This is the second article written for the Summer 1997 NRICH
Online Maths Video-conference linking schools in London and
Norfolk. This article discusses some possible answers to some of
the problems posed in the article "
Geometry and Gravity 1 ".
1. Question on curvature
|


|
Positive
curvature : |
$C < 2 \pi r$ |
| Flat : |
$C = 2 \pi r$ |
| Negative
curvature : |
$C > 2 \pi r$ |
| Positive
curvature : |
more gap than
overlap |
| Flat : |
equal amounts |
| Negative
curvature : |
less gap than
overlap. |
2. Task with geodesics

The diagram depicts a
tabletop viewed from above on which there is a pyramid with a
square base. The geodesic starts on the tabletop, meets and
climbs the pyramid, and eventually returns to the tabletop. We
define the deflection angle $d$ of the geodesic as shown in the
diagram.
What you should have found in your investigations:
- the nearer the path goes to the top vertex the greater $d$
is,
- the steeper the pyramid the greater $d$ is,
- there is a complicated dependence on the angle of approach
and where the geodesic first hits the pyramid. (These two
factors together determine how near the line goes to the top
vertex.)
To be more specific, consider these possible cases:
number of triangles
crossed by geodesic |
DEFLECTION
equilateral triangles (this is
obviously a special case of the more general isosceles
case with $z=60^{\circ}$)
|
DEFLECTION
isosceles triangles

|
|
2
|
$30^{\circ}$ |
$(2z-90)^{\circ}$ |
|
3
|
$60^{\circ}$ |
$(4z-180) ^{\circ}$ |
|
4
|
$90^{\circ}$ |
$(6z-270) ^{\circ}$ |
To see how we can prove this, let's look at the situation when
the geodesic crosses just two of the isosceles triangles.
Remember that when they are flattened the geodesic is just a
straight line but, to find the deflection in the plane, we have
to put the pyramid up again in its initial position.

Let the geodesic be at angle $x$ to side $A B$.
Then angle $B D E=x$
angle $B E D=180^{\circ}-x-z$
angle $F E O=180^{\circ}-x-z$.
Using the property that the external angle equals the sum of
the interior opposite angles, angle $B F
E=(180^{\circ}-x-z)-z$, so the geodesic makes angle
$180^{\circ}-x-2z$ to side $B C$.

Now stand up the pyramid in
its original position.

At $F$ we see that
$d=90^{\circ}-x-(180^{\circ}-x- 2z)=2z-90^{\circ}$
You can try proving the formulae when the geodesic crosses 3 or 4
triangles.
We see that the more triangles the geodesic crosses the greater
$d$ is, and the nearer the line goes to the top vertex, the more
triangles it crosses. This is rule (i) above.
Also the greater the value of $z$, the greater $d$ is, and the
steeper the pyramid the greater the value of $z$. This is rule
(ii) above.
Notice that given that the geodesic crosses a certain number of
triangles, the deflection does not depend on $x$. However the
number of triangles crossed does depend on $x$ and on the length
$A D$. For the equilateral case the conditions, with all the edge
lengths of one unit, are:
The geodesic crosses only two triangles if $A
D+\frac{\sqrt{3}}{2}\cot x > \frac{3}{2}$.
The geodesic crosses only three triangles if $A
D+\frac{\sqrt{3}}{2}\cot x < \frac{3}{2}$ and $A
D+\sqrt{3}\cot x> 1$.
The geodesic crosses only four triangles if $A D+\sqrt{3}\cot
x< 1$ and $A D+\frac{\sqrt{3}}{2}\cot x>
\frac{1}{2}$,
so this is quite complicated.
**
This experiment models the deflection of an object in the
gravitational field of another object, e.g. the deflection of a
comet in the gravitational field of the sun. The vertex of the
pyramid represents the sun, a heavy object producing curvature
around it. The geodesic is the path of the comet. If it is far
away (so that the geodesic never touches the pyramid) there is no
deflection. If it passes very close to the vertex of the pyramid
the deflection is large.
**
3. Final problem
You should have found the following values of $E$:
For any triangulation (regular or irregular) of a sphere,
$E=720^{\circ}$.
For any triangulation of a doughnut, $E=0$.
For any triangulation of a doughnut with two holes,
$E=-720^{\circ}$.
e.g. For a cube: gap angle at each vertex $=360^{\circ}-(3\times
90^{\circ}) =90^{\circ}$, number of vertices $=8$ so $E=8\times
90^{\circ}=720^{\circ}$.

For a doughnut a simple triangulation looks
like the sketch.
There are two types of vertices.
At type $A$, the gap angle $=360^{\circ}-(2\times
90^{\circ}+2\times 45 ^{\circ})=90^{\circ}$. At type $B$, the
gap angle $=360^{\circ}-(2\times 135^{\circ}+2\times
90^{\circ})=-90^{\circ}$. There are 8 of each type of vertex so
$E=8\times 90^{\circ}+8\times(-90^{\circ})=0$.

We can stick together two one-holed
doughnuts to give a doughnut with two holes, as seen from above
in the sketch. There are now 8 type $A$ vertices, 16 type $B$
vertices and 4 type $C$ vertices, where the gap angle is 0.
$E=8\times 90^{\circ}+16\times (-90^{\circ})=-720^{\circ}$.
The general rule is $E=720^{\circ}\times(1-h)$ where $h$ is the
number of holes (or handles). This does not depend on the
triangulation, it is a topological invariant. To see how the
proof might go, consider a triangle with angles $a$, $b$, $c$.
What is the contribution to $E$? Now replace it by 3 triangles
with angles $(a_1, b_1, d_1)$, $(b_2, c_2, d_2)$, $(c_3, a_3,
d_3)$. What is the contribution of this to $E$?
The theorem is the Gauss- Bonnet theorem which says that the
total curvature of a 2-dimensional closed surface depends only on
its topology.