Physics of the human body [draft ed.] 127-127-128-1

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Physics of the Human Body

a

Table of Contents Preface

i

Review of Newtonian Physics

1

Linear Momentum Work and energy Power Extended bodies

1 1 2 2

Center of mass Rotation Moment of inertia

2 3 5

Applications Firearms Automobile collisions

Dimensional analysis and scaling laws Dimensional analysis Surface waves in deep water

Allometric scaling Strength to weight ratio Structural strength Terminal velocity How high can you jump? Basal metabolism Lifetimes Brain size

Musculoskeletal system Skeletons of land verterbrates Muscle Mechanical (dis)advantage Virtual work

Hill’s Law Basics of locomotion Walking Scaling arguments Timing Walking power

Transition to running Running power Other considerations Maximum speed Animals that cannot run

8 8 9

13 13 14

15 15 16 17 17 18 19 19

23 23 23 24 24

25 26 26 28 29 30

32 33 34 34 34

Fluid mechanics and living organisms

37

Euler’s equation Conservation of fluid

37 38

The Navier-Stokes equation Reynolds’ number Steady Flow in pipes Impedance Pulsatile flow in arteries

Bioenergetics Energy input to living organisms Energy demand of organs Heating and cooling Appendix—Planck’s blackbody radiation law

Some ideas from thermodynamics The First Law of Thermodynamics Equation of state Thermodynamic relations Heat engines and Carnot efficiency The Second Law of Thermodynamics Statistical Mechanics

39 40 40 42 42

45 45 46 47 49

53 53 54 54 56 59 59

Boyle’s Law Maxwell-Boltzmann distribution

59 60

Applications of Maxwell-Boltzmann distributions

62

Liquid-vapor equilibrium Reaction rates Dissolved gases The bends

Diffusive processes and Brownian motion Random walk model of diffusion

62 62 63 64

65 65

The mean and variance of the position

66

Stochastic differential equation Thermal “force” and osmotic pressure Brownian motion and diffusion Diffusion across a synapse How a cell eats

67 69 70 70 70

The properties of water Static properties Solvent properties Solubility of H2O in H2O Viscosity Specific heat Heat of vaporization Vapor pressure

73 73 73 74 74 75 76 77

b

Physics of the Human Body

Excitable cells Electrocardiography Neurons Sensory cells Membrane potentials Energetics of muscle tissue

Sound, hearing and the human voice Sound propagation in gases Hearing and the ear Response to sound Frequency response of the ear Spatial discrimination of sounds The human voice

The human visual system Simple lenses Problems with lenses Problems with vision Light → nerve impulses Vision and the brain

79 79 82 84 84 85

87 87 90 92 93 93 95

99 99 101 102 103 105

Effects of radiation on living organisms

107

Background Effects of penetrating radiations Passage of radiation through matter Attenuation of radiation in matter Radiation exposure and dosage units Effects of radiation exposure Appendix

107 108 110 113 114 115 118

Physics of diagnostic methods Auscultation Blood pressure Ultrasound imaging Magnetic resonance imaging X-rays and computerized tomography Radioactive tracers

119 119 120 121 122 125 126

Mathematical review Review of Trigonometry Areas and Volumes Integration Differentiation The fundamental theorem of calculus Functions of several variables Ordinary differential equations Vectors Vector analysis Basic ideas of probability theory

127 127 128 128 129 130 130 131 134 135 137

Physics of the Human Body

i Preface

Preface Before the 19th Century, most natural philosophers distinguished living- from non-living objects. Living objects were supposed animated by a healthy dollop of “life force”, whereas inanimate matter lacked this essence. Dying thus implied losing the life force1. The dualistic point of view led people to worry about what happened to the animating principle after death. Since nothing definite was known, thinkers felt free to propose notions of an afterlife that persist in the dominant religious beliefs of the present day. The stunning successes of the scientific approach led thinkers to question the foundations of religious dogma. Galileo challenged the idea that the Earth was the center of the universe (without defying the Church’s teachings about the immortality of the soul, Heaven and Hell, and so forth). But the development of the precise sciences of physics and chemistry made clear that minerals and living tissues were composed of the same elements. Eventually a chemical compound—urea— heretofore found only in living tissues, was synthesized entirely from inorganic materials2. This discovery effectively discredited the idea that inorganic and organic compounds differed in some fundamental, unknown way. Other scientific discoveries furthered the idea that life was a process governed by physical law. I mention specifically the circulation of blood, discovered by William Harvey in 1628; and the measurement of the speed of the

1. 2. 3. 4.

impulse in a frog’s sciatic nerve (27 m/s) by Hermann Helmholtz, in 1850. By the late 19th and early 20th Centuries, most scientists had accepted the materialistic view: all the processes we call “life” obey the same laws of physics and chemistry as processes involving inanimate matter. Some of the evidence supporting materialism was physical: the law of energy conservation was proposed in connection with biological investigations. Some was chemical: oxygen input equals (bound) oxygen output; also oxygen input measures total energy output (heat + work).3 Finally, Darwin and others realized that life has a history—as we would expect if it is a physical process. We call this history the theory of evolution. In the materialistic view we regard living organisms as enormously complex collections of matter, capable of powering certain chemical reactions with free energy4 from external sources. These reactions include synthesizing the structural elements of the organism itself, as well as generating new autonomous organisms of the same pattern, i.e., reproduction. Self-replication is the key distinction between living organisms and inanimate matter. Our mental model of a living organism is therefore a kind of machine. In this sense, living organisms are the most complicated machines we know of, and therefore in many ways not especially amenable to the methods of physics.

Theories of a dichotomy between living and dead matter are collectively known as dualism. Friedrich Wöhler synthesized urea in 1828. “L’etat n’a pas besoin des savants.” (The state does not need wise men.)—Robespierre, refusing clemency for Lavoisier. The term “free energy” comes from thermodynamics. It has a specific technical meaning and in no way implies the absence of cost.

ii So we must understand at the outset that the ways we can apply physics to the understanding of any life form—much less to the human body—will be strictly limited in scope. That is, we can use physics to understand some aspects of life, but not others. Here is an example of what I mean: many years ago the mathematician Alan Turing5 tried to understand the process of morphogenesis6 during the development of an embryo, using a simple physical model7. He was excited to discover that a sphere of cells (the blastula), when it reaches a certain size, becomes unstable with respect to dimpling inward. Since that is just what a blastula does (dimples in on opposite sides, with the dimples then joining to form a tube, i.e. higher organisms are topologically akin to doughnuts), this seemed a major breakthrough.

However, this is not actually the way Mother Nature does it: at a certain point in the blastula’s growth, one cell (probably the oldest cell, whose timer goes off first) sends out a long fiber into the hollow interior. The fiber has a sticky end which encounters a cell on the opposite side of the sphere and sticks firmly to it. The emitting cell then reels 5. 6. 7. 8.

in the fiber like a fisherman, pulling the opposite sides of the sphere inward and transforming the sphere into a doughnut. In other words, when a crucial developmental event takes place, rather than leaving matters to chance (as would be implied if the inward dimpling resulted from structural instability), a living organism employs elaborate molecular machinery that ensures the necessary outcome with high probability. Since the analysis of complex machinery is more the province of the engineer than of the physicist, we shall have little to say about such details, except to note that this is how things work. What physics can usefully contribute here is a discussion of the general principles governing such molecular machines, so that we can understand their potentialities and their limitations. An example of what physics can do is Erwin Schrödinger’s analysis8 of why quantum mechanics is required to understand genetics. Although he was ignorant of DNA, his ideas are quite modern. In particular Schrödinger notes that the stability of the genetic code—the ability to pass information essentially unchanged from generation to generation—results from the fact that bound atoms and molecules have a discrete distribution of energy levels, rather than a continuous one. Thus at low enough temperatures the structural stabilty defeats the thermal tendency toward disorder. Finally, it is important to recall that physics is an experimental science. Despite our emphasis on theoretical physics, we shall be careful to confront our ideas with actual measurements.

Yes, that Alan Turing, the man who broke the Enigma code in WWII, developed a mathematical theory of computers, and was hounded to the point of suicide by the British Secret Service. ...meaning “the development of form”. A.M. Turing, “The Chemical Basis of Morphogenesis”, Phil. Trans. Roy. Soc. (London) B 237 (1952) 37. E. Schrödinger, What is Life (Cambridge University Press, Cambridge, UK, 2000—first publication 1944).

Physics of the Human Body Chapter 1:

Review of Newtonian Physics

1

Review of Newtonian Physics 1.

Linear Momentum

Newtonian physics describes the motion of pointlike masses via Newton’s Second Law of Motion, →

→

→

F = m

dv dp ≡ dt dt

→

where F is the force acting on an object of mass m, and →

df

A car of mass 1000 kg moving at 30 km/hr in the x direction collides with a car of mass 2000 kg moving at 20 km/hr in the −y direction. What is the velocity of the resulting mess, ignoring friction with the road and assuming the cars stick together after the collision? Solution:

→

p = mv

The momentum before equals the momentum after (no external forces!)

is the momentum. For a collection of objects, whose masses are m k , k = 1, 2, … , n we have d → d Ptot ≡ dt dt

Example

n

∑ k=1

n

→

pk =

→

→

∑ Fk ≡ Ftot . k=1

vx →  1000 ⋅ 30  = 3000 ⋅   ≡ pout pin =   v 2000 ⋅ (−20)  y Solving we see immediately that →

 10  vout =   km ⁄ hr . −16.33 End example →

→

In particular, if all the forcesFk sum to zero1, the →

time derivative of the total momentum Ptot vanishes, meaning that the total momentum is constant. This is the content of Newton’s First Law of Motion,

2.

Work and energy

Next we may define the work done by a force on a moving body as the force component in the direction of motion, times the distance moved. For →

small (vector) displacements dx we therefore have “In the absence of external forces, an object at rest remains at rest, and an object in uniform motion remains in that state of motion.”

Another name for this principle is Conservation of Momentum.

1.

…in a vectorial sense.

df →

→

df

dW = F ⋅ dx = Fx dx + Fy dy + Fz dz . →

If we replace F by Newton’s Second Law, and note that for a moving object the displacement in time dt is

2

Chapter 1:

→

Solution:

→

dx = v dt we may integrate

∫ dW =

→

∫ F ⋅ dx = →

∫m

→

→ → dv → ⋅ dx ≡ m ∫ dv ⋅ v dt

→ → ∫ d 12m v ⋅ v ≡ 12 m v2 ,   thereby finding that the work done in accelerating

=

→

an object from zero velocity to velocity v is W =

1 2

→

1

→

mv⋅v ≡

2

m v2 .

The latter expression is called the kinetic energy of a moving object because it is strictly associated with the object’s motion.

3.

From the expression →

dW = F ⋅ dx

for work we note that the energy consumed per unit time in exerting a force upon a moving body is df

A 5% grade rises 5 feet for each 100 feet of forward progress. Thus after walking an hour (3600 seconds) the man has risen about 282 meters, thereby doing total mechanical work W = 100 kg × 9.8 m ⁄ sec2 × 282 m = 276360 Joules . His power consumption from climbing is therefore ∆P = 276360 J ⁄ 3600 sec = 77 Watts , hence his total is almost 200 Watts. End of Example

4. Extended bodies Center of mass

Power →

Review of Newtonian Physics

→

→ dx → → dW = F⋅ ≡ F⋅v . P = dt dt

The rate of energy expenditure per unit time is called power. Examples of power consumption are found in propulsion problems—we exert a certain force to keep a vehicle or ship moving at constant speed against various resistive forces such as air resistance, viscosity, friction, wave generation and the like, that would tend to slow it down. Example What is the total mechanical power required by a man of mass 100 kg, to walk up a 5% grade at 3.5 mi/hr, given that walking on the level at that speed requires 120 Watts?

Up until now we have been pretending that masses are mathematical points, with no extension. Any motion of an extended body—which for now we shall consider rigid—can be decomposed into two independent motions: displacement of the center of mass, and rotation about the center of mass. We begin by defining the center of mass: an extended body can be thought of as comprised of N individual pointlike particles, each of mass m k . The center of mass is a point in space (it need not be located within the body) defined in terms of the positions of the individual locations of its components: N →

df

Xcm =

∑

→

m k xk

k=0 N

∑

mk

k=1 →

The importance of Xcm is that the force of gravitation on an object behaves as though all the mass comprising it were concentrated at the center of mass. Moreover, if the sum of all the external forces acting on each of the N constituents does not

Physics of the Human Body Chapter 1:

Review of Newtonian Physics

vanish, the equation of motion describing the displacement of an extended object is →

→

Ftot = M

d2Xcm dt2

,

where N

M =

∑ mk. k=1

Rotation

Because displacements and rotations are independent we may treat rotations as though a given object’s center of mass were at rest. Now for this to continue to be so, the net force acting on the body must vanish. One way this can happen, however, is for equal and opposite forces to act on different parts of the object, as shown below:

3

same number. We say that the scalar product is invariant with respect to rotation of coordinates. The proof of this statement is so is so simple and clever that I cannot resist including it. Consider the scalar product of a vector with itself: →

→

a ⋅ a = axax + ayay + azaz From the Pythagorean Theorem we see that this is just the square of the length of the vector, which is obviously unchanged by a rotation. Since the (squared) length of a vector is invariant, so must be the difference of the (squared) lengths of two vectors, or any constant multiple thereof: that is, 1 4

→

→

→

→

→

→

→

→

a + b ⋅ a + b − a − b ⋅ a − b         → → ≡ a⋅b is invariant under rotations. A final remark about the scalar product: if we denote the length of a vector by →

→

→

 a⋅a ≡ a a = √ then the scalar product has the value →

→

a ⋅ b = a b cosθ where θ is the angle between the two vectors.

Naturally we expect the unbalanced forces shown to cause the object to rotate about its center in the clockwise direction, and indeed that is what will happen. In order to describe rotations in terms of vectors we must define a new way to multiply two vectors. Presumably the reader is already familiar with the scalar, or “dot” product of two vectors: →

→

df

a ⋅ b = axbx + ayby + azbz ,

that we used in defining work and power. This is called a scalar product because the result is just a number; moreover, if the coordinate system is rotated the scalar product (computed in terms of components along the new coordinate axes) is the

The new way to multiply two vectors is called the vector product (or sometimes “cross” product) and ∧

is defined as follows: if x is the vector of unit length along the x-axis, and similarly for y and z, then we define df

→ ∧ ∧ a × b = x aybz − azby + y azbx − axbz +     →

∧ + z axby − aybx .   That is, when we vector-multiply two vectors the result is a vector, rather than a scalar.

4

Chapter 1:

Lz = x py − y px .

Exercise Can you show from the above definition that the unit vectors of a right-handed coordinate system obey the cyclic relations ∧ ∧ ∧ x×y = z

∧ ∧ ∧ z×x = y, →

a×a ≡ 0,

and since the velocity points in the same direction

and finally, that the vector multiplication depends on the order: →

→

→

that for any vector a ,

→

→

d → dr → → dp L = ×p + r× . dt dt dt

dr → = v dt

→

→

Thus the derivative can be calculated using the product rule of differentiation, as though the multiplicands were functions rather than vectors:

(We have to remember to preserve the order of the factors when differentiating.) Now we may note that

∧ ∧ ∧ y×z = x

→

Review of Newtonian Physics

→

a×b ≡ −b×a ?

→

→

dp F = dt →

(In other words, when we reverse the order of vector multiplication, the resulting vector reverses its direction.)

so we can now say

Another important fact about vector products is that their lengths can be computed as

where N is the torque.

→

→

as the momentum, the cross product v × p is zero. On the other hand, Newton’s Second Law says

→ → → d → L = r×F = N , dt →

→

a × b = a b sinθ   where again θ is the angle between the two vectors.

→

Now, let r be the (instantaneous) position of a point mass m, measured in a coordinate system whose origin will be the center of rotation, and let →

p be its instantaneous momentum. Then we can define a new vector, called the angular momentum, as → df →

→

L = r×p .

If we now add up all the angular momentum vectors for all the constituents of an extended object, we see that its time derivative is equal to the sum of all the torques acting on the body:

d → L = dt total

n

∑

→

→

→

rk × Fk = Ntotal .

k=1

Thus, in the absence of torques2 the total angular momentum of a body remains constant in time. This fact is called conservation of angular momentum.

→

Now let us calculate the derivative of L with respect to time. If we go back to the original definition of a cross product, we see that each component of the product involves products of the components of the multiplicands, for example

Finally, let us consider a rigid3 object that is rotating about an axis. Imagine the two-dimensional Cartesian coordinate system shown below has been rotated slightly about an axis perpendicular

Physics of the Human Body Chapter 1:

Review of Newtonian Physics

to the plane (z-axis) to form a new frame, as shown below.

5

formation from body-fixed to space-fixed coordinates: →

→

→

→

r = r ′ + δϕ × r ′ .

The algebraic sign of rotation conventionally fol-

!

: if the right lows the “right-hand rule” thumb points along the rotation axis, the fingers curl in the direction of positive rotation. If the position vector describes a moving particle, after a small time interval δt the position (in laboratory coordinates) becomes The coordinates of a point (x, y) in the original frame and those of the same point, (x′ , y′) in the new frame are related by x′  xcos(δϕ) + ysin(δϕ)  y′ = −xsin(δϕ) + ycos(δϕ)    

 cos(δϕ) sin(δϕ) x ≡   −sin(δϕ) cos(δϕ) y  1 δϕ x ≈  . −δϕ 1  y In vectorial notation this becomes

→

→

→

where the angular velocity vector Ω is df

→ ∧ δϕ n = Ω δt .

Expanding to first order in δt, and noting that in coordinates attached to a rigid body the constitu→

ents do not move, i.e. v ′ = 0, we find →

→

→

→

→

r(t) + v δt = r ′(t) + Ω δt × r ′(t)

or →

→

→

v = Ω × r′.

∧ → r ′ = r − δϕ z × r ,

→

→

→

r(t+δt) = r ′(t+δt) + Ω δt × r ′(t+δt) ,

→

→

or, solving for r , ∧ → r = r ′ + δϕ z × r ′ .

→

→

We imagine the primed coordinates are fixed in the rotating object and that the unprimed ones refer to coordinates in a laboratory. Although in this case we are rotating about the z-axis, there is nothing special about that choice. We thus define a rotation vector df

→ ∧ δϕ = δϕ n ,

∧ where n is a unit vector along the axis of rotation. →

In terms of δϕ we rewrite the infinitesimal trans-

5.

Moment of inertia

We now substitute back into Newton’s Second Law as applied to the rotation of an extended body. For each constituent, → → → d → d →  r ′ × mv  = m r ′ ×  Ω × r ′     dt  dt  →

= N

If we add up the contributions from all the constituents and all the torques, we find

6

Chapter 1: . . . Ix x Ω x + Ix y Ω y + Ix z Ω z = N x

. . . Iy x Ω x + Iy y Ω y + Iy z Ω z = N y . . . Iz x Ω x + Iz y Ω y + Iz z Ω z = N z

Review of Newtonian Physics

Consider a simple pendulum, consisting of a massless rod of length R, suspended from a pivot at one end, with a mass m on the other end, as shownbelow.

Here the tableau of coefficients Ixx Ixy Ixz   I = Iyx Iyy Iyz Izx Izy Izz    is something called the inertia tensor, and we have used Newton’s dot ( ⋅ ) notation to indicate time derivatives. That is, . dΩ Ω ≡ . dt In this course we will not be concerned with the off-diagonal elements of the inertia tensor. The diagonal elements are defined by n

Ixx =

∑ k=1

mk y2k + z2k   

∑ k=1

mk x2k + z2k   

∑ k=1

and ∧ F = −mg y

Hence the torque is → ∧ N = −mgR sinθ z

n

Izz =

∧ ∧ r = Rsinθ x − Rcosθ y

→

→

n

Iyy =

The force acting on the pendulum is gravity; taking the axis of rotation to pass through the point of suspension, we have

mk x2k + y2k   

where the coordinates refer to the body-fixed axes.

Examples In simple cases the axis of rotation and the direction of the torque coincide, so the torque causes angular acceleration. That is, suppose the torque is in the z-direction. Assuming the z-axis is one of the principal axes of the system, the equations of motion then reduce to . Izz Ωz = Nz .

and the equation of motion is I

d2θ = −mgR sinθ d t2

where I = mR2 . For small angles this is just the equation of an harmonic oscillator of angular frequency 1⁄2

g ω =   R Now suppose that instead of a mass on the end of a massless rod, the pendulum consisted of a uniform thin rod, of length R and of overall mass m. Then as we see below,

Physics of the Human Body Chapter 1:

Review of Newtonian Physics

7

d2θ 1 1 mR2 2 = mgR sinθ 2 3 dt or 3g d2θ = sinθ . 2 2R dt We see that the solutions of such equations grow in time, rather than oscillating sinusoidally (as with the case where the pendulum is hung below its pivot). the gravitational force now acts at the midpoint (center of mass) of the rod, and the torque is therefore half of the previous value. Another change is the moment of inertia. Since the previous pendulum consisted of a point mass m a distance R from the pivot, we obtained I = mR2. Now, however, we must add the contribution from many constituents. To do this we consider a small slice of the rod, of mass dy m; R

dm =

the contribution of this slice is d I = dm x2 + y2 ≡ y2 dm ,   so adding them all up (i.e. by evaluating the integral) we get R

m R 1 dy y2 = mR2 . I = ∫ dI = ∫ R 3 0 0 That is, because much of the mass is closer to the axis of rotation in this case, the moment of inertia is smaller than for the simple pendulum. As a third example, consider an inverted pendulum, such as a meter stick balanced on its end. In this case the torque changes sign (because the position of the center of mass is above the point of suspension rather than below) but the moment of inertia remains the same. For a uniform rod, the equation of motion now becomes

Example In the preceding equation let us re-scale the time variable, t → λτ, to make it dimensionless (see the following chapter). The equation then becomes

1 d2θ d2θ 3g = = sinθ 2 2 2 2 R d (λτ) λ dτ  2R ⁄ 3g  , we find so if we choose λ = √ d2θ = sinθ . d τ2 This equation can be integrated once by multiplying with the integrating factor dθ ⁄ dτ, giving 2

1 d  dθ  dθ dθ d2θ ⋅ 2 = = sinθ   2 dτ  dτ  dτ dτ dτ = −

dcosθ . dτ

Integrating once and solving for dθ ⁄ dτ we have 1⁄2 dθ = ± 2( cosθi − cosθ) dτ  

where we assume the meter stick starts at initial angle θi with zero initial angular velocity. The time to fall from θi to θf > θi is thus 1⁄2

R ∆t =    3g 

θf

−1⁄2

∫θ dθ cosθi − cosθ i

.

8

Chapter 1:

For θi = 2o and θf = 20o we evaluate the integral to get ∆ t ≈ 1.25 π √  R ⁄ 3g  . End of Example

Review of Newtonian Physics

The time of passage is

We now look at some more detailed applications of Newtonian mechanics to problems of interest in the study of living organisms.

1 2

Firearms

The bullet from a 0.357 Magnum pistol weighs about 200 grains (7000 grains to the pound) and carries a muzzle (kinetic) energy of 500 ft-lb. If it strikes and passes through a typical human body (assumed 1 foot thick) and loses about 10% of its speed thereby, 1. How much energy is lost? 2. What is the average force experienced by the person who has been struck? 3. How long does the bullet take to pass through? Beginning with Question 1, we see that mv2 −

1 2

m(v − δv)2 ≈m v δv

δv = 0.2 E = 100 ft−lb . ≈ 2E v The average force is given by _ df 1 ∆x ∆E 100 ft−lb F(x) dx = = F = 1 ft ∆x ∫0 ∆x = 100 lbs .

2.

dx ∆x ≈ , 0.95 v v(x)

mv2 = 500 ft−lb

or v =

1 2

∫0

where we replace v(x) by its average (assuming it decreases linearly with distance). We solve the for the speed v from

6. Applications

∆E =

∆x

∆t =

32000 √  m

ft/sec

where m is the weight in pounds2. Since the bullet weighs 1/35 of a pound, the speed is 1058 ft/sec. This means it takes about 0.001 sec (1 millisecond) to pass through the body. Exercise A rapidly moving object (like a bullet) passing through a fluid (such as air or water), displaces the matter it must pass through. If the cross-sectional area of the object is A and the mass density of the medium is ρ, then the mass displaced in time dt is dm = ρAv dt where v is the object’s speed. The displaced matter 1 2

gains some fraction C of the speed v so that the net momentum loss per unit time is df

dp 1 = − CρAv2 = Ffric . 2 dt Thus the equation of motion of the object is (neglecting other forces such as gravity) m

dv 1 = CρAv2 2 dt

or

Do you understand this? Note that to convert mass to weight we must multiply by g, which in these (English) units is 32 ft/sec2.

Physics of the Human Body Chapter 1:

Review of Newtonian Physics

noting that

dv dv dx dv = = v , dx dx dt dt

Let us first analyze what happens when a car crashes into a large fixed, stationary object:

we have v

9

L−X

dv = −Γv2. dx

We can solve this to find dv ≡ dlnv = −Γdx   v or v = vi e−Γx . Given these facts, use the fact that the density of living tissue is about that of water, to estimate the fractional loss of speed of a bullet of diameter 0.357 inch, passing through a body 1 foot thick. Assume the geometrical factor C is 0.3 . (20%) End of Exercise

Automobile collisions

When automobiles crash into stationary objects or each other, the forces in the collision are so large they dwarf such forces as tire friction with the pavement. Thus during the impact, we may treat the car(s) as though it (they) were on a frictionless surface. Automobiles are also designed so that collisions are as inelastic as possible. The force required to crush a car to a depth x is approximately3 Fcrush ≈ mω2 x ; that is, the crushed portion behaves something like a one-way spring. Here m is the car’s mass, and the angular frequency ω has the approximate value 20 sec–1.

3.

We denote the position of the cm by X and the depth of crush by x. If L is the distance from the cm to the front of the car, then the depth of crush (for times after the front has contacted the wall) x = X+L (note that at t = 0 , X(t) = −L ). The equation of motion for X(t) is therefore m

d2X = −mω2 (X + L) 2 dt

m

d2x = −mω2 x . 2 dt

or

The solution to this equation is . x(0) x(t) = x(0) cos(ω t) + sin(ω t) ω

≡

v0 ω

sin(ω t) .

We see that the car comes to rest (that is, . x(t) = 0 ) when ωt =

π . 2

The depth of crush at that instant is

That is, the spring constant is proportional to the mass because all cars are made of the same materials.

10

Chapter 1:

xmax

v0 = . ω

The time for the car to come to rest is π , t = 2ω and the maximum acceleration is .. Xmax = − v0 ω . For example, if the speed is 30 mi/hr when the car hits the wall, the maximum deceleration is .. Xmax = −44 ft/sec ×20 sec−1 ≈ −27 g . Manifestly, this is nothing to sneeze at. The average deceleration is this multiplied by 2 ⁄ π, or 17 g. The presence of an air bag/safety harness is therefore a great comfort! About 75% of automobile accidents are two-car, rather than single-car collisions. Assuming two cars of different masses crash frontally, as shown,

Review of Newtonian Physics

In a frame of reference moving with velocity Vcm , the two vehicles appear to have equal and opposite momenta: m1 m2 v − v2 p′1 = m1 v1 − Vcm = m1 + m2  1    m1 m2 p′2 = m2 v2 − Vcm = v − v1 m1 + m2  2    Calling v = v1 − v2 and m =

m1 m2 m1 + m2

we may write . m v = F2,1 where F2,1 is the force exerted on car 1 by car 2 . Now the cars manifestly exert equal and opposite forces on each other (Newton’s Third Law) hence F1,2 = −F2,1 . Thus the equations of motion of the cars are equivalent.

C (t)

the key difference between this situation and the one-car collision is that, since we neglect friction with the pavement, we have an inelastic collision of two objects on a frictionless surface so momentum is conserved. Thus we may define the velocity4 of the joint center of mass, df

Vcm =

m1 v1 + m2 v2 ; m1 + m2

Next we must determine what F2,1 actually is. If X1 and X2 are respectively the positions of the centers of mass of each of the two cars, the position of their joint center of mass is Xcm =

m1X1 + m2X2 . m1 + m2

The depths of crush on the two cars may then be written x1 = C (t) − X1 − L1 x2 = C (t) − X2 + L2 ,

4.

…we do not bother with arrows since the motion takes place in one dimension.

Physics of the Human Body Chapter 1:

Review of Newtonian Physics

where we have called the point of contact C (t) . Then since the forces are F2,1 = −m1 ω2 x1 F1,2 = −m2 ω2 x2 Newton’s 3rd Law gives us 0 = m1 x1 + m2 x2 = M C (t) − Xcm(t) + ∆   where ∆ = m2L2 − m1L1 , and M = m1 + m2 .

Expressing the.. coordinates in terms of the relative acceleration X we now find maximum decelerations

.. m2 = − v ω X1 M 0  max .. m1 = v ω. X2 M 0  max The consequences of these results in terms of severity of injury are as follows. Suppose a large car ( m1 ) hits a small one. Then the deceleration— hence the force—experienced by its occupants varies roughly like m2 ⁄ m1 . Rates of injury seem to follow this law approximately, as shown in the graph below.

Thus C (t) moves at the same speed as the overall center of mass, i.e. . . C (t) = Xcm ≡ Vcm . That is, in the co-moving frame the point of contact is stationary5. We may thus take the point of contact as the origin of coordinates in the co-moving frame. The equation of motion for the relative coordinate, X = X1 − X2 , is then .. mX = −mω2 X − X0   and the problem now reduces to the one-car collision, with solution

X(t) = X0 +

v0 ω

sin(ω t) .

The relative motion ceases in the collision time t =

π , 2ω

as before.

5.

11

This would not necessarily be true if the force required to crush to a given depth were not approximately harmonic and proportional to the overall mass of the vehicle.

12

Chapter 1:

Review of Newtonian Physics

Physics of the Human Body Chapter 2 Dimensional analysis and scaling laws

13

Dimensional analysis and scaling laws 1.

Dimensional analysis

One of the simplest, yet most powerful, tools in the physicist’s bag of tricks is dimensional analysis1. All quantities of physical interest have dimensions that can be expressed in terms of three fundamental quantities: mass (M), length (L) and time (T). We express the dimensionality of a quantity by enclosing it in square brackets. Thus, for velocity (change of position divided by change of time) we write [v] = LT −1 . Similarly acceleration (change of velocity divided by change of time) has dimensionality  dv  [a] =   = LT −2 .  dt  From Newton’s Second law of Motion, →

→

F = ma , we see the dimensions of force are [F] = M ⋅ [a] ≡ MLT −2 ; and since work (energy) is force times distance, [E] = [F] ⋅ L ≡ ML2T −2 . Dimensional analysis has three important applications: 1. We can avoid algebraic errors by requiring the dimensions of the quantities on two sides of an equation or inequality to be consistent.

2. We can reduce the number of independent parameters in a calculation by re-expressing the problem in terms of relations between dimensionless quantities (pure numbers). Thus, consider the problem of a physical pendulum, whose equation of motion in the absence of driving torques is I

d2θ + mgl sinθ = 0 ; dt2

Let us redefine the time in terms of a dimensionless variable τ and a quantity ω with dimensions of frequency (note θ is already dimensionless because it is defined as a ratio of lengths): t =

τ . ω

Then the original equation becomes Iω2 d2θ + sinθ = 0 . mgl dτ2 If we now choose ω so that Iω2 = 1 mgl our new dimensionless equation is d2θ + sinθ = 0 . dτ2 We can make sure we have not made an algebraic error by checking the dimensional consistency of the relation Iω2 = 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:

 Iω2  2 −2 −1 −1 2 −1   = ML T M L T L mgl  

= M0L0T0 = constant . Hence the expression is dimensionally consistent and contains no obvious algebraic errors2.

3. We can “solve” certain physical problems without actually doing the detailed calculations needed for a complete solution. Thus, suppose we want an expression for the frequency ν of a simple pendulum. The frequency can only depend on the dimensional parameters of the system: the mass m of the bob, the length λ of the string, and the acceleration g of gravity. We suppose the answer to be some pure number (such as π or √2) times a product of the parameters each raised to some power: ν = constant × mα λβ gγ . Now require the dimensional consistencyof both sides by putting square brackets around all dimensional quantities3: α

β

γ

[ν] = [m] [λ] [g]

which can be rewritten

Dimensional analysis

The solution is α = 0 , β = −1⁄2 , γ = 1⁄2 . The fact that the exponent of M is zero means that the period does not depend on the mass of the pendulum bob4. Hence our formula for the pendulum’s frequency becomes 1⁄2

g  ν = constant ×   , l a well-known formula. (In fact, for a pendulum started at zero angular velocity from initial angle θ0 the constant is (for small θ0)  √2  θ0 dθ  ∫ 4  0 √  cosθ − cosθ 0 

−1

≈

1 2π

but dimensional analysis does not reveal this—to learn its value one must actually solve the differential equation of motion.) Another point worth making in this context: if there is more than one dimensional quantity of each genre—for example the radius r of the bob, the mass µ of the string—then the above result must be multiplied by an unknown function of dimensionless ratios: 1⁄2

g ν =   F(µ ⁄ m, r ⁄ l, … ) . l

γ

L M0 L0 T−1 = Mα Lβ  2  ≡ Mα Lβ + γ T−2γ . T  Exponents of like dimensional quantities on both sides of this equation must agree. In general this yields three equations: 0 = α 0 = β+γ −1 = −2γ .

2. 3. 4.

Surface waves in deep water

We can use dimensional analysis to determine the speed of surface waves on deep water. The quantities in the problem are the wavelength λ, the density ρ of the fluid, and the acceleration of gravity, since the forces are again gravitational. The dimensional equation is v = constant × λα ρβ gγ .

Of course this test will not catch errors of sign or purely numerical factors such as 2 or π. We drop the constant because it has no dimensions. This is characteristic of problems involving the gravitational force.

Physics of the Human Body Chapter 2 Dimensional analysis and scaling laws

We are tacitly assuming here that the water depth is so great compared with the wavelength as to be effectively infinite; and that the viscous forces may be ignored. Converting this to dimensional terms we have L1 T−1 M0 = Lα Mβ L−3β Lγ T−2γ ; enforcing dimensional consistency gives, as before, three equations: 1 = α −3β + γ −1 = −2γ 0 = β. The solution is γ = 1⁄2 , α = 1⁄2 , β = 0 . Thus there is no dependence on the density of the fluid; instead the wave speed is

15

Strength to weight ratio

To understand this, consider first two cubes. The first is 1 meter on a side, the second 2 meters on a side. The volume of the first is obviously 1 m3. That of the second, however, is 8 m3. The same holds true of spheres: a sphere of radius 2 m has 8 times the volume of one with a 1 m radius. In fact, the same is true of the volumes of similar objects of any shape: if one is twice as large in any dimension—length, width, thickness—its volume will be 8 times that of the smaller object. Suppose the two objects are made of the same kind of stuff: for example, gold. Since the density of a

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 constant.

2.

Allometric scaling

Insects have nothing to fear from gravity. No fall in the Earth’s gravitational field can kill an insect. A mouse can fall down a 1000 foot deep mine shaft and suffer a minor shock, but no lasting injury, when it hits. A rat, on the other hand, might well die; a man would certainly die; and a horse would splash. Are mere bugs and mice so much sturdier than larger animals, or is something else taking place here? Galileo Galilei, the great Florentine scientist, was the first to point out that similar objects, no matter what their shapes, can be related by scaling 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.

material does not change with size or shape, the object that is twice as tall will have eight times the mass. If my height were doubled to 12 feet and all my other dimensions were also doubled, then my weight would be nearly 1 ton. Similarly, were I shrunk to three feet in height, with my other dimensions kept in proportion, I would weigh about 32 pounds. Exercise Given the above information, what do I weigh?

Galileo Galilei, Dialogue Concerning Two New Sciences (1638).

16

Chapter 2:

Structural strength

The strength of tendons and bones increases as their cross-sectional area—that is, roughly as the squares of their linear dimensions. Similarly, the force a muscle can exert in tension also scales as L2. Bone, like any other structural material, has strength proportional to its cross-sectional area. If an animal is scaled up in size keeping its bones in proportion, then at twice the linear size the strength, relative to its weight, of its bones is only half as great as those of the smaller animal. More precisely, the strength to weight ratio of geometri−1 cally similar animals scales as L−1 or M ⁄3. This poses a problem for the larger animal—it has to more carefully avoid bangs and bumps than its smaller counterpart, for fear of breaking bones or tearing tendons6.

Large animals—rhinoceri, elephants, dinosaurs— solved this evolutionary problem by changing their proportions. Their bones are proportionately much thicker than those of smaller animals. And

Allometric scaling

velop, hence there is a limit to the size of land animals. This seems to have been reached in the largest dinosaurs, which may have weighed 60 tons. Some schools of thought have believed such large sizes were possible only for quasi-aquatic animals, who could have been partly supported by the water of the lakes they waded in. Suppose the average density of bone is three times that of water (the density of most rocks is about that, e.g.). Let x =

Vbone Abone Lbone = ; V V

then the cross-sectional area of the bone (structural strength, that is) must be proportional to the weight of the animal: σmax Abone = m g = g 3ρ xV + (1 − x)ρ V   where σmax is the maximum stress a bone can bear. Moreover, Lbone = λ V ⁄3 , 1

since the overall length of the bones in the skeleton must be proportional to the typical linear dimension of the animal. Thus σmax Abone Lbone ρgλL V

df

= x

Lmax = 1 + 2x L

hence −1

Galileo’s depiction of the bones of light and heavy animals. (From Dialogue on Two New Sciences.)

so are their muscles, whose driving power is proportional to their cross sections. But there is a limit to the thickness that bones or muscles can de-

6.

mbone 3 L  L  1 − = 3x = ,  m 2 Lmax Lmax   where Lmax is a constant with dimensions of length. For small animals, therefore, the ratio of bone mass to overall mass increases linearly with the linear dimension of the animal. Eventually, however, just to keep the animal from collapsing into a puddle the animal must become all bone—in other words

This pertains to football players as well. The greater size of today’s players implies a greater rate of injury and a shorter playing career.

Physics of the Human Body Chapter 2 Dimensional analysis and scaling laws

it has no room for muscles or organs. The result is shown below.

or

17

[v] = LT−1M0 = gα mβ Γγ Aγ   1⁄2

m g v = constant ×    ΓA 

1

∝ L ⁄2.

We can reach the same conclusion using Newton’s Second Law of motion FTot = W + FAir = m a or Whales live in water, hence are not supported by immense bones but rather by buoyancy. Their bones are mostly levers that muscles are attached to. The largest whale ever caught would have weighed about 350-400 tons (length 110 feet). When a whale beaches accidentally or because of disease, it quickly suffocates under its own weight because its supporting structure is inadequate in the absence of buoyancy. The ability of whales to bypass the structural restrictions of self-support was also noted by Galileo5.

Terminal velocity

Now, what about a falling mouse, horse or elephant? At low speeds viscosity dominates; but at higher speeds the air’s resistance to movement through it is proportional to cross-sectional area. But whereas an elephant-sized mouse would be some 100 times taller than a mouse, and its cross section therefore 1002 = 10,000 times greater, its mass would be 1003 = 106 times as large, hence its ratio of cross-section to mass would be 100 times less. How does this affect a free fall? The force of gravity is W = −m g , and that of air resistance is 2

FAir = ΓA v

where Γ is a constant proportional to the air density. Dimensional analysis would then say

dv Γ − A v2 = −g . m dt This equation can be integrated by separation of variables: dv = dt λv − g 2

where λ = Γ A ⁄ m , giving 1⁄2

 mg  v(t) = −   tanh(t ⁄ τ) Γ A where τ = √  g Γ A ⁄ m is a characteristic time (that could also be found by dimensional analysis). Initially the gravitational force accelerates the ob. ject downward with acceleration v = −g. As it picks up speed, air resistance becomes important, . so the net force decreases and the acceleration v falls to zero. When this happens, the object remains at a constant speed, the terminal velocity. This is shown on the following page. The asymptotic speed of impact increases with the ratio of weight to surface area, which means it increases as the square root of the linear size of the animal: vterm ∝ L ⁄2 . 1

How high can you jump?

To jump a height h an animal must expend energy E = mgh .

18

Chapter 2:

Allometric scaling

vice versa7. Paradoxical as this may sound, it is indeed correct. A flea can jump about 20 cm straight up, and a human about 60 cm. While the (human) world record in the (running) high jump exceeds 230 cm, such jumps are accomplished by converting forward motion into vertical motion8, as well as by adroitly timed movements of arms and legs, that allow the athlete’s center of mass to remain well below the bar at all times.

Basal metabolism To accomplish this it must exert a force F over a distance D so that F ⋅ D ~ mgh. Muscular forces scale as the cross-sectional area of the muscles involved, i.e. as L2 where as usual, L is the linear size of the animal. Obviously D, the distance the muscle contracts, also scales like L. The mass scales as volume, L3, so we find h ~

L3 × constant L3

—the height of the jump is independent of the animal’s size! Or to put it another way, scaling suggests that a flea can jump as high as a man, and

What determines the ultimate length a whale can grow to? Several factors combine to limit its size (which may be the largest possible size for any mammal): the need for oxygen and food increases as the cube of its length, but the generation of energy to sustain life also means the production of waste heat. The surface of a whale must be sleek so it can swim rapidly. This means it can not have projecting radiator fins (other than its swimming apparatus, of course). But then the whale’s ability to get rid of its excess body heat, even in very cold water, is limited by the surface area, which increases only as the square of its length. Basal metabolism of mammals (that is, the minimum rate of energy generation of an organism) has long been known to scale empirically as df

B =

3⁄4 dQ = const. ( Mass) . dt

The origin of this relation, graphed on the following page, sometimes called Kleiber’s Law, has recently been explained by West, et al.9 in terms of optimizing the pumping efficiency for fluid flow in the circulatory and pulmonary systems of mammals. They note that the terminus of a capillary or alveolar duct must necessarily be of constant size, independent of animal mass. Since the arterial 7. 8. 9.

The first person to analyze this seeming paradox was Giovanni Alfonso Borelli (1608-1679) in his book De motu animalium. Indeed, the jumper is using his leg as a sort of pole-vaulting pole. G.B. West, J.H. Brown and B.J. Enquist, Science 276 (1997) 122.

Physics of the Human Body Chapter 2 Dimensional analysis and scaling laws

19

coolant in a radiator), its flesh actually cooks within its jacket of blubber because the residual metabolic heat production has no way to escape. The temperature rises, therefore (the onset of decay from bacterial action accelerates this process). Some of the old-time whalers apparently enjoyed whale meat “cooked” in this fashion11!

Lifetimes

network and bronchial systems each have a treelike structure, and since the sub sections of the tree are self-similar (“fractal”) what determines the size of the entry—the aorta or the trachea, respectively—is the ratio of branch diameter to branch length, and the fact that the branches are (almost) always bifurcations10. When this ratio is chosen to minimize resistance to flow, hence pumping power, Klieber’s Law emerges. The largest whales are certainly at the ragged edge, maintaining a precarious balance between energy production and heat dissipation. When a whale dies (by being killed by hunters, e.g.) and its heart stops circulating the blood (which acts like the

Suppose the heart- and blood volumes both scale proportionally with the volume of the animal. The dV rate of circulation, , must be proportional to the dt basal metabolism, so we can say the frequency at which the heart beats is dVblood 3 = c ⋅ M ⁄4 . dt

f ⋅ Vheart =

But since the mass is proportional to the volume we have a scaling law f ~ M

−1⁄4

.

If we suppose that each animal’s heart beats a fixed number of times during its lifetime, N = f ⋅ τ = constant , we see that lifetimes should scale as

τ ~ M ⁄4 . 1

As the graph to the left shows, this “law” is satisfied very well empirically.

Brain size

Interestingly, the brain mass also scales as body mass to the 3⁄4 power, Mbrain ~ M ⁄4 3

(see figure on following page). The data fall on a band with slope 3⁄4 on a log-log graph, with primates occupying the upper edge of 10. The physics of fluids determines that bifurcation is better than, say, trifurcation. 11. Herman Melville, Moby Dick.

20

Chapter 2:

the band, and modern humans represented by a point well above the band. That is, Homo sapiens has a substantially larger ratio of brain to body mass than any other species.

Two interesting questions arise from this empirical relation. First, why—like Klieber’s Law for basal metabolism—is the power 3⁄4 ? And second, why is the human brain so much larger (1.5−2× larger) than the empirical scaling would predict? Much of the brain’s computational power is devoted to muscular control. An elephant’s trunk, e.g. has 6 major muscle groups divided into about 105 individually controllable muscle units. Its brain weighs 3.6–5.4 kg. For comparison, there are only 639 muscles in the human body, and human brains weigh about 1.3 kg. Kleiber’s Law for basal metabolism followed from the self-similar character of arterial or bronchial networks9. Similarly, the density of neuromuscular junctions is presumably constant; however to optimize the power demand of the brain segment that controls the musculoskeletal system, requires the number of neurons per unit of muscle to diminish

Allometric scaling

with body size. This optimization can be effected via bifurcation of axons—that is, the neural network forms a self-similar bifurcating tree like the arterial network, where the constraints on transmission of nerve impulses determine the scale of bifurcations. Among mammals, apes, elephants and 3 whales have brains larger than the M ⁄4 fit would predict. When we examine the elephant brain we see it has massive temporal lobes and huge sections devoted to controlling the trunk muscles. The temporal lobes provide the proverbial “elephant’s memory”12; and the sections controlling the trunk musculature have to be enormous because this is the animal’s most vital organ13. Not much brain is left over for abstract reasoning, and elephants do not seem to have such ability. The dolphin and orca (killer whale) are fast swimmers, and also possess extremely sophisticated sonar. They achieve their speed by constantly adjusting thousands of subcutaneous muscles that cancel out turbulence—and thereby greatly reduce drag. The dolphin brain therefore devotes much of its computatonal power to the sensory and feedback control elements of the drag-reducing and sonar systems. Though their brains are moderately larger than humans’, dolphins’ intelligence ranks somewhere between wolves and chimpanzees14. Since the central nervous system is a great consumer of energy15, we conclude that evolutionary parsimony gives animals the smallest possible brains consistent with species survival. This accounts for the brains (or lack thereof) of the rest

12. …needed to recall the locations of oases and water holes along the migration paths that elephant herds must follow. 13. Elephants with injured trunks soon die. 14. I feel impelled to add, however, that the killer whale has always impressed me as having human-level intelligence. Studies of their behavior in the wild reveal a level of planning and forethought that are unmatched by any animal other than human beings. 15. Neurons require operating power 10× the average of other cells; moreover, the energy needed to create a zygote’s central nervous system places severe demands on the mother during gestation.

Physics of the Human Body Chapter 2 Dimensional analysis and scaling laws

of the animal kingdom, but does not really explain why humans took the odd evolutionary path that led to intelligence. Perhaps that path was mandated by the development of the human hand, with all that implies.

21

22

Chapter 2:

Allometric scaling

Physics of the Human Body Chapter 3:

23 Musculoskeletal system

Musculoskeletal system This chapter analyzes the gross aspects of the skeleton, muscles and tendons as a system of articulated levers operated by ropes, pulleys and contractile units. The object of this system is to exert forces of various kinds on the external world—for example to move the animal (locomotion) or to procure food.

1.

2.

Muscle

Muscle tissue consists of basic contractile units called sarcomeres. The sarcomeres are attached end to end, with the demarcations marked by

Skeletons of land verterbrates

The verterbrate skeleton consists of bone1, a com-

posite material consisting of bone cells, collagen (a fibrous protein arranged in long strands), and inorganic rod-like crystals of Ca10(PO4)6(OH)2, perhaps 50 Å in diameter and 200 to 2,000 Å long. The quantitative properties of bone are given in the Table below:

Properties of Bone Type of Stress Ultimate Strength (× 108 Nt ⁄ m2) Compression Tension Bending Young’s modulus

1.

1.5 1.2 — 1.5 2.1 — 2.2 171 — 185

Z-membranes. At the microscopic level, the sarcomere contracts because the cross-bridges on the myosin fibers ratchet along the actin fibers. We discuss this in more detail in Chapter X, Bioenergetics. The sarcomere can move only one way: it contracts. Thus (contractile) muscular force must be balanced by a weight, a spring, or an opposing muscle, in order that the fibers can be pulled back to their initial uncontracted state. The important things to remember about muscles are 1. they can exert a maximum stress of 3×105 Nt/m2; and 2. they can contract at most 20–25% of their overall length (maximum strain=0.2–0.25).

Data on the properties of bone were taken from the article: “bone”, Encyclopædia Britannica Online (Copyright © 1994-2001 Encyclopædia Britannica, Inc.).

24

Chapter 3:

3.

Mechanical (dis)advantage

Mechanical (dis)advantage

We now analyze the muscular contraction force needed to lift a given weight. To do this we must take into account the articulated bones, which act somewhat like scissors jacks. Because muscle can contact at most 25% of its length, the arrangement of vertebrate skeletal levers actuated by muscular contraction sacrifices mechanical advantage for range of motion. A secondary result is that the “output” achieves absolute speeds much greater than those of the muscular contractions. A second constraint on the evolutionary optimization of organisms is the fact that muscles can only exert force while contracting. To make a muscle return to its uncontracted state it is necessary to pair it with a muscle that, when contracting, stretches the opposing muscle2. That is, all limbic muscles occur in pairs, called flexors and extensors.

hence does work T δ l. Since the tendon attaches below the knee, we set

Virtual work

To analyze the levers and muscles comprising a limb, we apply the principle of virtual work3. For example, suppose the leg shown to the right raises the weight W by a distance δh, and thereby performs work W δh. In so doing, the angle of the femur (thighbone) from the vertical changes from θ to θ − δθ. Taking the femur and tibia (shinbone) to be the same length L, and the muscles and tendons to have the (greater) length l, we see from the (Pythagorean theorem) that 2

2

2

l = L + 2Ldsinθ + d

where d is the offset of the muscle attachments. The muscle exerts a force T over the distance δl 2. 3.

W δh ≈ 2T δ l and noting that

δh = −2Lsinθ δθ , at last find 1⁄2

T = W tanθ 1 + 2Γsinθ + Γ2   where Γ is the ratio L⁄d . For small angles from the vertical, very little tension is needed to raise or

This is not universal—exoskeletal animals like spiders extend their muscles using a hydraulic system, whereas certain other muscles compress springy tissues that expand when the muscles relax. See, e.g., H. Goldstein, Classical Mechanics, 2nd ed. (Addison-Wesley Publishing Co., Reading, MA, 1980), p. 17ff.

Physics of the Human Body Chapter 3:

25 Musculoskeletal system

lower the weight. However, for angles beyond 45o the force rises very rapidly indeed. For my leg the ratio Γ is about 3.3, hence at an angle of 45o the mechanical (dis)advantage (at 45o) is about 4. Thus the tension must equal 4 times the weight to be lifted. But at an angle of 60o the tension rises to 7 times the weight. This is why exercise physiologists warn us not to perform knee bends at angles exceeding 45o—we can damage our knee ligaments at higher angles!

Let us estimate how much I can bench press. My upper arm has a circumference of 15 inches or 38 cm. Setting this to 2πr I find the radius of my upper arm muscles (biceps and triceps) to be r ≈ 6 cm. Their total cross-section is thus some 114 cm2. The cross-section of the upper arm bone (humerus) is perhaps 3 cm2, so we may take the area of the triceps to be about 74 cm2 (about 2⁄3 the difference). Multiply by the maximum stress, 3×105 Nt/m2, to get a net force of 2220 Nt per arm muscle. From the preceding analysis we see that the net weight I can lift (using my triceps alone) is W = Tmax

d cotθ ; l o

with L=32 cm and d=7 cm, and with θ=45 , we get

2220 Nt ≈ 189 lb . 5.3

This is very close to the amount I can currently bench press (10 repetitions), about 180 lb.

4. 5. 6.

Hill’s Law

Many of the ideas in this section are taken from the excellent book by C.J. Pennycuick.4 The physiologist A.V. Hill proposed the following empirical relationship5 betweenthe absolute speed of muscular contraction and the force being exerted by the muscle: v = v0

Fmax − F F0 + F

where Fmax, F0 and v0 are empirically determined parameters. It turns out to be convenient to express Hill’s equation in terms of the stress,

Example

W ≈ 2×

4.

df

σ =

F , A

and the strain rate df

ψ =

v 1 δx ≡ . L δt L

We can express the stress in terms of the strain rate, σ =

σmaxψ0 − σ0ψ ψ0 + ψ

and use the result to calculate, for a given strain rate, the power output per unit volume of muscle tissue: P = σψ. Different types of muscle tissue contract at different maximum rates 6. The so-called “fast” muscles are (relatively) anærobic in their metabolism. (That is, they use a chemical reaction that does not require free oxygen to generate their energy.) Anærobic muscles contain fewer mitochondria (the sites of oxidative phosphorylation) and lower density of myoglobin (myoglobin is related to hæmoglobin; it is used by muscle cells to store

C.J. Pennycuick, Newton Rules Biology (Oxford U. Press, Oxford, 1992). A.V. Hill, Proc. Roy. Soc. Ser. B 126 (1938) 136-195; see also Science Progress 38 (1950) 209-230. We are speaking here of striated rather than smooth muscle tissue. The latter behaves very differently and is not involved in locomotion.

26

Chapter 3:

oxygen), than the “slow” muscles. In other words, “fast” muscle tissue is usually paler in color than “slow” muscle tissue. The (qualitative) specific power curves for the two types of muscles are shown in the figure below. We see that the maximum power is developed by fast

5.

Basics of locomotion

Basics of locomotion

We shall concentrate here primarily on human locomotion, although the general principles apply to all land animals. First, we note that all locomotion requires cyclic motions of the legs: after one period the pattern repeats. We define three parameters of locomotion. The speed v is obvious; the stride length s is the distance between successive placements of the same foot; and the stepping frequency f is the inverse of the time interval ∆t between successive placements of the same foot: f = 1 ⁄ ∆t . Clearly these parameters are related by7 v =

muscles at a higher strain rate. Another factor that influences the ratio of fast to slow muscle fibers in a given muscle group is their endurance. Anærobic metabolism (which all cells are capable of, to some extent) generates lactic acid which must be removed by the usual transport mechanisms lest it interfere with muscle function. Thus, the ærobic muscles are for sustained effort at moderate speeds, whereas the anærobic ones are for emergency speed in short bursts. We conclude that there is an optimum speed for each type of muscle, and that evolutionary selection will lead to the optimal adaptation of muscle types to the demands made on them.

7. 8. 9.

s ≡ sf. ∆t

If an animal must increase its speed of locomotion, it can only do so by increasing its stepping frequency, its stride length, or both. Broadly speaking, only two types of gait are observed in locomotion: walking and running8. The distinction is that in walking, the animal always has at least one foot on the ground; whereas at some point in running all the animal’s feet are off the ground. This distinction profoundly affects the animal’s performance.

Walking

According to D’Arcy Thompson9, animals walk faster by increasing their stride length, or as he put it, “stepping out”. In fact, a simple experiment on a treadmill demonstrates the falsity of this claim: the graph on p. 27 below presents data from the author’s walking and running on a treadmill. It is

This is exactly the same as the relation between frequency, wavelength and propagation speed. The elephant’s amble (or shamble) is a walk, whereas the trot, canter and gallop are variations of running gaits distinguished by the order in which the legs move. D’Arcy Wentworth Thompson, On Growth and Form (Cambridge U. Press, New York, 1961), p. 29. (This is an abridged and annotated version of the 1917 edition.)

Physics of the Human Body Chapter 3:

27 Musculoskeletal system

point of efficiency. (You can try this if you wish, but be careful!) Another viewpoint treats the leg as a rigid pole with a mass on its end. When the foot is planted, the pole rotates about the point of contact. The initial angular momentum of a mass m moving horizontally at speed v, a vertical distance l cosθ0 above the ground, is easy to see that for a wide range of walking speeds my stepping frequency is proportional to my speed. (This experiment has been repeated by my students, with essentially the same results.) That is, I increase my walking speed by increasing my pace, not my stride length. When we look for an explanation of this observation, we realize that nothing else is possible. In walking the legs are always bearing the animal’s weight. The diagram below relates the forces along

J = mv l cosθ0 . Once the end of the pole is planted in the ground, the mass rotates about the point of contact. The angular momentum is then . J = m l2 θ since the moment of inertia is m l2. We now use conservation of angular momentum to say that . J = mv l cosθ0 = m l2 θ . The equation of motion of an inverted pendulum of the form shown below is .. m l2 θ = mgl sinθ

the leg bones to the opening angle of the legs. As we can see, the force increases rapidly as the opening angle increases. That is, the forces that must be exerted by the muscles, simply to keep the animal from sprawling, must increase beyond the

from which we can see, by integrating once10 with . dθ that the integrating factor θ = dt . . 2 2 θ + 2ω cosθ = θ20 + 2ω2 cosθ0 = const. (This is also the equation of conservation of energy, once the rotation begins.11)

10. See any standard reference on mechanics, such as Goldstein (op. cit.) or ordinary differential equations.

28

Chapter 3:

Basics of locomotion

In order that the pole vaulter complete his vault, the angular velocity must not vanish before the angle decreases to 0 (top of the arc—note: angles are usually taken to be positive in the counterclockwise direction). This means that . 2g θ20 ≥ 1 − cosθ0 , l  

leg angles roughly constant, meaning that the ratio of stride length to leg length remains essentially independent of walking speed. The conclusion is that the best way for a walking animal to increase its speed is to increase its stepping frequency.

or, imposing conservation of energy,

Scaling arguments

2

v ≥ secθ0 secθ0 − 1 . 2gl   That is, if the angle between the legs is too large the walker has difficulty completing his stride. Of course the above argument is only schematic— realistic walking involves bending the knee and a pelvic rotation (because the legs are actually angled to keep the body’s center of mass over the feet). We also use the rotation of the foot about the ankle joint to prolong the time of contact with the ground while our center of mass rises and falls. Thus the center of mass need not rise as much as it would with rigid legs. Nevertheless, the constraints of statics and dynamics tend to keep the

R.M. Alexander12 has graphed the (dimensionless) ratio of stride length to leg length, vs. the dimensionless Froude number df

ϕ =

v2 gl

for a variety of animals. One such graph appears below. The line through the data represents a regression (least-squares) fit. Also note that both the horizontal and vertical scales are logarithmic (equal distances represent factors of 10). That is, Professor Alexander is saying that he expects to discover a relation in the form of a power law, s = f(ϕ) ~ α ϕβ , l

11. We note that energy is not conserved in the initial collision of the pole with the pivot point—if it were, the vaulter and pole would recoil! 12. R.M. Alexander, “How dinosaurs ran”, Scientific American April, 1991, p. 130ff.

Physics of the Human Body Chapter 3:

29 Musculoskeletal system

where α and β are constants. A rough argument that might justify such an expectation goes as follows: since v = sf and since, for a pendulum (inverted or otherwise) a characteristic frequency is  g  1⁄2 f ∝   , l we see that 1⁄2

 v2  s ∝   = ϕ0.5 . l  gl  In fact, the slope of the line at the fastest gaits does approach 0.5 (meaning that the exponent β is close to 1⁄2). The first comment that we can make about the attempt to fit the data with a regression curve is that the data below ϕ ≈ 0.5 (roughly where the gait changes from walking to running) should not scale at all. That is, since geometry limits the leg s opening angle to a certain range, the ratio is more l or less independent of speed, hence of ϕ. We can also argue that if locomotion is akin to vaulting, the angle θ0 has to obey the inequality 2   θ0 ≤ cos−1  , 1 + 2 ϕ Γ     1 + √ where Γ is a geometrical constant. The time for the pole vaulter to rotate from angle θ0 to −θ0 can be computed by separating the variables in the equation of motion: we find ω∆t =

θ0

∫−θ

dθ Λ + 2cosθ0 − cosθ    0

where Λ is the dimensionless parameter . θ20 ⁄ ω2 = ϕ ⁄ cos2θ0 . 13. op. cit., p. 51.

−1⁄2

Here is a table of ϕ, twice the stepping frequency (in terms of ω), and the initial leg angle, that illustrates this point: 2f ⁄ ω θ0 (deg) ϕ

( Γ = 1) 0.1 0.399781 17.36 0.2 0.398596 23.64 0.3 0.397565 27.99 0.4 0.410874 30.00 0.5 0.439127 30.00 0.6 0.462887 30.00 0.7 0.483315 30.00 5.0 0.727090 30.00 10.0 0.790407 30.00 20.0 0.839795 30.00 We see 2f ⁄ ω is nearly independent of ϕ, over a wide range of ϕ. In other words, the data shown in Alexander’s graph on the preceding page are better fit with two straight lines, one that is horizontal (exponent 0) and one with slope (exponent) 0.5 .

Timing

We now take a closer look at the stepping frequency. Pennycuick13 has measured the stepping

30

Chapter 3:

frequencies of various African mammals, and de−1 termined that they scale as l ⁄2, as shown in the graph below. This is, of course, what would be expected if locomotion were pendulum-like. However, consider the frequency for an animal of leg-length 1 m: from the graph it should be 0.9 Hz, whereas from the table above we extract a somewhat smaller number at the walk, namely about 0.6 Hz (note we must 1⁄2

g  multiply by ω =   ≈ 3.1 Hz and divide by 2). L The same animal runs at a frequency of nearly 2 Hz. The predicted frequency of the pendulum model is 1.13 Hz, barely half that observed14. What are we to conclude from this disagreement? There are several alternatives: 1. The omitted details, such as bending the leg or damping, are so important as to materially change the numerical results. 2. The timing is not primarily determined by the kinematics of an inverted pendulum, but rather by some other aspect of the problem, such as the forces exerted by the animal’s muscles. I tend to favor the second possibility. The notion that an animal tends to walk “in resonance” with its pendulum motion (or at some other natural frequency) arose from the intuition that a swing goes higher if we push in time with it—that is, if we drive the system at resonance. This point of view misleads, however. While it is true that we can most easily store energy at resonance, the energy dissipated per cycle is just the integral of dissipative force times distance:

14. These results are insensitive to Γ. 15. “After this, therefore on account of this.”

Basics of locomotion

→

→ ∆E = o∫ F ⋅ dx

where the integration follows one complete cycle (the integral does not vanish since the frictional force always opposes the motion, i.e. the integrand is negative). Since this is independent of the driving force, we expect the power dissipated in alternately contracting and relaxing muscles to be proportional to the contraction frequency. In other words, there is no particular advantage, insofar as minimizing energy loss to friction is concerned, in driving the musculoskeletal system at a resonant frequency. Therefore in my opinion the scaling of stepping frequency as the inverse 0.5 power of the leg length, observed by Pennycuick and others, results from some other aspect of the problem. Or in other words, to conclude from this empirical scaling law that locomotion is best represented as a swing metronome is an example of the logical fallacy known as post hoc, ergo propter hoc15. To understand what is going on, then, we must return to our understanding of muscle tissue. As we saw, each type of muscle has an optimal contraction speed, or in dimensionless terms, an optimal strain rate ψ . Presumably what we are observing is that strain rates, optimized for the entirety of an animal’s lifestyle to maximize the −1 probability of species survival, scale as L ⁄2 . Whether we can model the factors that lead to such optimization sufficiently well to explain the empirical scaling law, is another question.

Walking power

We have seen by direct measurement (p. 27) that walking gaits maintain constant stride length and vary speed by varying the stepping frequency. We

Physics of the Human Body Chapter 3:

31 Musculoskeletal system

now attempt to estimate the mechanical power consumption of walking and compare it with data. From the inelasticity of the collision of a leg with the ground, assuming the correctness of the polevault model, we see that the mechanical energy loss per step is ∆E =

1 2 2 mv sin θ0 . 2

The power input, to keep a constant pace is therefore P =

∆E 1 mv3 mv3 = mv2 2f sin2θ0 ≈ = , 4s 8l 2 ∆t o

where we have taken θ0 = 30 . For a 100 Kg person walking at 3.5 mi/hr (1.6 m/sec) with leg length 1 m, we find a (mechanical) power loss of 51 Watts. Let us compare this with the actual power consumption, taken from the table shown below16

Energy cost of walking at various speeds Weight (lb)

mi/hr 100 120 140 160 180 200 220 2.0 2.5 3.0 3.5 4.0 4.5 5.0 6.0 7.0

65 80 62 74 60 72 59 71 59 70 69 82 77 92 86 99 96 111

93 88 83 83 81 97 108 114 128

105 100 95 93 94 110 123 130 146

120 112 108 107 105 122 138 147 165

133 124 120 119 118 138 154 167 187

145 138 132 130 129 151 169 190 212

According to this table, the rate of energy consumption in this exercise is 130 Kcal/mi. If we convert this to power, we obtain 1.3×105 ⋅ 4.2 ⋅ 3.5 joules ≈ 530 watts , 3600 sec a factor of 10 greater than our estimate. We must ask wherein lies the discrepancy. First, as we shall see when we study some thermodynamics, the efficiency of conversion of chemical (food) energy to mechanical work is only 20%-25% in humans. This accounts for a factor of, say, 4 or 5. But then we have also neglected internal friction in our book-keeping, as well as losses from imperfect restitutionby the(springy)tendons.Thesedissipative effects could easily amount to another 50 watts of mechanical power consumption, which would account for everything. Finally, it is not clear whether the above value of 130 Kcal/mi includes basal metabolism (which amounts to some 50-75 W) for that person. Our simple model predicted that the energy cost to transport a unit mass a unit distance should be independent of mass and should vary as the square17 of the speed. The data from the table do indeed predict a mass-independent cost, and do

16. …cited on an exercise website, http://walking.about.com/sports/walking/library/how/blhowcalburn.htm 17. The power per unit mass varies like v3 so the energy per unit distance varies like one less power of v.

32

Chapter 3:

seem to increase asymptotically as v2. However, the data exhibit a dip—that is, there seems to be a speed at which the efficiency of converting food into transport is maximum. What leads to this minimum? We recall from the figure on p. 26 that each type of muscle has an optimal contraction speed, at which it produces maximum mechanical power output. Thus we should expect, for each animal, an optimum speed of locomotion. Of course the minimum is much flatter than the maximum in the graph of power vs. strain rate, since the inherent power requirement of walking is growing as v3.

Transition to running

animal on the wall of his cave. No one, prior to the sequential photographs18 of Eadweard Muybridge (1830-1904) would have been so bold as to hazard the opinion that the horse becomes fully airborne. This is shown in the second and third frames of the figure below.

6. Transition to running When we analyze running gaits (trot, canter, gallop) we see that at some point in the cycle the animal’s feet are entirely free of the ground. That is, running consists of a series of forward bounds. Among bipedal animals like humans and ratite birds (ostrich, emu, etc.) this is true at every step. It is also true of the cheetah, which leaps twice each stride. In heavier quadrupeds such as horses, rhinoceri and giraffes, the animal is airborne only once per complete stride. Presumably larger animals must take some extra steps per stride to correct their balance and adjust their takeoff position. Precise observations of the gaits of running animals were not possible until near the end of the 19th Century. The persistance of human vision precluded early observers from being able to determine, for example, whether a horse keeps three, two or one feet on the ground while galloping. Controversy has raged over the precise number, probably since the first painter drew a galloping

The question now arises, “Why does an animal bother to run?” Manifestly it is to move faster than a walk will accomplish. But for the physicist the question is why running works better than simply moving the legs faster. We recall that the speed of locomotion is given by v = sf and that in walking s remains fixed and f increases proportionately to speed, for the geometric reasons we have discussed. In running, the opposite happens: the stepping frequency remains nearly constant and the stride length is proportional to the speed. In fact, the stride length is the distance of the jump19, s =

2vx vy , g

where the velocity vector at takeoff is

18. Muybridge’s work, financed by the railroad baron Leland Stanford, led eventually to Edison’s kinetiscope, thence to the motion picture, and all the consequences thereof. 19. The student will recognize this as the formula for the range of a projectile.

Physics of the Human Body Chapter 3:

33 Musculoskeletal system

vx v =   v  y and as usual x is the horizontal and y the vertical direction. If the height h of jump is a constant, then the stride length is proportional to the speed, and this explains Alexander’s results for running: →

1⁄2

1⁄2

s  h v = 2     ∝ ϕ0.5, l  l   gl  where h is the maximum height of the jump. (It may also explain the spread in Alexander’s results, since h ⁄ l is certainly not the same for all species.)

The upshot is that running power increses only as the first power of speed. Margaria, et al.20 have measured the energy cost of walking and running as a function of speed, using trained athletes as test subjects. Their data appear in the graph below. We 20

15

2

Run 10

Walk

5

0

5

10

15

20

Km / hr

Running power

Finally, we look at the power requirement of running. By definition, gvx ∆E P λ + λ−1 = = (1 − η)  4  m ∆t where λ =

vy . vx

Here η is the fraction of kinetic energy that can be stored in the stretching of tendons and restored as kinetic energy for the next stride. The minimum value of λ + λ−1 is 2, but to avoid excessive stress on takeoff and landing, most animals tend toward values of λ much smaller than unity. In retrospect it should not surprise us that running power may be written P = ζ mg vx where ζ is a dimensionless constant. Clearly the average force an animal exerts in running at constant speed must be proportional to its weight, mg; since power is force times speed, voila!

note that as we expect from our previous results, walking power increases (at least as fast) as v3 whereas running power is linear in v. Therefore, at some speed vcrit the power cost for walking will exceed that for running, and the animal will change gaits. From the graph I have extracted the empirical relation (in MKS units) P = 0.6 + 4.6 vx . m If we suppose 75% of the energy is stored and that the power is minimized, we find the coefficient of vx to be 0.125 g = 1.225 J/kg-m, which, when multiplied by 4 to convert to total energy usage, gives 4.9—close enough to the empirical value of 4.6. The constant term, 0.6, must be related to basal metabolism. If a 70 Kg person eats 2000 Kcal/day to maintain body weight, then this is about 1.4 watts/kg. It is close enough to the 0.6 W/kg in the above formula that we may conclude they are the same (especially since the latter figure is suspect).

20. R. Margaria, P. Cerretelli, P. Aghemo and G. Sassi, J. Appl. Physiol. 18 (1963) 367-370.

34

Chapter 3:

Other considerations

I stated that the power cost of walking grows at least as fast as v3; in fact, because the stride length in walking is fixed, and because the stepping frequency is limited by the maximum contraction frequency of the leg muscles, the chemical power requirement becomes infinite at some walking speed. A better way to say this is that the rate of energy consumption is inversely proportional to a factor that vanishes at a finite maximum strain rate of muscle tissue. The reason for this is that the chemical energy to make a myosin cross-bridge switch from one actin site to another is constant. Hence the chemical energy needed to contract a muscle a certain distance is independent of the strain rate. However, empirically, the mechanical power output is zero at zero strain rate and also at the maximum strain rate. This is just Hill’s Law discussed in §4 above. Hence to get the same mechanical power output when the strain rate is higher than optimum the body recruits more muscle fibers. This process ends (short of infinity!) when all the available muscle is in use.

Maximum speed

The transition from walking to running therefore is an economy measure, that reduces the rate of increase of power with speed. The upper limit on speed may be imposed by two factors: 1. The maximum power output the organism can sustain. Since 75-80% of the energy produced must be dissipated as heat, the strain on an exothermic animal’s cooling system can become severe. For example, assuming the power requirement to be similar to that of humans, an animal running at 5 m sec–1 generates power at the rate of 23 Watts Kg–1. If the heat is to be dissipated by water evaporation, the rate of evaporation will be about 0.008 gm sec–1. For a 70 Kg animal this amounts to water loss of 4 Kg in the course of a 42 Km run. It is thus not surprising that Phidippides (who bore the news of victory from Marathon to Athens in 490 BC) died at the end of his run, nor that modern

Transition to running

marathon runners carry liquid with them and drink as they run. 2. The maximum stress the animal’s tendons and ligaments can sustain. Each time an animal reverses its vertical direction it must subject itself to substantial vertical acceleration—several g’s, in fact. The harder the running surface, the greater the shock. (This is the origin of “shin splints”, wherein the tendons of the lower leg separate painfully from the bone.) A human sprinter has a top speed of about 23 mi/hr. The horse, giraffe, buffalo and rhinocerous can gallop at about 30 mi/hr. Cheetahs and gazelles are said to be capable of 60 mi/hr bursts of speed. To achieve such speeds an animal must be capable of leaping higher than the norm—and cheetahs and gazelles are indeed excellent leapers. They are also relatively light, and are built on a particularly springy plan, hence they are less liable to injure themselves on takeoff and landing. Whether it is stress or heat dissipation that most limits the top speeds of running mammals, however, is not definitely known. The cheetah’s ability to sustain a sprint extends to perhaps 100 or so meters. The effort clearly exhausts them, as much video footage of their hunts confirms.

Animals that cannot run

As we have seen, the time a running animal can be airborne is thang =

2vy g

The average acceleration an animal experiences in changing from landing to takeoff is of order 2g

hmax ∆h

=

v2y ∆h

=

v2y γl

where ∆h is the stopping distance, which we suppose to be a fraction γ of the leg-length. The maximum (momentary) acceleration humans can sustain without serious injury is about 5 g.

Physics of the Human Body Chapter 3:

35 Musculoskeletal system

Since strength-to-weight ratio scales as M ⁄3, an elephant’s maximum sustainable acceleration should be a bit over 1 g. That is, an elephant has no safety margin for anything as athletic as jumping. In consequence, elephants must not leave the ground during locomotion, consequently their ability to run may be primarily limited by their ability to withstand stress. −1

Of course this may also be a sort of chicken-andegg problem. Since the maximum height of a jump does not scale with body size, and since, empirically, the stepping frequency scales inversely as  l , an elephant or larger animal may not be able √ to run because it has literally “run out of time”—it cannot get its legs back and forth in the available time, so it does not try to. It walks—or rather, shambles21.

21. But don’t make an elephant angry—it can shamble along at 15 mi/hr, a lot faster than anyone but a trained runner can go!

36

Chapter 3:

Transition to running

Physics of the Human Body Chapter 4: Fluid mechanics and living organisms

37

Fluid mechanics and living organisms In this chapter we discuss the basic laws of fluid flow as they apply to life processes at various size scales. For example, fluid dynamics at low Reynolds’ number dominates the universe of unicellular organisms: bacteria and protozoa. Fluid flow—both uniform and pulsatile—is the dominant process by which both air and blood circulate in land animals, including humans.

1.

Euler’s equation

We consider the form of Newton’s second law appropriate to a fluid. The mass of a small volume of fluid is

In the absence of viscous forces the equation of motion is therefore →

 ∂v → →  → ρ  + (v ⋅∇)v  = − ∇p + ρ g .   ∂t This is sometimes known as Euler’s equation (L. Euler, 1755).

Example We first apply Euler’s equation to two problems in hydrostatics. In hydrostatics the fluid is not mov→ ing, hence v = 0 and we obtain →

−∇p + ρ g = 0 .

∆m = ρ ∆V ; its acceleration is therefore determined by the forces acting upon it, →

→

→  ∂v → →  dv ≡ ρ ∆V  + (v ⋅∇)v  = ∆F . ∆m dt  ∂t 

The forces are generally of two kinds: 1. external, long-range forces—gravity, electromagnetism; 2. internal forces—especially pressure and viscosity.

→ ∧ Defining g = −g z we easily see that

p(x,y,z;t) ≡ p(z) so that dp = −ρ g . dz We now have two cases to consider:

. 1. The fluid is incompressible (ρ = 0).

2. The fluid is a perfect gas: ρ =

Thus, e.g., if the fluid is in a gravitational field of

µ p, RT

→

local acceleration g , → → ∆F = ρ∆V g . On the other hand, consider a pressure gradient in—say—the x-direction, as shown to the right. Clearly the net x-component of force is ∂p   ∆V ; ∆Fx = p(x) − p(x+dx)A ≈ − ∂x  

p(x)

p(x+dx)

Fx = A p(x) − p(x+dx)  

38

Chapter 4:

where R = 8.3 J oK−1 gm−mol−1 is the gas constant, T the absolute temperature, and µ the grammolecular mass. In Case 1 we may easily integrate the equation to get p(z) = p(0) − ρgz z  = p0 1 −  z0   where, if p0 is 1 atmosphere (= 0.76 m ρHg g ) and the fluid is water, the height z0 in which the pressure changes by 1 atm is z0 =

ρHg ρH O

× 0.76 m ≈ 10.3 m ≈ 34 ft .

mass (where ρ is the mass density). If a fluid is flowing with a velocity vx in the x-direction (say), then in time dt we expect dN = n vx dt dy dz particles to be transported across an imaginary surface of area dA = dy dz . This leads to the concept of the flux vector →

that describes the transport of particles across the →

surface. The dimensionality of j is L−2 T−1, that is number per unit area per unit time. Correspondingly we can define the flux vector of mass transport (dimensionality M L−2 T−1) →

p(z) = p0 e−µ g z ⁄ RT . The scale height of the atmosphere is then z0 =

8.3 × 273 ≈ 8 km . 0.029 × 9.8

End of example

→

J = ρv .

In Case 2, we may write

or

→

j = nv

2

µg 1 dp = − RT p dz

Fluid mechanics and living organisms

Now imagine a box with edges dx, dy and dz in the res pe c tive directions. We erect a vector of un it length pointing outward from each face. The surface integral of any vector over the surface of this box is defined as the sum, over all six faces, of the component of the vector along the outward normal. Thus for the flux →

j we have1

2.

Conservation of fluid

A volume dV = dx dy dz of fluid contains dN = n dV molecules (where n is the number density) and dm = ρ dV

1.

∫∫

box

→

→

j ⋅ dS = −jx(x) + jx(x + dx) dy dz   + −jy(y) + jy(y + dy) dx dz  

+ −jz(z) + jz(z + dz) dx dy .   This represents the rate at which the particles flow out of the box, that is,

The notation jx(x) stands for jx(x,y,z,t); similarly, jx(x+dx) means jx(x+dx,y,z,t).

Physics of the Human Body Chapter 4: Fluid mechanics and living organisms

∫∫

→

→

j ⋅ dS = −

box

d ∂n dV . (n dx dy dz) = − dt ∂t

If we then compare both sides (by expanding the small differences −jx(x) + jx(x + dx) to first order) we have −

df

→ ∂n ∂ ∂ ∂ = jx + jy + jz = ∇ ⋅ j . ∂x ∂y ∂z ∂t

Note we have defined a new differential operator, the divergence of a vector, as →

→

df

d iv j ≡ ∇ ⋅ j =

∂ ∂ ∂ jx + jy + j . ∂x ∂y ∂z z

39

In real life there is no such thing as an incompressible fluid, or solid, for that matter. If one squeezes hard enough, anything is guaranteed to compress. However, when atoms or molecules are touching (or nearly so) as in the case of liquids and solids, the energy needed to decrease even slightly the volume of a given number of molecules is enormous. Hence for practical purposes, over a considerable range of pressure, we can neglect any change in density. When we assume the density is constant, we see that conservation implies that the divergence of the velocity vector vanishes, →

∇⋅ v = 0.

The equation of continuity, or equation of number conservation is therefore → ∂n + ∇⋅j = 0. ∂t

The additional force per unit volume of fluid then →

takes the form η∇2 v where η is the viscosity. Its dimensionality is ML−1T−1 . The Euler equation then takes the form (ignoring body forces such as gravitation) →

Suppose the quantity being conserved is something other than particle number—for example, energy, electric charge, mass, or even probability. The conservation laws for such quantities are identical in form, with the number density and number flux replaced with the appropriate density (energy, →

charge, mass, probability) and the flux vector j replaced by the corresponding current density.

 ∂v → →  → ρ  + (v ⋅∇)v  = −∇p + η∇2 v .   ∂t This is known as the Navier-Stokes equation. The physical meaning of viscosity can be understood in terms of the force exerted on a fixed plate by a parallel moving plate, if there is a layer of fluid in between, as shown below:

v

3.

Liquid

The Navier-Stokes equation

We now consider viscous fluids. Viscosity arises from the transfer of momentum between a fluid and a solid in relative motion, or between adjacent relatively moving layers of fluid. The derivation of the expression for internal forces resulting from viscosity in the most general case is somewhat advanced and would take us too far afield. Thus we give only the end result in the specialized (and simplified) case of an isotropic, incompressible liquid.

Force on plate

The velocity of the fluid has a uniform gradient in the vertical (y) direction, and the force acting on the fixed plate is Fx df ∂vx = η . Aplate ∂y

40

4.

Chapter 4:

Reynolds’ number

Consider an object moving through a fluid. A guide to whether viscosity is important is the ratio of inertial force to viscous force. The inertial force is just, in crude dimensional terms, ma = ρ L3

v2 = ρL2v2 . L

What about an organism the size of a bacterium? Here the length is 2 µ ≡ 2×10−6 m, the speed is 30 µ/s and ρ and η are as before. Reynolds’s number is thus 6×10−5 hence a germ’s life is dominated by viscosity. If a spherical bacterium moving at the above speed stops turning its propeller, we find3 1 dv 9η dv ⁄ dt ≡ = − , v dx 2ρa2

The viscous force is ηA

∂ux ∂y

= η L2

v = ηLv . L

the coasting distance is x =

Their (dimensionless) ratio is therefore df

ρav , R = η where a is a typical linear dimension. This ratio is called Reynolds’s number. If R is large, viscosity is a minor effect; whereas if R is much smaller than unity, viscosity dominates inertia.

Fluid mechanics and living organisms

2ρa2 v ≈ 0.1 Å . 9η 0

Since the size of an atom is about 1 Å = 10−8 cm , this is somewhat like our trying to move around by swimming through concrete.

Example Consider various objects moving in water. The viscosity of water is easiest to remember in cgs units: η = 0.01 gm/cm/sec. Also easy to remember for water is its density: ρ = 1 gm/cm3. A typical speed for a rowboat is 1-2 miles per hour, or 50–100 cm/sec. A rowboat is about 3 m long and 1 m wide, hence R ≈ 1 × (100 − 300) × (50 − 100) ⁄ 0.01 = (0.5 − 3.0) ×106 . That is, Reynolds’s number is enormously larger than 1 so viscosity is irrelevant2 for rowboats, submarines, porpoises, whales etc. 2.

3.

5.

Steady Flow in pipes

We now use the Navier-Stokes equation to study a case of obvious importance in the physics of the human body: flow of a viscous fluid through a long cylindrical pipe. First we tackle steady (time-independent) flow, then we shall examine pulsatile flow. Let us take the direction of flow to be the z-direction. In principle the velocity could have eddies and vortices, even if its average tendency is along the direction of flow. However, the conservation equation

In that case, one might well ask, what causes the resistance to propelling a boat through water? It turns out that a displacement hull continuously generates surface waves that carry away energy. A simple calculation shows that the power needed is proportional to Mv3/L where M is the ship’s mass, L its water line length, and v its speed. Here we have used Stokes’s law, combined with Newton’s second law and the formula for the volume of a sphere. See, e.g., E.M. Purcell, “Life at Low Reynolds Number”, American J. Physics 45 (1977) 3-11.

Physics of the Human Body Chapter 4: Fluid mechanics and living organisms

→

∇⋅v =

1 ∂ 1 ∂ ∂ vz + vϕ + rv  = 0 r ∂ϕ r ∂r  r ∂z

in cylindrical coordinates provides some extra information. First of all, the rotational symmetry of the pipe guarantees there is no ϕ dependence. Next, the flow is steady, hence the z-component → of v cannot vary along the pipe. But suppose the radial component, vr , is non-zero. From fluid conservation we find rvr = constant = A ; clearly, if A ≠ 0, vr → ∞ as r → 0 , that is, the velocity blows up at the center of the pipe. Since that is obviously nonsense, we must have A ≡ 0 and hence vr ≡ 0. There is no radial flow. Although there might still be a screw-like rotational flow along the pipe, the Navier-Stokes equation tells us that it has no effect in steady flow. That is, the term  ∂ 1 ∂ ∂→ + vz  v v ⋅ ∇ v ≡ vr + vϕ   r ∂r ∂ϕ ∂z   manifestly vanishes. →

→

Before attacking the Navier-Stokes equation per se, we first examine the balance of forces. From the definition of viscosity in terms of velocity gradients the force exerted on an annulus of fluid of length

41

 ∂vz(r+dr) ∂vz(r)  dFz = 2πLη (r+dr) − r  ∂r ∂r   ∂  ∂vz  = 2π L η r  dr ; ∂ r ∂ r  however the force driving the annulus is the pressure differential acting on the ends, dfz = ∆p 2π r dr . Setting the sum to zero (because the fluid is unaccelerated) we have ∆p λ ∂  ∂vz  r = − r. r  = − η ∂ r ∂ r  ηL Returning to the Navier-Stokes equation, we note that since the velocity has components only in the z-direction, ∧

→

v = vz(r) z , ∂vz   ∂vz ∂p 2 + vz . ρ  ≡ 0 = η∇ vz − ∂t ∂z ∂z   →

Since there is no radial component of v , we see ∂p =0, ∂r i.e., p = p(z). But it is also clear that ∇2 vz =

L

1 ∂  ∂vz  r  r ∂r  ∂r 

is a function of r alone, as

∂p is a function of z alone. ∂z

Therefore, 1 ∂p λ = constant = − , η η ∂z and so λ ∂  ∂vz  r  = − r, ∂r  ∂r  η L, lying between radii r and r+dr, by the fluid outside- and inside it is

r

∂vz

λ = −   r2 + A , ∂r  2η 

42

Chapter 4:

Fluid mechanics and living organisms

pressure differential to be like a voltage, then the Poiseuille resistance of a pipe is

λ vz = −   r2 + A log r + B .  4η 

Z = Manifestly, vz(r = 0) is finite, hence A=0. On the other hand, the velocity falls to zero at the wall of the pipe, or vz =

8ηL . π ρ R4

This notion is useful in studying how to optimize the branching of pipes in terms of the total power required.

λ  2 2 R −r .  4η 

Finally, since

7.

∂p = −λ , ∂z we can solve to find z z  p(z) = p0 − λz ≡ p0 1 −  + pL , L L  where L is the length of the pipe. The result of solving the Navier-Stokes equation is identical to what we got from setting the total force (viscous plus pressure) to zero. We can now calculate how much mass flows →

through the pipe per unit time. The flux is ρv ; integrating over the cross-section of the pipe we therefore have R

dM = 2π ∫ ρ vz(r) r dr dt 0

=

≡

2πλρ 4η

1 4 1 4  2 R − 4 R   

π(p0 − pL) ρR4 8ηL

.

This is known as Poiseuille’s Law.

6. Impedance By analogy with Ohm’s law we can define an impedance for fluid flow through a pipe. If we take the mass flow to be like electric current and the

Pulsatile flow in arteries

We now consider non-steady flow through a cylindrical tube. Assuming the flow is entirely in the z-direction and that the velocity depends only on radial distance and time, the Navier-Stokes equation →

 ∂v → →  → ρ  + (v ⋅∇)v  = −∇p + η∇2 v , ∂t   reduces to 1 ∂p 1 ∂  ∂v  ρ ∂v = − + . r r ∂r  ∂r  η ∂t η ∂z ∂p = Re −λ eiωt where λ is now Let us suppose   ∂z complex and the angular frequency ω is known. (That is, we analyze the flow for simple harmonic time dependence.) Then we also set v(r, t) = u(r) eiωt and obtain  i3ρω  1 d  du  −λ r + .  u =   r dr  dr  η  η  The solution is u(r) =

λ + A J0 (κ r) , iρω

where J0 is the regular Bessel function of zero’th order and 1⁄2

κ = −iρω ⁄ η  .   Since u(r=R) = 0 we may determine the constant A to get

Physics of the Human Body Chapter 4: Fluid mechanics and living organisms

u(r) =

43

J0(κ r)  λ  1 −  . iρω  J0(κ R) 

Thus vz(r) = Re u(r) eiωt   =

 M(s) λω  sinω t + θ(s) − θ(S) sin ω t −   η  M(S) 

where s = κr S = κR, and we have written the Kelvin function4 J0 x √−i  = M(x) eiθ(x) .   The fluid flow through the pipe can be obtained by integrating ρ v over the area of the pipe, as before. This is facilitated by the relation5 d xJ (x) = xJ0(x) . dx  1  This procedure leads to a complex dynamical impedance for the cylindrical pipe, analogous to the situation in AC circuits. To determine the flow for the actual time-dependence of the pressure (which is certainly not harmonic, but is periodic) we expand the pressure function in Fourier6 series. That is, if the pressure looks like the function shown above right, that is, a periodic function

φ(t) = Re ∑ λn e2πint ⁄ T  .    n →

→

Since we have neglected the v ⋅ ∇ v term, the   Navier-Stokes equation is linear. Hence the different Fourier components superpose, and we may write vz(r, t) = Re ∑ un(r) eiωnt   n  where each Fourier component un(r) of velocity is obtained from the previous formula for u(r)by re2π . placing λ with λn and ω with ωn = T The most important thing to note is that the velocity profile across the artery (shown on the next page) is no longer parabolic, but much flatter—thus the rate of transport is proportional to R2 rather than R4. This has a large effect on optimization of branching for minimum power usage.

−∇p = − φ(t) , φ(t + T) = φ(t) , we may expand in Fourier series

4. 5. 6.

See, e.g., Handbook of Mathematical Functions, ed. by M. Abramowitz and I. Stegun (National Bureau of Standards, Washington, DC, 1964), p. 379 ff. Ibid., p. 361. Consult any standard text on applied mathematics, such as Mathews and Walker, Mathematical Methods of Physics.

44

Chapter 4:

Fluid mechanics and living organisms

Comparing square and circular pipes of equal cross-sectional areas we find the central flow speeds in the ratio vL⁄2, L⁄2 !  v(r = 0)

≈ 0.93

"

and the rates of mass transport in the ratio (dm ⁄ dt)! (dm ⁄ dt)"

=

512 1 5 ∑ 2 2 π m, n odd (mn) (m + n2)

≈ 0.88 . It should not surprise us to find that for pipes of equal area, the circular cross section poses the least resistance. *for advanced students

Example* Another interesting case is a pipe of rectangular cross section. Here the equation of steady flow is ∂2vz ∂x

2

+

∂2vz ∂y

= −

2

λ η

whose solution may be written (take the origin in the x-y plane at a corner of the rectangle) vz =

∑

amn sin

m, n

mπx nπy sin . Lx Ly

For this solution the fluid flow rate is Lx Ly dm = ρ ∫ dx ∫ dy vz(x, y) dt 0 0

= ρ

4LxLy 2

π

∑

m, n odd

amn . mn

We use the Navier-Stokes equation, together with the orthogonality of the functions sin (2n+1)θ to   determine amn : amn

16 = 4 π mn

−1

 m2 n2  + .  2 2 L L x y  

Physics of the Human Body Chapter 5

45 Bioenergetics

Bioenergetics In Chapter 8 (Thermodynamics), we find the maximum theoretical efficiency of a heat engine, in turning heat to work1, to be ηCarnot =

T> − T< . T>

Since the upper temperature (body internal temperature) is about 310 oK and the exterior temperature is—in cold weather—about 273 oK, the absolute maximum efficiency is about 12%. Including the effects of heat leaks, friction and the need to have a reasonable power-to-weight ratio, the expected efficiency would be no better than half that (and worse on hot days!). However, from measurement of work output and food input, we discover the human being is about 25% efficient in converting food energy to work. The conclusion is inescapable: living organisms are not heat engines.

1.

Energy input to living organisms

How does a living organism get its energy? As with all other things biological, Nature provides a small machine (actually, a series of small machines). In this case it is an energy-storing molecule called adenosine tri-phosphate (ATP) that is constructed in the cellular energy factory (the mitochondrion, a tubular organelle present in all cells) using the energy from oxidation of sugar (glucose). The ATP

1. 2. 3.

molecule gives up one of its phosphate groups to a molecule that needs to be promoted to a higher energy state, thereby becoming adenosine bi-phosphate (ADP). About 11.5 kcal/mol can be provided by this maneuver, or about 0.5 eV per reaction2. The oxidation of food provides energy as shown in the table below:

Energy output from various foods Food Lipid (fat) Protein Alcohol Sugar Carbohydrate

Kcal/ gm(food) 9.3 4.0 7.1 3.8 4.1

L(CO2)/ Kcal/ gm(food) L(O2) 1.39 4.7 0.75 4.5 0.97 4.9 0.74 5.1 0.81 5.0

The interesting column is the energy ouput per liter of oxygen consumed: the numbers are almost constant. This is the reason why oxygen consumption can be used to measure basal metabolism, the rate of energy consumption needed to just sustain the processes of life, with no significant muscular or cognitive activity taking place3.

In principle stochiometric combustion of food with oxygen is 100% efficient in converting chemical energy to heat, so we neglect inefficiencies in the burners of heat engines. An electron volt is the energy gained by letting an electron move across a potential difference of one volt. Since the electron charge is 1.6×10-19 Coulombs, and since 1 Coulomb × 1 Volt = 1 Joule, 1eV is 1.6×1019 J. As we shall see subsequently, the brain consumes significant amounts of energy during periods of intense concentration.

46

2.

Physics of the Human Body Energy demand of organs

Energy demand of organs

We can easily estimate the power required to run the human heart. Basically we use the enthalpy, which for a moving fluid is h =

1 ρ v2 + p + ρgz . 2

The stream of blood leaving the heart has net pressure increment 60 mm (of Hg) or about 8×104 dyne/cm2. When blood returns to the heart through the vena cava, its pressure is is basically the diastolic (gauge) pressure of the body and the speed of flow is roughly the same as that of the exiting blood. The height is the same so there is no change in potential energy. Therefore the amount of work done per unit time must be P = ∆p

dV . dt

The volume of the ventricle is about 100 cm3 so assuming 60 beats per minute (that is, one contraction per second) we get P ≈ 8×104 × 100 ≈ 107erg ⁄ sec = 1 Watt . Larger hearts, faster beats or higher blood pressures can double this. If we multiply by 4 or 5 to account for the inefficiency of converting chemical energy to work we find about 5-7 Watts or 90-120 Calories per day is needed to keep the heart pumping. This is only a few percent of our basal metabolic power demand.

What about the energy demand of kidneys? The thermodynamic potential of a concentration difference across a permeable membrane4 is  c2  ∆E = RT ln    c1 

4.

so that if the rate of transfer of material (in moles dν , the power needed is per unit time) is dt P =

 c2  dν RT ln   . dt  c1 

The logarithm of concentration ratio is a number of order 1; the average flow through the kidneys is 125 cm3/sec, or about 1/8 of a liter of fluid per second. Assuming the material that must be either held back (or eliminated into the bladder) is about 0.1 molar in concentration, we get dν ≈ 0.0125 gm−mol ⁄ sec . dt Thus the average power demand is about 30 W, or about 660 Cal/day. That is, the energy required to operate our kidneys is a non-trivial fraction of the total energy budget of the human body. Suppose we apply a different method for estimating energy requirements, based on the measured blood flow to various organs in the resting human being.

That is, the energy needed to maintain a concentration gradient against the process of diffusion in the opposite direction.

Physics of the Human Body Chapter 5

47 Bioenergetics

If we assume the energy requirements of each organ system in the table below is proportional to the blood volume that circulates through it, then we find the results shown in the third column of the table:

Flow* (ml/min)

Metabolism (Kcal/day)

Liver, intestines, spleen Kidneys Brain Heart Skeletal muscles Skin

1400

480

1100 750 250 1200 500

380 260 90 410 170

Other organs Total

600 5800

210 2000

System

*

taken from McDonald’s Blood Flow in Arteries, 4th ed. (Oxford U. Press, Inc. New York, 1988) Table 2.4, p. 36.

We see that our estimate for the heart (based on work output) is very close to that based on blood flow through the coronary arteries feeding the heart muscle. Our estimate for the kidneys was somewhat high, but considering that the concentration ratio c2 ⁄ c1 and the absolute concentration of dissolved substances were only “guess-timates”, the agreement seems remarkably good. The energy demand of the resting brain, 12 W or 260 Kcal/day seems rather small. Measurements of

5. 6. 7.

blood flow to various parts of the brain5, as well as of its heat output, yield comparable numbers for energy demand. However, when the brain is most active, as during periods of intense concentration on difficult problems (for example, the homework in this course), its demand rises 20-fold.

3.

Heating and cooling

There are three mechanisms of heat loss by the body: a) radiation; b) conduction; c)convection/evaporation. Bodies radiate and absorb electromagnetic energy (black-body radiation) according to the StefanBoltzmann law, P = σ A T4 where A is the radiating area, T the absolute temperature (in oK), and the Stefan-Boltzmann constant σ is6 σ =

2π5 k4 15 h3c2

= 5.7×10−8 Watts ⁄ m2 ⁄ oK4 . A perfect (“black”) radiator can be shown theoretically to be a perfect absorber7. Thus if an object of area A and temperature T2 is situated with an enclosure whose temperature is T1, it will radiate energy at a rate

N.A. Lasssen, D.H. Ingvar and E. Skinhøj, “Brain function and blood flow”, Scientific American, Oct. 1978, p. 62. One of the early triumphs of quantum mechanics was the calculation of σ from first principles, in terms of the speed of light, c, Boltzmann’s constant k and Planck’s constant h. Because this derivation is important, it is repeated in the Appendix to this chapter. Otherwise the Second Law of Thermodynamics will be violated.

48

Physics of the Human Body Heating and cooling

Pout = σAT42 and absorb at the rate Pin = σAT41 . Since the surface area of a typical human body is 1.5 m2, the net rate of radiation loss from a naked body at surface temperature 37 oC, into a room at 70 oF (21 oC) is

The constant κ is the thermal conductivity, often given in calories per centimeter per unit time, per degree Celsius. For fat the value in these units is 5×10−4, so assuming the thickness of the fatty layer is 1 cm and the area is 1.5 m2 as before, we find that a naked body loses heat by conduction in a room at 70 oF, at the rate

P ≈ 5×10−4 × 1.5×104 Pnet = σA T42 − T41  

= 500 Watts .

= 5.7×10 × 1.5 × 310 − 294   −8

4

4

. . ≈ 150 Watts It is therefore not surprising that we can feel a “chill” radiating from a cold surface such as an open refrigerator or a large window on a cold day. For example, suppose the surface is at 0 oC instead of room temperature. The additional radiated power is ∆P ≈ 5.7×10−8 ×1.5 × 2944 − 2734   = 164 W . In this case our net radiation rate more than doubles (because the thermal radiation absorption from our surroundings is greatly decreased) which we interpret as a sudden chill. What about thermal conduction across the fatty layer under the skin? Heat is conducted one-dimensionally through a uniform material according to Newton’s law of heat conductivity jx = −κ

8. 9.

∂T . ∂x

cal 310 − 294 × o 1 cm− K

That is, conduction is much more important than radiation as a mechanism by which we lose heat. Finally, if the air is moving we can lose heat a lot faster, by convection. This is why even a mild breeze leads to a substantial “wind chill” on a cold day. And of course any form of evaporation (perspiring) removes large amounts of heat because of th e large h eat of vaporization of water (540 cal/gm). Note that if we were unclothed, we would need to eat about 13,000 Cal daily just to maintain our core temperature against radiation and conduction into a rather warm environment. Most animals avoid this problem by maintaining some form of insulation (blubber, feathers or fur). Our closest hominid relatives (chimpanzees) have considerable body fur, despite living in a much warmer (tropical or sub-tropical) climate, to provide insulation against the night chill8. The absence of human body hair, by and large9, points to the fact that because of our adaptive strategies—clothing and fire—we have not needed fur for at least a half-million years.

They also huddle together for warmth. There are wide variations in pilosity (hairiness), known from antiquity: “My brother Esau is an hairy man, but I am a smooth man.”

Physics of the Human Body Chapter 5 Appendix–Planck’s blackbody radiation law

49

Appendix–Planck’s blackbody radiation law Blackbody radiation is by definition the electromagnetic radiation emitted by a perfectly “black” object—a perfect absorber of radiation (and consequently, by the Second Law of thermodynamics, a perfect emitter)—at a definite temperature T. A very good approximation to a blackbody is a container with a small hole in it, as shown below:

Quantum mechanics began with Max Planck’s successful derivation of the spectrum of this radiation by means of one remarkable new idea: the quantum hypothesis. The object of this Appendix is to give the highlights of Planck’s results, in modern language.

Electromagnetic radiation in free space is described by two of Maxwell’s equations10: →

1 ∂B ∇×E = − c ∂t →

→

1 ∂E ∇×B = . c ∂t →

If the container is a good thermal conductor the heat source will maintain it at a constant temperature; what comes out of the hole is radiation with a spectrum like that shown below:

Taking the curl of the first and the time derivative of the second, and using the identity →

→

→

∇ × ∇ × E  ≡ ∇ ∇ ⋅ E  − ∇2 E     →

we find (in the absence of charge, ∇ ⋅ E = 0 ) →

→ 1 ∂2E − ∇2 E = 0 ; 2 2 c ∂t

that is, the electric vector satisfies a wave equation. Suppose we now try plane wave solutions of the form →

→

→ →

E ( x, t) = E (t) eik ⋅ x ; →

→

if we choose the wave vector k suitably, this solution to the wave equation can be made to satisfy appropriate boundary conditions11 at the sides of

10. We use Gaussian units rather than SI to exhibit explicitly the speed of light, c. 11. —for example, periodic boundary conditions.

50

Physics of the Human Body Heating and cooling

a box. That is, we have the solutions appropriate to a box of electromagnetic radiation.

This means the total energy in a finite box is infinite.

Now with this form of the solution, the wave equation reduces to

What about the frequency spectrum of energy in the box? If we integrate over angles only, and note that

→

→ d2E (t) + k2c2 E (t) = 0 2 dt

k2 dk =

which is the equation for a simple harmonic oscillator of angular frequency

ω = kc .

we find that if the harmonic oscillators behave classically, their energy density in the frequency range [ν, ν + dν] is df

In classical physics, a harmonic oscillator can have any energy at all. As we have seen in our discussion of Brownian motion, a (damped) classical harmonic oscillator in thermodynamic equilibrium with a heat bath at temperature T has average energy kT. Thus to find the energy of electromagnetic radiation in a box of volume

Ω = Lx Ly Lz we need to count the number of oscillators and multiply by the average energy of each: E = Nosc kT , where ∞

∞

Nosc = 2 ∑

nx = −∞

∑

ny = −∞

∞

∑

→ 2

nz = −∞

Ω 8π3

∫ d3k .

That is, for each mode there are two polarizations, the wave numbers appropriate to periodic boundary conditions are kx =

2πnx , etc. Lx

and we take the limit as the dimensions of the box become large (which permits replacing the sum by an integral). 12. …due to Rayleigh, among others. 13. W. Wien, Annalen der Physik 58 (1896) 662.

1 2 (2π)3 2 ω dω ≡ ν dν c3 c3

dU =

8π kBT 2 dE dλ = ν dν ≡ 8π kBT 4 . 3 Ω λ c

However, while the blackbody radiation spectrum looks like this formula12 at long wavelengths, at short wavelengths it resembles the observed spectrum not at all. At short wavelengths, as Planck knew, the radiation spectrum is well fit by the formula13 dU = α e−β ⁄ λT

dλ λ5

where α and β are constants. He therefore guessed a formula that had the correct short- and long wavelength behavior—and that, incidentally, agreed very well with all the data. Planck’s formula was dU =

8πh ν3 dν c3 ehν ⁄ kBT − 1

where h is a new constant of nature, now called Planck’s constant. If Planck had left matters there, he still would be famous for proposing the correct formula describing blackbody radiation. However, he felt impelled to find some kind of underlying reason for the formula to work.

Physics of the Human Body Chapter 5 Appendix–Planck’s blackbody radiation law

So he worked backward. The number of oscillators per unit volume, per unit frequency, was certainly as given above, dn =

8π kBT 3

c

ν2 dν .

Thus what must be wrong was the equipartition assumption, that gave an oscillator at temperature T the average energy 〈ε〉 = kBT, independent of frequency. Planck’s formula clearly indicates that the average energy depends on frequency; 〈εν〉 =

hν hν ⁄ kBT

e

−1

;

we can rewrite this as ehν ⁄ kBT − 1 hν  ≡  〈εν〉 = hν ⁄ k T 2 e B −1 ehν ⁄ kBT − 1   hν

∞

∑ nhν e−nβhν ≡

n=0

∞

∑

e−nβhν

n=0

where we have set β = 1 ⁄ kBT . This is just the Boltzmann formula for the average energy of a system at temperature T, where the possible energies do not take on continuous values but rather are restricted to a discrete set of values

51

Planck also realized immediately several implications of his formula: first, it permits us to calculate the the intensity of emitted radiation from a blackbody—that is, to obtain the Stefan Boltzmann constant, in terms of other quantities. The total energy density of radiation in a box is U =

∞

∫0 dU

=

=

8πh c3

∞

ν3 dν

∫0 eβhν − 1

∞ 3 4 x dx 8πh k T .  ∫ B 3   0 ex − 1 c

The integral can be evaluated in closed form, for example by expanding, integrating term-by-term, and comparing with the Reimann zeta function; or else by contour integration techniques. The result is π4 ⁄ 15 . To get the power radiated per unit area, convert the energy density to a flux (pretending it is a fluid): →

v. j = U v ≡ Uc ^ →

Averaging the normal component over a halfspace gives the net radiated flux as →

→

〈 j 〉 = U v ≡ Uc ⁄ 4 so the result for the Stefan-Boltzmann constant is σS−B =

2π5 hk4B 15c2

.

ε(n) ν = nhν, n = 0, 1, 2, … That is, the energies are restricted to integer multiples of a fundamental energy, hν. This hypothesis, central to the blackbody radiation formula, was called the quantum hypothesis14 by Planck.

14. …from the Latin word, meaning “how much”.

Planck also realized that fitting his formula to the blackbody radiation measurements yields both h a nd kB ; from the latter one can calculate Avogadro’s number from the formula NA =

R kB

52

Physics of the Human Body Heating and cooling

as well as the charge on the electron (F is the Faraday, about 2.9×1014 esu, or 96,500 Coulombs, per gram-mol): e =

F . NA

Physics of the Human Body Chapter 6 Some ideas from thermodynamics

53

Some ideas from thermodynamics The branch of physics called thermodynamics deals with the laws governing the behavior of physical systems at finite temperature. These laws are phenomenological, in the sense that they represent empirical relations between various observed quantities. These quantities include pressure, temperature, density and other measurable parameters that define the state of a given system. Thus, in discussing magnetic systems, e.g., the variables might include temperature, magnetization and external magnetic field; in discussing elastic systems we might want to use temperature, stress and strain as defining variables. Some useful references for the student who wants to pursue the subject further are 1. F.W. Sears, Thermodynamics, 2nd ed. (AddisonWesley Publishing Co., Reading, MA, 1963). 2. K. Huang, Statistical Mechanics (John Wiley and Sons, Inc., New York, 1963). 3. R.P. Feynman, Statistical Mechanics (W.A. Benjamin, Inc., Reading, MA, 1972).

1.

The First Law of Thermodynamics

We imagine a system such as a gas, where temperature, pressure and density are the appropriate system-defining variables. Then conservation of energy states that the following differential relationship must hold: dU = dQ − dW

(1)

where dQ represents (a small amount of) heat flowing into the system from some heat source; dW the mechanical work done by the system, and dU the change in internal energy. The quantity U depends, by assumption, only on the present state of the system and not on the history of how the system got into that state. In

that case, dU must be, in the mathematical sense, an exact differential. To understand what this means, consider the work done by sliding a heavy block up an inclined plane. There is a force acting downhill (gravity) as well as friction. Obviously the work done traversing the wiggly path up the hill is greater than that along the straight path, since friction always acts opposite to the direction of motion. That is, the part of the work arising from overcoming gravity is independent of path (because gravity is a conservative force) but that used to overcome friction is path-dependent (because friction is a dissipative force). In other words, the work output depends on how the job is done, so dW is not an exact differential. Similarly, the heat input depends on the details of how the heat enters the system—for example, it can depend on the rate at which heat is supplied— so dQ is also not an exact differential. But experience indicates that the internal energy of a system does not depend on how it got there, hence we postulate that dU is exact. Now suppose this postulate were wrong: then conclusions we can deduce from it would at some point disagree with experience. So far no such cases have arisen; moreover, the microscopic theory of thermodynamics (that is, statistical mechanics) suggests strongly that dU is exact.

54

2.

Physics of the Human Body Equation of state

Equation of state

We can use different sets of variables to define the state of a thermodynamic system. For example we can use temperature and pressure, pressure and volume, or temperature and volume. The reason we have only two independent variables is that the three variables p, V, T are related by an equation of state which takes the form f(p, V, T) = 0 .

relationships. That is, imagine it depends on pressure and volume: U ≡ U(p, V) so that  ∂U   ∂U  dU =   dp +   dV .  ∂p V  ∂V p

(5)

(2)

For example one typical equation of state is the perfect gas law, pV = νRT

(3)

where R is the perfect gas constant (numerical value ≈ 8.3 J ⁄ oK ⁄ gm−mol) and ν is the number of gram-moles present. Another well-known equation of state is the Van der Waals equation, p + an2 1 − nV0 = n kBT (4)    where n is the number-density of the molecules, a is a constant representing an average (attractive) inter-molecular interaction energy, and V0 is proportional to the volume of a molecule (that is, if there are N molecules the available volume is V − N V0 rather than V). Also kB = R ⁄ NA is Boltzmann’s constant. The Van der Waals equation of state represents an attempt to take into account two physical effects that cause actual gases to deviate from the behavior of an ideal gas: intermolecular forces, and the finite size of molecules. Although not a precise representation of any gas (except possibly over a limited range of the variables) Eq. (4) is useful in understanding qualitatively the behavior of gases in phase transitions.

Now imagine that we go from internal energy U1 = U(p1, V1) to U2 = U(p2, V2) via two different paths as shown above. Since the result is independent of path, it must be true that

∫

Γ

dU =

∫

Γ

f(p, V) dp + g(p, V) dV = 0 ,  

where the integral is taken around the closed path Γ. Now, by noting that if the result is true for a large path it must also be true for the small square path shown below, we find −g(p + dp)dV −f(V)dp

f(V+dV)dp g(p)dV

3.

Thermodynamic relations

Because the internal energy differential dU is (by hypothesis) perfect, we immediately derive certain

Physics of the Human Body Chapter 6 Some ideas from thermodynamics

 ∆Q   ∂H  cp =   ≡    ∆T p  ∂T V

g(p,V) − g(p+dp, V) dV +   or

+ f(p, V+dV) − f(p, V) dp = 0  

55

(7)

where the enthalpy H is defined as H = U + pV . If a thermally isolated ideal gas is allowed to expand slowly into a vacuum, as shown below,

∂ f ∂ g  ∂V − ∂p  dp dV = 0 ,   or finally ∂  ∂U  ∂  ∂U  = .   ∂p  ∂V  ∂V  ∂p  V p In other words, for dU to be an exact differential the order of its second derivatives with respect to p and V is immaterial. (This result is just Stokes’s Theorem in two dimensions.) Since the the work output can be written dW = pdV , we can express the heat input into a system in the alternate forms   ∂U   ∂U  dQ =   dp + p +   ∂V  ∂p V   p

  dV (5a) 

  ∂V   ∂U   dQ = p   +    dT + ∂T  ∂T     p p

(5b)

  ∂V   ∂U   + p   +    dp ∂p  ∂p     T T  ∂U    ∂U   dQ =   dT + p +    dV . (5c) ∂V  ∂T V   T  The specific heat at constant volume is defined by  ∆Q   ∂U  cV =   ≡   ,  ∆T V  ∂T V

(6)

where the last relation follows from letting dV=0 above. The specific heat at constant pressure is defined as

Joule’s free-expansion experiment it is found that the temperature does not change. On the other hand the volume may change substantially. Since the system is thermally isolated the change in heat must be zero. On the other hand, no work is done in the expansion since no force is exerted on anything. Therefore we deduce that the internal energy of an ideal gas must be unchanged in the experiment; that is, the internal energy of an ideal gas depends neither on volume nor on pressure, but solely on temperature. From the above experimental result we further deduce that the internal energy of the ideal gas has the form U = cV T + constant .

(8)

Since the origin of the energy scale is arbitrary, we may set the constant to zero (that is, it is unmeasureable).

56

Physics of the Human Body Heat engines and Carnot efficiency

For an ideal gas, since pV = νRT, we see the enthalpy is also a function of temperature only, H = U + pV ≡ cV + R T ;   that is, the molar specific heats cp and cV are related by cp = cV + R .

4.

Heat engines and Carnot efficiency

The first law of thermodynamics is really nothing more than the law of conservation of energy, taking into account that heat is a form of energy. The key difference between heat and other forms of energy, however, becomes manifest in the process of conversion from one form of energy to another. Electrical energy—in the form of charge on a capacitor or a current in a wire—can in principle1 be converted 100% into mechanical energy (work), chemical energy (charging a battery, or electrolysis, e.g.), electromagnetic energy (radio, light), or heat (radiant heater, oven, etc.). For that reason we say that electricity is “high grade” energy. Similarly chemical energy can be converted into electrical energy (battery, fuel cell), or into mechanical work, with high efficiency. Heat is different. If we want to convert heat into mechanical work or electricity using engines or engine-driven generators, the fundamental nature of heat imposes a limitation on the maximum efficiency that can be achieved with the most perfect machine conceivable. This limitation was discovered by the French engineer Sadi Carnot. Carnot’s argument was based on a heat engine using a perfect gas as its working fluid. The first part of the argument is the assumption that perpetual motion, in the form of an engine that cre-

1.

ates more energy than it consumes, is impossible. Thus if we consider heat engines, we are entitled to imagine one that has no heat leaks, friction or other ways in which energy can be used unproductively. The most perfect kind of heat engine is one that begins in a certain state (pressure, volume), takes heat Q1 from a source at temperature T1, produces some work W ≤ Q1 , and if there is any leftover heat, df

Q2 = Q1 − W , that has not been converted to work, dumps that into a heat sink at temperature T2 . Then the system is restored to its original state and the engine is in a position to repeat these actions. The key to getting the most work possible for a given heat input is to make all parts of the cycle reversible. That is, when heat is moved from the source to the engine, it is done with the engine at the same temperature as the source, and so slowly that one can reverse the engine and put the heat back with no loss or change of state. This is illustrated below. In the first step, a cylinder of gas at temperature T1 is placed in contact with

the source and slowly expanded, extracting heat

Practical considerations—friction, resistance, etc.—restrict the achievable conversion efficiencies to less than 100%, of course.

Physics of the Human Body Chapter 6 Some ideas from thermodynamics

and doing work. The cylinder is then removed from the source and insulated so no heat can flow in or out. The piston is further expanded, doing more work and allowing the temperature to fall to T2 . The cylinder is then placed in contact with the sink and the gas compressed so that heat flows into the sink. (Work has to be put in in order to compress the gas.) Finally, the cylinder is removed from the sink and insulated so no heat flows in or out, and the gas compressed still further until the temperature has come back to T1 . The gas is now back in its original state. How do we know the system returns to its original state at the end of a cycle? The answer is that we must show it is possible to construct a cycle that does precisely this. If the working fluid is a perfect gas we may find relations for the two types of expansion (or compression). When the cylinder is in contact with the heat reservoir, we say a change of volume is isothermal. When it is isolated, the change of volume is adiabatic. In mathematical terms, isothermal means Carnot cycle Isothermal Adiabatic

Pressure

1

2

3

4

Volume

dT = 0 whereas adiabatic means dQ = 0 . In an isothermal process (we assume we are working with 1 mole of gas) pV = RT = constant , whereas in an adiabatic process,

57

dU ≡ cV dT = −p dV . Applying the perfect gas equation of state we have cV dT = −R dT + V dp or with dT =

1 d(pV) R

we have (η − 1) V dp + η p dV = 0

(9)

or cp dV dp + = 0, cV V p and −cp ⁄ cV

V p = p0   V  0 Since df

γ =

.

(10)

cp > 1, cV

the curve of pressure vs. volume is steeper than that of an isotherm passing through the same point. Thus on the diagram to the left we see the initial isothermal expansion (from point 1 to point 2) lies on a curve that falls more slowly than the subsequent adiabatic expansion (from point 2 to point 3). Similarly, when the gas is compressed isothermally (point 3 to point 4) the pressure rises less steeply than in the subsequent adiabatic compression (4 to 1) that returns to the initial point. The possibility of returning to the initial state then depends on being able to return to the initial volume and pressure (point 1 on the diagram). The equation of state guarantees that the temperature will then be the initial temperature, T1 . Moreover, since the internal energy of an ideal gas depends only on temperature, U will also have returned to its initial value. A solution that looks like the figure is possible if the compression ratio λ satisfies

58

Physics of the Human Body Heat engines and Carnot efficiency

1 ⁄ (γ − 1)

Vmax  T1  λ = >   Vmin  T2  Then it is easy to see that

.

(11)

 T2  V3 = λVmin    T1  and

η =

1 ⁄ (γ − 1)

Vmin < V3 < Vmax and Vmin < V1 < Vmax . What about the net work done by the engine in this cycle? It is clearly the area contained within the intersecting curves, W =

1

2

∫2 p dV + ∫3 p dV + ∫4 p dV + ∫1 p dV 1 ⁄ (γ − 1)

. = RT1 − T2 ln λ T2 ⁄ T1        It is worth noting that the areas under the adiabatic parts of the curves, W34 =

4

∫3 p dV

and W12 =

.  

Hence the Carnot efficiency is

 T1  V1 = Vmin    T2  so that

4

1 ⁄ (γ − 1)

df

Q1 = W23 = RT 1 ln λ T2 ⁄ T1   

1 ⁄ (γ − 1)

3

Now to get the efficiency we need to evaluate the energy extracted from the heat source. This is just the area

2

∫1 p dV

cancel exactly. This should not be surprising since on these segments of the cycle, ∆Q = 0, and therefore W34 = cV T1 − T2   and W12 = cV T2 − T1 ≡ −W34 .  

T1 − T2 ∆W = . T1 Q1

Carnot’s proof that this is the maximum possible efficiency of any heat engine is based on the assumption that a machine that creates work from nothing (that is, a perpetuum mobile of the first kind) is impossible. Suppose some other kind of engine could do better than a Carnot cycle engine that uses an ideal gas. Then we could use it to produce energy, and run the Carnot engine backward to dump heat back into the heat source. That is, if engine X can produce work ∆WX > ∆WCarnot from the same heat input Q1, then running the Carnot engine backwards uses work ∆WCarnot to put the heat Q1 back, leaving the source and the sink in their initial conditions, but with a net output of work, ∆WX − ∆WCarnot > 0 , having been produced. This argument of Carnot’s is so general and clever that it has been applied in many other circumstances. For example, Einstein used it to show that all forms of energy content must contribute to the weight of an object according to g ∆m = g

∆E c2

—for if one form of energy did not contribute to weight, one could in principle use pulleys, ropes and an energy converter to make a perpetual motion machine.

Physics of the Human Body Chapter 6 Some ideas from thermodynamics

5.

The Second Law of Thermodynamics

Carnot’s discussion of the efficiency of heat engines led naturally to the concept of entropy, although it was not until the end of the 19th Century that Boltzmann was able to give a physical interpretation of entropy in terms of degree of order and probability. Now in the Carnot engine, we note that the heat taken out of the reservoir at T1 is Q1 = W23

1 ⁄ (γ − 1)

= RT 1 ln λ T2 ⁄ T1   

  

and that dumped in the sink at temperature T2 is 1 ⁄ (γ − 1)

Q2 = W41 = RT 2 ln λ T2 ⁄ T1   

;  

that is, Q2 Q1 = . T1 T2

(12)

Like many results in physics, Eq. 12 is more general than the arguments used to derive it. If we imagine a lot of heat reservoirs (at different temperatures) we could add up all the heats extracted from them, divided by their temperatures, to obtain the integral df

S =

∫

dQ . T

2. 3.

dQ T

(14)

is sometimes called the second law of thermodynamics. One of its chief consequences is that in a closed system the entropy always increases or remains the same.

6. Statistical Mechanics Thermodynamics is somewhat of a misnomer—it deals with equilibrium relations rather than timedependent ones !2 .Thetheory consistsoftwolaws: the First Law, which is just conservation of energy; and the Second Law, which posits a new thermodynamic function, entropy, that (in a closed system) always increases or remains the same. The remainder of thermodynamics consists of empirical relations (equation of state, specific heat) and relationships that can be derived from them.

Statistical mechanics is an attempt to derive the laws of thermodynamics from a microscopic description at the atomic level. That is, historically statistical mechanics accepted the reality of atoms and molecules, even though its early practitioners did not know how exactly big atoms were, and had no means by which to observe them directly!3 .

(13)

It turns out that S is independent of the path the system takes in getting from state 1 to state 2, hence it is another thermodynamic function like the internal energy, U, that depends only on the state of the system. This new thermodynamic function is called entropy. The differential relation

dS =

59

Boyle’s Law

Daniel Bernoulli seems to have been the first to use the statistical approach to obtain a practical result, Boyle’s Law, which states that at constant temperature the product of pressure and volume of a gas is constant: pV = constant .

(15)

…that is, it should really be called thermostatics. Of course technological progress has somewhat altered this—we can now isolate and work with individual atoms, as well as visualize the atomic granularity of the surfaces of solids.

60

Physics of the Human Body Statistical Mechanics

N

N

i=1

i=1

m m 1 v2x (i) ≡ N ⋅ ∑ v2x (i) Fx = ∑ L L N Nm 2 〈vx 〉 , L

=

(19)

where 〈v2x 〉 is the mean-square velocity component in the x-direction. But there is nothing special about one direction or another, so we may write 〈v2x 〉 ≡ His (vastly oversimplified) picture of a gas consisted of a box of point masses whizzing around randomly, and colliding elastically with the walls of the box. The particles do not interact with each other. We shall imagine a box in the form of a rectangular parallelepiped of length L and crosssectional area A, as shown below!4 . A single molecule collides with the face at the right end of the box (say, lying in the y-z plane) and reverses the x component of its velocity. The momentum transferred to the wall is thus ∆px = 2mvx .

(16)

The time between such collisions is 2L ∆t = vx

(17)

1 2 1 〈vx + v2y + v2z 〉 = 〈v2〉 . 3 3

Now, finally, the force per unit area (that is, the pressure) acting on the rightmost face is df

p =

Fx N 2 m 2 N2 = 〈v 〉 ≡ 〈ε〉 . (20) A LA 3 2 V 3

Not only does Eq. 20 express Boyle’s Law, it even tells us (once we extend Boyle’s Law to the ideal gas law) how the average kinetic energy of a molecule is related to the absolute temperature: RT =

N2 〈ε〉 . ν 3

(21)

If we take one mole of molecules, then ν = 1 and N = NA so we have 〈ε〉 =

df

3 R 3 T = kBT . 2 NA 2

(22)

so the average force exerted by this molecule is mv2x = . fx = L ∆T ∆px

(18)

If there are N such molecules whizzing about, the total force they exert on the rightmost face of the box is

4.

Maxwell-Boltzmann distribution

When we study matter from the atomic viewpoint we can ask more detailed questions than those permitted by thermodynamics. For example we might ask what fraction of molecules of a perfect gas, at temperature T, have velocities between v and v + dv .

The shape of the container is immaterial. A derivation for a spherical container may be found in G.B. Benedek and F.M.H. Villars, Physics with Illustrative Examples from Medicine and Biology: Statistical Physics (Springer Verlag, New York, 2000) pp. 250-252.

Physics of the Human Body Chapter 6 Some ideas from thermodynamics

James Clerk Maxwell first derived the probability distribution of molecular velocities in a dilute gas in thermal equilibrium. He argued that if the temperature is constant, then the probability that a molecule has x-component of velocity in the range [vx , vx + dvx] must be the same function as the probability that it has y-component in the range [vy , vy + dvy]. That is, the motions in orthogonal directions (and their corresponding probabilities) are statistically independent. The total probability function therefore has the form5 → pv d3v = f(vx) f(vy) f(vz) dvx dvy dvz . (23)   →

Moreover, p( v ) can depend only on the squared → → → magnitude, |v|2 ≡ v ⋅ v , of the velocity, since it must be6 invariant under rotations and reflections of the coordinates. Thus we have p(v2x + v2y + v2z ) = f(v2x ) f(v2y ) f(v2z ) . Now suppose we rotate the coordinates so that v2y = v2z = 0 ; 2

2

p(v ) = A f(v ) , or ln f(v2x + v2y + v2z ) + 2 ln A = ln f(v2x ) + ln f(v2y ) + ln f(v2z ) . It is easy to show that the only function for which this is true is f(u) = −α u + Γ where α is an undetermined constant and Γ =

1 2

ln A2

is a normalization constant that we determine from the condition that the total probability that a particle has some velocity is 1. Therefore the velocity distribution function is 2

p(v2) = A3 e−α v . To determine the constant α we need to calculate the average kinetic energy, since the kinetic theory of gases, as we just saw, relates the average kinetic energy of a molecule to temperature via 1 2

〈 mv2 〉 =

3 k T; 2 B

therefore, since df

〈v2〉 =

2

∫ d3v v2 e−αv ∫ d3v e−αv

6.

=

3 , 2α

we find 1 2

α = m ⁄ kBT ,

p(v2) = A3 e−ε ⁄ kBT ,

(24)

where ε is the kinetic energy.

Ludwig Boltzmann noticed that Maxwell’s velocity distribution has the form (24), and saw that the law for the variation of gas density with altitude in a uniform gravitational field, ρ(z) = ρ(0) e−mgz ⁄ kBT

(25)

was the same law, with kinetic energy replaced with potential energy. He therefore proposed a probability distribution law governing the energies of a complicated system: p(E) = g e−E ⁄ kT ,

5.

2

or in other words,

then it becomes clear that (let A = f(0) ) 2

61

(26)

The sparseness of the Latin alphabet leads us to use the symbol p to mean both probability, as in this section; and pressure, as elsewhere. …that is, by symmetry. This is an illustration of the enormous power of symmetry in theoretical physics.

62

Physics of the Human Body Applications of Maxwell-Boltzmann distributions

of which Maxwell’s result was a special case.

7.

Applications of Maxwell-Boltzmann distributions

We now consider three applications of the distribution law, Eq. 26, each with biological implications.

Liquid-vapor equilibrium

The first concerns the equilibrium of a liquid and a gas in contact, say water and water vapor. We note the well known fact that energy is required to convert a molecule of water from the liquid to the gaseous phase. This is basically the binding energy of water molecules to the body of liquid. In thermodynamics it is called the chemical potential. In general the calculation of chemical potentials from first principles—that is, from a knowledge of the microscopic forces acting between water molecules, and using the laws of quantum mechanics— is a difficult task that has only been performed for some simple systems. So we shall assume it is a given (slowly varying) function of temperature and pressure. We call this energy µ.

Reaction rates

A second application of Boltzmann’s Law concerns activation and reaction rates. Suppose that before two molecules can undergo a chemical reaction (that ultimately releases energy, to be sure!) they have to approach each other within a certain distance. If there is a repulsive potential between

Then according to Boltzmann, the probability for a molecule to be in a state of energy µ relative to the liquid state—that is to be gas rathor than liquid—is proportional to e−µ ⁄ kBT and we may write the partial pressure of vapor molecules as pvap = p0 e−µ ⁄ kBT where p0 is a constant that, like µ, could in principle be calculated from fundamental considerations, but has not been because of the difficulty of the calculation. That is, we regard µ and p0 as empirical fitting parameters. If we take the logarithm of the vapor pressure and plot it against 1 ⁄ T we should obtain a straight line of negative slope, −µ ⁄ kBT ; as the figure at the above right shows, this is indeed what we find.

them, as shown below, their motion must surmount the barrier before they can get close enough to react. The height of the barrier is the “activation energy” ∆E; not surprisingly, the fraction of molecules with energies equal to or greater than this energy varies like f ≈ e − ∆E ⁄ kBT.

Physics of the Human Body Chapter 6 Some ideas from thermodynamics

Thus we expect the reaction rate to increase very rapidly with temperature, when kBT