Physics of the Human Body 13 Chapter 2 Dimensional analysis and scaling laws Dimensional analysis and scaling laws 1. Dimensional analysis 2. We can reduce the number of independent One of the simplest, yet most powerful, tools in the parameters in a calculation by re-expressing the physicist’s bag of tricks is dimensional analysis1. All problem in terms of relations between dimen- quantities of physical interest have dimensions sionless quantities (pure numbers). that can be expressed in terms of three fundamen- Thus, consider the problem of a physical pendu- tal quantities: mass (M), length (L) and time (T). lum, whose equation of motion in the absence of driving torques is We express the dimensionality of a quantity by d2θ I + mgl sinθ = 0 ; enclosing it in square brackets. Thus, for velocity dt2 (change of position divided by change of time) we write Let us redefine the time in terms of a dimensionless variable τ and a quantity ω with dimensions of −1 [v] = LT . frequency (note θ is already dimensionless because Similarly acceleration (change of velocity divided it is defined as a ratio of lengths): by change of time) has dimensionality τ = t ω . dv −2 [a] = = LT . dt Then the original equation becomes From Newton’s Second law of Motion, Iω2 d2θ → + sinθ = 0 . → mgl τ2 F = ma , d ω we see the dimensions of force are If we now choose so that = ⋅ ≡ −2 Iω2 [F] M [a] MLT ; = 1 mgl and since work (energy) is force times distance, − our new dimensionless equation is [E] = [F] ⋅ L ≡ ML2T 2 . d2θ + sinθ = 0 . dτ2 Dimensional analysis has three important applica- We can make sure we have not made an algebraic tions: error by checking the dimensional consistency of 1. We can avoid algebraic errors by requiring the the relation dimensions of the quantities on two sides of an Iω2 equation or inequality to be consistent. = 1 : mgl − Since [I] = ML2 and [g] = LT 2 we have 1. In biological and physiological applications dimensional analysis is often called allometric scaling. 14 Chapter 2: Dimensional analysis α = β = −1⁄ γ = 1⁄ 2 The solution is 0 , 2 , 2 . The fact Iω − − − − = ML2 T 2 M 1 L 1T2 L 1 that the exponent of M is zero means that the mgl period does not depend on the mass of the pendu- lum bob4. Hence our formula for the pendulum’s = M0L0T0 = constant . frequency becomes Hence the expression is dimensionally consistent 1⁄2 and contains no obvious algebraic errors2. g ν = constant × , l a well-known formula. (In fact, for a pendulum 3. We can “solve” certain physical problems with- started at zero angular velocity from initial angle out actually doing the detailed calculations θ θ 0 the constant is (for small 0) needed for a complete solution. − θ 1 Thus, suppose we want an expression for the fre- √2 0 dθ 1 ∫ ≈ quency ν of a simple pendulum. The frequency can 4 0 √θ − θ 2π cos cos 0 only depend on the dimensional parameters of the system: the mass m of the bob, the length λ of the but dimensional analysis does not reveal this—to string, and the acceleration g of gravity. We sup- learn its value one must actually solve the differ- pose the answer to be some pure number (such as ential equation of motion.) π or √2) times a product of the parameters each Another point worth making in this context: if raised to some power: there is more than one dimensional quantity of α β γ each genre—for example the radius r of the bob, ν = constant × m λ g . the mass µ of the string—then the above result Now require the dimensional consistencyof both must be multiplied by an unknown function of sides by putting square brackets around all dimen- dimensionless ratios: sional quantities3: 1⁄2 α β γ g [ν] = [m] [λ] [g] ν = F(µ ⁄ m, r ⁄ l, … ) . l which can be rewritten γ 0 0 −1 = α β L ≡ α β + γ −2γ M L T M L 2 M L T . Surface waves in deep water T We can use dimensional analysis to determine the Exponents of like dimensional quantities on both speed of surface waves on deep water. The quanti- sides of this equation must agree. In general this ties in the problem are the wavelength λ, the yields three equations: density ρ of the fluid, and the acceleration of 0 = α gravity, since the forces are again gravitational. The dimensional equation is 0 = β + γ − = − γ α β γ 1 2 . v = constant × λ ρ g . 2. Of course this test will not catch errors of sign or purely numerical factors such as 2 or π. 3. We drop the constant because it has no dimensions. 4. This is characteristic of problems involving the gravitational force. Physics of the Human Body 15 Chapter 2 Dimensional analysis and scaling laws We are tacitly assuming here that the water depth Strength to weight ratio is so great compared with the wavelength as to be To understand this, consider first two cubes. The effectively infinite; and that the viscous forces may first is 1 meter on a side, the second 2 meters on a be ignored. Converting this to dimensional terms side. The volume of the first is obviously 1 m3. we have That of the second, however, is 8 m3. The same − α β − β γ − γ holds true of spheres: a sphere of radius 2 m has 8 L1 T 1 M0 = L M L 3 L T 2 ; times the volume of one with a 1 m radius. enforcing dimensional consistency gives, as before, three equations: In fact, the same is true of the volumes of similar objects of any shape: if one is twice as large in any 1 = α −3β + γ dimension—length, width, thickness—its volume −1 = −2γ will be 8 times that of the smaller object. 0 = β . Suppose the two objects are made of the same kind γ = 1⁄ α = 1⁄ β = The solution is 2 , 2 , 0 . Thus there of stuff: for example, gold. Since the density of a is no dependence on the density of the fluid; instead the wave speed is v = constant × √ gλ . Again this agrees with the result one obtains by solving the partial differential equations of fluid mechanics, up to the unknown numerical con- stant. 2. Allometric scaling Insects have nothing to fear from gravity. No fall in the Earth’s gravitational field can kill an insect. material does not change with size or shape, the A mouse can fall down a 1000 foot deep mine shaft object that is twice as tall will have eight times the and suffer a minor shock, but no lasting injury, mass. If my height were doubled to 12 feet and all when it hits. A rat, on the other hand, might well my other dimensions were also doubled, then my die; a man would certainly die; and a horse would weight would be nearly 1 ton. Similarly, were I splash. shrunk to three feet in height, with my other dimensions kept in proportion, I would weigh Are mere bugs and mice so much sturdier than about 32 pounds. larger animals, or is something else taking place here? Galileo Galilei, the great Florentine scientist, Exercise was the first to point out that similar objects, no matter what their shapes, can be related by scaling Given the above information, what do I weigh? laws5. These laws explain why an elephant cannot look like an enlarged mouse, why a giant human necessarily has certain problems, and why there are ultimate limits to the sizes of animals, plants and structures. 5. Galileo Galilei, Dialogue Concerning Two New Sciences (1638). 16 Chapter 2: Allometric scaling Structural strength velop, hence there is a limit to the size of land The strength of tendons and bones increases as animals. This seems to have been reached in the their cross-sectional area—that is, roughly as the largest dinosaurs, which may have weighed 60 squares of their linear dimensions. Similarly, the tons. Some schools of thought have believed such force a muscle can exert in tension also scales as large sizes were possible only for quasi-aquatic ani- L2. Bone, like any other structural material, has mals, who could have been partly supported by the strength proportional to its cross-sectional area. If water of the lakes they waded in. an animal is scaled up in size keeping its bones in Suppose the average density of bone is three times proportion, then at twice the linear size the that of water (the density of most rocks is about strength, relative to its weight, of its bones is only that, e.g.). Let half as great as those of the smaller animal. More precisely, the strength to weight ratio of geometri- Vbone Abone Lbone x = = ; − −1⁄ cally similar animals scales as L 1 or M 3. V V then the cross-sectional area of the bone (struc- This poses a problem for the larger animal—it has tural strength, that is) must be proportional to the to more carefully avoid bangs and bumps than its weight of the animal: smaller counterpart, for fear of breaking bones or tearing tendons6. σ A = m g = g 3ρ xV + (1 − x)ρ V max bone σ where max is the maximum stress a bone can bear. Large animals—rhinoceri, elephants, dinosaurs— Moreover, solved this evolutionary problem by changing their = λ 1⁄3 Lbone V , proportions. Their bones are proportionately much thicker than those of smaller animals.