Why a folded pizza slice stays stiff
Tonight's idea
Try it tonight, with pizza or with a sheet of paper held by one edge. Why does it work?
A curve's curvature at a point is \(1/r\), where \(r\) is the radius of the circle that fits it best there. A straight line has curvature 0. At a point on a surface, cut straight down into the surface in every direction. Each cut is a curve. Count its curvature as positive if it bends toward you and negative if it bends away. Call the biggest and smallest values \(k_1\) and \(k_2\).
Gaussian curvature. \(K = k_1 k_2\). Flat sheet: \(0 \times 0 = 0\). Can of radius \(r\): \(\tfrac1r \times 0 = 0\). Ball of radius \(R\): \(\tfrac1R \times \tfrac1R = \tfrac{1}{R^2}\). Saddle: negative.
So \(K = 0\) does not mean flat. It means that at least one of \(k_1\) and \(k_2\) is zero.
In 1827 Gauss proved what he called his Remarkable Theorem: bend a surface any way you like, without stretching, squashing or tearing it, and \(K\) stays the same at every point.
Watch one: measure a ball's curvature without leaving it
Be an ant that lives on a ball of radius \(R\). It cannot cut into the ball or see \(R\). It can still walk the straightest possible paths, and measure angles and areas. Here is a fact you may use: on a surface with the same \(K\) everywhere, a triangle whose sides are straightest paths has
\(\text{angle sum} - \pi = K \times \text{area}\), with the angles in radians.
The ant walks from the North Pole down a meridian to the equator, a quarter of the way round the equator, and back up a meridian to the pole. All three sides are great circles, the straightest paths on a ball. Each angle is \(90°\), so the angle sum is \(\tfrac{3\pi}{2}\) and the extra is \(\tfrac{3\pi}{2} - \pi = \tfrac{\pi}{2}\). Eight such triangles tile the ball, so the area is \(\tfrac18 \times 4\pi R^2 = \tfrac{\pi R^2}{2}\). Then \(K = \text{extra angle} \div \text{area} = \tfrac{\pi}{2} \div \tfrac{\pi R^2}{2} = \tfrac{1}{R^2}\).
That matches the box above, yet it was found from inside the surface. Bending without stretching changes no length, angle or area measured within the surface, so the ant's \(K\) cannot change. That is the idea behind Gauss's theorem.
Your turn
- Find \(K\), with its sign, at a point of each. (a) A flat tabletop. (b) The side of a can of radius 3 cm. (c) A ball of radius 11 cm. (d) A saddle whose most upward-bending cut fits a circle of radius 5 cm and whose most downward-bending cut fits a circle of radius 10 cm. (e) Which of the four can you not make by bending paper without stretching or tearing it? Why?
- The ant walks another triangle on a ball of radius \(R\): North Pole, down a meridian to the equator, through \(60°\) of longitude along the equator, and back up to the pole. (a) Find its three angles and its extra angle in radians. (b) What fraction of the ball is it? Write its area in terms of \(R\). (c) Compute extra angle \(\div\) area. (d) Model the Earth as a ball with \(R = 6371\) km (good, not exact). Find the area in km².
- (a) An orange has radius 4 cm. Find \(K\) on its peel, and explain why a piece of peel cannot be pressed flat without stretching or tearing. (b) Why can no flat map of part of a ball show every distance to one scale? (c) A paper label peels off a can and lies flat. Why does that not break the theorem? (d) Draw a straight-sided triangle on paper and roll it into a tube. Find its angle sum now.
- Model the slice as a thin sheet that bends but cannot stretch (an idealisation), held level by the crust, left and right alike. By symmetry, on the centre line \(k_1\) and \(k_2\) belong to the cut along the slice (crust to tip) and the cut across it, so \(K = k_{\text{along}} k_{\text{across}}\) there. (a) Find \(K\) for the slice lying flat. (b) Held flat by the crust, it droops. Which curvature is now not zero? Is the theorem broken? (c) Fold the crust into a U, so \(k_{\text{across}}\) is not zero on the centre line. What does the theorem force, and what does it say about the centre line? (d) A real folded slice can still droop near the tip. Give two ways it escapes the model.
Stretch: all the curvature of a ball
On a ball of radius \(R\), draw the equator and two full circles through both poles, crossing there at an angle \(\theta\) with \(0° \lt \theta \lt 180°\). They cut the ball into 8 triangles. (a) With \(\theta = 90°\), add up the extra angles of all 8. (b) Show that the total is the same for every \(\theta\). (c) Use \(K \times \text{area}\) to explain why it does not depend on \(R\). (d) A deeper theorem says that \(K\), added up over the whole of any smooth closed surface without holes, such as an egg, comes to this same number, and for a ring doughnut to 0. A doughnut's outer rim curves like a ball. What must happen around the hole?
Answers
- (a) \(0 \times 0 = 0\). (b) \(\tfrac13 \times 0 = 0\). (c) \(\tfrac1{11} \times \tfrac1{11} = \tfrac1{121} \approx 0.0083\) per cm². Seen from outside, both curvatures are negative and the product is the same. (d) The two cuts bend opposite ways, so \(K = \tfrac15 \times \left(-\tfrac1{10}\right) = -\tfrac1{50} = -0.02\) per cm². (e) The ball and the saddle. Flat paper has \(K = 0\), and bending without stretching keeps \(K = 0\) at every point, so paper can never match a surface where \(K\) is not zero. The tabletop and the can both have \(K = 0\), and paper does make both.
- (a) Meridians cross the equator at right angles, and the angle at the pole is the difference in longitude. So the angles are \(90°, 90°, 60°\), the sum is \(240° = \tfrac{4\pi}{3}\), and the extra is \(\tfrac{4\pi}{3} - \pi = \tfrac{\pi}{3}\). (b) It is \(\tfrac{60}{360} = \tfrac16\) of the northern half, so \(\tfrac1{12}\) of the ball: area \(\tfrac1{12} \times 4\pi R^2 = \tfrac{\pi R^2}{3}\). (c) \(\tfrac{\pi}{3} \div \tfrac{\pi R^2}{3} = \tfrac{1}{R^2}\), the same \(K\) as before. (d) \(\tfrac{\pi}{3} \times 6371^2 \approx 4.25 \times 10^7\) km², about 42.5 million km².
- (a) \(K = \tfrac14 \times \tfrac14 = \tfrac1{16} = 0.0625\) per cm². A flat piece has \(K = 0\). Pressing the peel flat without stretching, squashing or tearing would keep \(K = \tfrac1{16}\), so it cannot be done: the peel must stretch or split. (b) Suppose a map got every distance right. Enlarged to full size and laid on the ball, it would be a bending without stretching, so \(K\) would have to match. But the map has \(K = 0\) and the ball has \(K = \tfrac{1}{R^2}\). So every flat map of the Earth, modelled as a ball, distorts some distances. (c) The side of the can has \(K = \tfrac1r \times 0 = 0\), the same as flat paper, so the theorem has no objection. (d) Still \(180°\). Rolling is bending without stretching, so no angle on the paper changes. That agrees with the ant's rule, because the sides are still straightest paths and on the tube \(K = 0\), so the extra is \(0 \times \text{area} = 0\).
- (a) \(K = 0 \times 0 = 0\). (b) \(k_{\text{along}}\): the centre line now curves down. \(k_{\text{across}}\) is still 0, so \(K = k_{\text{along}} \times 0 = 0\). Nothing is broken, and that is why a flat slice is free to droop. (c) \(K\) must still be 0, and \(k_{\text{across}}\) is not 0, so \(k_{\text{along}} = 0\) at every point of the centre line. The centre line is the along cut, so it is a curve with curvature 0 everywhere: a straight line. The slice cannot droop. To droop it would need both curvatures at once, and that takes \(K \neq 0\). (d) The U can flatten out toward the tip, and where \(k_{\text{across}}\) is back to 0 the slice may bend. Or the dough can stretch, tear or crease, which the model ruled out.
Stretch. (a) Each is the worked example's triangle with extra \(\tfrac{\pi}{2}\), so the total is \(8 \times \tfrac{\pi}{2} = 4\pi\). (b) Every triangle has one corner at a pole and two right angles on the equator, so its extra equals its angle at the pole. The four angles round each pole are \(\theta\), \(\pi - \theta\), \(\theta\), \(\pi - \theta\), which add to \(2\pi\). Two poles give \(4\pi\). (c) The total extra is \(K\) times the total area: \(\tfrac{1}{R^2} \times 4\pi R^2 = 4\pi\). A small ball is more curved but has less area, and the two cancel exactly. (d) The total must be 0, so the positive \(K\) on the outside has to be cancelled by negative \(K\) somewhere. Around the hole the surface is a saddle: it curves one way going round the hole and the opposite way going through it.