The Blooming Greek Astronomy

Abstract

In an atmosphere in which mathematics and geometry were burgeoning, ancient Greek astronomers naturally took every opportunity to put forth brand new theories once they believed they had unraveled the mystery of nature from their nightly observation of the heavens.

明月幾時有?把酒問青天。

不知天上宮闕,今夕是何年?

How rare the moon, so round and bright

With cup in hand, I ask the blue sky,

“I do not know in the celestial sphere

What name this festive night goes by?”

From Shuidiao Getou by Song dynasty poet Su Shi (1037–1101)

Translated by Lin Yutang (1895–1976)

Appendices

Appendix 2.1: Principle of Minimum Energy: Cases and Proofs

To explain why Mother Nature has selected these polyhedra (the Platonic solids) as the shapes for the crystallization of matter, scientists today have proposed the theory of the “natural tendency to require the least amount of energy,” which states that among all the available shapes, the energy required to form these polyhedra are minimum. A popular (two-dimensional) example that supports this theory is the regular hexagonal shape of the honeycomb. The reason for the formation of this shape stems from the fact that for a given length, a regular hexagon with a perimeter equal to that length has the largest enclosed area among all regular polygons that tile the plane (without leaving gaps). Therefore, for a given amount of propolis (bee glue) produced by honey bees and the same amount of labor (or time), a honeycomb cell with a regular hexagonal shape will provide the largest amount of space for the bees. The proof of a regular hexagon having the “same perimeter but maximum area” is as follows.

• 1. Regular hexagon (Fig. 2.9: right):

  Fig. 2.9

  

  Three different types of regular tiling: square/regular quadrilateral (left); equilateral triangle (center); regular hexagon (right)

  Each side is shared by 2 hexagons, and each hexagon has 6 sides, so a side corresponds to one-third of a hexagon. Suppose each side of the hexagon is of length R. Since a regular hexagon has an area of $\frac{3\sqrt 3 }{2}R^{2}$, each side corresponds to an area of:

  $$\frac{1}{3} \times \frac{3\sqrt 3 }{2}R^{2} = \frac{\sqrt 3 }{2}R^{2}$$

  (2.6)

• 2. Regular quadrilateral or tetragon (square; Fig. 2.9: left):

  Each side is shared by 2 squares, and each square has 4 sides, so a side corresponds to one-half of a square. Suppose each side of the square is of length R. Since a square has an area of $R^{2}$, each side corresponds to an area of:

  $$\frac{1}{2} \times R^{2} = \frac{1}{2}R^{2}$$

  (2.7)

• 3. Equilateral triangle (Fig. 2.9: center):

  Each side is shared by 2 triangles, and each triangle has 3 sides, so a side corresponds to two-thirds of a triangle. Suppose each side of the triangle is of length R. Since an equilateral triangle has an area of $\frac{\sqrt 3 }{4}R^{2}$, each side corresponds to an area of:

  $$\frac{2}{3} \times \frac{\sqrt 3 }{4}R^{2} = \frac{\sqrt 3 }{6}R$$

  (2.8)

From the three proofs above, it is clear that the regular polygons can be ranked by the criterion “area per unit length of perimeter” as follows, in descending order: (a) regular hexagon; (b) square (regular quadrilateral); (c) equilateral triangle.

If we increase the number of sides of the regular polygon to 8 (an octagon) or 12 (a dodecagon), the tiling will leave gaps and much of the area would be wasted and fail to satisfy the minimum energy requirement. This highly efficient hexagonal shape is adopted not only by bees. Tens of thousands of columns of the igneous rock basalt, formed from underwater eruptions of magma and found in Northern Ireland, the northern coast of Germany and the Pescadores Islands of Taiwan, also have hexagonal cross-sections. Efficiency is a mysterious force of nature. Whether it is the living bees or the inanimate basalt columns, the nature tends to opt for the most efficient geometric shape in the formation of physical structures.

Another geometric shape considered the most efficient is the circle, or its three-dimensional counterpart, the sphere. They are found in abundance in the Universe, from objects as large as planets to those as tiny as water droplets or phytoplankton, all of which are spherical. Therefore, to form a soap bubble with the least amount of soapy water,¹⁹ the shape of the bubble is bound to be spherical. A sphere is therefore regarded as the most efficient surface. If we join two soap bubbles of equal size together, the common wall separating them will be a flat surface as a result of surface tension. If six bubbles of identical sizes are joined in a symmetrical fashion, then the space in the middle will form a regular hexahedron (i.e. a cube). If twelve bubbles are joined symmetrically, the shape in the middle will be a regular dodecahedron, with all twelve sides being regular pentagons of the same size. This shape of the space that forms in the middle of the bubbles represents that which minimizes its volume.

As an application of regular polyhedra, the dice for tabletop games and gambling have been designed with the principle of minimum energy. When a die in the shape of a regular polyhedron is rolled on a table, the probability of each face turning up is identical. In addition, a rolled regular polyhedral die takes the least amount of time to come to rest compared with other non-regular polyhedral dice. This shortens the total amount of time required to determine the results of a game, as well as saving gamblers’ mental and physical efforts. Although saving energy is not the factor gamblers are most concerned about, saving time is one way for casino owners to maximize their income. Therefore, using regular polyhedron-shaped dice in gambling is a win-win for both the casinos and the gamblers, and this has endured over a thousand years uninterrupted or with little change.

The idea that all matter is made up of Platonic solids is exemplified in today’s vast computer-animated film industry, where multimedia figures are created by animators as computer memory capacity and computing power continue to grow beyond leaps and bounds. This technology was first implemented by engineers at the Boeing Company in the development of flight simulators. The software engineers who designed the simulators employed millions of triangles to represent the complex shapes of terrestrial surfaces that appeared in the background of simulated flight scenarios. This technology was later used by the computer animation movie studio Pixar to create animated feature films, where dozens of human characters and various animate and inanimate objects were rendered with tens of millions of basic geometric shapes. Capable of adapting in response to changes in the environment, these animated entities are transformed into life-like characters, creating stunning visual effects and revolutionizing the film industry of the 21st century.

Appendix 2.2: Derivation of Equations for Planetary Retrograde Motion

The retrograde motions²⁰ of planets baffled generations of ancient astronomers, who proposed a number of theoretical models to try to explain the phenomenon. Here we analyze their causes from a modern perspective and perform a few simple calculations.

When observing celestial objects in the sky from Earth, we not only see the Sun and the Moon but also a few what are known as “wandering stars,” or planets, which move relative to a fixed background of distant stars. A planet normally moves westward relative to the fixed stars, and this is called prograde (or direct) motion. Occasionally, however, it would appear to move eastward relative to the fixed stars, which is referred to as retrograde motion. This rather strange planetary movement was a great mystery in astronomy for a long time until geocentrism was abandoned in favor of the heliocentric (Sun-centered) model, which offers a much simpler way to explain retrograde motions.

Consider the planetary system of Fig. 2.10. P is a planet that orbits the Sun and is superior to Earth (i.e., Earth’s orbit lies inside that of P). According to Kepler’s third law of planetary motion, Earth’s angular speed is faster than that of P. For example, suppose Earth has completed half of its orbit around the sun, P may have completed only one quarter of its orbit. In the diagram below, five pairs of E and P’s relative positions are shown: When Earth is at the E1 position and P at P1, P’s projection onto the celestial sphere is A1; when Earth and P are at the positions $E2$ and the $P2$, respectively, P’s projection is $A2$; and so on. For an observer on Earth, P’s trajectory, as projected onto the distant background, appears to be $A1 \to A2 \to A3 \to A4 \to A5$, where the path $A1 \to A2$ is prograde (direct), the part $A2 \to A3 \to A4$ is retrograde, and the path $A4 \to A5$ resumes the original prograde direction.

Fig. 2.10

Planetary retrograde motion

We will now try to compute the time required for a planet to complete a retrograde motion relative to Earth. Three assumptions are required:

• 1. The masses of planet P and Earth are negligible (compared with the Sun), and the Sun can be regarded as a fixed point.

• 2. Planet P has the same orbital plane as Earth’s and travels in a circular orbit (instead of an ellipse, which is more tractable mathematically).

• 3. Gravitational forces on P, Earth and the Sun from other Solar System bodies are ignored.

Next, we consider a simple model as shown in Fig. 2.11. Here we have a Cartesian coordinate system with the Sun S located at the origin. Let E be Earth and P be the planet under observation, and suppose the radii of their respective circular orbits around the Sun are $R_{1}$ and $R_{2}$. Let E and P have angular orbital speeds of $\omega_{1}$ and $\omega_{2}$, respectively. Select $t = 0$ as the time when planet P is in opposition to the Sun (i.e., the Sun, Earth and P are perfectly aligned in a straight line, with Earth between the other two bodies). Now all three bodies are located on the x-axis.

Fig. 2.11

Computing the various positions of a planet’s retrograde trajectory

After a while, at time t, Earth’s coordinates can be expressed as

$$\left( {x_{E} ,y_{E} } \right) = \left( {R_{1} \cos \omega_{1} t,R_{1} \sin \omega_{1} t} \right)$$

(2.9)

Likewise, planet P’s coordinates $\left( {x_{M} ,y_{M} } \right)$ are

$$\left( {x_{M} ,y_{M} } \right) = \left( {R_{2} \cos \omega_{2} t,R_{2} \sin \omega_{2} t} \right)$$

(2.10)

We then define the bearing of P with respect to E to be

$$\lambda = \tan^{ - 1} \left( {\frac{{y_{M} - y_{E} }}{{x_{M} - x_{E} }}} \right)$$

(2.11)

Notice that whether P is orbiting in a prograde or retrograde direction at time t depends on the sign of $d\lambda /dt$. If $d\lambda /dt > 0$, the movement is prograde, and if it is negative, the motion is retrograde. Substituting (2.9) and (2.10) into (2.11) we have

$$\tan \lambda = \frac{{y_{M} - y_{E} }}{{x_{M} - x_{E} }} = \frac{{R_{2} \sin \omega_{2} t - R_{1} \sin \omega_{1} t}}{{R_{2} \cos \omega_{2} t - R_{1} \cos \omega_{1} t}}$$

(2.12)

Now we take the derivative of (2.12) with respect to t to get

$$\sec^{2} \lambda \frac{d\lambda }{dt} = \frac{{\omega_{1} R_{1}^{2} + \omega_{2} R_{2}^{2} - \left( {\omega_{1} + \omega_{2} } \right)R_{1} R_{2} \cos \left( {\omega_{1} - \omega_{2} } \right)t}}{{\left( {R_{2} \cos \omega_{2} t - R_{1} \cos \omega_{1} t} \right)^{2} }}$$

(2.13)

Note that at $t = 0$ (when the Sun, Earth and P are in syzygy), as Earth’s angular orbital speed is higher, P will move in a retrograde direction.

Next, we will compute the time $t > 0$ when $d\lambda /dt$ becomes zero. Both sides of Eq. (2.13) vanishes when $t = t_{S}$ satisfies the following equation:

$$\omega_{1} R_{1}^{2} + \omega_{2} R_{2}^{2} = \left( {\omega_{1} + \omega_{2} } \right)R_{1} R_{2} \cos \left( {\omega_{1} - \omega_{2} } \right)t_{S}$$

(2.14)

or

$$t_{S} = \frac{{\cos^{ - 1} \left[ {\frac{{\omega_{1} R_{1}^{2} + \omega_{2} R_{2}^{2} }}{{\left( {\omega_{1} + \omega_{2} } \right)R_{1} R_{2} }}} \right]}}{{\omega_{1} - \omega_{2} }}$$

(2.15)

Since $t = 0$ is precisely the time when P is halfway through its retrograde movement (as $t > 0$ and $t < 0$ are perfectly symmetric), the amount of time for P to complete a retrograde cycle is

$$T = 2t_{S} = \frac{{2\cos^{ - 1} \left[ {\frac{{\omega_{1} R_{1}^{2} + \omega_{2} R_{2}^{2} }}{{\left( {\omega_{1} + \omega_{2} } \right)R_{1} R_{2} }}} \right]}}{{\omega_{1} - \omega_{2} }}$$

(2.16)

We can apply the above equation to compute the retrograde constants of the planets in the Solar System relative to Earth. Substituting each planet’s orbital data into the equation, we get:

• The number of days in Mars’ retrogradation is 72 days for each synodic period of 25.6 months.

• The number of days in Jupiter’s retrogradation is 121 days for each synodic period of 13.1 months.

• The number of days in Saturn’s retrogradation is 138 days for each synodic period of 12.4 months.

• The number of days in Uranus’ retrogradation is 151 days for each synodic period of 12.15 months.

• The number of days in Neptune’s retrogradation is 158 days for each synodic period of 12.07 months.

The above results are close to those obtained from actual observations. The above data indicates that the farther away a planet is from Earth, the shorter the time interval between its two consecutive retrograde motions. This interval is the time interval for Earth to be located between the Sun and the planet in question for two consecutive times. For its relationship with the sidereal period (time required for a planet to complete one cycle around the Sun), refer to (3.3) in Chap. 3.