SECTION ONE -MATERIAL TRANSPORT

  ...spontaneous “events of physical and chiefly of chemical nature are going on continuously which aim at a balance of the existing potentials of energy. Since life requires the continuation of these potentials of energy, work must be performed continuously for the prevention or reversion of these spontaneous changes.”

                                                                                                   (Otto Meyerhof, 1924)

 

I. INTRODUCTION.

All physiological functions involve some form of transport of matter - into, out of, or from one place to another within the organism. Two modalities of such transport are found: 1) one in which the organism expends metabolically derived energy (performs work) to accomplish the transport, as when the contractions of the cardiac muscle transport blood from the veins to the arteries at a substantially higher pressure, and 2) one in which the energy dissipated is the potential energy accumulated in part of the organism as a consequence of its work, as when the blood spontaneously flows through the blood vessels from the large arteries, where it is at high pressure, to the veins, at lower pressure. This is identical to the behavior of a car on a hill. The car will roll down the hill spontaneously. Using the energy provided by the oxidation of the fuel, its engine is capable of doing work on the car by driving it uphill. If the car has zero velocity when reaching the top, therefore having zero kinetic energy, the potential energy thusly gained by the car is derived from the work done by the engine.

It is part of our experience that whenever a system has acquired some potential energy, it has also acquired the tendency to lose it by having some of its parts move towards a position such that its potential energy is reduced. In fact, the presence of potential energy is recognized by the existence of such tendency. The car on top of the hill will roll down unless some constraint prevents it from doing that. So, any process that reduces the potential energy of a system is spontaneous and will happen unless actively prevented; any process that increases the potential energy of a system is non-spontaneous and will not happen unless special conditions apply. It is possible to drive a non-spontaneous process by coupling it with a second process during which the loss of potential energy of the system is larger than its gain in the first. The engine will drive the car uphill because the fuel loses more potential energy during oxidation than the car gains in being brought to the top of the hill. The balance shows up as heat (and kinetic energy, if its velocity is not zero upon reaching the top.. Another exceptional condition is found when the body has a velocity in a direction in which it gains potential energy by spending its kinetic energy. For example, when a ball is thrown upwards.

Section One-Material Transport deals with blood flow in the circulatory system, air flow in the respiratory system, and diffusional flow across plasma membranes. In this Introduction we will review a few basic ideas in mechanics and develop some central concepts that later become essential to the analysis of these different phenomena as particular cases of a single general kind of process. The selection of those three process not only reflects their obvious physiological importance, but also the fact that those are the ones that more completely  lend themselves to this type of analysis

a. Newton’s Second Law

Dynamics is the part of mechanics that studies the relationships between movement and its causes. Newton’s second law of mechanics, the familiar F =M · a, establishes the foundations of dynamics by stating that the acceleration imparted to a free body (a = F/M) is proportional to the force applied to it, and the constant of proportionality is the inverse of a property of the body called its mass. Indeed, this law causally links movement to a  force by defining force as the reason for any change in the state of rest or movement of a body.

Consequently, if a free body is at rest with respect to a set of coordinates as a frame of reference and a force is applied to it, the body will experience the acceleration  a = F / M in the direction of the force and its velocity in the same direction will constantly increase as long as the force remains applied. Since the initial velocity with respect to the selected frame of reference is zero, the velocity at any time t will be  v = a · t and the distance traveled s = ½ a · t2 .This is the situation encountered by a space vehicle far enough from all celestial bodies as to be essentially free from gravitational attractions while maintaining a constant position with respect to some sidereal frame of reference. If a rocket in the vehicle is fired and kept on for some time, the force generated by the rocket will accelerate the vehicle. The resultant velocity will continuously increase until the rocket is turned off, at which point the velocity will have acquired a value that will be maintained  indefinitely.

 

A. MOVEMENT IN A VISCOUS MEDIUM

The increase in kinetic energy of the above ship , Ek = ½M · v2   where v is the final velocity, will be equal to the work done by the applied force  W = F · s. (PROBLEM 1: Using the relationships given in Ia., show that E k = W when the potential energy remains constant). When the sum of the kinetic and potential energies of the body being considered (the ship in this case) is constant, i.e., when the  mechanical energy is conserved, the force involved is said to be conservative. The gravitational force is an example of  conservative forces as will be shown later.

a. Gravitation

Gravitation is the property of mass that requires that any two masses will exert on each other  an attractive force. The intensity of this force is given by  

             EQUATION   I. 1

where FG is the gravitational force on the body (its weight) in newtons (N), ME is the mass of body E in kilograms, MB  is the mass of body B, d  is the distance between the centers of mass of the two bodies in meters, and k is a proportionality constant with units N·m2·kg-2. If body E is the Earth, ME is a constant, and the distance d is also a constant for objects placed at the same height from the surface at one location. Therefore, the attractive force will be directly proportional to the mass of the object, i.e., , where K = (k · ME) /d2. (PROBLEM 2: Is the gravitational force stronger in South Florida than in Nepal?  If you used a typical doctor’s scale to measure your weight, would the result of the measurements in South Florida and in Nepal be equal?)   For objects that are sufficiently close to the surface, their distances to the center of mass of the planet are essentially the same, since their differences are insignificant compared to the total distances. Strictly speaking, though, the intensity of the gravitational force varies for the same object depending of where on Earth it is placed, since the Earth is not a perfect sphere, the altitude of different points on its surface will affect its distance from the center, and the vertical component of the centripetal pseudo-force varies with latitude. When comparing the gravitational force on different objects located at the same place on Earth, it is useful to characterize the place by the value of the constant K, with units of N·kg-1, which is in fact the value of the intensity of the force experienced by a mass of 1 kg located in the same place and, simultaneously, the acceleration imposed on any free mass at that place and distance from the center of the Earth. In other words, , where is the acceleration due to gravity at a particular distance from the center of mass of the Earth and at a given latitude. Its value for most points on the surface is approximately 9.8 m/sec2. (Confirm that the units of K and those of g are the same). This is, therefore, a particular case of Newton’s second law when applied to objects exposed to the gravitational force of the Earth, when  F = M · a = M · g. (PROBLEM 3:  Demonstrate that if a cannon ball and a marble are dropped simultaneously from the same height on the surface of the Moon, they will hit the ground at exactly the same time. Who discovered this?).

b. The Gravitational Force is Conservative

Consider a body A of mass MA  located at a distance D-1 from the surface of the Earth. This distance is  small enough to place the body under the predominant influence of the Earth’s  gravitational field. The gravitational force acting on it is    FG = g A MA.

In order to maintain that distance, the body must be under the influence of another force FO equal to and directly opposed to FG. Let FO be the weight of another object B of mass MB = MA  placed initially at a distance d-1 from the surface and connected to the first body by an inextensible cord over a frictionless pulley (see FIG I.1).   If, by reducing the mass of B by an infinitesimal amount, FO is diminished by an also infinitesimal amount *F, the body will experience a very small net force FN = FG-(FO-*F)=*F.  The infinitesimal acceleration imposed by this Text Box: infinitesimal force will in time generate a very small velocity and A will move down as B moves up. Let the movement proceed until the distances are D.2 and d.2, respectively. Since the cord is inextensible,     D.2 - D.1 = - (d.2 - d.1) .  The potential energy change for A is   )UA = FG (D.2 - D.1),   whereas the potential energy change for B is    )UB = FO (d.2 - d.1), since FO has been reduced by only a negligible amount. Clearly, )UA = - )UB or, in words, the potential energy lost by A is equal to the potential energy gained by B and the total potential energy of the system remains constant. In this situation the velocity acquired by the bodies is so small that their kinetic energy is negligible. So, the mechanical energy has been conserved.

Imagine now that a force is applied to a body on the surface of the Moon. The body will simultaneously be under the influence of two forces: the applied force and the Moon’s gravitational force. If the resultant movement of the body brings it to a higher altitude, its potential energy will have increased in proportion to the increase in altitude as indicated above. This increase in potential energy is equal to the work done by the force against gravity, or, in equivalent terms, to the negative work done by the gravitational force as the body gains altitude. If now the applied force is removed, gravitation will decelerate any remaining upward motion and the body will then reach maximal altitude and start to fall down with increasing velocity. Its kinetic energy at the time it hits the ground,  EK = ½ Mv2,  will be equal to the work done by the gravitational force on the object as its mass is accelerated, and, therefore, equal to the change in its potential energy. This was shown in PROBLEM 1. Keep in mind that work is energy in transit. When the gravitational force does work on the object by accelerating it, the system’s mechanical energy remains constant because potential energy is quantitatively being transformed into kinetic energy, i.e., the process is conservative and gravitational force is a conservative force. When the object hits the ground, all the kinetic energy is lost and transferred to the surroundings in forms that will eventually decay to heat.

Notice that, at the beginning of this paragraph, the direction of the movement of the object was not specified. It could have been straight up, following a slanted path or any conceivable trajectory. Still, the work done by the gravitational force, and consequently the change in potential energy of the system, depends only on the change in vertical distance to the surface because this is the direction of the force. This is a property of conservative forces. (PROBLEM 4: If a bullet with a mass of 10 g is fired from the ground in a vertical direction at a velocity of 980 m. s-1, calculate how high it will go, what potential energy it will gain and what will be its velocity and kinetic energy when it hits the ground. Repeat the calculations for the case in which the bullet is fired at an angle of 45o with the horizontal. Ignore the effects of the Earth’s rotation and atmosphere.)

The reason why the  example just analyzed refers to a process conducted on the Moon instead of the Earth is the Earth’s atmosphere, which would exert an influence similar to that of friction in the pulley of  FIG. I.1.

c. Friction and Viscosity

When the surface of one body slides over that of another, each body exerts a frictional force on the other. When one body is held in place, the direction of the frictional force acting on the moving body is opposite to that of the force that drives the movement. Consequently, the frictional force opposes the movement by reducing the net force acting on the moving body.

Viscosity is internal friction in a fluid. The internal friction generated by the movement of a solid body immersed in a fluid results in a viscous force given by Stokes’ law as being proportional to the viscosity of the fluid and to the velocity of the body relative to the fluid:  Ffr = k · v · 0, and directed opposite to the movement.  If a metallic sphere is deposited at the top of a tall cylinder full of motor oil, the gravitational force will accelerate it. As the downward velocity of the sphere increases, the viscous  force, with an upward direction, also increases. The net force acting on the sphere is the sum of the gravitational force, which is constant, and the viscous force, which is increasing from an initial value of zero at rest. As the downward velocity increases, the intensity of the upward viscous force grows, the net force is reduced and the downward acceleration becomes smaller. Soon the net force and the acceleration become zero and the velocity remains constant until the sphere hits the bottom of the cylinder.  Symbolically,

FG + Ffr = 0 = FG - k·v ·0.             EQUATION   I.2

Since the final velocity of the sphere does not reach the value it would have in the absence of the fluid, the kinetic energy acquired is also less. Then, the potential energy lost by virtue of having lost altitude in a gravitational field is larger than the kinetic energy gained. In the example of the body descending and ascending at the Moon presented in I.A.b., the potential energy was fully transformed into kinetic energy and vice versa because there was no viscous fluid present. If the body was moving upward under the influence of some undefined force, the gravitational force was doing negative work as the height increased, and that negative work was being stored as potential energy. When the body is allowed to fall, that same amount of work (now positive) is done by the gravitational force in accelerating the body and imparting it with a kinetic energy equal in magnitude to the lost potential energy. In the presence of a viscous fluid, part of the work done by the gravitational force on a falling body is equal to the negative work done by the viscous force, which, unlike the negative work done by the gravitational force, cannot be stored as potential energy and will be transformed into heat. The viscous force is, therefore, not conservative.

 

B. DRIVE TOWARD EQUILIBRIUM

At the beginning of this Introduction it was stated that any process by which the potential energy of a mechanical system is reduced will happen spontaneously. A mass in a gravitational field is said to have potential energy associated with its position because, if allowed, it will spontaneously move toward the position at which that tendency to move disappears: the position of equilibrium. In classical mechanics, the  tendency to restore equilibrium is represented by a force responsible for the movement towards equilibrium, force whose intensity is related to the magnitude of the potential energy by   F = - grad U,   where U is potential energy and grad is gradient.

a. Gradient

When the magnitude of a variable changes in space, the magnitude and direction of its steepest slope with respect to the three spatial coordinates is its gradient. This may be greatly simplified in many cases by choosing the direction of the x coordinate along the direction of the gradient. Then, using the above expression for the force

            EQUATION   I. 3

(Forces, distances, gradients, velocities and accelerations are vectors, but we will continue using here a scalar notation.)

Text Box: In FIG I.2, a sphere has been placed on the top left hand corner of an inclined plane. The plane is placed in a system of cartesian coordinates such that A is the height h of the plane taken along the y axis and the distance B is taken along the x axis. When the sphere is released, it will roll straight down following the continuous arrow, not the diagonal dashed line, and the force F responsible for that movement will have, therefore, the same direction. If we chose the direction of x along the diagonal D, dU/dx  would be  proportional to the slope of the dashed line, which is equal to A/D. On the other hand, if we chose x to have the direction of B, dU/dx would be proportional to the slope of the continuous arrow, which is equal to A/B and has a larger absolute value than A/D. (Notice that we say ‘absolute value’ because both slopes are negative). Since the continuous line is the steepest, its slope is the gradient of height and, when multiplied by FG will yield the gradient of gravitational potential energy. If we had two inclined planes with the same height but different angle, the sphere will roll faster down the one with the steepest angle. This means that the acceleration and, therefore the force, are also greater, which indicates a larger potential energy gradient. 

The change in potential energy of the sphere due to the gravitational field in descending the distance h as it rolls down the plane is )U = -FG·h,  and   )U/)x = - )(FG · h)/)x  describes the change in potential energy as the sphere descends while increasing the value of its x coordinate.  If the distance h is much smaller than the distance to the center of mass of the Earth, the change in gravitational force with h (and, therefore, with x, since the path down the plane is a function of h vs. x) is insignificant, so,   )U/)x = - FG · )h/)x.   Then, using the notation of calculus,

     EQUATION  I. 4

Text Box: FIG I.3 shows a plot of U vs. x. The continuous line applies when x is taken in the direction of the gradient and the dashed line applies when x is taken along the diagonal D in FIG I.2. These are straight lines because, as mentioned before, if FG is considered constant U is proportional to height and, in the inclined plane, the height is proportional to x. If FG is allowed to increase with x, the curves would be convex. The figure also shows that the force is positive (points in the same direction as x) while the slope of U(x) is negative, since U (and h) decrease as x increases. So, the slope, and therefore the gradient, have a negative value while the force has a positive value. This requires the negative sign in  F = - dU/dx and in       F = - FG · dh/dx.

Text Box: The force F responsible for the movement of the sphere down the plane must be a consequence of the gravitational force FG. In FIG I.4, FG is decomposed into F and a force normal to the inclined plane, which is canceled by the reaction from the plane. That leaves the sphere under the influence of only F.  Because the triangle formed by A, B and C, and the triangle formed by F and FG are similar,   F/ FG = A/C ,   which means that the force that drives the movement down the plane is smaller than the gravitational force by the same factor as the vertical distance A is smaller than the actual path C. An important derived result is that   F · C = FG · A , which means that the work done by F along the distance C is equal to the work done by FG down the distance A. Therefore, the potential energy lost by the sphere in going down along C is equal to the potential energy lost falling down vertically or, in fact, following any other path that takes it to lose height by an amount equal to A. (PROBLEM 5: Justify the mechanical advantage of the inclined plane for lifting heavy weights to a certain height. Is the energy spent diminished by its use?)

b. Potential and Field

We have seen that the change in potential energy of a mass in a gravitational field is given by  )U = -FG · )h and that the gravitational force is FG = k m1 m2 / r2. Here, values of )h are very small increments of r. The force can also be expressed as FG = (k m1 / r2 ) · m2 , where the quantity in parentheses is force per unit mass, with units newtons·Kg-1. This is an intensive quantity E, named field, that may be used to characterize different points in a region of space in terms of the intensity (and direction) of the gravitational force experienced by a one Kg mass placed at those locations. Alternatively, the field may be seen as a function of position in space, or point function. Then, the force experienced by a body is the product of the field at its position times its mass: FP = EP ·M. This is equivalent to the substitution done in Equation 1 by which  F = K ·M, where, later, K was substituted by g. Now, we are substituting it by E. All of these are just different names for the same thing, as may be easily shown by dimensional analysis.

In a similar manner, another property may be defined by expressing  )U = - {(k M1 / r2) · )h} ·M2. The quantity between the curly braces is the potential energy per unit mass, with units of joules · Kg-1. This is also an intensive quantity A, named potential, that may be used to characterize a point in space in terms of the potential energy of a one Kg mass placed at that location. Then, the potential energy of a body at a given location is the product of the value of the potential at that position times the body’s mass. The same relationship that exists between potential energy and force exists between potential and  field:  )A = - E · )h, or, using the notation of calculus, d A /dh = - E. In an inclined plane, where h is a function of x,  d A /dx = (d A / dh) · (d h /dx) = - E · (d h/dx), which are similar to Equation I.3.    (PROBLEM 6: A mass of 10 Kg is moved a distance of 50 m from point A (10 m above sea level) to point B (20 m above sea level) in a region where g = 9.8m · s-2.  What is the intensity of the gravitational field in that region, what is the difference in potential between A and B, and what is the gain in potential energy of the 10 Kg mass?).

The concepts of potential and field, as those of energy and force, are general and applicable to any system in which potential energy is stored, be it electrical, hydraulic, gravitational or chemical. Their relationships may be better grasped by observing the following diagram.

                            

 

Text Box: c. Movement in a viscous medium: Flux

In I. A. c., the example of a metallic sphere falling through a viscous fluid was used to discuss movement in a viscous medium and the role of dissipative frictional forces. It was established that the net force acting on the sphere eventually becomes zero as the opposing frictional force grows with its velocity. Symbolically,   FG + Ffr = 0 = FG - k · v · 0   (Equation I.2) .

 

Imagine a uniform suspension of many small metal spheres with a small mass, say 1 gram, in a viscous fluid (see FIG. I.5). The tube that contains the suspension is placed in a gravitational field so that its axis is parallel to the direction of the field. The spheres will start flowing downward and soon acquire a constant velocity. 

In the last section it was explained that any field, including the gravitational field, is equal to the negative gradient of the potential. Then, if the x axis is aligned with the direction of the field,  the gravitational force may be expressed as    - (d A / dx) ·M, which leads to

- (d A / dx) ·M - k · v · 0 = 0

when a constant velocity is reached. Solving for v,

                v =  -M / (k0) · (d A /dx),               EQUATION  I. 5

where v is the constant velocity acquired by each sphere. If the spheres did not have the exact same weight or if their interactions with the surrounding fluid differed, it would still be possible to define an average velocity which could be used in Equation I.5.

Planes A and B are normal to the axis of the cylinder and have an area of S m2. Then, the number n of particles that cross one of them, say B, in one second per unit cross sectional area of the plane is called the flux J of those particles: J = n @ S . Let us select A and B so that their distance is equal to the distance d traveled by the particles in one second (their velocity). Then, the number of particles contained in the volume defined by A and B will cross B in one second. This volume is equal to the area S times the distance d,  the number of particles is  S @ d @ c,  and J is equal to the velocity, given in meters per second, times the concentration of particles in the fluid, given as number of particles per cubic meter:   J = S @ v  @ c / S = v @c.  The units of J are  m · sec-1 · m-3 = sec-1 · m-2 , since n is a pure number. This definition of flux is quite general and flexible. It can be easily adjusted to express the flow of different materials by changing the definition of concentration from number of particles per unit volume to any other unit that consists of a fixed number of particles, as a mole, for example. In this case, the concentration could be expressed in moles per cubic meter and the flux would be given in moles per second per square meter. If the material that is flowing is an incompressible fluid, and not particles suspended or dissolved in it, the notion of concentration loses significance. The particles that flow are the fluid molecules themselves and the number of such particles is determined by the volume. Then,  the flux is the unit of volume  per second per unit area.  In the case now under consideration of the suspension of metallic spheres, an appropriate unit would be the kilogram, since mass is so intimately related to the nature of the forces acting in this system. Then, the concentration is given in kilograms per cubic meter and the flux in kilograms per second per square meter. So,

 J = v · c (m · sec-1 · Kg · m-3)

And, by substituting the value of v from Equation I.5

J  =  - M · (k0)-1 · d A /dx · M · vol-1     EQUATION  I. 6

The units of k and 0 are, respectively,  meters and N·sec·m-2. (PROBLEM 7: Check that the units of 0 are N·sec·m-2 if k is measured in meters. Also, show that the dimension of M·(k0)-1 is velocity per unit field).

The velocity may be expressed in terms of a property of the system, the mobility   u = v / E or velocity per unit field, which is a function of properties of the particles and of the fluid in which the particles move. This mobility is, of course, equal to M·(k0)-1 in Equations I.5 and I.6, since    v = u · E, and E = - d A/ dx. Making the appropriate substitution,

           J = - u · E · c = - u · (d A / dx) · c          EQUATION  I.6.a    

which is a general flux equation derived using the principles of the theory of irreversible processes. All processes involving friction must be irreversible, since frictional forces are not conservative. As just mentioned, the factor u = M/(k0) has the units of velocity per unit field, which suggests a conductance of the medium, while its inverse k0/M represents the medium’s resistance to movement. The derivative of the potential with respect to distance, which in this case refers to height, is the strength of the gravitational field. As mentioned earlier, it must be preceded by a negative sign because the directions of the field and of the potential gradient are opposite.

 

C.  FORCES OTHER THAN GRAVITATIONAL

Gravitation is a fact of life that affects many of our functions either by facilitating or obstructing them. In fact, many are the examples of structures and processes in our bodies that clearly have evolved in response to gravity, and many functions happen somewhat differently in the horizontal position and while standing. Furthermore, there are significant physiological adaptations that take place in near zero gravity, as studies on astronauts have evidenced. Still, of the many modes of transport of matter that happen in association with physiological functions, none is exclusively driven by gravitational forces. The reason for having initiated our discussion of general principles using gravitation as the example is that it seemed more intuitively familiar.

Those physiological transports that lend themselves to a neat physical treatment may be separated in two groups: convective flows and diffusional flows, and the forces responsible for the transport are originated by  potential energy gradients other than gravitational. We will address the following processes: blood flow in the circulatory system, air flow in the respiratory system, diffusion of electrically neutral molecules across the plasma membrane, and diffusion of ions also across the plasma membrane. All will be treated as passive (spontaneous) flows that entail the loss of some form of potential energy, that are driven by a force given by the negative gradient of that potential energy, and that involve the generation of a form of frictional force that opposes the movement. These flows are all described by equations similar to EQ. I. 6.a   with the adjustments required to account for the particular ways in which we represent the corresponding potential energies. Attention will also be given to the active processes by which metabolic energy is transformed into those forms of potential energy that drive the spontaneous material transports.

 

 

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