To Explain the World: The Discovery of Modern Science

by Steven Weinberg

  There is a fundamental mathematical theorem (not made explicit until the early nineteenth century) that virtually any disturbance (that is, any disturbance having a sufficiently smooth dependence on distance along the wave) can be expressed as a sum of sine waves with various wavelengths. (This is known as “Fourier analysis.”)

  Each individual sine wave exhibits a characteristic oscillation in time, as well as in distance along the wave’s direction of motion. If the wave is traveling with velocity v, then in time t it travels a distance vt. The number of wavelengths that pass a fixed point in time t will thus be vt/λ, so the number of cycles per second at a given point in which the amplitude and rate of change both keep going back to the same value is v/λ. This is known as the frequency, denoted by the symbol ν (nu), so ν = v/λ. The velocity of a wave of vibration of a string is close to a constant, depending on the string tension and mass, but nearly independent of its wavelength or its amplitude, so for these waves (as for light) the frequency is simply inversely proportional to the wavelength.

  Now consider a string of some musical instrument, with length L. The amplitude of the wave must vanish at the ends of the string, where the string is held fixed. This condition limits the wavelengths of the individual sine waves that can contribute to the total amplitude of the string’s vibration. We have noted that the distance between points where the amplitude of any sine wave vanishes can be any whole number of half wavelengths. Thus the wave on a string that is fixed at both ends must contain a whole number N of half wavelengths, so that L = Nλ/2. That is, the only possible wavelengths are λ = 2L/N, with N = 1, 2, 3, etc., and so the only possible frequencies are*

  ν = vN/2L

  The lowest frequency, for the case N = 1, is v/2L; all the higher frequencies, for N = 2, N = 3, etc., are known as the “overtones.” For instance, the lowest frequency of the middle C string of any instrument is 261.63 cycles per second, but it also vibrates at 523.26 cycles per second, 784.89 cycles per second, and so on. The intensities of the different overtones make the difference in the qualities of the sounds from different musical instruments.

  Now, suppose that vibrations are set up in two strings that have different lengths L1 and L2, but are otherwise identical, and in particular have the same wave velocity v. In time t the modes of vibration of the lowest frequency of the first and second strings will go through n1 = ν1t = vt/2L1 and n2 = ν2t = vt/2L2 cycles or fractions of cycles, respectively. The ratio is

  n1/n2 = L2/L1

  Thus, in order for the lowest vibrations of both strings each to go through whole numbers of cycles in the same time, the quantity L2/L1 must be a ratio of whole numbers—that is, a rational number. (In this case, in the same time each overtone of each string will also go through a whole number of cycles.) The sound produced by the two strings will thus repeat itself, just as if a single string had been plucked. This seems to contribute to the pleasantness of the sound.

  For instance, if L2/L1 = 1/2, then the vibration of string 2 of lowest frequency will go through two complete cycles for every complete cycle of the corresponding vibration of string 1. In this case, we say that the notes produced by the two strings are an octave apart. All of the different C keys on the piano keyboard produce frequencies that are octaves apart. If L2/L1 = 2/3, the two strings make a chord called a fifth. For example, if one string produces middle C, at 261.63 cycles per second, then another string that is 2/3 as long will produce middle G, at a frequency 3/2 × 261.63 = 392.45 cycles per second.* If L2/L1 = 3/4, the chord is called a fourth.

  The other reason for the pleasantness of these chords has to do with the overtones. In order for the N1th overtone of string 1 to have the same frequency as the N2th overtone of string 2, we must have vN1/2L1 = vN2/2L2, and so

  L2/L1 = N2/N1

  Again, the ratio of the lengths is a rational number, though for a different reason. But if this ratio is an irrational number, like π or the square root of 2, then the overtones of the two strings can never match, though the frequencies of high overtones will come arbitrarily close. This apparently sounds terrible.

  4. The Pythagorean Theorem

  The so-called Pythagorean theorem is the most famous result of plane geometry. Although it is believed to be due to a member of the school of Pythagoras, possibly Archytas, the details of its origin are unknown. What follows is the simplest proof, one that makes use of the notion of proportionality commonly used in Greek mathematics.

  Consider a triangle with corner points A, B, and P, with the angle at P a right angle. The theorem states that the area of a square whose side is AB (the hypotenuse of the triangle) equals the sum of the areas of squares whose sides are the other two sides of the triangle, AP and BP. In modern algebraic terms, we can think of AB, AP, and BP as numerical quantities, equal to the lengths of these sides, and state the theorem as

  AB2 = AP2 + BP2

  The trick in the proof is to draw a line from P to the hypotenuse AB, which intersects the hypotenuse at a right angle, say at point C. (See Figure 2.) This divides triangle ABP into two smaller right triangles, APC and BPC. It is easy to see that both of these smaller triangles are similar to triangle ABP—that is, all their corresponding angles are equal. If we take the angles at the corners A and B to be α (alpha) and β (beta), then triangle ABP has angles α, β, and 90°, so α + β + 90° = 180°. Triangle APC has two angles equal to α and 90°, so to make the sum of the angles 180° its third angle must be β. Likewise, triangle BPC has two angles equal to β and 90°, so its third angle must be α.

  Because these triangles are all similar, their corresponding sides are proportional. That is, AC must be in the same proportion to the hypotenuse AP of triangle ACP that AP has to the hypotenuse AB of the original triangle ABP, and BC must be in the same proportion to BP that BP has to AB. We can put this in more convenient algebraic terms as a statement about ratios of the lengths AC, AP, etc.:

  It follows immediately that AP2 = AC × AB, and BP2 = BC × AB. Adding these two equations gives

  AP2 + BP2 = (AC + BC) × AB

  But AC + BC = AB, so this is the result that was to be proved.

  Figure 2. Proof of the Pythagorean theorem. This theorem states that the sum of the areas of two squares whose sides are AP and BP equals the area of a square whose sides are the hypotenuse AB. To prove the theorem, a line is drawn from P to a point C, which is chosen so that this line is perpendicular to the line from A to B.

  5. Irrational Numbers

  The only numbers that were familiar to early Greek mathematicians were rational. These are numbers that are either whole numbers, like 1, 2, 3, etc., or ratios of whole numbers, like 1/2, 2/3, etc. If the ratio of the lengths of two lines is a rational number, the lines were said to be “commensurable”—for instance, if the ratio is 3/5, then five times one line has the same length as three times the other. It was therefore shocking to learn that not all lines were commensurable. In particular, in a right isosceles triangle, the hypotenuse is incommensurable with either of the two equal sides. In modern terms, since according to the theorem of Pythagoras the square of the hypotenuse of such a triangle equals twice the square of either of the two equal sides, the length of the hypotenuse equals the length of either of the other sides times the square root of 2, so this amounts to the statement that the square root of 2 is not a rational number. The proof given by Euclid in Book X of the Elements consists of assuming the opposite, in modern terms that there is a rational number whose square is 2, and then deriving an absurdity.

  Suppose that a rational number p/q (with p and q whole numbers) has a square equal to 2:

  (p/q)2 = 2

  There will then be an infinity of such pairs of numbers, found by multiplying any given p and q by any equal whole numbers, but let us take p and q to be the smallest whole numbers for which (p/q)2 = 2. It follows from this equation that

  p2 = 2q2

  This shows that p2 is an even number, but the product of any two odd numbers is odd, so p must be even. That is, we
can write p = 2pʹ, where pʹ is a whole number. But then

  q2 = 2pʹ2

  so by the same reasoning as before, q is even, and can therefore be written as q = 2qʹ, where qʹ is a whole number. But then p/q = pʹ/qʹ, so

  (pʹ/qʹ)2 = 2

  with pʹ and qʹ whole numbers that are respectively half p and q, contradicting the definition of p and q as the smallest whole numbers for which (p/q)2 = 2. Thus the original assumption, that there are whole numbers p and q for which (p/q)2 = 2, leads to a contradiction, and is therefore impossible.

  This theorem has an obvious extension: any number like 3, 5, 6, etc. that is not itself the square of a whole number cannot be the square of a rational number. For instance, if 3 = (p/q)2, with p and q the smallest whole numbers for which this holds, then p2 = 3q2, but this is impossible unless p = 3pʹ for some whole number pʹ, but then q2 = 3pʹ2, so q = 3qʹ for some whole number q, so 3 = (pʹ/qʹ)2, contradicting the statement that p and q are the smallest whole numbers for which p2 = 3q2. Thus the square roots of 3, 5, 6 . . . are all irrational.

  In modern mathematics we accept the existence of irrational numbers, such as the number denoted whose square is 2. The decimal expansion of such numbers goes on forever, without ending or repeating; for example, = 1.414215562. . . . The numbers of rational and irrational numbers are both infinite, but in a sense there are far more irrational than rational numbers, for the rational numbers can be listed in an infinite sequence that includes any given rational number:

  1, 2, 1/2, 3, 1/3, 2/3, 3/2, 4, 1/4, 3/4, 4/3, . . .

  while no such list of all irrational numbers is possible.

  6. Terminal Velocity

  To understand how observations of falling bodies might have led Aristotle to his ideas about motion, we can make use of a physical principle unknown to Aristotle, Newton’s second law of motion. This principle tells us that the acceleration a of a body (the rate at which its speed increases) equals the total force F acting on the body divided by the body’s mass m:

  a = F/m

  There are two main forces that act on a body falling through the air. One is the force of gravity, which is proportional to the body’s mass:

  Fgrav = mg

  Here g is a constant independent of the nature of the falling body. It equals the acceleration of a falling body that is subject only to gravity, and has the value 32 feet/second per second on and near the Earth’s surface. The other force is the resistance of the air. This is a quantity f(v) proportional to the density of the air, which increases with velocity and also depends on the body’s shape and size, but does not depend on its mass:

  Fair = −f(v)

  A minus sign is put in this formula for the force of air resistance because we are thinking of acceleration in a downward direction, and for a falling body the force of air resistance acts upward, so with this minus sign in the formula, f(v) is positive. For instance, for a body falling through a sufficiently viscous fluid, the air resistance is proportional to velocity

  f(v) = kv

  with k a positive constant that depends on the body’s size and shape. For a meteor or a missile entering the thin air of the upper atmosphere, we have instead

  f(v) = Kv2

  with K another positive constant.

  Using the formulas for these forces in the total force F = Fgrav + Fair and using the result in Newton’s law, we have

  a = g − f(v)/m

  When a body is first released, its velocity vanishes, so there is no air resistance, and its acceleration downward is just g. As time passes its velocity increases, and air resistance begins to reduce its acceleration. Eventually the velocity approaches a value where the term – f(v)/m just cancels the term g in the formula for acceleration, and the acceleration becomes negligible. This is the terminal velocity, defined as the solution of the equation:

  f(vterminal) = gm

  Aristotle never spoke of terminal velocity, but the velocity given by this formula has some of the same properties that he attributed to the velocity of falling bodies. Since f(v) is an increasing function of v, the terminal velocity increases with the mass m. In the special case where f(v) = kv, the terminal velocity is simply proportional to the mass and inversely proportional to the air resistance:

  vterminal = gm/k

  But these are not general properties of the velocity of falling bodies; heavy bodies do not reach terminal velocity until they have fallen for a long time.

  7. Falling Drops

  Strato observed that falling drops get farther and farther apart as they fall, and concluded from this that these drops accelerate downward. If one drop has fallen farther than another, then it has been falling longer, and if the drops are separating, then the one that is falling longer must also be falling faster, showing that its fall is accelerating. Though Strato did not know it, the acceleration is constant, and as we shall see, this results in a separation between drops that is proportional to the time elapsed.

  As mentioned in Technical Note 6, if air resistance is neglected, then the acceleration downward of any falling body is a constant g, which in the neighborhood of the Earth’s surface has the value 32 feet/second per second. If a body falls from rest, then after a time interval τ (tau) its velocity downward will be gτ. Hence if drops 1 and 2 fall from rest from the same downspout at times t1 and t2, then at a later time t the speed downward of these drops with be v1 = g(t – t1) and v2 = g(t – t2), respectively. The difference in their speeds will therefore be

  v1 − v2 = g(t − t1) − g(t − t2) = g(t2 − t1)

  Although both v1 and v2 are increasing with time, their difference is independent of the time t, so the separation s between the drops simply increases in proportion to the time:

  s = (v1 − v2)t = gt(t1 − t2)

  For instance, if the second drop leaves the downspout a tenth of a second after the first drop, then after half a second the drops will be 32 × 1/2 × 1/10 = 1.6 feet apart.

  8. Reflection

  The derivation of the law of reflection by Hero of Alexandria was one of the earliest examples of the mathematical deduction of a physical principle from a deeper, more general, principle. Suppose an observer at point A sees the reflection in a mirror of an object at point B. If the observer sees the image of the object at a point P on the mirror, the light ray must have traveled from B to P and then to A. (Hero probably would have said that the light traveled from the observer at A to the mirror and then to the object at B, as if the eye reached out to touch the object, but this makes no difference in the argument below.) The problem of reflection is: where on the mirror is P?

  To answer this question, Hero assumed that light always takes the shortest possible path. In the case of reflection, this implies that P should be located so that the total length of the path from B to P and then to A is the shortest path that goes from B to anywhere on the mirror and then to A. From this, he concluded that angle θi (thetai) between the mirror and the incident ray (the line from B to the mirror) equals angle θr between the mirror and the reflected ray (the line from the mirror to A).

  Here is the proof of the equal-angles rule. Draw a line perpendicular to the mirror from B to a point Bʹ that is as far behind the mirror as B is in front of the mirror. (See Figure 3.) Suppose that this line intersects the mirror at point C. Sides BʹC and CP of right triangle BʹCP have the same lengths as sides BC and CP of right triangle BCP, so the hypotenuses BʹP and BP of these two triangles must also have the same length. The total distance traveled by the light ray from B to P and then to A is therefore the same as the distance that would be traveled by the light ray if it went from Bʹ to P and then to A. The shortest distance between points Bʹ and A is a straight line, so the path that minimizes the total distance between the object and the observer is the one for which P is on the straight line between Bʹ and A. When two straight lines intersect, the angles on opposite sides of the intersection point are equal, so angle θ between line BʹP and the mirror equals angle θr between the reflected ray and the mirror.
But because the two right triangles BʹCP and BCP have the same sides, angle θ must also equal angle θi between the incident ray BP and the mirror. So, since both θi and θr are equal to θ, they are equal to each other. This is the fundamental equal-angles rule that determines the location P on the mirror of the image of the object.

  Figure 3. Proof of Hero’s theorem. This theorem states that the shortest path from an object at B to the mirror and then to an eye at A is one for which angles θi and θr are equal. The solid lines marked with arrows represent the path of a light ray; the horizontal line is the mirror; and the dashed line is a line perpendicular to the mirror that runs from B to a point B' on the other side of the mirror at an equal distance from it.

  9. Floating and Submerged Bodies

  In his great work On Floating Bodies, Archimedes assumed that if bodies are floating or suspended in water in such a way that equal areas at equal depths in the water are pressed down by different weights, then the water and the bodies will move until all equal areas at any given depth are pressed down by the same weight. From this assumption, he derived general consequences about both floating and submerged bodies, some of which were even of practical importance.