Planes, Spheres and Pseudospheres — Greg Egan
by Greg Egan
A sheet of paper or a chunk of cloth can lie completely flat on a desk, however it will also be rolled
up right into a cylinder or a cone, and twisted in sure methods, with out being stretched or sheared.
Figures drawn on the fabric when it lies flat retain their form, within the sense that distances measured inside the materials itself
are unchanged by the way in which it occurs to sit down in threedimensional house.
The geometry of strange paper is flat, Euclidean geometry, however what concerning the equal for
spherical geometry (which might be one thing like a chunk of orange peel),
or for hyperbolic geometry?
The purpose of this web page is to research the chances for all three circumstances: given
a chunk of an idealised twodimensional materials whose geometry is Euclidean, spherical, or hyperbolic, what shapes can it undertake in threedimensional Euclidean house?
Assumptions
 Our idealised materials is assumed to be infinitesimally skinny, and completely versatile.
For instance, a flat sheet may very well be rolled right into a cylinder whose radius is as small as we like, so
lengthy because it’s not zero. However we are going to not permit the fabric to include sharp folds, the place the normal to the floor undergoes a discontinuous bounce from one facet of the fold to the opposite.  The fabric is assumed to be completely inextensible.
So nevertheless a lot we bend it, it doesn’t stretch or shear: all distances alongside curves drawn
on the fabric, and all angles between such curves the place they cross, stay unchanged.  We’ll generally discuss options the place the fabric would possibly intersect itself in threedimensional
house, as a result of probably the most mathematically easy statements about basic options received’t rule
that out.  We’ll assume that every one the mathematical features describing the way in which the fabric is embedded in
threedimensional house are “sufficiently easy”: at any time when we have to take a spinoff of one thing,
that spinoff exists and is steady, besides probably on the boundary of the fabric.
Gaussian Curvature
There are three sorts of twodimensional house which have fixed Gaussian curvature, Okay:
[There are also spaces that consist of pieces of these spaces, with boundaries, or pieces where all or some ofthe boundaries have been joined together.]
Gaussian curvature, Okay, is a quantity that quantifies the way and extent during which the geometry of a twodimensional
house departs from Euclidean geometry. It may be outlined in many various methods, however one strategy is to have a look at the geometry of a circle: the set of factors that lie at a set distance from a given level, the place we have to measure
this alongside the geodesic curves
that minimise the space between factors.
On the floor of a sphere with radius a, we measure distances alongside
great circles,
and a circle of radius r centred on the pole in spherical coordinates will lie at an angle of
θ = r/a radians from the pole. A simple calculation offers the
circumference and space of such a circle:
Circumference(r) = 2 π a sin(r/a)
Space(r) = 2 π a^{2} (1 – cos(r/a))
Generally, a twodimensional house might need totally different curvature at totally different
factors, however we will nonetheless characterise the native geometry by inspecting the limiting behaviour because the radius r of the
circle turns into very small. If we take the primary two nonzero phrases of the
Taylor series
for the formulation above, we get:
Circumference(r) ≈ 2 π r – (π/(3 a^{2})) r^{3}
Space(r) ≈ π r^{2} – (π/(12 a^{2})) r^{4}
We are able to quantify the extent to which these formulation deviate from the same old Euclidean formulation for circumference and space, 2 π r and π r^{2}, by defining the Gaussian curvature Okay to be
both:
Okay = (3/π) lim_{r → 0} (2 π r – Circumference(r)) / r^{3}
or:
Okay = (12/π) lim_{r → 0} (π r ^{2} – Space(r)) / r^{4}
For the case of a sphere of radius a, each definitions yield:
Okay = 1 / a^{2}
One other method of characterising Gaussian curvature is by way of the notion of angular extra.
For any triangle within the aircraft whose sides are straight line segments, the sum of the
three inside angles on the vertices is precisely π. On a sphere of radius a, the sum of the
three angles for
a triangle whose sides are geodesics
is determined by the world of the triangle, A. Particularly:
Sumofangles(A) = π + A/a^{2} = π + Okay A
We received’t show this, however it’s straightforward to examine for some easy circumstances.
For instance, a triangle with one vertex on the north pole, and two which can be 90 levels aside on the equator,
may have an space oneeighth that of the entire sphere, or 4πa^{2}/8 = πa^{2}/2,
and since all three angles are rightangles, the sum of its angles might be 3π/2, in settlement
with the system. The native worth of the curvature Okay might be obtained within the limiting case of
a small triangle:
Okay = lim_{A→0} (Sumofangles(A) – π) / A
We now have used a sphere to show how geometry is affected by optimistic Gaussian curvature, however
precisely the identical mathematical relationships maintain in surfaces with damaging curvature.
So when Okay < 0, the circumference and space of a circle are bigger than these of a circle of the identical
radius within the aircraft, and the sum of the three angles of a triangle whose sides are geodesics might be
lower than π.
The outline we’re given of a twodimensional house will normally are available in both of two kinds. We may be
instructed how the house has been embedded
as a floor in threedimensional Euclidean house, just like the sphere we now have simply described.
Or we may be given an outline of the floor’s intrinsic geometry, by way of a
metric.
The appendix to this net web page explains how the Gaussian curvature is calculated in each these circumstances.
For a extra complete therapy, there are a lot of good textbooks,
similar to Lipschutz,^{[1]} that give an in depth account of “the classical principle of surfaces,” which offers with the geometry of twodimensional areas by way of surfaces embedded in threedimensional
house.
Intrinsically Flat Surfaces
There are 4 sorts of surfaces embedded in threedimensional Euclidean house which can be intrinsically flat, with a continuing Gaussian curvature of zero:
 planes
 cylinders
 cones
 tangent surfaces
A few of these classes might be seen as both particular circumstances or limiting circumstances of others; for instance, cylinders are like cones whose apex is moved infinitely distant.
Planes
That planes are flat is not only apparent, it’s the case the place, if our measure of intrinsic curvature gave any
different consequence than zero, we might throw it away and search for a brand new one. To examine that
we don’t want to do this, notice that we will parameterise any aircraft with the system:
x(u, v) = P + u a + v b
the place P is a degree on the aircraft and a and b are two mounted, linearly independent vectors which can be parallel to the aircraft. All three second derivatives of x(u, v) with respect to the
parameters are zero, so the second fundamental form, S, of the floor, outlined within the appendix,
might be zero in every single place, making its determinant zero, whereas the first fundamental form, g, may have a nonzero determinant as a result of a and b are linearly unbiased.
So the Gaussian curvature, Okay = det(S) / det(g), might be zero.
Cylinders
Given a curve c(s) that lies in a aircraft, the cylinder
generated by c(s) is the floor traced out by all of the traces perpendicular to the aircraft that go via factors on c(s).
If the curve is closed, the cylinder might be finite in a single course, and if the curve is selfintersecting,
the cylinder might be a selfintersecting floor, but when the curve is infinitely lengthy and by no means intersects itself,
the cylinder might be infinite in all instructions, and might be an embedding of your complete Euclidean aircraft.
Suppose the parameter s for the curve is its arc length, and let p be a unitlength regular
to the aircraft of the curve (we are going to reserve n for the traditional vector to the cylinder itself). Then we will parameterise the cylinder as:
x(u, v) = c(u) + v p
Following the strategies described within the appendix, we now have:
x_{u} = c‘(u)
x_{v} = p
n = x_{u} × x_{v} / x_{u} × x_{v} = c‘(u) × p
x_{uu} = c”(u)
x_{uv} = 0
x_{vv} = 0
g(u, v) =
x_{u} · x_{u} x_{u} · x_{v} x_{u} · x_{v} x_{v} · x_{v} = S(u, v) =
x_{uu} · n x_{uv} · n x_{uv} · n x_{vv} · n = Okay = det(S) / det(g) = 0 / 1 = 0
There’s a simple technique to assemble an
isometry (a onetoone
distancepreserving operate) between these cylinders and all or a part of the Euclidean aircraft: we map the purpose with
coordinates (u, v) on the cylinder to the purpose (x, y) within the traditional
Cartesian coordinates on the aircraft, setting:
x = u
y = v
The metric in u, v coordinates is precisely the identical because the Euclidean metric, so this
map preserves distances and angles measured inside the embedded floor.
If the curve c(s) by no means intersects itself, within the case of an infinitely lengthy curve we get an isometry between your complete Euclidean aircraft and the cylinder. For finite curves, or if we select to restrict the vary of the
v coordinate, the isometry is with a chunk of the aircraft.
If the cylinder is generated by a closed curve, the topology is now not that of a chunk of the aircraft, however
that of a chunk with two reverse edges joined collectively. Regionally, the excellence is just not essential, and the
Gaussian curvature stays zero in every single place, however clearly a round cylinder is just not globally isometric to
a rectangle, or a strip of finite width, as a result of some factors on the other edges find yourself nearer to one another
within the cylinder.
Cones
We’ll outline a cone because the floor traced out by all of the traces that go via a
curve c(s) and a few mounted level A, the apex of the cone, which doesn’t lie on the curve.
Not like the cylinder, we won’t require c(s) to be planar, however we are going to impose the situation
that the tangent to the curve by no means factors instantly in direction of the apex. We are able to parameterise this floor as:
x(u, v) = A + v (c(u) – A)
The unique curve is given by v = 1, and the apex by v = 0.
We then have, for v ≠ 0, and assuming c(s) is parameterised by arc size:
x_{u} = v c‘(u)
x_{v} = c(u) – A
n = x_{u} × x_{v} / x_{u} × x_{v} = v c‘(u) × (c(u) – A) / v c‘(u) × (c(u) – A)
x_{uu} = v c”(u)
x_{uv} = c‘(u)
x_{vv} = 0
g(u, v) =
x_{u} · x_{u} x_{u} · x_{v} x_{u} · x_{v} x_{v} · x_{v} =
v^{2} v c‘(u) · (c(u) – A) v c‘(u) · (c(u) – A) c(u) – A^{2} S(u, v) =
x_{uu} · n x_{uv} · n x_{uv} · n x_{vv} · n = Okay = det(S) / det(g) = 0 / (v^{2} (c(u) – A^{2} – (c‘(u) · (c(u) – A))^{2})) = 0
Generally, the apex, v = 0, won’t include a neighbourhood that appears like a part of a aircraft, so it must be
excluded, and the surfaces with v < 0 and v > 0 aren’t linked to one another. Relying on the precise form of the curve and the relative location of the apex, cones would possibly or would possibly
not be selfintersecting elsewhere.
Probably the most acquainted instance of a cone is a proper round cone, generated by a circle and an apex that lies instantly above the centre of the circle. The portion of the aircraft that’s embedded as a
proper round cone is all the time a wedge with an angle lower than 2π. Nonetheless, cones generated by
sufficiently meandering curves can come from wedges of any angle, together with angles better than or equal to 2π.
If the wedge angle is precisely 2π, as within the second illustration initially of this part,
then the cone would possibly or won’t rely as an embedding of your complete Euclidean aircraft, relying on how strict
we wish to be concerning the smoothness of the embedding on the apex. If the wedge angle is better than 2π,
as within the third illustration, then the
cone can’t be an embedding of only one copy of the Euclidean aircraft
(within the illustration,
an additional piece that comes from a second copy of the aircraft is colored inexperienced). However it is going to nonetheless be
a wonderfully good intrinsically flat house, as long as we exclude the apex.
To assemble an isometry between a cone described within the (u, v)
coordinates we used within the definition, and a part of the aircraft (or perhaps elements of multiple aircraft),
it’s best if we work in polar coordinates (r, θ). An isometry is then given by:
r = v c(u) – A
θ = ∫_{0}^{u} (1/c(t) – A) √[1 – ((c(t) – A) · c‘(t))^{2} / c(t) – A^{2}] dt
The integral for θ quantities to including up the space we journey alongside the curve perpendicular to a line in direction of the apex, divided by the space to the apex, to present us the angle we now have travelled “round” the apex to achieve the coordinate worth u.
For instance, if c(s)
is a circle of radius R and the apex lies at a distance H instantly above the centre of the circle,
we now have:
c(s) = (R cos(s/R), R sin(s/R), 0)
A = (0, 0, H)c(s) – A = (R cos(s/R), R sin(s/R), –H) = √(R^{2} + H^{2})
r = v c(u) – A = v √(R^{2} + H^{2})
θ = ∫_{0}^{u} (1/c(t) – A) √[1 – ((c(t) – A) · c‘(t))^{2} / c(t) – A^{2}] dt
= (1/√(R^{2} + H^{2})) ∫_{0}^{u} √[1 – ((R cos(t/R), R sin(t/R), –H) · (–sin(t/R), cos(t/R), 0))^{2} / (R^{2} + H^{2})] dt
= u/√(R^{2} + H^{2})
The u coordinate will vary from 0 to 2πR, so θ will vary from 0 to
2πR/√(R^{2} + H^{2}).
This confirms that any proper round cone has a wedge angle of lower than 2π.
However for different cones, if θ exceeds 2π we might want to get hold of the rest of the floor from
a number of further copies of the aircraft.
Tangent surfaces
Suppose we now have a curve c(s), parameterised by arc size s, that
has
no factors of inflection:
that’s, no factors the place the unitlength tangent to the curve, c‘(s), has a charge of change of zero as we transfer alongside the curve. This is similar as saying that for all s, c”(s) ≠ 0.
The tangent surface of c(s) is the floor consisting of all traces which can be tangents to the curve.
[In some literature, this is called the tangent developable; the term
“developable”
refers to any intrinsically flat surface.] We are able to parameterise the tangent floor as:
x(u, v) = c(u) + v c‘(u)
As a result of c(s) is parameterised by arc size, c‘(s) has a continuing size
of 1, and its charge of change, c”(s), is all the time orthogonal to it, since it could actually solely change course,
not size. We then have, for v ≠ 0:
x_{u} = c‘(u) + v c”(u)
x_{v} = c‘(u)
n = x_{u} × x_{v} / x_{u} × x_{v} = v c”(u) × c‘(u) / v c”(u) × c‘(u)
x_{uu} = c”(u) + v c”'(u)
x_{uv} = c”(u)
x_{vv} = 0
g(u, v) =
x_{u} · x_{u} x_{u} · x_{v} x_{u} · x_{v} x_{v} · x_{v} = S(u, v) =
x_{uu} · n x_{uv} · n x_{uv} · n x_{vv} · n = Okay = det(S) / det(g) = 0 / (v^{2} c”(u)^{2}) = 0
It won’t be instantly apparent from our definitions, however we do not get hold of a single,
easy floor if we permit the v coordinate to tackle each optimistic and damaging values.
The primary determine initially of this part makes this clear; the yellow and crimson surfaces meet alongside the
producing curve, however there’s a cusp the place they arrive collectively, relatively than a easy transition.
To search out an isometry between a tangent floor and a part of the aircraft, we are going to deal with the
explicit instance of the tangent floor to a helix, h(s), with radius a
and pitch managed by b:
We outline a parameter c = √(a^{2} + b^{2})
h(s) = (a cos(s/c), a sin(s/c), b s/c)
h‘(s) = (–a/c sin(s/c), a/c cos(s/c), b/c)
h”(s) = (–a/c^{2} cos(s/c), –a/c^{2} sin(s/c), 0)
h”'(s) = (a/c^{3} sin(s/c), –a/c^{3} cos(s/c), 0)x(u, v) = h(u) + v h‘(u)
= (a cos(u/c), a sin(u/c), b u/c) +
v (–a/c sin(u/c), a/c cos(u/c), b/c)x_{u} = h‘(u) + v h”(u)
= (–a/c sin(u/c), a/c cos(u/c), b/c) + v (–a/c^{2} cos(u/c), –a/c^{2} sin(u/c), 0)
x_{v} = h‘(u)
= (–a/c sin(u/c), a/c cos(u/c), b/c)
n = v h”(u) × h‘(u) / v h”(u) × h‘(u)
= signal(v) (–b/c sin(u/c), b/c cos(u/c), –a/c)
x_{uu} = h”(u) + v h”'(u)
= (–a/c^{2} cos(u/c), –a/c^{2} sin(u/c), 0) + v (a/c^{3} sin(u/c), –a/c^{3} cos(u/c), 0)
x_{uv} = h”(u)
= (–a/c^{2} cos(u/c), –a/c^{2} sin(u/c), 0)
x_{vv} = 0
g(u, v) = = S(u, v) = =
–signal(v) a b v / c^{4} 0 0 0
To assemble an isometry, we are going to first establish a handy set of
geodesics
on the tangent floor, which might want to correspond to straight traces within the aircraft.
To start out, we are going to notice that any straight line (within the sense of being straight in
threedimensional Euclidean house) in any easily embedded floor is a geodesic.^{[4]} So the traces of mounted u and ranging v in a tangent
floor are all geodesics.
To search out different geodesics, we are going to make use of the truth that the metric for the
helix tangent floor is totally unbiased of the u coordinate. It follows
that altering u by a set quantity in every single place maps the floor into itself isometrically;
that’s, it’s a symmetry of the floor, which corresponds to a screwlike movement, the place we rotate
the floor across the axis of the helix whereas additionally shifting all the things alongside the axis.
An extra consequence of that is that, alongside any geodesic G(s), the dot product between
the tangent to the geodesic, G‘(s), and the vector area that corresponds to a uniform
change within the u coordinate, which is simply (1,0) in (u, v) coordinates,
is fixed. This consequence is called Killing’s theorem, and it’s described in a bit extra
element here [this article is associated with my novel Incandescence, but there
is no need to know anything about that book to follow the discussion of the geometry].
So, Killing’s theorem offers us, for a geodesic:
G(s) = (u_{G}(s), v_{G}(s))
with tangent vector:
G‘(s) = (u_{G}‘(s), v_{G}‘(s))
a conserved amount C alongside the geodesic:
G‘(s)^{T} g(G(s)) (1,0) = C
(1 + a^{2} v_{G}(s)^{2} / c^{4}) u_{G}‘(s) + v_{G}‘(s) = C
We are able to examine this for the geodesics we already learn about, the traces of fixed u and ranging v:
u_{G}(s) = u_{0}
v_{G}(s) = su_{G}‘(s) = 0
v_{G}‘(s) = 1(1 + a^{2} v_{G}(s)^{2} / c^{4}) u_{G}‘(s) + v_{G}‘(s)
= 1
If the geodesic is parameterised by arc size, its tangent may have a size of 1 in every single place,
and we will use that to resolve for
u_{G}‘(s) by way of v_{G}(s) and v_{G}‘(s).
G‘(s)^{T} g(G(s)) G‘(s) = 1
(1 + a^{2} v_{G}(s)^{2} / c^{4})
u_{G}‘(s)^{2} + 2 u_{G}‘(s) v_{G}‘(s) + v_{G}‘(s)^{2} = 1
u_{G}‘(s) =
(–v_{G}‘(s) ±
√[1 + (a^{2}/c^{4}) v_{G}(s)^{2} (1 – v_{G}‘(s)^{2})]) / (1 + a^{2} v_{G}(s)^{2} / c^{4})
With this consequence, we now have for the conserved amount alongside the geodesic:
(1 + a^{2} v_{G}(s)^{2} / c^{4}) u_{G}‘(s) + v_{G}‘(s) = C
±
√[1 + (a^{2}/c^{4}) v_{G}(s)^{2} (1 – v_{G}‘(s)^{2})] = C
1 + (a^{2}/c^{4}) v_{G}(s)^{2} (1 – v_{G}‘(s)^{2}) = C^{2}
Suppose we wish to discover a geodesic that meets one in every of our straightline geodesics of
fixed u sooner or later (u_{0}, v_{0}), and the
two geodesics are orthogonal to one another the place they meet, at s = 0. Then we are going to
need the tangent to the geodesic at s = 0 to be:
G‘(0) = (u_{G}‘(0), v_{G}‘(0))
= (c^{2} / (a v_{0}), –c^{2} / (a v_{0}))
since this tangent has unit size, and it’s orthogonal to (0,1), the tangent to the straightline geodesic:
G‘(0)^{T} g(u_{0}, v_{0}) G‘(0) = 1
G‘(0)^{T} g(u_{0}, v_{0}) (0,1) = 0
The preliminary worth this offers us for v_{G}‘(0) lets us compute the conserved
amount initially of the geodesic:
C^{2} = 1 + (a^{2}/c^{4}) v_{G}(s)^{2} (1 – v_{G}‘(s)^{2})
= (a^{2}/c^{4}) v_{0}^{2}
and the differential equation we have to clear up turns into:
1 + (a^{2}/c^{4}) v_{G}(s)^{2} (1 – v_{G}‘(s)^{2}) = (a^{2}/c^{4}) v_{0}^{2}
with the preliminary circumstances:
v_{G}(0) = v_{0}
v_{G}‘(0) = –c^{2} / (a v_{0})
That is solved by:
v_{G}(s) = √[s^{2} – 2 (c^{2}/a) s + v_{0}^{2}]
If we take a look at the factors within the (s, v_{0}) aircraft the place v_{G}(s) = 0, it seems they lie on a circle, with radius
c^{2}/a, and centre (c^{2}/a, 0).
This circle within the (s, v_{0}) aircraft corresponds to the helix within the tangent floor, and v_{G}(s) is exactly the space from (s, v_{0}) to this circle, measured alongside a tangent to the circle.
A line of fixed
v_{0} within the (s, v_{0}) aircraft
with v_{0} < c^{2}/a will intersect the circle,
and for values of s that lie contained in the circle, v_{G}(s) turns into imaginary. In
different phrases, these geodesics of fixed v_{0} attain the boundary of the tangent floor, the helix,
and are available to an finish there. However for
v_{0} > c^{2}/a these geodesics by no means hit the boundary. Equally, geodesics of fixed s and ranging v_{0} will solely hit the boundary
and are available to an finish if
0 < s < 2c^{2}/a.
If we substitute this resolution for v_{G}(s) again into our
unique equation for the conserved amount, earlier than we eradicated u_{G}‘(s),
we will now clear up that equation for u_{G}‘(s) and combine it to seek out u_{G}(s). The answer we get hold of is:
u_{G}(s) = u_{0} +
2 (c^{2}/a) arctan(s / (v_{0} + v_{G}(s)))
How does u_{G}(s) relate to the circle within the (s, v_{0}) aircraft that corresponds to the helix? For factors that lie on that circle, it should equal u_{0} plus the arc size alongside the circle, measured from the origin. Which means all through the (s, v_{0}) aircraft, the vary
of u values is restricted to the interval u_{0} ± π (c^{2}/a), and a single copy of the aircraft can solely
correspond to that portion of the tangent floor.
So the isometry we now have discovered maps areas of the helix tangent floor the place the u coordinate modifications by 2π (c^{2}/a) to copies of the aircraft with a disk of radius c^{2}/a reduce out of it. The second determine initially of this
part makes use of totally different colors to point out the totally different copies of the aircraft which can be mapped to the floor.
Be aware that the curvature of the producing helix, as a curve
in threedimensional Euclidean house:
κ = h”(s) = a/c^{2}
is exactly the curvature (the inverse of the radius, c^{2}/a) of the corresponding circle within the aircraft. That is definitely not true usually for the photographs of
curves below this isometry; for instance, most straight traces within the aircraft don’t map to straight traces in
threedimensional house. Nonetheless, it’s not a coincidence both; the producing curve
of any tangent floor all the time has the identical curvature in threedimensional Euclidean house because it has intrinsically
as a curve within the tangent floor, or equivalently, because the corresponding curve below an isometry with the Euclidean aircraft.
Dominated surfaces
All 4 sorts of intrinsically flat surfaces we now have described are examples of
ruled surfaces: surfaces which can be constructed completely from a household of straight traces.
However not each dominated floor is intrinsically flat. If we compute the Gaussian curvature of an arbitrary dominated
floor, what additional circumstances are required to make sure that Okay = 0?
Suppose we now have a curve, c(s),
parameterised by arc size, and at every level on the curve there’s a unitlength vector L(s)
that offers the course of the straight line passing via that time. The ensuing floor is parameterised
as:
x(u, v) = c(u) + v L(u)
We carry out the same old calculations to seek out the Gaussian curvature:
x_{u} = c‘(u) + v L‘(u)
x_{v} = L(u)
n = x_{u} × x_{v} / x_{u} × x_{v} = (c‘(u) + v L‘(u)) × L(u) / (c‘(u) + v L‘(u)) × L(u)
= (c‘(u) + v L‘(u)) × L(u) /
√(c‘(u) + v L‘(u)^{2} – (c‘(u) · L(u))^{2})
x_{uu} = c”(u) + v L”(u)
x_{uv} = L‘(u)
x_{vv} = 0
g(u, v) =
x_{u} · x_{u} x_{u} · x_{v} x_{u} · x_{v} x_{v} · x_{v} =
c‘(u) + v L‘(u)^{2} c‘(u) · L(u) c‘(u) · L(u) 1 S(u, v) =
x_{uu} · n x_{uv} · n x_{uv} · n x_{vv} · n =
(c”(u) + v L”(u)) · n L‘(u) · n L‘(u) · n 0 Okay = det(S) / det(g) =
–[L‘(u) · c‘(u) × L(u) / (c‘(u) + v L‘(u)^{2} – (c‘(u) · L(u))^{2})]^{2}
This expression for the curvature tells us that it can by no means be optimistic, since it’s the reverse of a squared realvalued amount. What’s extra, it can’t be equal to a fixed damaging worth, since
the one technique to take away the dependence on v would contain both setting L‘(u) = 0, or in any other case making the numerator zero, each of which might produce flat surfaces.
However the dominated floor might be intrinsically flat if, and provided that:
L‘(u) · c‘(u) × L(u) = 0
This
scalar triple
product will clearly be zero if a number of of the three vectors are zero. However it is going to even be zero if
any two of the vectors are parallel, or if all three vectors are coplanar.
 For a generalised cylinder (which incorporates the aircraft as a particular case), the instructions of the traces are fixed, so L‘(u) = 0.
 For a cone, L(u) is a scalar a number of of c(u) – A,
the place A is the apex, so by the product rule for derivatives, L‘(u) lies within the aircraft spanned by
L(u) and c‘(u).  For a tangent floor, L(u) = c‘(u).
Are there are some other prospects in addition to these 4 circumstances? We are able to splice totally different sorts of
floor collectively, e.g. we may be part of a cone and a cylinder, however regionally, is there some other technique to
meet this situation on the dominated floor that we haven’t already thought of?
The reply isn’t any; this checklist is full! Here’s a proof, taken from Lipschutz.^{[5]}
Proof that any dominated floor with Gaussian curvature zero have to be piecewise a aircraft/cylinder/cone/tangent floor
The situation that the scalar triple product
L‘(u) · c‘(u) × L(u) = 0
means there have to be three features, α(u), β(u), γ(u),
which show the linear dependence of the three vectors:
α(u) c‘(u) + β(u) L(u) + γ(u) L‘(u) = 0
α(u), β(u), γ(u) are by no means all equal to zero for a similar worth of u.
This have to be true for your complete vary of the u parameter related to the floor in query, however it’s potential that totally different subsets of {α(u), β(u), γ(u)}
might be zero on varied intervals of the parameter.
Case 1. Suppose α(u) = 0 on some interval for u. Then on that interval, the unique situation turns into:
β(u) L(u) + γ(u) L‘(u) = 0
β(u), γ(u) are by no means each equal to zero for a similar worth of u.
As a result of L(u) is a unit vector, its spinoff is all the time orthogonal to it, and we now have:
L(u) · (β(u) L(u) + γ(u) L‘(u)) =
β(u) = 0
γ(u) L‘(u) = 0, and γ(u) ≠ 0 as a result of β(u) = 0
So L‘(u) = 0
This implies L(u) is a continuing vector, and the portion of the floor akin to this
interval have to be a generalised cylinder (which incorporates the aircraft as a particular case).
Case 2. Suppose α(u) ≠ 0 on some interval for u. Then we will clear up the
unique equation for c‘(u):
c‘(u) = –β(u)/α(u) L(u) – γ(u)/α(u) L‘(u)
We’ll outline a brand new curve, C(u), as:
C(u) = c(u) + γ(u)/α(u) L(u)
Differentiating with respect to u, we now have:
C‘(u) = c‘(u) + d(γ(u)/α(u))/du L(u)
+ γ(u)/α(u) L‘(u)
= η(u) L(u)
the place we now have outlined η(u) as:
η(u) = d(γ(u)/α(u))/du – β(u)/α(u)
Case 2A. Suppose η(u) = 0 on some subinterval of the one we’re contemplating the place
α(u) ≠ 0. Then on that subinterval we now have:
C‘(u) = 0
C(u) = A, for some fixed vector A
c(u) = A – γ(u)/α(u) L(u)
x(u, v) = A + (v – γ(u)/α(u)) L(u)
Which means all of the traces that generate the floor go via A, and the floor should both be a generalised cone with apex A, or a part of a aircraft.
Case 2B. Suppose η(u) ≠ 0 on some subinterval. Then we now have:
L(u) = C‘(u) / η(u)
c(u) = C(u) – γ(u)/α(u) L(u)
= C(u) – γ(u)/[α(u) η(u)] C‘(u)
x(u, v) = c(u) + v L(u)
= C(u) + (v/η(u) – γ(u)/[α(u) η(u)]) C‘(u)
This implies the traces that generate the floor are all tangents to the curve C(u),
and the floor right here is a part of the tangent floor to C(u).
The one remaining query is whether or not there may be any technique to have an intrinsically flat floor that’s
not a dominated floor. The reply to that’s no;^{[6]} we received’t
reproduce the proof right here, however it formalises the intuitive concept that if, at each level, one of many principal curvatures
of the floor is zero, then if you happen to hold shifting throughout the floor within the corresponding course, you have to
be following a straight line in threedimensional house.
Rectifying developables
Given a easy, nonselfintersecting curve c(s) in threedimensional Euclidean house, is it potential to seek out an intrinsically flat floor during which that curve is a geodesic? Instinct means that the reply needs to be sure: a strip of (completely versatile) paper, slim sufficient that it received’t stumble upon itself in any tight locations on the curve, may very well be positioned in order that its centreline coincides with the curve.
The floor that meets this situation is called the rectifying developable for the curve.^{}[7] We all know from the earlier part that it is going to be a dominated floor, so we will
parameterise it as:
x(u, v) = c(u) + v L(u)
for some L(u) but to be decided, which we are going to write initially as:
L(u) = A(u) c‘(u) + B(u) c”(u)
+ D(u) c‘(u) × c”(u)
A standard vector to the floor at any level on the curve c(u) might be given by:
N(u) = L(u) × c‘(u)
= B(u) c”(u) × c‘(u) +
D(u) (c‘(u) × c”(u)) × c‘(u)
= B(u) c”(u) × c‘(u) +
D(u) c”(u)
Alongside a geodesic, the
tangent to the curve, c‘(u), its spinoff, c”(u), and the traditional vector to the floor, N(u), all lie in the identical aircraft.^{[3]} So to make c(u) a geodesic, we will need to have:
N(u) · (c‘(u) × c”(u)) = 0
which requires (assuming c”(u) is nonzero) that B(u) = 0. Then the situation for this dominated floor to be intrinsically flat is:
L‘(u) · c‘(u) × L(u) = 0
[A‘(u) c‘(u) + A(u) c”(u)
+ D‘(u) c‘(u) × c”(u)
+ D(u) c‘(u) × c”'(u)] · [c‘(u) ×
(A(u) c‘(u)
+ D(u) c‘(u) × c”(u))] = 0
[A‘(u) c‘(u) + A(u) c”(u)
+ D‘(u) c‘(u) × c”(u)
+ D(u) c‘(u) × c”'(u)] · [–D(u) c”(u)] = 0
A(u) c”(u)^{2} + D(u) (c‘(u) × c”'(u)) · c”(u) = 0
A(u) κ^{2}(u) – D(u) κ^{2}(u) τ(u) = 0
Right here κ(u) is the size of c”(u), which measures the curvature of c(u),
and τ(u) is the torsion of the curve, which measures the speed at which the aircraft of the curve is altering
its orientation as we transfer alongside the curve. It’s given by:
τ(s) = (c‘(s) × c”(s)) · c”'(s) / κ(s)^{2}
If we wish L(u) to be a unitlength vector, this situation might be glad by:
L(u) = (τ(u) c‘(u) + c‘(u) × c”(u))
/ √[τ(u)^{2} + κ(u)^{2}]
The numerator here’s a vector related to the curve, generally known as the
Darboux vector,
which describes the “angular velocity” of the orthogonal set of axes given by the unitlength tangent to
the curve, c‘(u), its spinoff, c”(u), and their cross product, as we transfer alongside the curve. (It’s common apply to additionally normalise the final two vectors to unit size, giving an orthonormal
body generally known as the FrenetSerret frame.)
So, if we take a easy, nonselfintersecting curve and assemble a dominated floor whose traces lie within the course of the Darboux vector at every level on the curve, the curve might be a geodesic
for that floor.
If c(u) is a straight line, the Darboux vector might be undefined, however we will use any aircraft that incorporates that line as a floor with the required properties.
The following easiest case could be a planar curve.
It will have zero torsion, so L(u) might be a set vector perpendicular to the aircraft of the curve,
giving us a cylinder. Equally, if we select a curve that could be a geodesic on a cone,
the rectifying developable might be that cone.
However generically, the floor we get from a curve might be a tangent floor, and we will explicitly establish
the curve whose tangents are the identical set of straight traces as we get from the Darboux vectors of the
unique curve. If we outline:
T(s) = c(s) + [κ(s)/(τ(s) κ'(s) – τ'(s) κ(s))] (τ(s) c‘(s) + c‘(s) × c”(s))
then it seems that T‘(s) is parallel to the Darboux vector for c(s),
so the tangent floor for T(s) is the rectifying developable for c(s).
Be aware that the parameter s right here is an arc size parameter for c(s) solely, and never for
T(s).
T(s) will represent an edge to the floor, simply
as any curve does for its personal tangent floor. The space measured alongside the ruling line from the purpose c(s) to the sting of the floor might be:
Distance to edge alongside ruling
=  κ(s) √[τ(s)^{2} + κ(s)^{2}] / (τ(s) κ'(s) – τ'(s) κ(s)) 
If we wish to know the orthogonal distance, measured inside the floor, between a degree on the sting, T(s), and the producing curve (which is a geodesic for the floor), we will make use of the truth that the sine of the angle between the ruling traces and
the curve is κ(s)/√[τ(s)^{2} + κ(s)^{2}], from which we get hold of:
Orthogonal distance from edge to prescribed geodesic
=  κ(s)^{2} / (τ(s) κ'(s) – τ'(s) κ(s)) 
=  1 / d(τ(s)/κ(s))/ds 
We are able to additionally use that system for the the sine of the angle between the ruling traces and
the curve to jot down an isometry between the rectifying developable and the aircraft, in Cartesian coordinates (x, y):
x = u
y = v (κ(s)/√[τ(s)^{2} + κ(s)^{2}])
If the curvature vector c”(u) is zero at an remoted level
u_{0}, the unitlength regular vector c”(u)/c”(u), which is undefined at
u_{0}, can both have the identical, or reverse, limiting values on both facet of u_{0}.
If the torsion additionally goes to zero on the similar level, the normalised Darboux vector we’re utilizing for L(u) can even swap course.
This doesn’t have an effect on the floor itself (for the reason that ruling traces, which observe the Darboux vector in each instructions, are unchanged) however to be able to make the (u, v) coordinates
steady throughout the change, we will multiply L(u) by an indication that’s chosen to be able to lengthen continuity as a lot as potential.
If the torsion does not go to zero at a degree the place the curvature does, then that marks a degree
the place c(s) intersects the sting curve T(s).
If we assemble the rectifying developable for a easy closed curve on which the curvature is by no means zero,
then the floor might be topologically a cylinder: a chunk of the aircraft whose reverse edges be part of up and not using a twist. But when there are an odd variety of factors of zero curvature and torsion the place the course of
the normalised curvature and Darboux vectors change discontinuously, the floor might be a Möbius strip.
Möbius strips
We are able to embed a Möbius strip isometrically in three dimensions by wrapping the strip round three cylinders, as within the first picture above. On this case,
the centreline is a piecewise operate that splices collectively three helices and three line segments, and the floor is spliced collectively from three cylindrical areas and three planar areas.
We are able to additionally embed a Möbius strip because the rectifying developable of a single easy curve,
as within the second picture above. The third picture exhibits what is going on with the FrenetSerret body.
The crimson vector would change course on the inflection level, I, if we merely outlined it as
c”(u)/c”(u), however we multiply it with an indication that makes it steady because it crosses that time. Nonetheless, because of that single change of signal, it now not agrees with its unique course
when it comes full circle; it must traverse the loop twice earlier than that occurs. This
is accompanied by the same change within the Darboux vector, so the ruling traces reverse orientation after
a single circuit across the loop, making the floor a Möbius strip.
The operate we now have chosen for the y coordinate of the curve would possibly look a bit unusual and arbitrary at first look:
y(t) = cos(t) + (2/5) cos(2t) + (1/15) cos(3t)
In truth, that is the best trigonometric operate
with a Taylor collection at π that could be a fixed plus a sixthorder time period:
y(π + ε) ≈ –2/3 + (1/3) ε^{6}
This produces an inflection level at π with appropriate behaviour for the curvature and torsion:
each go to zero, however the torsion should go to zero no less than as quickly because the curvature, to ensure that the width of the strip we will match across the centreline, 1 / d(τ(s)/κ(s))/ds, to stay nonzero.
As a result of we now have chosen to make the curve symmetrical below a
180° rotation across the yaxis, the torsion have to be an excellent operate of ε, so the bottom
order it could actually have is quadratic.
κ(π + ε) ≈ (4/(5√5)) ε
τ(π + ε) ≈ –(2/9) ε^{2}
Möbius strips and Klein’s bottles in 4 dimensions
Though this net web page is concentrated on surfaces embedded in threedimensional house, we are going to notice
that embeddings in 4 dimensions permit for extra prospects. Intrinsically flat surfaces in
4 dimensions needn’t be dominated surfaces, and a few topologies that can’t be embedded in three
dimensions as a result of the floor will unavoidably intersect itself might be embedded in 4 dimensions.
Take into account the next floor embedded in 4 dimensions, parameterised as:
x(u, v) = (cos u cos v, sin u cos v,
2 cos(u/2) sin v, 2 sin(u/2) sin v)
In 4 dimensions, we will nonetheless use the strategies described within the appendix to seek out the metric:
x_{u} = (–sin u cos v, cos u cos v,
–sin(u/2) sin v, cos(u/2) sin v)
x_{v} = (–cos u sin v, –sin u sin v,
2 cos(u/2) cos v, 2 sin(u/2) cos v)
g(u, v) =
x_{u} · x_{u} x_{u} · x_{v} x_{u} · x_{v} x_{v} · x_{v} =
Whereas we may proceed to compute the Gaussian curvature from the metric with the entire equipment of
Christoffel symbols and the Riemann curvature tensor described within the
second appendix, we don’t really have to do all that work to conclude that the floor is intrinsically
flat! The metric is already virtually within the type of the twodimensional Euclidean metric, aside from the
g_{22} part, which is a operate of the second parameter v, relatively than being 1.
But when we modified to a brand new parameter rather than v that measured arc size alongside every curve of
various v and fixed u, then g_{22} would turn into 1, and the metric could be exactly the Euclidean metric, since this alteration would don’t have any impact on the opposite elements.
What’s the topology of this floor? If we permit u to vary from 0 to 2π, and v
to vary from –V to V for some worth of V better than 0 and strictly much less
than π, then we will look at how the 2 ends, u = 0 and u = 2π, of the coordinate
rectangle are associated within the embedding:
x(0, v) = (cos v, 0, 2 sin v, 0)
x(2π, v) = (cos v, 0, –2 sin v, 0)
x(2π, –v) = (cos v, 0, 2 sin v, 0)
We see that the 2 ends coincide, however with the worth of the v coordinate negated, which quantities to a 180° twist. That is exactly
the topology of a Möbius strip.
What if we set V = π, widening the strip in order that v ranges from –π to π?
We then have the opposite two sides of the coordinate rectangle assembly:
x(u, –π) = (–cos u, –sin u, 0, 0)
x(u, π) = (–cos u, –sin u, 0, 0)
Since these sides meet up with no change in orientation, this exhibits that the floor has closed up
right into a Klein’s bottle. After all, it stays intrinsically flat, and there are not any selfintersections (apart from
the required assembly of the borders of the coordinate rectangle). The “tube” of the Klein’s bottle
is generated by taking an ellipse, given by the curve for various v at u = 0:
x(0, v) = (cos v, 0, 2 sin v, 0)
and making use of a steady household of fourdimensional rotations to it, by an angle of
u within the xy aircraft and an angle of u/2 within the zw aircraft.
The animation initially of this part exhibits a single, inflexible Möbius strip
with this embedding, however it’s projected down to a few dimensions utilizing the idea:
{(0, 0, cos α, sin α), (1, 0, 0, 0), (0, 1, 0, 0)}
with α cyling from 0 to 2π over the course of the animation.
To be clear, these projections to threedimensional house are not intrinsically flat.
Intrinsically Curved Surfaces
Surfaces of Revolution
We’ll begin our exploration of surfaces of fixed nonzero Gaussian curvature by searching for examples that may
be produced by taking a planar curve and rotating it round an axis. It will definitely not embody all of the surfaces
we wish to catalogue, ultimately, however the benefit of beginning right here is to seek out some examples whereas protecting the
arithmetic comparatively easy, as these surfaces of revolution might be described with a single operate of 1 variable.
One technique to parameterise a floor of revolution is to explain the z coordinate as a operate of the
radial cylindrical coordinate, ρ, i.e. the space from the axis. Whereas we may reverse this, and describe the
radius as a operate of z, or, to permit for utterly basic curves, describe each ρ and z as features of some
parameter, the strategy we now have chosen will make our calculations simpler, and the outcomes less complicated to specific.
So, we are going to parameterise our floor as:
x(φ, ρ) = (ρ cos φ, ρ sin φ, f(ρ))
the place the righthand facet listed here are strange Cartesian coordinates, x, y, z. We now have departed from our earlier apply of utilizing (u, v) because the names for the 2 floor coordinates, in favour of (φ, ρ), that are extra suggestive of the geometry, no less than for readers who’ve
been uncovered to the traditional selection of symbols for
cylindrical coordinate systems.
Following the same old recipe for computing the Gaussian curvature of an embedded floor (spelled out within the appendix), we now have:
x_{φ} = (–ρ sin φ, ρ cos φ, 0)
x_{ρ} = (cos φ, sin φ, f ‘(ρ))
n = x_{φ} × x_{ρ} / x_{φ} × x_{ρ} = (f ‘(ρ) cos φ, f ‘(ρ) sin φ, –1) / √[1 + f ‘(ρ)^{2}]x_{φφ} = (–ρ cos φ, –ρ sin φ, 0)
x_{φρ} = (–sin φ, cos φ, 0)
x_{ρρ} = (0, 0, f ”(ρ))
g(φ, ρ) =
x_{φ} · x_{φ} x_{φ} · x_{ρ} x_{φ} · x_{ρ} x_{ρ} · x_{ρ} = S(φ, ρ) =
x_{φφ} · n x_{φρ} · n x_{φρ} · n x_{ρρ} · n =
–ρ f ‘(ρ) / √[1 + f ‘(ρ)^{2}] 0 0 –f ”(ρ) / √[1 + f ‘(ρ)^{2}] Okay = det(S) / det(g) = f ‘(ρ) f ”(ρ) / [ρ (1 + f ‘(ρ)^{2})^{2})]
To provide a floor of fixed Gaussian curvature, Okay, we have to clear up the differential equation right here
for f(ρ). As a primary step, we will rewrite this as:
–1/(2 ρ) d(1/(1 + f ‘(ρ)^{2}))/dρ = Okay
That is straightfoward to resolve for f ‘(ρ):
d(1/(1 + f ‘(ρ)^{2}))/dρ = –2 Okay ρ
1/(1 + f ‘(ρ)^{2}) = –Okay ρ^{2} + c
f ‘(ρ) = ±√[1/(–K ρ^{2} + c) – 1]
Right here c is a continuing of integration. The values that c can take whereas the
amount contained in the sq. root stays nonnegative (for some nonempty vary of values for ρ) will depend upon the signal of Okay. The situation we have to fulfill is:
0 < –Okay ρ^{2} + c < 1
That’s to say, a parabola centred on ρ = 0 that both factors down (if Okay > 0) or up (if Okay < 0), and which has a most or minimal worth of c at ρ = 0, should lie between 0 and 1 for some
nonempty vary of values for ρ.
Suppose Okay = 1/a^{2}, i.e. the Gaussian curvature takes the optimistic worth that corresponds to a sphere of radius a. Then c have to be better than zero, or the downwardspointing parabola won’t ever tackle any optimistic values. It will likely be handy to set c = χ^{2}. The vary of values for ρ will then be:
If 0 < χ ≤ 1  0 < ρ < a χ  
If χ > 1  a √[χ^{2} – 1] < ρ < a χ 
For damaging curvature, say Okay = –1/a^{2}, we will need to have c lower than 1, or the
upwardspointing parabola won’t ever tackle any values lower than 1. On this case, we are going to set c = 1 – ξ^{2}, and the vary of values for ρ might be:
If 0 < ξ ≤ 1  0 < ρ < a ξ  
If ξ > 1  a √[ξ^{2} – 1] < ρ < a ξ 
We are able to combine our consequence for f ‘(ρ) to seek out f(ρ). For the particular case c = 1,
which is simply appropriate with optimistic curvature, so we are going to set Okay = 1/a^{2}:
Okay = 1/a^{2}
c = 1f ‘(ρ) = ±ρ / √[a^{2} – ρ^{2}]
f(ρ) = z_{eq} ∓ √[a^{2} – ρ^{2}]
That is simply the system for 2 quadrants of a circle of radius a, and the ensuing floor of revolution is a sphere of radius a, with the z coordinate of its equator given by the fixed
of integration z_{eq}.
For the particular case c = 0, which is simply appropriate with damaging curvature:
Tractroid
Okay = –1/a^{2}
c = 0f ‘(ρ) = ±√[a^{2} – ρ^{2}] / ρ
f(ρ) = z_{eq} ± (√[a^{2} – ρ^{2}] – a arccosh(a/ρ))
This operate known as a
tractrix,
and its floor of revolution, a tractroid, is probably the most wellknown instance of a
pseudosphere:
a floor of fixed damaging curvature.
If c ≠ 1 and c ≠ 0, we now have:
f(ρ) = z_{eq} ± √[(1 – c)/K] (E(arcsin(√[K/c] ρ)  c/(c – 1)) – e_{0})
If Okay > 0 e_{0} = E(c/(c – 1)) If Okay < 0 e_{0} = E(arcsin(√[(c – 1)/c])  c/(c – 1))
The operate E is an elliptic integral of the second kind, outlined as:
E(s  m) = ∫_{0}^{s} √[1 – m sin^{2} t] dt
E(m) = E(½π  m)
The picture on the correct exhibits crosssections via the surfaces of revolution of fixed
Gaussian curvature Okay = 1/a^{2}, for a wide range of equatorial radii.
Right here, c = χ^{2}, and χ is the utmost
worth of ρ/a for every curve.
These curves all meet the horizontal axis with vertical tangents, to allow them to be easily
prolonged by reflection on this axis; for instance, the curve for χ = 1 will yield an entire sphere. Nonetheless, the factors on the prime of the curve will (in all different circumstances in addition to
the sphere) have to be excluded from the floor, with the “poles” for χ < 1 having a neighborhood topology just like the tip of a cone, and the tops of the curves for χ > 1 giving rise to round boundaries.
We are able to compute the lengths of those curves:
L = ∫_{ρ1}^{ρ2} √[1 + f ‘(ρ)^{2}] dρ
= ∫_{ρ1}^{ρ2} 1/√[χ^{2} – (ρ/a)^{2}] dρ
= a arcsin(ρ/(a χ)) _{ρ1}^{ρ2}
If 0 < χ ≤ 1  L = (π/2) a  
If χ > 1  L = (π/2 – arctan(√[χ^{2} – 1])) a 
So for χ ≤ 1, this is similar because the meridian from the equator to the pole on a sphere of radius a, whereas for χ > 1 it’s shorter.
What concerning the floor space?
A = 2 π ∫_{ρ1}^{ρ2} ρ √[1 + f ‘(ρ)^{2}] dρ
= 2 π ∫_{ρ1}^{ρ2} ρ/√[χ^{2} – (ρ/a)^{2}] dρ
= –2 π a √[a^{2} χ^{2} – ρ^{2}] _{ρ1}^{ρ2}
If 0 < χ < 1  A = 2 π a^{2} χ  
If χ ≥ 1  A = 2 π a^{2} 
For χ < 1, that is lower than the hemisphere of a sphere of radius a, whereas for
χ ≥ 1 it’s precisely the identical.
Now let’s compute
geodesics on these surfaces. To start out, notice that the metric in (φ, ρ) coordinates might be discovered from
our preliminary calculations, as soon as we substitute the consequence for f ‘(ρ):
g(φ, ρ)  =  = 

As a result of the metric is unbiased of the φ coordinate, we will discover a household of geodesics by imposing the
requirement that every such curve has a conserved amount: the dot product of the unitlength tangent to the geodesic and the vector area that corresponds to a uniform enhance within the φ coordinate. We received’t
undergo the derivation, however the outcomes are simply checked. We declare that these curves are geodesics (for c ≠ 0):
φ_{G}(s) = φ_{0} + arctan((√[c/K] / ρ_{0}) tan(s √Okay)) / √c
ρ_{G}(s) = √[ρ_{0}^{2} + (c/K – ρ_{0}^{2}) sin^{2}(s √K)]
To confirm this, we compute the coordinates of the tangent to the geodesic:
φ_{G}‘(s) = ρ_{0} Okay
/ (ρ_{0}^{2} Okay + (c – ρ_{0}^{2} Okay) sin^{2}(s √Okay))
ρ_{G}‘(s) = (c – ρ_{0}^{2} Okay) sin(s √Okay)
/ √[c tan^{2}(s √K) + ρ_{0}^{2} K]
Alongside these geodesics, the metric turns into:
g(φ_{G}(s), ρ_{G}(s))  = 

and we now have:
(φ_{G}‘(s), ρ_{G}‘(s))^{T}
g(φ_{G}(s), ρ_{G}(s))
(φ_{G}‘(s), ρ_{G}‘(s)) = 1
(φ_{G}‘(s), ρ_{G}‘(s))^{T}
g(φ_{G}(s), ρ_{G}(s))
(1, 0) = ρ_{0}
This confirms that the tangent has a size of 1, and that its dot product with the φ coordinate
vector area is a conserved amount, the fixed ρ_{0}, that parameterises the household of geodesics.
For optimistic curvature we will write the geodesics as:
Okay = 1/a^{2}
c = χ^{2}Generic case (with s measured from level the place ρ = ρ_{0}):
φ_{G}(s) = φ_{0} + arctan((aχ/ρ_{0}) tan(s/a)) / χ
ρ_{G}(s) = √[ρ_{0}^{2} + (a^{2} χ^{2} – ρ_{0}^{2}) sin^{2}(s/a)]Meridians (with s measured from the equator):
φ_{G}(s) = φ_{0}
ρ_{G}(s) = a χ cos(s/a)
At s = 0, we now have (for each optimistic and damaging curvature):
φ_{G}(0) = φ_{0}
ρ_{G}(0) = ρ_{0}
φ_{G}‘(0) = 1/ρ_{0}
ρ_{G}‘(0) = 0
So at s = 0 the geodesic is tangent to a circle of latitude of radius ρ_{0},
and that is the minimal worth of the ρ coordinate for the geodesic. Nonetheless,
for surfaces with χ > 1, if we select a worth
for ρ_{0} that’s lower than the minimal worth for ρ on your complete floor, a √[χ^{2} – 1], we nonetheless get hold of a legitimate geodesic, however it solely exists for
values of s massive sufficient to present legitimate values for ρ.
For the optimistic curvature case, we now have:
φ_{G}((π/2) a) = φ_{0} + π/(2 χ)
ρ_{G}((π/2) a) = a χ
This tells us that after we journey a distance, (π/2) a,
that’s the similar because the equatortopole distance on a sphere of radius a,
we arrive on the equator (the utmost worth for ρ). The metric in (φ, ρ) coordinates
is singular on the equator, however we will calculate the angle between the geodesic and the equator
as a restrict:
cosine of angle between geodesic and circle of latitude
= dot product between tangent to geodesic and unit vector within the φ course
= (φ_{G}‘(s), ρ_{G}‘(s))^{T}
g(φ_{G}(s), ρ_{G}(s))
(1/ρ_{G}(s), 0)
= ρ_{0} / ρ_{G}(s)cosine of angle between geodesic and equator (ρ = a χ)
= ρ_{0} / (a χ)
By various ρ_{0}, we will assemble a household of geodesics that every one
go via the identical level on the equator at varied angles.
The animation above exhibits surfaces of revolution, all with the identical fixed optimistic curvature, as
the parameter χ cycles between 0.5 and 1.5. The 2 views present the identical floor from the entrance and the again.
The Earth has been mapped to those surfaces by a easy isometry:
when χ is lower than 1, a part of the equator of the Earth, and the total size of every meridian crossing it, is mapped to the equator and meridians on the floor; when χ is larger than 1, the total equator of the Earth,
and a part of the equator of a second copy of the Earth, and a part of every meridian crossing them, is mapped to the
equator and meridians on the floor. There’s usually a discontinuity within the map alongside the
midline of the view of the floor from behind, however this isn’t a boundary or singularity within the floor
itself. As we talked about earlier,
the poles for the surfaces with
χ lower than 1 are singular factors, in the same technique to the tip of a cone, and have to be excluded.
For χ better than 1 there are two round
boundaries the place ρ hits a nonzero minimal worth and the floor can’t be prolonged.
The crimson curves are geodesics, fanning out from a degree on the equator after which reconverging. The geodesics are solely drawn till they’ve gone midway across the axis, or till they’ve reconverged, whichever is larger.
No matter
the worth of χ, the geodesics all the time reconverge after the identical distance, π a, however for χ better than 1
among the geodesics hit the boundary of the floor and terminate there. Additionally,
the space π a will
now not correspond to half the size of the equator, and the second reconvergence, at a distance of 2π a, will now not happen on the level the place we began.
The equator is all the time a geodesic, however usually geodesics will solely kind closed loops of finite size
when χ is rational, since every portion of a geodesic that lies north or south of the equator will
span a longitude of π/χ. And even for rational values of χ, it’d take
a really massive variety of crossings of the equator earlier than these modifications in longitude sum to an integer a number of of 2π. This results in the conclusion that, usually, the shortest geodesic path from
a degree P (not on the equator) that wraps across the floor and returns to P will not
be a part of a finite loop, which in flip means it should return to P with its tangent pointing in a unique course than when it started. The angle between the tangent to the geodesic and the circle of
latitude that P lies on will need to have the identical cosine at any time when they cross, so if they don’t seem to be the identical angles,
they have to be opposites.
So, whereas the native geometry of those
surfaces is precisely like that of a sphere, the international properties aren’t the identical.
Bugs confined to a floor like this, however capable of journey distances similar to the size of the entire floor, may definitely distinguish it from a sphere by appropriate “experiments” with geodesics.
Now let’s think about the case of damaging curvature.
The picture on the correct exhibits crosssections via the surfaces of revolution of fixed
Gaussian curvature Okay = –1/a^{2}, for varied equatorial radii.
We now have set c = 1 – ξ^{2}, and ξ is the utmost
worth of ρ/a for every curve.
All of those curves meet the horizontal axis with a slope of zero, so not like the case with optimistic curvature,
the surfaces can’t be easily prolonged by reflection right here. For ξ < 1, the surfaces may have a conical
tip on the level of most z; for ξ = 1, the tractrix extends to infinity within the z course.
For ξ > 1, the curves all have a vertical tangent on the minimal radius, ρ = a √[ξ^{2} – 1], to allow them to be prolonged easily there by reflection in a horizontal line on the most z worth.
The lengths of those curves are:
L = ∫_{ρ1}^{ρ2} √[1 + f ‘(ρ)^{2}] dρ
= ∫_{ρ1}^{ρ2} 1/√[1 – ξ^{2} + (ρ/a)^{2}] dρ
= a arctanh(ρ/√[ρ^{2} + a^{2}(1 – ξ^{2})]) _{ρ1}^{ρ2}
If 0 < ξ ≤ 1  L = a arctanh(ξ)  
If ξ > 1  L = a arctanh(1/ξ) 
The world of the floor of revolution is:
A = 2 π ∫_{ρ1}^{ρ2} ρ √[1 + f ‘(ρ)^{2}] dρ
= 2 π ∫_{ρ1}^{ρ2} ρ/√[1 – ξ^{2} + (ρ/a)^{2}] dρ
= 2 π a √[ρ^{2} + a^{2} (1 – ξ^{2})] _{ρ1}^{ρ2}
If 0 < ξ < 1  A = 2 π a^{2} (1 – √[1 – ξ^{2}])  
If ξ ≥ 1  A = 2 π a^{2} 
For ξ < 1 that is lower than the hemisphere of a sphere of radius a, whereas for
ξ ≥ 1 it’s precisely the identical.
For damaging curvature we will write the geodesics as:
Okay = –1/a^{2}
c = 1 – ξ^{2}Generic case (with s measured from level the place ρ = ρ_{0}):
φ_{G}(s) = φ_{0} + arctan(√[1 – ξ^{2}] (a/ρ_{0}) tanh(s/a)) / √[1 – ξ^{2}]ρ_{G}(s) = √[ρ_{0}^{2} + (ρ_{0}^{2} + a^{2} (1 – ξ^{2})) sinh^{2}(s/a)]Asymptotes to waist (for ξ > 1 solely, with s measured from the equator):
φ_{G}(s) = φ_{0} + (½log(1 + (ξ^{2} – 1) exp(2s/a)) – log(ξ)) / √[ξ^{2} – 1]ρ_{G}(s) = a √[ξ^{2} – 1 + exp(–2s/a)]Meridians:
φ_{G}(s) = φ_{0}
 for ξ < 1 (with s measured from the pole):
ρ_{G}(s) = a √[1 – ξ^{2}] sinh(s/a) for ξ > 1 (with s measured from the “waist”):
ρ_{G}(s) = a √[ξ^{2} – 1] cosh(s/a)
We are able to additionally compute the particular case c = 0, ξ = 1, for geodesics on the tractroid:
Tractroid
Okay = –1/a^{2}
c = 0Generic case (with s measured from level the place ρ = ρ_{0}):
φ_{G}(s) = φ_{0} + (a/ρ_{0}) tanh(s/a)
ρ_{G}(s) = ρ_{0} cosh(s/a)Meridians (with s measured from the equator):
φ_{G}(s) = φ_{0}
ρ_{G}(s) = a exp(–s/a)
As with the positively curved surfaces, within the generic case these geodesics are tangent, at s = 0, to a circle of latitude of radius ρ_{0},
and that is the minimal worth of the ρ coordinate for the geodesic. If ξ > 1 and we select ρ_{0} to be lower than the minimal worth for ρ on your complete floor, a √[ξ^{2} – 1], we nonetheless get hold of a legitimate geodesic, however it solely exists for
values of s massive sufficient to present legitimate values for ρ.
Nonetheless, not like the optimistic curvature case, all geodesics on a given floor do not take the identical distance from their level of minimal ρ earlier than they attain the equator. And the equator itself is not a geodesic; its geodesic curvature (the curvature by way of the intrinsic geometry, relative to the geodesics, which
are thought of to be straight) is similar as its curvature in threedimensional Euclidean house: 1/(a ξ), the reciprocal of the radius of the equatorial circle.
For ξ ≤ 1, most geodesics will
begin and finish on the equator. Meridians will begin on the equator and finish on the pole, or for ξ = 1 will proceed to infinity.
For ξ > 1, the floor has a “waist,” a circle of minimal ρ the place ρ = a √[ξ^{2} – 1]. That is clearly a geodesic by symmetry, and if we insert this worth for ρ_{0} into the system for the generic ρ_{G}(s), it turns into fixed. Geodesics with ρ_{0} < a √[ξ^{2} – 1]
will cross the waist, and have endpoints on the 2 distinct copies of the equator. Geodesics with
ρ_{0} > a √[ξ^{2} – 1] will
by no means cross the waist, and can begin and finish on the identical copy of the equator. Geodesics with
ρ_{0} = a √[ξ^{2} – 1]
that don’t begin on the waist are asymptotic to it.
As earlier than, the cosine of the angle between a geodesic and any circle of latitude might be
equal to ρ_{0} / ρ_{G}(s),
so the cosine of the angle between a geodesic and the equator is ρ_{0} / (a ξ). For ξ > 1 the geodesics that cross the waist might be people who make angles better than arccos √[1 – 1/ξ^{2}] with the equator. If a geodesic leaves the equator at exactly this angle,
it is going to strategy the waist asymptotically, by no means reaching it but in addition by no means returning to the equator.
The animation above exhibits back and front views of surfaces of revolution with Okay = –1, with the parameter ξ starting from 1.10125 to 1.70728. These values don’t have any particular significance by way of the properties of the floor, however they had been chosen in order that the actual tiling of the hyperbolic
aircraft used as an example the floor wraps round neatly on the two endpoints. On this tiling, seven equilateral triangles meet at a degree, and the equilateral triangles are additional subdivided into six rightangled triangles.
This tiling can’t exist within the Euclidean aircraft, the place the angles at every vertex of an equilateral triangle
are all the time π/3 radians, however within the hyperbolic aircraft the sum of the angles of a triangle might be lower than π by an quantity that is determined by the world of the triangle, as we mentioned earlier.
For these equilateral triangles with angles of 2π/7 at every vertex,
for a complete of 6π/7, the world is π/7.
The entire space of every of the surfaces within the animation is 4π.
The isometries used to map the hyperbolic aircraft to those surfaces take a set geodesic within the hyperbolic aircraft
to the circle across the waist of the floor, after which map geodesics orthogonal to the primary one to meridians
on the floor.
The photographs on the left and beneath present some examples of those surfaces rotating, to present a clearer sense of how the negatively curved geometry matches into threedimensional Euclidean house.
The animation above exhibits surfaces with Okay = –1 and the parameter ξ starting from (√13)/7 ≈ 0.515 to (√45)/7 ≈ 0.958. Every floor corresponds to a round wedge of the hyperbolic aircraft, with a wedge angle of 2 π √[1 – ξ^{2}], a radius of arctanh(ξ), and a complete floor
space of 2 π (1 – √[1 – ξ^{2}]).
The wedge angles right here vary from 6 × 2π/7 to 2 × 2π/7, as might be seen by the numbers of triangles from the tiling that match across the pole.
The tractroid, with ξ = 1, might be seen as a limiting case of this household of conelike surfaces, the place the peak of the cone goes to infinity whereas the wedge angle goes to zero, in such a method that the entire space of the floor stays finite at 2π a^{2}. As a result of the space from the pole to the equator goes to infinity, any realworld
instance of the tractroid will simply be a portion of the entire floor, within the neighborhood of the equator relatively than the
pole.
The equator of the tractroid is just not a geodesic; it corresponds to a finite portion of an infinite curve within the hyperbolic
aircraft generally known as a horocycle,
which might be considered a sort of circle whose centre lies, not within the hyperbolic aircraft itself,
however “at infinity.” In hyperbolic geometry, a degree like this is called an
ideal point.
Within the Poincaré disk model of the hyperbolic aircraft, the aircraft is represented
by the inside of a disk, whereas the perfect factors lie on the boundary of the disk, and horocycles
are represented by circles which can be tangent to the boundary circle.
The meridians of the tractroid are geodesics, assembly the horocycle of the equator orthogonally,
and converging on a standard preferrred level.
As a result of there are not any pairs of geodesics from the sides in our tiling that converge on a really perfect level, there may be essentially
a visual “seam” within the sample alongside one meridian of the tractroid within the picture on the correct.
The picture on the left exhibits the outlines of areas of the hyperbolic aircraft, within the
Poincaré disk mannequin, which can be mapped to the negatively curved surfaces we now have illustrated.
The dashed white shapes are the 2 extremes of the wedges mapped to the conelike surfaces with ξ < 1.
The stable white line exhibits the area mapped to the tractroid. The big arc on the left, akin to the equator of the tractroid, is a part of a horocycle; the small arc on the correct, akin to the circle of latitude
the place we selected to truncate the tractroid, is a part of one other horocycle.
The black shapes are three areas mapped to
the hourglasslike surfaces with ξ > 1.
Surfaces Constructed From Helices
Within the earlier part, we constructed surfaces of revolution composed of circles of various radius, ρ,
with their z coordinate (their location alongside the axis of revolution) given by a operate, f(ρ),
which glad a differential equation that ensured fixed Gaussian curvature.
One technique to generalise this development, whereas nonetheless working with only one operate of a single variable, is to
exchange all of the circles with helices of a specified pitch, managed by a parameter b
that could be a fixed for every floor. The parameterisation of the floor then takes the shape:
x(φ, ρ) = (ρ cos φ, ρ sin φ, f(ρ) + b φ)
The surfaces of revolution correspond to b = 0. Computing the Gaussian curvature:
x_{φ} = (–ρ sin φ, ρ cos φ, b)
x_{ρ} = (cos φ, sin φ, f ‘(ρ))
n = x_{φ} × x_{ρ} / x_{φ} × x_{ρ} = (ρ f ‘(ρ) cos φ – b sin φ,
b cos φ + ρ f ‘(ρ) sin φ,
–ρ) / √[b^{2} + ρ^{2} (1 + f ‘(ρ)^{2})]x_{φφ} = (–ρ cos φ, –ρ sin φ, 0)
x_{φρ} = (–sin φ, cos φ, 0)
x_{ρρ} = (0, 0, f ”(ρ))
g(φ, ρ) =
x_{φ} · x_{φ} x_{φ} · x_{ρ} x_{φ} · x_{ρ} x_{ρ} · x_{ρ} =
ρ^{2} + b^{2} b f ‘(ρ) b f ‘(ρ) 1 + f ‘(ρ)^{2} S(φ, ρ) =
x_{φφ} · n x_{φρ} · n x_{φρ} · n x_{ρρ} · n =
–ρ^{2} f ‘(ρ) / √[b^{2} + ρ^{2} (1 + f ‘(ρ)^{2})] b / √[b^{2} + ρ^{2} (1 + f ‘(ρ)^{2})] b / √[b^{2} + ρ^{2} (1 + f ‘(ρ)^{2})] –ρ f ”(ρ) / √[b^{2} + ρ^{2} (1 + f ‘(ρ)^{2})] Okay = det(S) / det(g) = (ρ^{3} f ‘(ρ) f ”(ρ) – b^{2}) / (b^{2} + ρ^{2} (1 + f ‘(ρ)^{2}))^{2})
We are able to rewrite this differential equation in f as:
q(ρ) = (b^{2} + ρ^{2} (1 + f ‘(ρ)^{2}))^{–1}
–½ ρ q‘(ρ) – q(ρ) = Okay
which is solved by:
q(ρ) = c / ρ^{2} – Okay
f ‘(ρ) = ±√[1/(–K ρ^{2} + c) – (1 + b^{2}/ρ^{2})]
Right here c is a continuing of integration, and the consequence for f ‘(ρ) agrees with that for a floor of revolution if we set b = 0.
For the amount contained in the sq. root to be finite and nonnegative we want:
0 < –Okay ρ^{2} + c < ρ^{2} / (ρ^{2} + b^{2})
and ρ > 0
In all the things that follows, we are going to assume that b > 0.
For the case of optimistic curvature, we are going to set Okay = 1/a^{2} and
c = χ^{2}, since we want c > 0 to be able to have a nonempty vary for ρ:
½(d + √[d^{2} + 4 a^{2} b^{2} χ^{2}])
< ρ^{2} < (a χ)^{2}
the place d = a^{2} (χ^{2} – 1) – b^{2}
So we now have the identical most worth for ρ as for the positivecurvature surfaces of revolution, however now we all the time have
a nonzero minimal worth for ρ, whatever the worth of χ.
For damaging curvature, we are going to set Okay = –1/a^{2} and
c = 1 – ξ^{2}, since we want c < 1 to be able to have a nonempty vary for ρ:
If 0 < ξ ≤ 1 and
b < a (1 – √[1 – ξ^{2}])½(d – √[d^{2} + 4 a^{2} b^{2} (ξ^{2} – 1)]) < ρ^{2} < ½(d + √[d^{2} + 4 a^{2} b^{2} (ξ^{2} – 1)]) If ξ > 1 a^{2} (ξ^{2} – 1) < ρ^{2} < ½(d + √[d^{2} + 4 a^{2} b^{2} (ξ^{2} – 1)]) the place d = a^{2} ξ^{2} – b^{2}
Generally we may have a nonzero minimal worth for ρ, but when ξ = 1 the boundaries turn into:
If ξ = 1 and b < a 0 < ρ^{2} < a^{2} – b^{2}
We’ll combine f ‘(ρ) first for the particular case ξ = 1 (i.e. c = 0).
This yields a tractroid when b = 0, and actually the integral is similar
aside from altering the fixed a within the earlier system to √[a^{2} – b^{2}]:
Dini’s floor
Okay = –1/a^{2}
c = 0
b < a
ρ_{m} = √[a^{2} – b^{2}]f ‘(ρ) = ±√[ρ_{m}^{2} – ρ^{2}] / ρ
f(ρ) = z_{ρm} ± (√[ρ_{m}^{2} – ρ^{2}] – ρ_{m} arccosh(ρ_{m}/ρ))
The consequence we get hold of by substituting this resolution into our parameterisation of helical surfaces
is called Dini’s surface.
Be aware that the parameterisations for this household of surfaces within the literature can look very totally different from the one we now have
used. For instance, MathWorld offers the parameterisation:
(x, y, z) = (a cos u sin v, a sin u sin v,
a [cos v + log(tan ½v)] + b u)
which has a Gaussian curvature of –1/(a^{2} + b^{2}) relatively than
–1/a^{2}, and whereas u right here is similar as our φ, the coordinate v is
analogous to the colatitude θ in spherical polar coordinates, and our ρ coordinate is proportional to
sin v. But when we changed a right here with √[a^{2} – b^{2}] this is able to describe
precisely the identical floor as ours.
As a result of the metric for all of the surfaces constructed from helices is unbiased of the φ coordinate, because it was
for the surfaces of revolution, we once more have a conserved amount alongside every geodesic: the dot product
between the unitlength tangent to the geodesic and the vector (1,0), which represents a uniform change of φ. The distinction is that, this time, relatively than pointing round in a circle, the vector
for a change in φ factors alongside the helices from which the floor is constructed.
For Dini’s floor, the formulation for ρ within the geodesics take a really comparable kind
to that of geodesics on the tractroid, however the formulation for φ are significantly extra sophisticated.
We don’t actually have “meridians” anymore, however the geodesics might be grouped into
three courses: these which have a minimal ρ worth, ρ_{0} > 0, these the place ρ is simply bounded beneath by zero, and are of finite size, and people that are of infinite size and strategy ρ = 0 asymptotically.
Dini’s floor
Okay = –1/a^{2}
c = 0
b < a
ρ_{m} = √[a^{2} – b^{2}]Geodesics with decrease certain ρ_{0} on ρ (with s measured from the purpose the place ρ = ρ_{0}):
ρ_{G}(s) = ρ_{0} cosh(s/a)
φ_{G}(ρ) = φ_{0}
+ (ρ_{m}/b) log(
(ρ_{m} + √[ρ_{m}^{2} – ρ^{2}])
(ρ_{m} – √[ρ_{m}^{2} – ρ_{0}^{2}]) / (ρ ρ_{0}))
+ ½ (a/b) log(
(a – √[ρ_{m}^{2} – ρ^{2}])
(a + √[ρ_{m}^{2} – ρ_{0}^{2}])
(ρ_{0}^{2} + b (b – √[b^{2} + ρ_{0}^{2}]))
(ρ (ρ – √[ρ^{2} – ρ_{0}^{2}])
+ (b (b + √[b^{2} + ρ_{0}^{2}]))) /
[(a + √[ρ_{m}^{2} – ρ^{2}])
(a – √[ρ_{m}^{2} – ρ_{0}^{2}])
(ρ_{0}^{2} + b (b + √[b^{2} + ρ_{0}^{2}]))
(ρ (ρ – √[ρ^{2} – ρ_{0}^{2}])
+ (b (b – √[b^{2} + ρ_{0}^{2}])))] )
The angle between the geodesic and the helix
ρ = ρ_{m} is acos(√[b^{2} + ρ_{0}^{2}]/a).Geodesics the place ρ goes to zero in a finite distance (with s measured from the purpose the place ρ = ρ_{m}):
ρ_{G}(s) = ρ_{m} cosh(s/a) – √[a^{2} – γ^{2}] sinh(s/a)
φ_{G}(ρ) = φ_{0}
+ (ρ_{m}/b) arctanh(√[1 – (ρ/ρ_{m})^{2}])
– (a/b) arccoth(a/√[ρ_{m}^{2} – ρ^{2}])
– ½ (a/b) log(
(ρ_{m} √[a^{2} – γ^{2}] + b γ – a^{2})
(b (b + γ) + ρ (ρ – √[ρ^{2} + b^{2} – γ^{2}])) /
[(ρ_{m} √[a^{2} – γ^{2}] – b γ – a^{2})
(b (b – γ) + ρ (ρ – √[ρ^{2} + b^{2} – γ^{2}]))] )
Right here γ is a parameter, 0 < γ < b. The angle between the geodesic and the helix
ρ = ρ_{m} is acos(γ/a).Geodesics the place ρ goes to zero asymptotically (with s measured from the purpose the place ρ = ρ_{m}):
ρ_{G}(s) = ρ_{m} exp(–s/a)
φ_{G}(ρ) = φ_{0}
+ (a/b) log(ρ_{m}(a – √[ρ_{m}^{2} – ρ^{2}])/(a ρ))
+ (ρ_{m}/b) log((ρ_{m} + √[ρ_{m}^{2} – ρ^{2}])/ρ)
The angle between the geodesic and the helix
ρ = ρ_{m} is acos(b/a).
It may be a bit stunning, at first, to understand that there are geodesics on Dini’s floor that
journey between the biggest helix, with ρ = ρ_{m}, and
factors the place ρ = 0, in a finite distance. On the tractroid,
the equator lies at one finish of an infinitely lengthy funnel that takes an infinite distance to taper all the way down to
zero width. However though there’s a contribution to the z coordinate of Dini’s floor
that grows infinitely damaging as ρ approaches zero, these finitelength geodesics counterbalance
that with the contribution from φ approaching optimistic infinity, in order that they terminate at a finite worth for z.
We are able to measure the width of Dini’s floor by the size of the orthogonal geodesics between the biggest helix ρ = ρ_{m} and the road ρ = 0;
these are the geodesics the place we set γ = 0, and their size is
a arccosh(a/b). The floor might be continued infinitely far in
the optimistic and damaging φ instructions, alongside the helices.
We are able to assemble an isometry between Dini’s floor and an infinitely lengthy strip inside the hyperbolic
aircraft, bounded on one facet by a geodesic, and on the opposite by a
hypercycle,
a curve that lies at a set distance from that geodesic.
The hypercycle corresponds to the biggest helix of Dini’s floor, and the geodesic corresponds to
the road ρ = 0. Within the Poincaré disk mannequin, a hypercycle is represented
by a round arc that meets the boundary of the disk on the similar factors as its related geodesic, whereas the
geodesic itself is both a diameter of the disk, or a round arc that meets the boundary orthogonally.
The picture on the correct exhibits a finite piece of Dini’s floor, with a = 1, b = 1/4 and –3π ≤ φ ≤ 3π. The isometry that we’ve used right here
maps the xaxis of the Poincaré disk mannequin to the road ρ = 0, whereas the biggest helix on the floor corresponds to a round arc within the higher half of the disk that passes
via the factors (–1, 0), (0, √[(1–b)/(1+b)]) and (1, 0).
The tiling is similar one we used for all of the negativelycurved surfaces within the earlier part.
The SineGordon Equation
This part coming quickly …
Appendix: Computing Gaussian Curvature
Gaussian curvature of an embedded floor
We assume that the surfaces we’re coping with are described by mathematical
features whose second derivatives exist, and are steady, in every single place, besides maybe alongside their boundaries. So at any inside level P on the floor we will all the time approximate its form in a small area
utilizing a Taylor series. If we select the origin of our coordinates to lie on the purpose P, and select the zaxis
to be orthogonal to the floor, the Taylor collection as much as quadratic phrases in x and y might be written within the kind:
z(x, y) = ½ A x^{2} + B x y + ½ C y^{2}
As a result of we now have positioned the origin on the floor there isn’t any fixed time period, and since
we now have made the xy aircraft tangent to the floor there are not any linear phrases. These three numbers,
A, B, and C, are sufficient to explain the form in addition to we have to calculate its Gaussian curvature.
To do that, we are going to begin by inspecting the curves we get if we slice the floor with a aircraft
that incorporates the normal to the floor at P, as within the diagrams on the higher proper. Every such curve (blue within the diagram)
is both a parabola or a straight line:
z(ρ, θ) = ½ [A (cos θ)^{2} + 2 B cos θ sin θ + C (sin θ)^{2}] ρ^{2}
Right here θ is the angle the slicing aircraft makes with the xaxis, and ρ is the horizontal distance
measured from the zaxis.
We are able to quantify these parabolas’
curvature,
κ, by computing the speed at which a unitlength tangent to the curve is altering course.
For an ideal round arc of radius R, the curvature outlined this manner is simply 1/R.
For a parabola of the shape z = ½b x^{2},
it is the same as b at x = 0.
And the place the parabola we get hold of by slicing via our quadratic floor passes via the purpose P, the curvature
is the same as:
κ(θ) = A (cos θ)^{2} + 2 B cos θ sin θ + C (sin θ)^{2}
See Also
Within the diagrams
on the decrease proper, the reciprocal of the sq. root of absolutely the
worth of the curvature is plotted as the space from the origin for various values of θ,
with crimson for optimistic curvature and inexperienced for damaging curvature. These plots are
conic sections: there are crimson and inexperienced hyperbolas the place the curvature modifications signal, and a crimson
ellipse the place the curvature stays optimistic.
If we outline S_{0} to be the matrix:
then we will write the quadratic expression for z, and the curvature alongside the slice, as:
z(x, y) =
½ (x, y)^{T} S_{0} (x, y)
κ(θ) = (cos θ, sin θ)^{T} S_{0} (cos θ, sin θ)
The curvature will attain most and minimal values when the vector (cos θ, sin θ)
is an
eigenvector
of the matrix S_{0}, and the curvature itself would be the corresponding
eigenvalue.
The dashed traces within the plots present the instructions of the eigenvectors.
However we don’t want to seek out the eigenvectors to know that
the eigenvalues, κ_{1} and κ_{2}, would be the two roots of the characteristic polynomial of S_{0}:
det(S_{0} – κ I) = κ^{2} – (A + C) κ + (A C – B^{2}) = 0
And we don’t want to resolve this quadratic equation explicitly to know that the product of its two roots
is the same as the fixed time period within the quadratic:
Okay = κ_{1} κ_{2} = A C – B^{2} = det(S_{0})
We name κ_{1} and κ_{2} the principal regular curvatures of the floor
at level P, and we declare that their product is the Gaussian curvature, Okay, of the floor at P.
In truth, that is the unique definition of Gaussian curvature, and the commonest method of defining it
when speaking about embedded surfaces. It’s not laborious to see
that this may agree with our earlier definition within the case of a sphere, since each curve produced by a standard slice
via a sphere of radius a may have curvature 1/a, so the product of κ_{1} and κ_{2} alongside any two slices might be 1/a^{2}.
Proof that the circumference and principal curvature definitions of Gaussian curvature are equal
We are able to present that the definition of Okay because the product of
the principal curvatures κ_{1} and κ_{2} agrees, usually, with our earlier definition of Okay
by way of the circumference of a circle.
To proceed, we might want to know the geodesic curves on our quadratic floor, or relatively, we might want to approximate
them with polynomial curves, nicely sufficient to compute the limiting behaviour of the circumference of a small circle.
We’ll begin by making use of a “canonical illustration” of
a curve in threedimensional house with a given curvature, κ, charge of change of curvature dκ/ds, and
torsion, τ,
sooner or later. A curve with these properties might be approximated with three cubic polynomials
within the arc size, s, measured from the purpose:^{[2]}
G(s) =
(s – (κ^{2}/6) s^{3},
(κτ/6) s^{3},
(κ/2) s^{2} + ((dκ/ds)/6) s^{3})
This curve has a tangent at s = 0 that factors alongside the xaxis, whereas the instantaneous
charge of change of the tangent vector at s = 0 is a vector of size κ pointing alongside the zaxis.
The torsion, τ, is the speed at which the aircraft of the curve is rotating across the curve.
Be aware that s is just not exactly the arc size of this curve, but when we compute the squared size
of the tangent with respect to s, it is the same as 1 plus phrases of order 3 and 4 in s:
G‘(s)^{2}
= 1 + (dκ/ds) κ s^{3} + (((dκ/ds)^{2} + κ^{4} + κ^{2} τ^{2})/4) s^{4}
We wish to discover the geodesic curves that go via the origin at an angle of θ with the xaxis,
and whose curvature is given by the operate κ(θ) that we computed earlier. We are able to set dκ/ds
initially of the curve to zero by symmetry, set τ to a operate τ(θ) but to be decided,
after which rotate the entire curve G(s) across the zaxis so its tangent at s = 0 factors within the right course.
So as to make the curve approximation to a geodesic
for our quadratic floor, we are going to use the truth that alongside a geodesic, the
tangent to the curve, its spinoff, and the traditional vector to the floor all lie in the identical aircraft.^{[3]} If we name our rotated curve G_{θ}(s), we discover that to the bottom order in s:
(G_{θ}‘(s) × G_{θ}”(s)) ·
(G_{θ}‘(s) × dG_{θ}(s)/dθ)
≈
½ κ(θ) (κ'(θ) + 2 τ(θ)) s^{2}
Right here the second cross product, between the tangent to the curve and the speed of change of a degree on the curve with respect to θ, is used to approximate a vector orthogonal to the floor, for the reason that floor itself is (roughly) swept out
by various s and θ. For a geodesic, with G_{θ}‘(s), G_{θ}”(s) and the traditional to the floor all coplanar, the dot product right here would
be precisely zero, and we will make this approximation zero by setting:
τ(θ) = –½κ'(θ) = (A – C) cos θ sin θ + B ((sin θ)^{2} – (cos θ)^{2})
It’s not stunning that the torsion is proportional to the spinoff of the curvature, since
the curvature reaches its most and minimal values when the aircraft slicing the floor is a aircraft of symmetry,
and we might anticipate the geodesics there to be planar, i.e. with zero torsion.
Having chosen the torsion this manner, we then discover that as much as order s^{3}:
dG_{θ}(s)/dθ ≈ s – (1/6)(A C – B^{2}) s^{3}
which tells us that the circumference of a circle of radius r in our quadratic floor, centred on the origin,
might be:
Circumference(r) ≈ 2 π r – (π/3)(A C – B^{2}) r^{3}
and that
Okay = (3/π) lim_{r → 0} (2 π r – Circumference(r)) / r^{3}
= A C – B^{2}
in settlement with our earlier calculation that gave Okay because the product of the principal curvatures.
The pictures on the correct present geodesics via the origin, and circles across the origin, for quadratic surfaces
of optimistic (prime) and damaging (backside) Gaussian curvature.
Probably the most frequent methods to explain a floor embedded in threedimensional house is by specifying its
three Cartesian coordinates (x, y, z) as features of two parameters, say u and v:
x(u, v) = (x(u, v), y(u, v), z(u, v))
We now have seen learn how to compute the Gaussian curvature at a single level on a quadratic floor that we described in a fastidiously chosen Cartesian coordinate system, however it seems to not be a lot more durable to undertake the identical
computation to an arbitrary level on a parameterised floor (assuming, as all the time, that the related features
are sufficiently easy for all of the derivatives we have to exist).
We begin by defining two matrices which can be related
with every level on the floor x(u, v):
g(u, v) =
x_{u} · x_{u} x_{u} · x_{v} x_{u} · x_{v} x_{v} · x_{v} S(u, v) =
x_{uu} · n x_{uv} · n x_{uv} · n x_{vv} · n
the place the image · is the same old Euclidean
dot product
between vectors,
and we now have launched abbreviations for the next vectors at every level:
x_{u} = ∂_{u}x(u, v) is a tangent vector to the curve of various u
x_{v} = ∂_{v}x(u, v) is a tangent vector to the curve of various v
n = x_{u} × x_{v} / x_{u} × x_{v} is a unitlength regular to the floor
x_{uu} = ∂_{u,u}x(u, v) is the instantaneous charge of change of x_{u} alongside the curve of various u
x_{uv} = ∂_{u,v}x(u, v) = ∂_{v,u}x(u, v) is the instantaneous charge of change of x_{u} alongside the curve of various v (and vice versa)
x_{vv} = ∂_{v,v}x(u, v) is the instantaneous charge of change of x_{v} alongside the curve of various v
The primary of those matrices, g, known as the
metric tensor or the
first fundamental form,
and S is called the second fundamental form.
The metric, g, can be utilized to compute the dot product between two vectors, say a and b,
that lie within the tangent aircraft to the
floor, given their coordinates within the foundation {x_{u}, x_{v}}:
a = a^{u} x_{u} + a^{v} x_{v}
b = b^{u} x_{u} + b^{v} x_{v}
a · b = (a^{u}, a^{v})^{T} g (b^{u}, b^{v})
The second elementary kind, S, captures the identical sort of quadratic Taylor collection for the perpendicular distance
from the tangent aircraft to the floor that we described with the matrix S_{0}:
Perpendicular distance from tangent aircraft at P=x(u_{P},v_{P}), to a closeby level on the floor,
x(u,v)
≈ ½ (u – u_{P}, v – v_{P})^{T} S (u – u_{P}, v – v_{P})
In truth, S_{0} is exactly what S could be if we selected parameters for the floor that corresponded
to a Cartesian grid within the tangent aircraft at P.
However since we will get hold of the Gaussian curvature, Okay, because the determinant of S_{0}, we will additionally
get hold of it from the determinant of S, together with an element that accounts for the change of foundation between the vectors
{x_{u}, x_{v}}, related to the parameters u and v, and the
orthonormal basis
related to the Cartesian coordinates we used when discussing S_{0}.
Suppose the matrix M takes the coordinates of a vector in an orthonormal foundation for the tangent aircraft, and
offers us coordinates within the foundation {x_{u}, x_{v}}. Then
we will use the truth that g offers us the dot product by way of the latter coordinates to jot down:
a · b = (a^{u}, a^{v})^{T} g (b^{u}, b^{v}) = (M (a^{x}, a^{y}))^{T} g
(M (b^{x}, b^{y}))
= (a^{x}, a^{y})^{T} (M^{T} g M) (b^{x}, b^{y})
the place (a^{x}, a^{y}), (b^{x}, b^{y}) are the coordinates of the vectors a and b within the orthonormal foundation. However we even have:
a · b = (a^{x}, a^{y})^{T} I (b^{x}, b^{y})
the place I is the two×2 identification matrix. For this to carry for all pairs of vectors a and b, we will need to have:
M^{T} g M = I
from which it follows that:
det(M^{T}) det(g) det(M) = det(I)
det(M)^{2} = 1 / det(g)
However the identical change of foundation matrix M applies to S and S_{0}:
S_{0} = M^{T} S M
which lets us write the Gaussian curvature as:
Okay = det(S_{0}) = det(M)^{2} det(S) = det(S) / det(g)
We are able to examine this with a calculation for a sphere of radius a embedded in threedimensional house, with the same old polar
coordinates because the parameters:
x(u, v) = a (sin u cos v, sin u sin v, cos u)
x_{u} = a (cos u cos v, cos u sin v, –sin u)
x_{v} = a (–sin u sin v, sin u cos v, 0)
n = (sin u cos v, sin u sin v, cos u)
x_{uu} = a (–sin u cos v, –sin u sin v, –cos u)
x_{uv} = a (–cos u sin v, cos u cos v, 0)
x_{vv} = a (–sin u cos v, –sin u sin v, 0)
g(u, v) =
x_{u} · x_{u} x_{u} · x_{v} x_{u} · x_{v} x_{v} · x_{v} = S(u, v) =
x_{uu} · n x_{uv} · n x_{uv} · n x_{vv} · n = Okay = det(S) / det(g) = [a^{2} (sin u)^{2} ] / [a^{4} (sin u)^{2}] = 1/a^{2}
Gaussian curvature from a metric
We’ve seen learn how to compute the Gaussian curvature for a floor embedded in threedimensional Euclidean house,
however we all know, from our definition of the curvature by way of the geometry of a circle inside the house, that
Gaussian curvature is an intrinsic property of the twodimensional house itself. So we ought to have the ability to compute
it with out making use of any details about the embedding, and in conditions the place the house of curiosity merely isn’t embedded in any higherdimensional house in any respect.
The intrinsic geometry of an area is decided by its
metric, g.
We described how g might be computed for an embedded floor within the final part,
the place the geometry of the floor is induced by the geometry of the threedimensional house during which it sits,
however within the context of intrinsic geometry the metric is just acknowledged as a part of the definition of the house.
So our objective might be to compute the Gaussian curvature from the metric alone.
There’s a substantial physique of arithmetic coping with varied measures of intrinsic curvature in areas of
arbitrary dimension, a lot of which was first developed within the nineteenth century, and later discovered a job
within the twentieth century physics of Basic Relativity. I sketched among the important concepts wanted for GR
in a short article that covers the notions of
parallel transport and the
Riemann curvature tensor, and these ideas are nicely value finding out each
for his or her bodily functions and for a deeper understanding of the geometry itself. However for our functions all we
actually need is a system for the Gaussian curvature by way of the metric.
We’ll strategy this by looking for an equation that describes the geodesic curves in an area with a given metric, g(u, v).
Suppose we now have a curve G(λ) from the purpose P with coordinates (u_{P},v_{P})
to the purpose Q with coordinates (u_{Q},v_{Q}):
G(λ) = (u_{G}(λ), v_{G}(λ))
G(0) = (u_{G}(0), v_{G}(0)) = (u_{P},v_{P})
G(1) = (u_{G}(1), v_{G}(1)) = (u_{P},v_{P})
Right here u_{G}(λ) and v_{G}(λ) are no matter features give the coordinates of G(λ), and λ is a parameter that ranges from 0 to 1 alongside the curve.
If G(λ) is a geodesic, then any curve from P to Q might be longer than
G(λ).
Particularly, suppose we now have features u_{δ}(λ)
and v_{δ}(λ) that fulfill:
u_{δ}(0) = u_{δ}(1) = 0
v_{δ}(0) = v_{δ}(1) = 0
and we add multiples of those features to the coordinates of G(λ) to acquire a household of
curves from P to Q, with a parameter μ specifying the totally different curves inside this household:
c_{μ}(λ) = (u_{G}(λ) + μ u_{δ}(λ),
v_{G}(λ) + μ v_{δ}(λ))
c_{0}(λ) = G(λ)
The size of any explicit curve on this household might be given by:
L(μ) = ∫_{0}^{1} √[c_{μ}‘(λ)^{T}
g c_{μ}‘(λ)] dλ
For the sake of brevity we merely write g for the metric, however notice that we imply
g(c_{μ}(λ)), i.e. we should consider the metric, which is a operate of
the coordinates u and v, alongside the curve c_{μ}(λ), which makes
the metric a operate of each μ and λ.
(We launched the metric as a method of computing the dot product between vectors within the tangent aircraft to
a floor embedded in threedimensional house, however we haven’t talked concerning the exact that means of
a “vector” when the twodimensional curved house isn’t embedded in any bigger house.
For our current functions, we are going to merely notice that this manner of computing the size of a curve is
in step with the way in which we launched the metric. The dot product of any vector with itself is the squared size
of that vector, so the sq. root of the dot product with itself of
the vector tangent to a curve, with respect to the parameter λ, offers the speed of change ds/dλ of arc size s alongside the curve with respect to λ, and this integral will give the entire size of the curve.
The article on General Relativity talked about earlier delves into precisely
what a tangent vector means within the context of intrinsic geometry.)
If c_{0}(λ) = G(λ) is a geodesic, the operate
L(μ) that offers the size of every curve on this household may have a minimal at μ = 0.
We might anticipate L(μ) to be a differentiable operate, with a spinoff of zero at its minimal:
L‘(0) = 0
What’s extra, we anticipate this to be true no matter features we select for u_{δ}(λ)
and v_{δ}(λ), as long as they’re zero at their endpoints. This quantities to saying that
nevertheless we select a household of curves from P to Q that features the geodesic G(λ),
the geodesic will all the time be the shortest, no less than in comparison with different curves that solely differ from it by a small
quantity. (In areas that aren’t
simply connected,
there may be a number of geodesics that be part of the identical two factors, solely one in every of which would be the shortest.)
We wish to discover the spinoff of L(μ). Beginning with the derivatives with respect to μ of some items of the integral, we now have:
dc_{μ}(λ)/dμ = (u_{δ}(λ), v_{δ}(λ))
dc_{μ}‘(λ)/dμ = (u_{δ}‘(λ), v_{δ}‘(λ))dg/dμ = ∂g/∂u u_{δ}(λ)
+ ∂g/∂v v_{δ}(λ)d [c_{μ}‘(λ)^{T}
g c_{μ}‘(λ)] / dμ =
2 (u_{δ}‘(λ), v_{δ}‘(λ))^{T} g c_{μ}‘(λ) +
c_{μ}‘(λ)^{T}
(∂g/∂u u_{δ}(λ)
+ ∂g/∂v v_{δ}(λ)) c_{μ}‘(λ)L‘(μ) = ∫_{0}^{1} (d √[c_{μ}‘(λ)^{T}
g c_{μ}‘(λ)] / dμ) dλ
= ∫_{0}^{1} (d [c_{μ}‘(λ)^{T}
g c_{μ}‘(λ)] / dμ) / (2 √[c_{μ}‘(λ)^{T}
g c_{μ}‘(λ)]) dλ
We are able to apply the system for
integration by parts:
∫_{0}^{1} f ‘(λ) g(λ) dλ = f(1) g(1) – f(0) g(0) – ∫_{0}^{1} f(λ) g‘(λ) dλ
to phrases multiplied by u_{δ}‘(λ) or v_{δ}‘(λ),
to rewrite the final integral as:
L‘(μ) = ∫_{0}^{1} {
– (u_{δ}(λ), v_{δ}(λ))^{T}
d/dλ [g c_{μ}‘(λ) / √[c_{μ}‘(λ)^{T}
g c_{μ}‘(λ)]] + ½ c_{μ}‘(λ)^{T}
(∂g/∂u u_{δ}(λ)
+ ∂g/∂v v_{δ}(λ)) c_{μ}‘(λ) / √[c_{μ}‘(λ)^{T}
g c_{μ}‘(λ)] }
dλ
If we consider this at μ = 0, it needs to be zero no matter decisions we make for
u_{δ}(λ) and v_{δ}(λ), which might solely be true if each these features
seem within the integral multiplied by elements which can be zero for all values of λ. Recall that
c_{0} is the same as G, so we now have:
– (1, 0)^{T}
d/dλ [g G‘(λ) / √[G‘(λ)^{T}
g G‘(λ)]] + ½ G‘(λ)^{T}
(∂g/∂u) G‘(λ) / √[G‘(λ)^{T}
g G‘(λ)] = 0
– (0, 1)^{T}
d/dλ [g G‘(λ) / √[G‘(λ)^{T}
g G‘(λ)]] + ½ G‘(λ)^{T}
(∂g/∂v) G‘(λ) / √[G‘(λ)^{T}
g G‘(λ)] = 0
We are able to divide
each equations by √[G‘(λ)^{T}
g G‘(λ)], multiply by –1, after which use:
√[G‘(λ)^{T}
g G‘(λ)] = ds/dλ
1/ √[G‘(λ)^{T}
g G‘(λ)] = dλ/ds
to alter all derivatives with respect to λ into derivatives with respect to arc size, s.
We additionally redefine G, u_{G} and v_{G} to be features of
s relatively than of λ, to maintain the notation easy.
(1, 0)^{T}
d/ds [g G‘(s)] – ½ G‘(s)^{T}
(∂g/∂u) G‘(s) = 0
(0, 1)^{T}
d/ds [g G‘(s)] – ½ G‘(s)^{T}
(∂g/∂v) G‘(s) = 0
If we write x^{1} for u_{G}(s) and x^{2} for v_{G}(s),
write the elements of the metric as g_{ij},
and use the
Einstein summation convention
of summing over {1,2} for any repeated index in a product,
we will write
each these equations as:
g_{ij} d^{2}x^{j}/ds^{2}
+ [∂g_{ij}/∂x^{k}
– ½ ∂g_{jk}/∂x^{i}] (dx^{j}/ds)
(dx^{okay}/ds) = 0
the place the index i is both 1 or 2 for the 2 equations, and j and okay are summed over {1,2}.
The expression in brackets might be changed by one thing extra symmetrical, the place we take the common
of the primary time period and a model of it with j and okay swapped:
Γ_{ijk} = ½(∂g_{ij}/∂x^{okay} + ∂g_{iokay}/∂x^{j} – ∂g_{jokay}/∂x^{i})
That is not similar to the expression in brackets, however once we sum over j and okay within the equation
the sum seems to be precisely the identical, as a result of (dx^{j}/ds)
(dx^{okay}/ds) is symmetrical once we swap j and okay.
The equations can then be written as:
g_{ij} d^{2}x^{j}/ds^{2}
+ Γ_{ijk} (dx^{j}/ds)
(dx^{okay}/ds) = 0
The Γ_{ijk} are generally known as the
Christoffel symbols of the first kind. It could make no distinction to any of our calculations if we used the unique expression
in brackets as a substitute, however utilizing the Christoffel symbols is regular apply, and it could threat
confusion if we didn’t observe it.
The most typical model of the geodesic equations is:
d^{2}x^{i}/ds^{2}
+ Γ^{i}_{jk} (dx^{j}/ds)
(dx^{okay}/ds) = 0
the place we now have eliminated the metric multiplying the second spinoff time period by multiplying via by the
inverse of the metric, g^{–1}, whose elements are normally written as g^{ij}, and we
outline Christoffel symbols of the second kind as:
Γ^{i}_{jk} = g^{im} Γ_{mjk}
= ½ g^{im}
(∂g_{mj}/∂x^{okay} + ∂g_{mokay}/∂x^{j} – ∂g_{jokay}/∂x^{m})
To examine that what we’ve executed to date is right, let’s compute the Christoffel
symbols for the best curved house: a sphere of radius a, with the parameters
u and v because the colatitude (angle from the north pole) and longitude.
g(u, v) = Γ_{1}(u, v) = Γ_{2}(u, v) =
0 a^{2} sin u cos u a^{2} sin u cos u 0 g^{–1}(u, v) =
a^{–2} 0 0 a^{–2} (sin u)^{–2} Γ^{1}(u, v) = Γ^{2}(u, v) =
We’ll sneak again into threedimensional house and write the Cartesian coordinates
for a household of geodesics that go via the purpose (a, 0, 0) and make an angle
θ with the equator. To maneuver a distance s alongside one in every of these geodesics, we simply rotate by the angle
s/a from our start line in direction of a degree the place the geodesic reaches its most
latitude, (0, a cos(θ), a sin(θ)):
G_{xyz}(s) = cos(s/a) (a, 0, 0) + sin(s/a) (0, a cos(θ), a sin(θ)) = a (cos(s/a), cos(θ) sin(s/a), sin(θ) sin(s/a))
The colatitude and longitude are then given by:
u_{G}(s) = acos[sin(θ) sin(s/a)]v_{G}(s) = atan[cos(θ) tan(s/a)]
The primary derivatives with respect to s are:
u_{G}‘(s) =
–sin(θ) cos(s/a) / (a [1 – sin(θ)^{2} sin(s/a)^{2}]^{1/2})
v_{G}‘(s) =
cos(θ) / (a [1 – sin(θ)^{2} sin(s/a)^{2}])
and the second derivatives are:
u_{G}”(s) =
sin(θ) cos(θ)^{2} sin(s/a) /
(a^{2} [1 – sin(θ)^{2} sin(s/a)^{2}]^{3/2})
v_{G}”(s) =
2 cos(θ) sin(θ)^{2} sin(s/a) cos(s/a) / (a^{2} [1 – sin(θ)^{2} sin(s/a)^{2}]^{2})
Alongside these geodesics, the Christoffel symbols of the second variety turn into:
Γ^{1}(s) =
0 0 0 –sin(θ) sin(s/a) √[1 – sin(θ)^{2} sin(s/a)^{2}] Γ^{2}(s) =
0 sin(θ) sin(s/a) / √[1 – sin(θ)^{2} sin(s/a)^{2}] sin(θ) sin(s/a) / √[1 – sin(θ)^{2} sin(s/a)^{2}] 0
It’s not laborious to examine that:
u_{G}”(s) + Γ^{1}_{22}(s) v_{G}‘(s)^{2} = 0
v_{G}”(s) + 2 Γ^{2}_{12}(s) u_{G}‘(s) v_{G}‘(s) = 0
which confirms that u_{G}(s), v_{G}(s) fulfill
the geodesic equations.
Our purpose now’s to compute the Gaussian curvature for any metric, by discovering the coefficient of the s^{3}
time period within the circumference of a circle with a geodesic radius of s.
To do that, we are going to want a Taylor collection for the geodesics via a degree to thirdorder in s.
The geodesic equation offers us the second derivatives by way of the primary derivatives, and we will differentiate once more
to get the third derivatives.
d^{2}x^{i}/ds^{2} = –Γ^{i}_{jk} (dx^{j}/ds)
(dx^{okay}/ds)d^{3}x^{i}/ds^{3} = –∂Γ^{i}_{jk}/∂x^{m} (dx^{j}/ds)
(dx^{okay}/ds)
(dx^{m}/ds)
– 2 Γ^{i}_{jk} (d^{2}x^{j}/ds^{2})
(dx^{okay}/ds)= –∂Γ^{i}_{jk}/∂x^{m} (dx^{j}/ds)
(dx^{okay}/ds)
(dx^{m}/ds)
+ 2 Γ^{i}_{jk} Γ^{j}_{pq} (dx^{p}/ds)
(dx^{q}/ds)
(dx^{okay}/ds)= [2 Γ^{i}_{jk} Γ^{j}_{pq}
– ∂Γ^{i}_{pk}/∂x^{q}] (dx^{p}/ds)
(dx^{q}/ds)
(dx^{okay}/ds)
If we abbreviate the primary derivatives by writing:
X^{i} = [dx^{i}/ds]_{s = 0}
then the thirdorder Taylor collection for the geodesic might be written:
x^{i} ≈ x^{i}_{P}
+ X^{i} s
– ½ Γ^{i}_{jk} X^{j} X^{okay}
s^{2}
+ (1/6) [2 Γ^{i}_{jk} Γ^{j}_{pq}
– ∂Γ^{i}_{pk}/∂x^{q}] X^{p} X^{q} X^{okay}
s^{3}
the place the Christoffel symbols and their derivatives are additionally evaluated at s = 0.
By rotating the tangent to the geodesic initially, a degree that lies at a
distance of s alongside the curve will sweep out a circle of fixed
geodesic radius. We wish to discover the size of the tangent to that circle, with respect to
the angle of rotation of the geodesic’s preliminary tangent. However relatively than explicitly writing the
elements of the preliminary tangent, X^{i}, as features of a course angle θ,
we are going to use the truth that their derivatives with respect to θ have to be the elements of a unitlength vector
orthogonal to the tangent. In Euclidean coordinates, this is able to merely be a matter of noting
that the counterclockwise unit vector orthogonal to some unit vector (X, Y)
is (–Y, X), however as a result of we’re coping with basic coordinates, we have to
use a barely fancier system:
Z^{α} = –ε^{αβ} g_{βγ} X^{γ} / √det(g) the place ε =
The image ε right here known as the
LeviCivita symbol.
Given any tangent vector X whose elements are charges of change of the u, v coordinates, the vector Z might be orthogonal to it, and of the identical size.
Taking the spinoff of the geodesic with respect to the preliminary tangent elements X^{α} offers us:
dx^{i}/dX^{α} ≈
δ^{i}_{α} s
– Γ^{i}_{αokay} X^{okay}
s^{2}
+ (1/6) [4 Γ^{i}_{jq} Γ^{j}_{αp}
+ 2 Γ^{i}_{jα} Γ^{j}_{pq}
– 2 ∂Γ^{i}_{αq}/∂x^{p}
– ∂Γ^{i}_{pq}/∂x^{α}] X^{p} X^{q}
s^{3}
We then get hold of the tangent vector to the circle by multiplying by Z^{α}
(with the implied sum from the repeated index α):
dx^{i}/dθ ≈
Z^{ i} s
– Γ^{i}_{αokay} X^{okay}
Z^{α}
s^{2}
+ (1/6) [4 Γ^{i}_{jq} Γ^{j}_{αp}
+ 2 Γ^{i}_{jα} Γ^{j}_{pq}
– 2 ∂Γ^{i}_{αq}/∂x^{p}
– ∂Γ^{i}_{pq}/∂x^{α}] X^{p} X^{q} Z^{α}
s^{3}
We compute the squared size of this tangent to the circle with the metric, however we will’t merely use the
metric initially of the geodesic; we have to use a Taylor collection in s.
g_{ij}(s) ≈ g_{ij}(0) +
∂g_{ij}/∂x^{p} X^{p} s +
½
[∂^{2}g_{ij}/(∂x^{p} ∂x^{q})
– ∂g_{ij}/∂x^{r} Γ^{r}_{pq}] X^{p} X^{q} s^{2}dG/dθ^{2} = g_{ij}(s) dx^{i}/dθ dx^{j}/dθ
The bottomorder time period within the squared size is the secondorder time period in s, which is:
 the zerothorder time period within the metric multiplied by the firstorder time period from every copy of the tangent vector dx^{i}/dθ
The coefficient is 1, as a result of Z is a unitlength vector.
g_{ij}(0) Z^{ i} Z^{ j} s^{2} = s^{2}
The thirdorder time period is the sum of:
 the zerothorder time period within the metric multiplied by the firstorder time period from
one copy of the tangent vector and the secondorder time period from the opposite  the firstorder time period within the metric multiplied by the firstorder time period from
every copy of the tangent vector
Be aware that multiplying Christoffel symbols of the second variety
by the metric lowers the primary index and converts them again to Christoffel symbols of the primary variety.
As a result of the expression in brackets is antisymmetric (i.e. it modifications signal) when α and β
are exchanged, summing it with the symmetric expression Z^{α} Z^{β}
offers zero.
[∂g_{βα}/∂x^{k} – 2 g_{iβ}(0) Γ^{i}_{αk}] X^{okay}
Z^{α} Z^{β}
s^{3}
=
[∂g_{βα}/∂x^{k} – 2 Γ_{βαk}] X^{okay}
Z^{α} Z^{β}
s^{3}
=
[∂g_{αk}/∂x^{β}
–
∂g_{βk}/∂x^{α}] X^{okay}
Z^{α} Z^{β}
s^{3}
= 0
The fourthorder time period is the sum of:
 the zerothorder time period within the metric multiplied by the firstorder time period from one copy of the tangent
vector and the thirdorder time period from the opposite  the zerothorder time period within the metric multiplied by the secondorder time period from every copy of the tangent
vector  the firstorder time period within the metric multiplied by the firstorder time period from one copy of the tangent
vector and the secondorder time period from the opposite  the secondorder time period within the metric multiplied by the firstorder time period from every copy of the tangent
vector
[(1/3) g_{iβ}(0)
(4 Γ^{i}_{jq} Γ^{j}_{αp}
+ 2 Γ^{i}_{jα} Γ^{j}_{pq}
– 2 ∂Γ^{i}_{αq}/∂x^{p}
– ∂Γ^{i}_{pq}/∂x^{α})
+ g_{ij}(0)
Γ^{i}_{αp}
Γ^{j}_{βq}
– 2 ∂g_{αj}/∂x^{p}
Γ^{j}_{βq}
+ ½(∂^{2}g_{αβ}/(∂x^{p} ∂x^{q})
– ∂g_{αβ}/∂x^{j} Γ^{j}_{pq})] X^{p} X^{q} Z^{α} Z^{β}
s^{4}= [(1/3)
(4 Γ_{βjq} Γ^{j}_{αp}
+ 2 Γ_{βjα} Γ^{j}_{pq}
– 2 g_{iβ}(0) ∂Γ^{i}_{αq}/∂x^{p}
– g_{iβ}(0) ∂Γ^{i}_{pq}/∂x^{α})
+ (Γ_{jαp}
– 2 ∂g_{αj}/∂x^{p})
Γ^{j}_{βq}
+ ½(∂^{2}g_{αβ}/(∂x^{p} ∂x^{q})
– ∂g_{αβ}/∂x^{j} Γ^{j}_{pq})] X^{p} X^{q} Z^{α} Z^{β}
s^{4}= – (1/3) (Γ_{jαq} Γ^{j}_{βp}
– Γ_{jpq} Γ^{j}_{βα}
+ ∂Γ_{βpq}/∂x^{α}
– ∂Γ_{βαq}/∂x^{p})
X^{p} X^{q} Z^{α} Z^{β}
s^{4}
The mixture of Christoffel symbols and their derivatives that seems right here:
R_{βqαp} = Γ_{jαq} Γ^{j}_{βp}
– Γ_{jpq} Γ^{j}_{βα}
+ ∂Γ_{βpq}/∂x^{α}
– ∂Γ_{βαq}/∂x^{p}
is called the
Riemann curvature tensor. Given the 4 indices, in two dimensions it has 16 elements, however it’s so symmetrical
that 12 of them are zero, and the opposite 4 all have the identical absolute worth, so it’s basically
described by a single quantity. Particularly, swapping both the primary two indices (β ↔ q) or the final two indices (α ↔ p) modifications the signal, and swapping the primary two with the final two
(βq ↔ αp) leaves the worth unchanged.
We’ll outline the opposite tensor we summed this towards as S:
S^{βqαp}
= X^{p} X^{q} Z^{α} Z^{β}
= X^{p} X^{q}
ε^{αγ} g_{γδ} X^{δ}
ε^{βη} g_{ησ} X^{σ} / det(g)
Whereas S itself is determined by the vector X, if we add up 4 copies the place we swap indices and alter
indicators as follows, we discover that:
S^{βqαp}
– S^{qβαp}
– S^{βqpα}
+ S^{qβpα}
= ½ (X^{T} g X)^{2}
(H^{βqαp}
– H^{qβαp}
– H^{βqpα}
+ H^{qβpα})
the place H is outlined as:
H^{βqαp} = g^{pq} g^{αβ}
Since X is all the time chosen to be a unitlength vector, the issue on the righthand facet simply turns into ½. If we multiply each side of this equation by R_{βqαp}, we get:
R_{βqαp} S^{βqαp}
– R_{βqαp} S^{qβαp}
– R_{βqαp} S^{βqpα}
+ R_{βqαp} S^{qβpα}
= ½ (R_{βqαp} H^{βqαp}
– R_{βqαp} H^{qβαp}
– R_{βqαp} H^{βqpα}
+ R_{βqαp} H^{qβpα})
The second, third and fourth phrases on both sides of the equation now have totally different ordering of the indices
on the 2 tensors, however we will use the symmetries of the Riemann tensor to make its indices match the opposite
tensor in every case, whereas additionally eradicating all of the minus indicators:
R_{βqαp} S^{βqαp}
+ R_{qβαp} S^{qβαp}
+ R_{βqpα} S^{βqpα}
+ R_{qβpα} S^{qβpα}
= ½ (R_{βqαp} H^{βqαp}
+ R_{qβαp} H^{qβαp}
+ R_{βqpα} H^{βqpα}
+ R_{qβpα} H^{qβpα})
Now that the order of the indices match, if we sum over all values for all indices we simply get
4 similar sums on both sides of the equation:
4 R_{βqαp} S^{βqαp}
= 2 R_{βqαp} H^{βqαp}
So our fourth order time period turns into,
unbiased of X:
– (1/6) g^{pq} g^{αβ} R_{βqαp} s^{4}
Our calculation to date has been for the squared size of the tangent vector dx^{i}/dθ:
dx^{i}/dθ^{2} ≈ s^{2} – (1/6) g^{pq} g^{αβ}
R_{βqαp} s^{4}
however we wish the size itself:
dx^{i}/dθ ≈ s – (1/12) g^{pq} g^{αβ} R_{βqαp} s^{3}
It follows that the Gaussian curvature is given by:
Okay = ½g^{pq} g^{αβ} R_{βqαp}
For the case of a sphere, we have already got the Christoffel symbols, and it’s not too laborious to
discover the Riemann tensor, whose 4 nonzero elements are:
R_{1212} = R_{2121} = a^{2} (sin u)^{2}
R_{1221} = R_{2112} = –a^{2} (sin u)^{2}
The elements of the inverse metric are:
g^{11} = a^{–2}
g^{22} = a^{–2} (sin u)^{–2}
g^{12} = g^{21} = 0
So we now have:
Okay = ½(g^{22} g^{11} R_{1212}
+ g^{11} g^{22} R_{2121}
+ g^{12} g^{21} R_{1221}
+ g^{21} g^{12} R_{2112}) = 1/a^{2}
as anticipated.
References
[1] Idea and Issues of Differential Geometry, Martin Lipschutz, Schaum’s Define Collection, McGrawHill 1969. [2] Lipschutz, op. cit., web page 83. [3] Lipschutz, op. cit., web page 233, equation (11.3). [4] Lipschutz, op. cit., web page 234, Theorem 11.7. [5] Lipschutz, op. cit., web page 256, solved drawback 11.26. [6] Lipschutz, op. cit., web page 257, solved drawback 11.27. [7] Seguin B, Chen Yc, Fried E. 2021 “Bridging the hole between rectifying developables and tangent developables: a household of developable surfaces related to an area curve.” Proc. R. Soc. A 477: 20200617. https://doi.org/10.1098/rspa.2020.0617For those who hyperlink to this web page, please use this URL: https://www.gregegan.net/SCIENCE/PSP/PSP.html
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