descartes’ Theorem.
Acknowledgment
Support from the Vicerrectoŕıa de Investigación of the University of Costa Rica is acknowl-
edged.
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	Introduction
	The Analytical Problem (in two dimensions)
	Both Fixed Points Are Finite
	The Differential Equation
	The Solution of the Differential Equation
	One Fixed Point At Infinity; One Fixed Point Finite
	The Differential Equation
	The Solution of the Differential Equation
	Both Fixed Points are at Infinity
	Descartes' Theorem
	Drucker's Characterization of a Surface of Revolution
	Proof of Descartes' Theorem

  We give a new, elementary, purely analytical development of
\textsc{Descartes}' theorem that a smooth connected surface is a perfect
focusing lens if and only if it is a connected subset of the ovoid obtained by
revolving a cartesian oval around its axis of symmetry.

Introduction 1
2 The Analytical Problem (in two dimensions) 3
2.1 Both Fixed Points Are Finite . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2.1.1 The Differential Equation . . . . . . . . . . . . . . . . . . . . . . . . 4
2.1.2 The Solution of the Differential Equation . . . . . . . . . . . . . . . . 6
2.2 One Fixed Point At Infinity; One Fixed Point Finite . . . . . . . . . . . . . 7
2.2.1 The Differential Equation . . . . . . . . . . . . . . . . . . . . . . . . 7
2.2.2 The Solution of the Differential Equation . . . . . . . . . . . . . . . . 8
2.3 Both Fixed Points are at Infinity . . . . . . . . . . . . . . . . . . . . . . . . 9
3 Descartes’ Theorem 9
3.1 Drucker’s Characterization of a Surface of Revolution . . . . . . . . . . . . 9
3.2 Proof of Descartes’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
1 Introduction
Almost two thousand three hundred years ago, the hellenistic mathematician Diokles [4]
gave the first proof that a mirror in the shape of a paraboloid of revolution reflects all incident
light rays, which are parallel to its axis of symmetry, to a single point, which Kepler [8],
in 1604, called the focus.
http://arxiv.org/abs/0704.1059v4
After the advent of the calculus it was possible to prove that the only such reflecting
surface is generated by revolving a (proper or degenerate) parabola around its axis of sym-
metry. This is a very famous and well-known result, and is treated in many easily accessible
sources. See, for example, Spiegel [16].
All these proofs are based on Heron’s Law of Reflection, θI = θRf , where θRf is
the angle the reflected ray makes with the normal to the reflecting surface at the point of
incidence of the incoming ray and θI is the angle the incident ray makes with the normal.
One transforms Heron’s equation into the ordinary differential equation (ODE) of
the cross-section curve of the mirror. Drucker’s paper [5] would seem to be the final word
on the subject.
Unfortunately and surprisingly, the corresponding result for a lens, instead of a mirror,
is less well-known, at least among mathematicians (although [11] is pleasant attempt to alter
that). Yet the case of a lens, too, is quite fascinating and is treatable by elementary means.
The purpose of this paper is to remedy the situation and fill this gap.
Indeed, it all started in 1637, when Descartes [3] asked for the refractive analogue of
the parabolic mirror:
Which shape of lens will focus all rays from one radiant point source to one
single image point?
We will call such a lens a perfect lens .
Descartes discovered that the cross-section curve of the perfect lens, assumed to be
a surface of revolution, is a fourth degree curve known today as the cartesian oval . It
can be defined as the locus of points the ”weighted” sum of whose distances from two fixed
points is a constant:
d1 + nd2 = c (1.0.1)
where d1 and d2 are the distances from any point on the curve to the two fixed points, called
the foci, and n is a constant. If one focus is at the origin and the other is at the point (b, 0)
where b > 0 , the equation can be written:
(1 − n2)(x2 + y2) + 2n2bx + c2 − n2b2
= 4c2(x2 + y2) (1.0.2)
If n = ±1, the oval is the conic section :
x − b
c2−b2
) = 1 (1.0.3)
More information on cartesian ovals can be found in [14] and [15] and [18].
Descartes’ own treatment, which is not altogether easy to read (see [1]), shows that
the oval is a solution, but does not show that it is the only solution.
The only treatments of Descartes’ result that we have seen in the literature do not
appear in books on mathematics (!), but rather on optics (see Hecht [7] and Klein [9])
and use Fermat’s Principle: A light ray traverses the path between two points which takes
the least time.
A non-trivial computation, based on the calculus of variations, shows that the time the
light ray takes to go from the radiant point to the image point is constant for every point
of the cross-section curve of the perfect lens (since if the time were different in two points of
the curve, it would not be minimal), and therefore its equation is that of the cartesian oval.
Moreover such treatments make physical assumptions about the velocity of light in differ-
ent media, while, as our treatment will show, the problem is really one in pure mathematics.
We have not seen any treatment of the subject which is founded purely on Snell’s law
of refraction, which describes the relationship between the angle of incidence and the angle
of refraction when light passes the boundary between two isotropic media (media in which
the path of a light ray is a straight line). The law states:
If θI is the angle the incident ray makes with the normal to the boundary at the
point of refraction, and if θR is the angle the refracted ray makes with the normal,
then at all points of the boundary the ratio
sin θI
sin θR
where n, called the index of refraction, is constant .
Such a treatment of Descartes’ theorem would seem desireable, since it is the immediate
generalization of the corresponding treatment of the perfect reflective mirror .
In this paper, we will present a new, self-contained, elementary, purely analytical proof,
based on Snell’s Law and Drucker’s paper [5], of the following complete form of Descartes’
theorem:
Theorem 1. (Descartes’ Theorem) A smooth connected surface is a perfect lens if and
only if it is a connected subset of the ovoid obtained by revolving a cartesian oval around its
axis of symmetry.
2 The Analytical Problem (in two dimensions)
We begin by solving the following two-dimensional purely analytical problem:
It is required to find the equation, f(x, y) = 0, of a smooth connected curve, C,
for which the straight lines from two fixed points cut the normal in two angles
whose sines are in constant ratio.
Please note the absence of physical modeling. The problem is purely mathematical, as its
its solution.
We will find and solve an ordinary differential equation (ODE) for which the equa-
tion of the curve is the general solution. The ODE, in fact, will be a restatement of Snell’s
Definition 1. We call any curve C that solves the problem a perfect two-dimensional
lens with respect to the points F and F ′.
2.1 Both Fixed Points Are Finite
2.1.1 The Differential Equation
We assume a cartesian coordinate system in the xy-plane.
Let the two fixed points be O(0, 0) and B(b, 0) with b > 0 (Here is we use the assumption
that F and F ′ are finite and distinct). Let P(x, y) be a variable point on the curve f(x, y) = 0.
We assume P is in the first quadrant and we assume that the curve is concave downwards
at P (that is, if y(x) is the function defined implicitly by the equation f(x, y) = 0, then
y′′(x0) < 0) . Let l1 be the length of the line segment OP and let l2 be the length of PB.
Let MN be the normal to the curve where N is on the concave side of the curve and P is
between M and N. Let θ1 := MP̂O, the angle that OP forms with the normal MN , and let
θ2 := BP̂N , the angle that BP forms with the normal MN . Let PT be the tangent line
to the curve at P where T is the point on the x-axis where the tangent line crosses it. Let
φ := P T̂B, the angle, measured counter clockwise, the tangent line forms with the x-axis.
We will use the geometry of the figure to obtain formulas for sin θ1 and sin θ2 in terms
of x, y, and the derivative, y′. When we substitute these expressions into Snell’s Law, we
obtain the desired differential equation for the curve f(x, y) = 0.
By the law of cosines applied twice to the △OPB
cos PÔB =
b2 + l2
cos PB̂O :=
b2 + l2
cos PÔB = cos(θ1 + φ − 90) cosPB̂O = cos(90 − θ2 − φ)
= sin(θ1 + φ) = sin(θ2 + φ)
So, we obtain our fundamental formulas:
sin(θ1 + φ) =
b2 + l2
sin(θ2 + φ) :=
b2 + l2
(2.1.1)
Moreover, it is evident that
< θ1 + φ < π 0 6 θ2 + φ <
(2.1.2)
b2 + l2
b2 + l2
2b2 − 2bx
b − x
Therefore, equations (2.1.1) become
sin(θ1 + φ) =
sin(θ2 + φ) =
b − x
By (2.1.1) and (2.1.2) and the definition of the arcsin function, we obtain
π − (θ1 + φ) = arcsin
θ2 + φ = arcsin
b − x
and therefore
θ1 = (π − φ) − arcsin
θ2 = arcsin
b − x
whence,
sin θ1 = sin(π − φ) cos
arcsin
− cos(π − φ) sin
arcsin
= sin φ
+ cos φ
1 + y′2
1 + y′2
sin θ2 = cos φ sin
arcsin
b − x
− sin φ cos
arcsin
b − x
1 + y′2
b − x
1 + y′2
b − x
1 + y′2
b − x
(1 + y′2)
Now, our assumption is that Snell’s Law holds, i.e., that
sin θ1
sin θ2
where n is a constant, holds for every point P (x, y) of the curve. Substituting our two
formulas for sin θ1 and sin θ2 into this equation gives us the equation:
1 + y′2
1 + y′2
1 + y′2
b − x
(1 + y′2)
= n (2.1.3)
Solving this equation (2.1.3) for y′ we obtain the differential equation of the curve:
b − x
) (2.1.4)
2.1.2 The Solution of the Differential Equation
We use the “arrow” notation. “P ⇒ Q” means “the proposition P (logically) implies the
proposition Q.”
(2.1.4) ⇒
b − x
yy′ + x
nyy′ − n(b − x)
yy′ + x
x2 + y2
+ n ·
−(b − x) + yy′
(b − x)2 + y2
2yy′ + 2x
x2 + y2
+ n ·
−2(b − x) + 2yy′
(b − x)2 + y2
x2 + y2 + n
(b − x)2 + y2
x2 + y2 + n
(b − x)2 + y2 = c
for some (arbitrary) constant c. We have therefore proved:
Theorem 2. The general solution for the differential equation (2.1.4) of the perfect two-
dimensional lens, C, with respect to the points F = (0, 0) and F ′ = (b, 0), where b > 0, is
given by the equation:
x2 + y2 + n
(b − x)2 + y2 = c. (2.1.5)
As we saw, this is the equation (1.0.1) of a cartesian oval with foci at the points (0, 0)
and (b, 0).
We have assumed that b is finite in this analysis, i.e., that the two foci are a finite distance
apart.
Now we consider the limiting cases where one or both foci are “at infinity.” We will see
that we obtain proper or degenerate conic sections for these cases.
2.2 One Fixed Point At Infinity; One Fixed Point Finite
2.2.1 The Differential Equation
We will slightly alter the treatment for the case of two finite foci. To do so, we begin with
the following:
Definition 2. A point at infinity is specified by means of a line through the origin. The
line joining P to a point at infinity is the line through P parallel to the given line.
Points at infinity are not considered to be on C.
We assume that the fixed point F is at −∞ along the x-axis and that the fixed point F ′
is at the point (b, 0) of the x-axis, where b > 0.
Intuitively, this means that a beam of light from −∞, parallel to the x-axis, is brought to
a point focus at (b, 0) by a single refracting curve, f(x, y) = 0, of index n.
The line joining P to the point at infinity is the line parallel to the x-axis through P . θ1
is the angle the horizontal line through P (x, y) makes with the normal while θ2 is the angle
PF makes with the normal. Finally, l be the length of PF .
Then, the earlier derivation of the ODE is applicable. We need only observe that
θ1 + φ =
So, substituting our two new formulas for sin θ1 and sin θ2 into Snell’s Law gives us, instead
of (2.1.3), the new equation:
1 + y′2
1 + y′2
b − x
(1 + y′2)
After some rearrangement, we obtain the differential equation of the curve:
(b − x) − yy′
(b − x)2 + y2
· n = 0. (2.2.1)
2.2.2 The Solution of the Differential Equation
The ODE (2.2.1) can be solved by the same computations as we did for the ODE (2.1.4)
which lead us to the
Theorem 3. The general solution for the differential equation (2.2.1) of the perfect two-
dimensional lens, C, with the radiant point at −∞ is given by the equation:
x + n
(b − x)2 + y2 = c. (2.2.2)
where c is an arbitrary constant.
We observe that the equation (2.2.2) has the following interesting interpretation. The
equation (2.2.2) says that the ratio of the distance of the point P from the line x =
c − bn
1 − n
its distance from the point (b, 0) is the constant ±n, and thererefore, by the focus-directrix
definition, is a conic section.
This theorem takes a more elegant form if we assume that the curve C passes through the
origin. Then, the constant c = nb and, after rationalizing (2.2.2), we obtain ([10], problem
B-10, Chapter 20):
Theorem 4. The general solution for the differential equation (2.2.1) of the perfect two-
dimensional lens, C, is a conic section whose focus is the point where the light is focused
and whose excentricity is the reciprocal of the index of refraction.
1. If n2 6= 1, C given by the equation:
x − nb
) = 1 (2.2.3)
Therefore C is an ellipse if n2 > 1 or an hiperbola if n2 < 1, either one of which is
centered at
n + 1
2. If n = 1, then C is the segment of the x-axis given by 0 6 x 6 b.
3. If n = −1, then C is the parabola
y2 = 4bx (2.2.4)
The reader should compare this result with that of the form of the perfect reflecting
mirror already cited in [5]. If n < 0, then we get reflection instead of refraction.
Maesumi [11] used Fermat’s Principle to treat this case in a very elegant paper, al-
though his definition of the index of refraction is the reciprocal of our (standard) one.
2.3 Both Fixed Points are at Infinity
Keeping the notation of the case of the radiant point at −∞, we assume that the refracted
rays form a parallel beam in the direction such that
θ2 + φ = Constant,
but, this means that
sin(θ2 + φ) =
b − x
(b − x)2 + y2
where C is some constant. But the condition that C goes through the origin means that
C = 1,
and rationalizing the resulting equation we obtain:
Theorem 5. If both fixed points are at infinity, then the perfect lens C has the equation:
x = 0 (2.3.1)
That is, it is the vertical y-axis.
3 Descartes’ Theorem
3.1 Drucker’s Characterization of a Surface of Revolution
In 1992 [5] Drucker published a very interesting paper in which he treated the problem
of finding all perfect mirrors, i.e., mirrors which reflect all rays issuing from one radiant
point to one image point.
After showing that the two dimensional curve with the perfect reflecting property is a
proper or degenerate conic section, he (implicitly) proved the following characterization of a
surface of revolution. Drucker, himself, did not state it explicitly.
Theorem 6. Let F and F ′ be two fixed points. If, for each point P of the smooth connected
surface S the normal ~N at P lies in the subspace spanned by the vectors
FP and
F ′P , then
S is a surface of revolution whose axis of revolution is the line through F and F ′.
Proof. We offer a new proof of Drucker’s theorem. It is based on an idea in Salmon
[15] which goes back to Monge [12].
Since, by definition, the normal MN is in the subspace spanned by FP and F ′P , it is in
the plane of △FPF ′.
If MN is always parallel to FF ′, then S is a plane which is perpendicular to FF ′. We
exclude this degenerate case for the rest of the argument. (See (2.3.1)).
Therefore MN is not always parallel to FF ′. Thus, the infinite line MN intersects FF ′
at some point. This is the characteristic property of the surface S.
Let (α, β, γ) be a point on FF ′ and let (l, m, n) be the line’s direction numbers where we
assume l · m · n 6= 0. The corollaries deal with the case where one or more coefficients are
equal to zero. Then, the equation of the line FF ′ is
x − α
y − β
z − γ
= t (3.1.1)
where t is the common value of the three fractions.
F (x, y, z) = 0 (3.1.2)
be the equation of S, where F is a continuously differentiable function of x, y, and z in some
open set R, and let (x0, y0, z0) be the point P on S..
Since MN is normal to S at P , its equation is:
x − x0
Fx(x0, y0, z0)
y − y0
Fy(x0, y0, z0)
z − z0
Fz(x0, y0, z0)
= T (3.1.3)
where T is the common value of the three fractions, and where Fx(x0, y0, z0) ≡
evaluated
in (x0, y0, z0), and where the other denominators have a similar interpretation. We assume
that all three denominators are different from zero. The corollaries deal with the cases where
the denominators are equal to zero.
Solving equations (3.1.1) and (3.1.3) for x, y, and z, and then equating the values ob-
tained, we get the following homogeneous linear system for the unknowns t, T , and 1:
lt − Fx(x0, y0, z0)T + (α − x0) · 1 = 0
mt − Fy(x0, y0, z0)T + (β − y0) · 1 = 0
nt − Fz(x0, y0, z0)T + (γ − z0) · 1 = 0
The analytical condition that this system have a nontrivial solution, which it does by as-
sumption, is that the determinant of their coefficients vanish:
∣∣∣∣∣∣
Fx(x0, y0, z0) Fy(x0, y0, z0) Fz(x0, y0, z0)
l m n
x0 − α y0 − β z0 − γ
∣∣∣∣∣∣
= 0 (3.1.4)
The determinant on the left-hand side of (3.1.4) is (one half of ) the Jacobian of the three
functions
Ω := F (x, y, z), u := lx + my + nz, v := (x − α)2 + (y − β)2 + (z − γ)2, (3.1.5)
evaluated at the point P of S.
But the point P is totally arbitrary, which means the Jacobian (3.1.4) vanishes in a full
neighborhood of P , since F is a continuously differentiable function of x, y, and z in some
open set R. According to a classical theorem (see Buck [2], Goursat [6], Osgood [13],
and Taylor [17], if the Jacobian of the three functions vanishes identically, then three
functions are functionally dependent.
That means that there is a function, Ω(u, v), of the two variables u and v, defined and
continuously differentiable in a neighborhood of the point (u0, v0), where
u0 := lx0 + my0 + nz0, v0 := (x0 − α)2 + (y0 − β)2 + (z0 − γ)2
for which the equation
F (x, y, z) = Ω
lx + my + nz, (x − α)2 + (y − β)2 + (z − γ)2
(3.1.6)
holds identically in a neighborhood of (x0, y0, z0).
Now, the equation
lx + my + nz = u (3.1.7)
represents a plane which cuts the line FF ′ (represented by (3.1.1)) perpendicularly, while
the equation
(x − α)2 + (y − β)2 + (z − γ)2 = v (3.1.8)
represents a sphere of radius
v and with center (α, β, γ) on the line FF ′.
The points (x, y, z) which are on the plane, (3.1.7), and on the sphere, (3.1.8), simulta-
neously, are on their circle of intersection and this circle has its center on the line FF ′.
Therefore, the equation (3.1.2) of S, i.e., Ω(u, v) = 0, represents a surface generated
by a circle of variable radius whose center moves along the line FF ′ and whose plane is
perpendicular to that line.
Thus, every planar transverse section of S, perpendicular to FF ′, consists of one or more
circles whose centers are on the line FF ′.
That is, S is a surface of revolution with axis FF ′.
This completes the proof of Drucker’s theorem.
Corollary 1. If the z-axis is the axis of revolution, we may take the origin as the point
(α, β, γ), and the equation (3.1.2) becomes
F (x, y, z) = Ω
z, x2 + y2 + z2
(3.1.9)
There are similar simplifications in (3.1.6) if we take the other coordinate axes as the
axis of revolution.
Corollary 2. If Fx(x0, y0, z0) ≡ 0, then the normal is everywhere perpendicular to the x −
axis and the equation (3.1.2) becomes the cylinder of revolution:
F (x, y, z) = Ω
my + nz, (y − β)2 + (z − γ)2
(3.1.10)
There are similar simplifications if the other components of the normal are zero.
Remark 1. In order to apply the theorem on functional dependence which we used in
the above proof, we have to make sure that we comply with all the hypotheses. The only
one, which we did not explicitly state in the body of the proof is, using the notations of
(3.1.7) and (3.1.8), is that at least one of the three jacobians
∂(u, v)
∂(x, y)
∂(u, v)
∂(y, z)
∂(u, v)
∂(x, z)
, (3.1.11)
is different from zero at (x0, y0, z0).
We claim that even more is true in our case. We will prove that at least two of the
jacobians (3.1.11) are different from zero.
Suppose, to the contrary, that at least two of them are equal to zero, say
∂(u, v)
∂(x, y)
∂(u, v)
∂(y, z)
= 0 (3.1.12)
This leads to
x − α
y − β
y − β
z − γ
, (3.1.13)
respectively. By (3.1.13)
x − α
y − β
z − γ
(3.1.14)
which is the equation of the axis FF ′. But, this means that S is just the straight line axis,
which is excluded by the hypothesis that S is a smooth surface. Therefore, at least two of
the jacobians (3.1.11) are different from zero and the theorem on functional dependence is
applicable.
Remark 2. The proof shows that the characteristic property of a surface S of revolution is
that the normal to any point of S intersects the axis of revolution.
3.2 Proof of Descartes’ Theorem
We adapt Drucker’s definition
Definition 3. Let S be a smooth connected surface and let F and F ′ be points not in S. We
say that S is a perfect lens relative to F and F ′ if, for each point P in S:
1. the normal ~N at P lies in the subspace spanned by the vectors
FP and
F ′P , and
2. the sines of the angles which
FP and
F ′P form with that normal are in constant
ratio for every point P in S.
By condition 2 of the definition, the cross-section of S sliced out by the xy-plane is a
plane curve C which is a perfect two-dimensional lens relative to F and F ′.
That means that C is either (part of) a cartesian oval, or (part of) a conic section,
or a degenerate case of either one.
Therefore, by condition 1 and Drucker’s Theorem, a three dimensional perfect lens S
is (part of) a surface of revolution with axis FF ′ obtained by rotating a two-dimensional
perfect lens S around it.
This completes the proof of Descartes’ Theorem.
Acknowledgment
Support from the Vicerrectoŕıa de Investigación of the University of Costa Rica is acknowl-
edged.
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	Introduction
	The Analytical Problem (in two dimensions)
	Both Fixed Points Are Finite
	The Differential Equation
	The Solution of the Differential Equation
	One Fixed Point At Infinity; One Fixed Point Finite
	The Differential Equation
	The Solution of the Differential Equation
	Both Fixed Points are at Infinity
	Descartes' Theorem
	Drucker's Characterization of a Surface of Revolution
	Proof of Descartes' Theorem
