Classical 1

Thisdocumentis based on Chapter 1 of the Foundations of for Chemists, by G.A.D. Ritchie and D.S. Sivia, published by the Oxford University Press (2000) in the Oxford Chemistry Primers series (vol. 93).

1.1 Introduction

The Oxford English Dictionary describes Physics as ‘the scientific study of the properties and interactions of matter and ’. An important aspect of this study is a desire to understand a wide range of disparate phenomena in terms (1642–1727) became Lucasian Pro- of a few basic rules or ‘laws of nature’. One of the first people to achieve a fessor at Cambridge at the age of 27, and major success in this endeavour was Sir , who was able to explain stayed at the university for 30 years before why apples fell to the ground and how orbited the sun within a single becoming Master of the Royal Mint. While his contributions to Mathematics and Physics of the universal law of gravitation. We begin our discussion of Physics, were far-reaching, his greatest interests were therefore, with Newton’s on and ; this topics is usually alchemy and the occult. called classical or Newtonian mechanics.

1.2 Newton's laws of motion

1.2.1 The first law A body remains at rest, or moves at a constant in a straight line, unless acted upon by a . The first law provides a definition of force: that which causes an object to change its speed or direction of motion. In lay terms, a push or a pull. It’s not too difficult to imagine that a force exerted along the line of motion will result in a change of speed, whereas a correspondingnudge in a perpendic- ular direction will alter the course of travel. This dependence on the orientation of the force tells us that it is a vector quantity, being specified by both a mag- nitude and a direction; by contrast, is a scalar quantity, having only the former attribute of ‘size’ (or how heavy or something is). The statement of the first law can be made more succinct by rephrasing it in terms of a uni- form , where velocity is a vector pertaining to speed (a scalar) in a given direction. Explicitly, 10 metres per (m/s or ms−1) is a speed but 10ms−1 due east is a velocity. We will not go into a detailed discussion of vectors, or any other topic in , in this text (referring the reader instead to volumes 77 and 82, by Sivia and Rawlings, in the Oxford Chemistry Primers series), but will try to give a brief reminder at opportune . 2 Classical mechanics

Magnitudes, directions and basic vector algebra A vector, F say, is represented graphically by an arrow where the length of the line corresponds to its magnitude, |F|, and the pointed-end indicates the direction. Alge- braically, vector quantities are often differentiated from scalar ones through the use of a bold font. If the vector lies in the x–y plane (so that it can easily be drawn on a piece of paper), then it can be decomposed, or resolved, into two components that are perpendicular to each other. Using the x (or horizontal) and y (or vertical) direc- tions as our reference, we can express F as (Fx , Fy) where Fx and Fy are scalars; specifically, if F points at an θ (anti-clockwise) with respect to the x-axis, then elementary trigonometry shows that

Fx = |F| cos θ and Fy = |F| sin θ .

Using the identity sin2θ + cos2θ ≡ 1, or Pythagoras’ theorem, it follows that the square of the magnitude of F is given by

2 2 2 |F| = Fx + Fy .

A=B ⇔ Ax =Bx ,Ay =By , ... Two vectors are said to be equal if they have the same magnitude and direction; this is equivalent to requiring each of the corresponding components to be equal. Vectors are added together graphically by sequentially placing the flat-end of one arrow representation at the location of another’s apex, so that the sum is the resultant from start to finish; algebraically, this amounts to separately adding- up the various components of the contributing vectors. Subtraction is just a special case of , with the negative vector having the same magnitude but opposite direction as its positive counterpart. Algebraically, then,

C = A ± B ⇒ Cx = Ax ± Bx , Cy = Ay ± By , ...

The multiplication and division of a vector by a scalar is straightforward, resulting in only a change of magnitude but not direction (to within a sign reversal, if the scalar is negative). In terms of components,

C = µA ⇒ Cx = µAx , Cy = µAy , ...

The product of two vector is more involved, and we defer its discussion to appropriate points in this chapter. Division by a vector is not defined, however, and this operation must never appear in our calculations.

Newton’s first law of motion can be paraphrased by saying that ‘a force is required to change an object’s velocity’. A stationary body is a special case when the magnitude of the velocity, or speed, happens to be zero. Indeed, its study forms a separate branch of mechanics called (as opposed to ); it plays a central role in working out the strength of the materials required to prevent a bridge from collapsing, or the frictional roughness of a surfaced required to stop a ladder from slipping, and so on. According to the first law, of course, the net force on an object must be zero if it is to remain stationary; this forms the basis of the topic of statics. 1.2 Newton’s laws of motion 3

1.2.2 The second law The force acting on a body is directly proportional to its rate of change of . The second law is a quantitative statement about the strength, and influence, of a force. To appreciate it, however, we must first understand what is meant by momentum. In everyday parlance, we sometimes talk about a movement developing a momentum of its own; the implication being that it is a trend that is becoming increasingly difficult to argue with or stop. A simple physical picture that Exercise 1.1 Given that the mechan- conjures-up the same sense is that of a car, or a truck, careering down a hill ical wavelength, λ , associated with a parti- due to brake failure. It’s not hard to imagine that the effort required to control cle having a momentum of magnitude p is the run-away vehicle will be greater the faster it’s going or the heavier it is; λ = h/p, where h is Planck’s constant, what is the speed of a neutron used in consequently, the definition of the momentum of an object as the product of experiments to study structure on an atomic its mass and velocity does not seem unreasonable. Like the velocity v, and in length scale (i.e. λ ≈ 0.1 nm)? contrast to the mass m, the momentum p is also a vector:

p = mv , (1.1) so that the direction of the (linear) momentum is along the (instantaneous) line of motion. Returning to Newton’s second law of motion, we can now state this mathe- matically as d F (mv) , (1.2) ∝ dt where F is the applied force and d/d t is the differential meaning d dv dm ‘the rate of change, with respect to (t), of’. If the mass of the object re- (mv)= m + v dt dt dt mains constant during the motion, so that dm/dt =0, then eqn (1.2) simplifies slightly to the more familiar form

F = ma , (1.3) where dv dvx dvy a = , (1.4) ax = , ay = , ... dt dt dt the rate of change of velocity with time, is the . We should note that the proportionality of eqn (1.2) has been replaced by an equality in going to eqn (1.3) by assuming that the force, mass and velocity are given in suit- Exercise 1.2 Is an moving in a cir- able (SI) units: newtons (N), kilograms (kg) and metres per second (ms−1) cular at a constant speed accelerating? respectively. Early on in our school life, we usually learn Newton’s second law of motion as ‘force = mass acceleration’. The discussion above shows that while this is generally true, we× must remember that force and acceleration are vectors. As such, eqn (1.3) yields separate relationships for each of its components: dv dv F = m x , F = m y , ... (1.5) x dt y dt We should also bear in mind the proviso of an , and add a term v dm/dt to the right-hand side of eqn (1.3) if dm/dt =0 (as for the case of a rocket where fuel burning is significant). 6 4 Classical mechanics

Newton’s second law of motion tells us that, for the same applied force, the magnitude of the acceleration experienced by a body is inversely proportional to its mass. That’s why motorcycleswith modestly powerful engines can easily overtake cars, or why racing cars are made from the lightest (but structurally strong) materials available. Incidentally, the change in the momentum of an object is called the and, by integration of eqn (1.2), is readily shown to be equal to Fdt :

R t2 t2 v1 = v(t1) Fdt = [mv ] = m(v2 v1) , (1.6) t1 − tZ1 where we have assumed that the mass does not change in the writing the term on the far right-hand side. If the force itself happens to be constant, then eqn (1.6) reduces to ‘force time = mass change-of-velocity’. To return a tennis ball so that it approaches× our opponent× at very high speed, for example, we need either a brief but powerfulcontactwith the racket or a less vigorous stroke that maintains contact between the racket-head and ball for a longer time.

1.2.3 The third law To every there is an equal and opposite . The third law is a statement about the nature of forces, in that they mediate a mutual interaction between two objects. If the earth is held in its orbit around the sun by the latter’s gravitational attraction pulling it towards the centre of the solar system, then the sun experi- ences a (gravitational) pull of equal magnitude towards the earth; the influence of this force on the sun is much smaller, of course, because of the enormous difference in their (a factor of more than one hundred thousand). Sim- ilarly, the forward propulsion of a bullet fired from a rifle is accompanied by a backwards recoil of the gun.

1.3 Newton's second law in action

To see how Newton’s laws of motion can be to used to understand the dynam- ical behaviour of the world around us in a quantitative manner, let us consider a of examples. The first is quaint, but very pedagogical, and involves the of a . The second entails oscillatory behaviour, and is of greater scientific relevance.

1.3.1 The trajectory of a cannon ball Suppose that a shell is fired out to sea from the battlements of a coastal fort that are at a height H metres above sea-level. If u is the speed of the projectile on leaving the gun-barrel, which is inclined at an angle θ to the horizontal, how far will it go before it hits the water? To deal with this problem we first need to set up a , or ref- erence framework, within which to do the calculation. Following convention, 1.3 Newton’s second law in action 5

Exercise 1.3 If the neutron in Exercise 1.1 enters a long, straight, horizontal and evacu- ated guide tube of radius 1 cm, how far along the guide will it have moved before it first makes contact with the bottom wall assum- ing an alignment of the initial velocity with the central axis?

Fig. 1.1 The x-y coordinate framework for the trajectory of a shell fired from a coastal fort. we will let x and y denote the (rightwards) horizontal and (upwards) vertical displacements respectively; for convenience, the origin (0, 0) is best chosen to be at the coastal-base of the fort. The situation is illustrated in Fig. 1.1. Ignoring minor factors such as air resistance, the only force acting on the projectile during its flight is : a downwards force of magnitude mg, where g =9.8 Nkg−1 is the strength of earth’s gravitational pull close to the surface. As such, the x and y components of F are

Fx = 0 and Fy = mg . − Their substitution into eqn (1.5), and a division by m, leads to two first-order for the velocity of the shell: dv dv x = 0 and y = g . (1.7) dt dt − These are said to be the for the projectile, and follow from the application of Newton’s second law of motion in the horizontal and vertical directions respectively. They represent the case where there is no acceleration in the horizontal direction and a constant acceleration vertically. An integra- tion of eqn (1.7), with respect to t, yields the time-dependence of the x and y components of the velocity v:

vx = α and vy = gt + β , − where α and β are constants. If we let t =0 be the starting-time of the trajec- tory, when vx = u cos θ and vy = u sin θ, then

vx = u cos θ and vy = u sin θ g t . (1.8) − These expressions can readily be turned into a pair of first-order differential equation for the location of the projectile, 6 Classical mechanics

dx dy = u cos θ and = u sin θ g t , (1.9) dt dt −

since velocity is the rate of change of , r =(x,y), with time:

dx dy dr vx = , vy = , ... v = . (1.10) dt dt dt An integration of eqn (1.9), with respect to t, yields

x = (u cos θ)t + α′ and y = (u sin θ)t 1 gt2 + β′ , − 2 where α′ and β′ are constants. As x =0 and y = H when t =0, α′ =0 and β′ =H. Hence,

x = (u cos θ)t and y = H + (u sin θ)t 1 gt2 (1.11) − 2 defines the trajectory of the shell as a function of time (after launch and before it hits the sea). The substitution of t = x/(u cos θ) into the expression for y reveals the path to be parabolic. Since sea-level corresponds to y =0 in our coordinate system, the time taken for the projectile to reach the surface of the water, T , is given by the solution to the quadratic equation

H + (u sin θ)T 1 gT 2 = 0 . − 2 Although this is formally satisfied by two values of T , only the one with the positive square-root makes physical sense (as t>0):

u sin θ + u2 sin2 θ +2gH T = . p g The range of the shell, R, is then equal to the x-coordinate in eqn (1.11) when t=T . Namely,

u cos θ R = u sin θ + u2 sin2 θ +2gH . (1.12) g  q  If required, the angle of inclination for the maximum range can be calculated by solving the equation dR/dθ =0. While the differentiation is straightfor- ward, if somewhat messy, finding the root(s) of the resultant non-linear equa- tion is not easy in general. The special case of H =0 can be done analytically, however, when eqn (1.12) simplifies to

u2 sin2θ sin2θ = 2 sinθ cos θ R = . g

This yields a maximum range of u2/g when θ =45◦. 1.3 Newton’s second law in action 7

1.3.2 As a second example of Newton’s second law in action, let’s consider the be- haviour of an that is loosely bound to a substrate. The situation can be modeled as a of mass m (kg) that is attached to a planar surface by a (weak) of stiffness k (Nm−1). While several types of motion are possible, for simplicity, we will only consider small perturbations of the atom, from its equilibrium position, along the line of the bond; this is illustrated in Fig. 1.2(a). For an ideal spring, the restoring force, F , is proportional to its extension or compression, x, so that F = k x , (1.13) − where k is called the spring constant; this is known as Hooke’s law. Using the fact that the acceleration of the mass, a, can be expressed as the second of its displacement with respect to time,

dv d dx d2x a = = = , dt dt dt dt2   Newton’s second law of motion for this one-dimensional problem, F = ma, leads to the differential equation d2x k x = m . (1.14) − dt2 This equation of motion has the general solution d2x x = A sin(ωt + φ) , (1.15) = − ω2 x dt2 which can be verified by substitution in eqn (1.14), where the angular fre- quency of the sinusoidal oscillation, ω, is given by

k ω = (1.16) rm

Fig. 1.2 A simple harmonic example: (a) the restoring force, −kx, given a small displacement, x, from the equilibrium position; (b) the resultant sinusoidal oscillation, with frequency ω. 8 Classical mechanics

and has (SI) units of radians per second. The determination of the (arbitrary) constants A and φ, known as the amplitude and phase respectively, requires further information, or boundary conditions. A spectroscopic measurement of ω tells us about the stiffness of the related atomic bond. The frequency of an oscillatory system is often given in terms of cycles per second, or Hertz (Hz), rather than rads−1. Denoting the latter by f, it is easily related to ω through ω = 2πf , (1.17)

since there are 2π radians in one complete turn (or 360◦).

1.4 Conservation of momentum

In Section 1.2.1, we noted that an object moved with a uniform velocity if there was no force acting on it. More formally, a time- of the second law of motion in eqn (1.2) shows that

p = mv = constant

if F =0. While this is not particularly useful for an isolated body, it becomes an important property once we realise that the total momentum of a collection of interacting is always conserved if there is no net external force acting on them. To see this, consider a system of N particles; we will label each with a th subscript, so that the i one has momentum . If Fi denotes the external th th force acting on the i particle, and Fij is the force exerted by the j on i, then Newton’s second law tells us that the ith component of the system obeys the equation of motion

dpi F1j = F12 + F13 + · · · + F1N F + F = , i ij dt Xj6=1 j=i X6 where the Σ-term constitutes the sum of all the internal interactions. Adding up the N such equation, for i =1, 2, 3,...,N, we obtain

N N N N dpi Fij = F1j + · · · + FNj F + F = . i ij dt Xi=1 Xj6=i Xj6=1 jX6=N i=1 i=1 j=i i=1 X X X6 X According to Newton’s third law, Fij = Fji. All the terms cancel out in pairs in the double-summation, therefore, and− the equation of motion reduces to

N N d F = p , (1.18) i dt i i=1 i=1 ! X X where we have used the linearity of the differential operator, d/dt, to write the sum of the as the derivative of the sum on the right-hand side. If the system is isolated, so that Fi =0, then an integration of eqn (1.18) with respect to time yields P 1.5 Work done, energy and 9

Fig. 1.3 (a) The of a general two-body collision, and (b) a simple special case.


mi vi = constant , (1.19) pi = mi vi i=1 X th where mi and vi are the mass and velocity of the i particle. In other words, the total (linear) momentum is conserved. As an explicit illustration of eqn (1.19), suppose that two particles, with masses m1 and m2, collide with incoming of u1 and u2 and separate with outgoing velocities v1 and v2; the situation is sketched in Fig. 1.3(a). The total initial momentum is m1u1 + m2 u2 , whereas the outgoing combination is m1v1+m2 v2. According to the conservation of momentum, therefore, one constraint on their motion is given by

m1u1 + m2 u2 = m1v1 + m2 v2 . (1.20)

This vector equation is shorthand for several simultaneous relationships, of course, since it is satisfied by every spatial component of the velocities (along the x,y,... directions). A particularly simple case of the above scenario occurs when particle 2 is initially at rest (u2 =0) and they both stick together on impact, or coalesce, so that v1 = v2 = v, as illustrated in Fig.1.3(b). Then, eqn (1.20) simplifies to

m1u1 = (m1 + m2) v .

That is to say, the initial and final velocities are in the same direction (i.e. a straight-line collision), but the onward-going speed is reduced by a factor of m1/(m1 +m2).

1.5 Work done, energy and power 1.5.1 Work done Lifting a heavy object is a difficult task, and entails a great deal of effort; we’d be inclined to say that it was hard work! In , the term ‘work’ has a very specific meaning: it is a measure of the distance moved in opposition to a 10 Classical mechanics

force. If the force F had a constant magnitude, and acted in a fixed direction, and we moved a distance L in a straight line directly against it, then the work done by us is defined to be F L newton metres (Nm) or joules (J); in other words, ‘force distance’. Conversely,| | if we were displaced by L along the direction of F,× then F L would be the work done by the force. When we are making| | the effort, there is no particular reason why F and the displacement vector L should be parallel (or antiparallel) to each other. If their orientations differ by an angle θ, but are otherwise acting in a uniform manner, then the work done by us, W, is given by W = F L cos θ . (1.21) − | | | | If W is negative, then W is the work that is done on us by the force. The presence of the cosine in| eqn| (1.21) can be understood in one of two equiva- lent ways: (i) the component of the displacement in the direction of the force is L cos θ, by elementary trigonometry; and, similarly, (ii) F cos θ is the component| | of the force along the displacement-path. | | In general, neither the displacement nor the direction of the force will lie along a fixed direction; indeed, the magnitude of F may also vary with posi- tion. The work done can then be calculated by dividing up the path into many small straight line segments, with displacements δL, over which the force is locally uniform, and adding up all the tiny contributions δW F δL cos θ . (1.22) ≈ − | | | | Taking the infinitesimal limit in the summation, and using the succinct notation of a scalar (or dot) product,

F • dr = F dr cos θ W = lim δW = F • dr (1.23) | | | | |δL|→0 − path X pathZ where we have used dr instead of dL on the right-hand side for consistency with the conventional notation F=F(r).

The scalar, or dot, product of two vectors The simplest way in which two vectors, A and B, can be multiplied together is called a ‘scalar’ or ‘dot’ product. The former indicates that the result of such a multiplication is a scalar quantity, whereas the latter describes the symbol used to denote this type of product. It is defined geometrically as the product of the lengths of the two vectors times the cosine of the angle between them,

A •B = |A| |B| cos θ , (1.24)

and is calculated algebraically through the sum of the products of corresponding com- ponents: A •B = Ax Bx + Ay By + · · · , (1.25)

where A=(Ax ,Ay , ... ) and B =(Bx , By , ... ). 1.5 Work done, energy and power 11

1.5.2 Potential energy is literally the potential of an object, or a system, to do work by virtue of its position or state of being. Water at the top of a high waterfall has a lot of potential energy, for example, since it is capable of driving a blade to generate electricity. Similarly, a spring stores potential energy when it is stretched or compressed which becomes available (for catapulting things) when the tension is released. To get a quantitative feel for the concept, let us calculate the potential energy stored in an ideal spring. As noted in Section 1.3, this can be used to model the behaviour of a chemical bond when subjected to a collinear perturbation. With reference to Fig. 1.2(a) and eqn (1.22), the small amount of work done, δW, in increasing the extension or compression of a spring of stiffness k from x to x + δx is given by δW kxδx. ≈ Summing up these tiny contributions in the limit of δx 0, the total amount of work needed to change the length of the spring from→ its relaxed state by a distance X is evaluated through the integral

x=X 2 X x x 1 2 W = k x d = k = 2 kX . 2 0 xZ=0   Taking this to be the amount of energy stored in the distorted spring, the po- tential energy V at displacement x is:

1 2 V = 2 k x . (1.26) It varies quadratically, whereas the the restoring force is linear. As a second example, consider two isolated ions with charges q and Q coulombs (C) that are separated by a distance r. According to Coulomb’s law, there is a mutual repulsion between them which is proportional to the product of the charges and inversely proportional to the square of their separation:

qQ F = 2 , (1.27) 4πǫo r

−12 2 −2 −1 where ǫo is a constant (8.854 10 C m N ) called the permittivity of free . A negative value of×F indicates that the force is actually attractive, and will be the case when q and Q have opposite signs. The inverse-square fall off in eqn (1.27) is illustrated in Fig.1.4(a). Taking the potential energy of this system to be zero when the two charges are infinitely far apart, V is equal to the work done in bringing them together from r = to r =R. Following eqns (1.23) and (1.27), that is ∞

r=R ∞ r ∞ r qQ d 1 1 V = F d = 2 . − = − 4πǫo r » r –R R r=Z∞ RZ 12 Classical mechanics

Fig. 1.4 The variation of the (mutually repulsive) force, F , and potential energy, V , with separa- tion r between two isolated (point-like) ions with charges q and Q.

Evaluating the integral, the potential energy at a separation of r is seen to be

qQ V = , (1.28) 4πǫo r

and is illustrated in Fig. 1.4(b). The potential energy increases with distance for charges of opposite sign, since work has to be done to pull them apart, and decreases with separation for charges of the same sign, as they naturally repel each other. Incidentally, an anion with a charge equivalent to one electron has q = 1.602 10−19 C. − × 1.5.3 Kinetic energy (KE) is the energy associated with motion. Although we might guess that it is a function of the mass and speed of an object, KE=f(m,v), what is the exact relationship? The formula for the kinetic energy can be derived by calculating the work done by a force F when a mass m is accelerated from rest to speed vo. Since the force need not be uniform, we will consider infinitesimally small incre- ments and add up their contributions as in eqn (1.23):


KE = F • dr . (1.29) v=0Z Over a short period of time, dt, the force and the resultant displacement, dr, can be related to the instantaneous velocity, v, through dv F = m and dr = v dt . dt This enables the integrand in eqn (1.29) to be written as

F • dr = m v • dv , 1.5 Work done, energy and power 13 where we have implicitly used the chain rule of differentiation and the com- • • mutative property of a . A direct connection between F • dr and A B = B A the speed can now be made by noting that

2 v•v v • v v = v = √v•v dv = d( )=2 d , | | which follows from the product rule of differentiation. In other words, the integral of eqn (1.29) simplifies to

v=vo vo m 2 m 2 1 2 KE = dv = v = 2 mvo . 2 2 0 v=0Z h i Hence, the kinetic energy (in joules, J) of a mass m (kg) moving at a speed v (m s−1) is foundto be

p2 KE = 1 mv2 = , (1.30) 2 2m where p = mv is the modulus of the momentum in eqn (1.1).

1.5.4 One of the most useful devices in Physics for carrying out quantitative calcu- lations are ‘conservation laws’. We met an example of this in Section 1.4, the conservation of linear momentum, which was derived from Newton’s second and third laws of motion. Perhaps the most fundamentalof all such rules, or as- sumptions based on experience, is that of the conservation of energy: Energy may be transformed from one form to another, but it cannot be created or destroyed. In other words, the total energy of a system is conserved. As a specific example, let us consider the simple harmonic case of Section 1.3. Substituting for the displacement x from eqn (1.15) into the formula for the corresponding potential energy (PE) in eqn (1.26),

1 2 2 PE = 2 kA sin (ωt + φ) . (1.31) Likewise, the substitution of the time-derivative of x in eqn (1.15) for v in eqn (1.30) leads to

dx 2 KE = 1 m = 1 mω2A2 cos2(ωt + φ) , 2 dt 2   which can be simplified to

1 2 2 KE = 2 kA cos (ωt + φ) . (1.32) with eqn (1.16). Although the potential and kinetic vary sinusoidally with time, their sum remains constant:

1 2 Total energy = PE +KE = 2 kA , (1.33) where we have implicitly used the identity sin2θ + cos2θ 1. The temporal variation of eqns (1.31)–(1.33)is illustrated in Fig.1.5, and shows≡ that any loss 14 Classical mechanics

Fig. 1.5 The temporal variation of the kinetic, potential and total energy of the simple harmonic system of Fig. 1.2, as per eqns (1.31)–(1.33).

in potential energy is matched by an equivalent gain in kinetic energy and vice versa. While the total energy of an isolated system is always conserved, there is Exercise 1.4 If two H+ ions traveling in op- often special interest in its kinetic energy. This is particularly true in collision posite directions at 300 m s−1 have a head-on collision, how close will they get before they phenomena, where the term elastic is used to denote the conservation of KE. If start to move apart? How would the situation the interaction of Fig.1.3(a) was elastic, for example, then the velocities would differ if one of the ions was initially at rest? satisfy the scalar constraint

2 2 2 2 m1 u1 + m2 u2 = m1 v1 + m2 v2 (1.34) | | | | | | | | in addition to the vector condition of eqn (1.20). The collision illustrated in Fig. 1.3(b) is inelastic since

2 2 m1 u1 > (m1 + m2) v | | | | from the conservation of momentum. The shortfall in kinetic energy cannot be recovered in this case as it becomes dispersed as internal energy (heat) in the putty-like encounter. The universal nature of the law of conservation of energy can be seen from the fact that it appears in several different guises in science. For example, the first law of , and the Born-Haber and Hess cycles in Chem- istry, are nothing more than a restatement of this principle.

1.5.5 Power Although the concept of energy tells us how much work is (or can be) done, it gives no indication of how quickly the relevant task is performed.This temporal aspect is encapsulated in the term power, which is defined to be the ‘rate of doing work’. In notation, the power P is related to the work done W through t2 dW P (t) = and W = P (t) dt . (1.35) dt tZ1 The average power is simply the total work done divided by the duration of the exercise. The standard unit for power is watts (W), which is equivalent to joules per second (Js−1). A measure more familiar from car specifications is ‘horsepower’, where 1hp = 746W. We note that the most common everyday 1.6 Fields, potentials and stability 15 unit of energy is a calorie (Cal), frequently misused for kCal (or 1000Cal), and is approximately equal to 4.2 J; one calorie is formally defined to be the energy required to raise the of 1 cm3 of water by 1◦C.

1.6 Fields, potentials and stability

In eqn (1.27), we encountered Coulomb’s law. It is an example of an ‘action- at-a-distance’ force, since it operates without direct contact between the two charges. In such contexts, it can be useful to introduce the concept of a ‘field of the force’. The idea is probably familiar, as the pattern made by iron-fillings scattered on a piece of paper above a bar gives a visual indication of the lines of the magnetic field. A small object, which is susceptible to effects of magnetism, will move along the field lines when placed in the vicinity of the bar magnet. There is, of course, a close link between a field and a force; it is essentially one of normalisation, or scaling. An electric field is defined to be the force per unit charge (NC−1), for example, and a gravitational field is the force per unit mass (Nkg−1). Thus an object with charge Q will experience a force

Felec = QE (1.36) when placed in an electric field E, and a mass m will experience of a force

Fgrav = mg (1.37) when in a gravitational field g. Just as it is sometimes useful to think in terms of a field rather than a force, so too is there an analogous quantity for potential energy; namely the potential, which corresponds to the PE per unit charge (JC−1) or per unit mass (Jkg−1) in the electrostatic and gravitational cases respectively. There is an intimate connection between fields and potentials, since force and work done are linked through the integral relationship of eqn (1.23). A consideration of eqns (1.13) and (1.26) provides explicit verification of a differential connection between the relevant force F and potential energy V , dV F = , − dx for an ideal spring. This is confirmed by the electrostatic example of eqns (1.27) and (1.28), from which it can be seen that dV F = . − dr A more detailed examination of the electrostatic case leads to the general vec- torial result

E = φ , (1.38) −∇ where φ is the and is a ‘vector differential operator’. The important thing to note about φ is∇ that it is a vector: its direction indicates ∇ 16 Classical mechanics

Fig. 1.6 The equipotentials, φ = constant, and electric field, E, shown in blue and red respec- tively, for three elementary cases: (a) a positive charge, (b) a negative charge and (c) parallel plates. The arrows indicate the direction in which a positive charge in the vicinity would move.

the direction in which the potential increases most rapidly, and its magnitude specifies how quickly it does so. The minus sign in eqn (1.38) tells us that objects move naturally from locations of high potential to the nearest lower ones. Some simple examples of electric fields and ‘equipotentials’ (lines, or surfaces, along which φ = constant) are shown in Fig. 1.6, and illustrate that the two are mutually perpendicular. There is a proviso to the discussion above, in that only ‘conservative’ forces have associated potentials. These are ones in which the work done in moving an object depends only on the start and end points, and not on the path taken between them. As such, potentials are a physical analogue of ‘state functions’ in Chemistry. The final topic we need to discuss in this section is that of ‘stability’. Sup- pose that a particle exists in a certain potential φ(r), where we have made the problem one-dimensional for simplicity. Are there any locations, or values of r, where it can remain at rest? According to Newton’s first law of motion, this can only happen if the net force is zero. Following eqns (1.36) and (1.38), for example, we need dφ/dr =0, or the stationary points of the potential. If the latter is a minimum, so that d2φ/dr 2 > 0, then it is a point of stable equilib- rium: the particle moves back to the stationary value of r if it is given a small nudge. Conversely, any perturbation will be amplified if d2φ/dr 2 <0 because such a maximum is a point of unstable equilibrium. The situation is illustrated graphically in Fig.1.7(a). The interaction between two neutral is often modelled in terms of a Lennard–Jones potential; its variation, as a function of their separation r, is plotted in Fig. 1.7(b). The point of stable equilibrium gives the natural bond length: if the atoms moves closer than this, the net force is repulsive; if separated farther, it’s attractive. For small perturbations from equilibrium, φ(r) can be approximated by a quadratic Taylors series expansion about the minimum; this is indicated in Fig.1.7(b) and, to within an additive constant, is equivalent to the simple harmonic potential of eqn (1.26).

1.7 Rotational motion

Perhaps the simplest and most common form of non straight line motion is one that follows a circular . The of the planets around the sun, or the 1.7 Rotational motion 17

Fig. 1.7 (a) A one-dimensional potential energy surface, φ(r), with points of stable and unstable equilibrium and a region of neutral stability. (b) The Lennard–Jones potential, which is often used to model the interaction of two neutral atoms as a function of their separation r; the dotted red line is the harmonic approximation for small perturbations from equilibrium. moon around the earth, for example, are very nearly circular (more generally elliptical); and the Bohr model of the atom had the moving around the nucleus along circular paths. In this section, we will consider such rotational, or angular, motion.

1.7.1 Centripetal forces According to Newton’s first law of motion, any deviation from travel in a straight line is indicative of a net force. In particular, requires a centrally acting, or centripetal, force that continually pulls the orbiting body towards a focal, or pivotal, point. In an astronomical setting, this is provided by gravity; and in the Bohr model, by the electrostatic attraction between the negatively charged electrons and the positive nucleus. As children we have all spun a bucket, or some other object, tied to our hand with a piece of string. The here is provided by the tension in the string. Our hand, however, feels an outward tug; but this is nothing more than Newton’s third law of motion in action.

1.7.2 and acceleration For simplicity, let us restrict ourselves to circular motion at a constant speed. This can either be specified in terms of an angular speed, ω, in rads−1, ora linear speed, v, in m s−1; the former being related to a frequency, f, in cycles per second through eqn (1.17). To see the connection between ω and v, con- sider an orbit of radius R. In a time t, the object will rotate through an angle θ = ωt and move a distance l = vt along an arc. Given also that θ = l/R , from the definition of radians,

l Rθ v = = = Rω . (1.39) t t This scalar relationship can be generalized to a vector one between the angular and linear velocities, ω and v, by using a vector (or cross) product: 18 Classical mechanics

The vector, or cross, product of two vectors The second way in which two vectors, A and B, can be multiplied together is called a ‘vector’ or ‘cross’ product. The former indicates that the result of such a multipli- cation is a vector quantity, whereas the latter describes the symbol used to denote this type of product. Geometrically, its modulus is the product of the lengths of the two vectors times the sine of the angle between them,

|A×B| = |A| |B| sin θ , (1.40)

and its direction is perpendicular to both and in the sense of a ‘right-hand screw’. That is to say, if the curl of the fingers on our right hand matches the sense of needed to go from A to B, then the out-stretched thumb indicates the direction of A×B. Algebraically, the components of the are calculated through the formula:

A×B = (Ay Bz − By Az ,Az Bx − Bz Ax ,Ax By − Bx Ay) , (1.41)

where A=(Ax ,Ay ,Az) and B =(Bx , By , Bz).

v = ω r , (1.42) × where the direction of ω is along the axis of rotation, in a ‘right-hand screw’ sense, ω = ω and r is any vector from the axis of rotation to the object in question.| The| modulus of this equation returns the simple connection between the in eqn (1.39), v = v = ω r = ω r sin α = ω R , | | | × | | | where α is the angle between ω and r, but it also captures the directionalaspect of the instantaneous motion. The acceleration of the orbiting body can be ascertained by differentiating eqn (1.42) with respect to time: dv dr a = = ω = ω v , (1.43) dt × dt × where we have implicitly assumed that dω/dt =0 and used the basic defini- tion v = dr/dt . Since v is tangential to the circular orbit and ω is axial, the centripetal nature of the acceleration is recovered from the direction of their cross product. The magnitude of the acceleration reduces to a = a = ω v = ω v | | | × | because ω and v are mutually perpendicular. Upon substitution for ω or v from eqn (1.39), v2 a = ω2R = (1.44) R where R is the radius of the circular motion. 1.7 Rotational motion 19

1.7.3 and Tight nuts and bolts are loosened by using a spanner. Very sticky cases require a stronger yank, or a wrench with a longer handle, or both. This suggests that the ‘turning force’ or torque, G, involves both the magnitude of the (linear) force used, F , and the distance from the pivot, R, at which it is applied;

G = F R , (1.45) perhaps? A more careful consideration of the situation leads to the definition of the torque vector, G, as

G = r F , (1.46) × where r is the location, relative to the pivot (or origin), at which the force, F, is applied. Its magnitude is consistent with eqn (1.45) when R is taken to be the perpendicular distance between the pivot and the force, and its direction is normal to the plane containing r and F and indicates the axis of the torque in a right-hand screw sense. Just as torque is the rotational analogue of a force, the angular momentum

L = r p (1.47) × is the rotational analogue of (linear) momentum, p. Indeed, they obey a rela- tionship that is equivalent to Newton’s second law of motion:

dL dp G = . (1.48) F = dt dt Torque and angular momentum are known as the ‘moments’ of the force and momentum respectively. In the absence of a torque, the angular momentum remains constant.

1.7.4 The of For an object of mass m moving in a circular orbit of radius R at speed v, the magnitude of the angular momentum about the centre of the motion reduces to

L = L = Rmv, (1.49) | | because the linear momentum, p = mv, is perpendicular to the instantaneous displacement, r, (of length R) from the origin. It can be written in terms of the angular speed, ω, by using eqn (1.39):

L = mR2 ω . (1.50) L = Iω

Similarly, the kinetic energy of the circulating  object is given by L2 KE = 1 mv2 = 1 mR2 ω2 . (1.51) KE = 1 Iω2 = 2 2 2 2I These formulae lead to the recognition of the quantity in brackets,

I = mR2 , (1.52) 20 Classical mechanics

as the rotational analogue of mass: the . For a set of point-masses that are all rotating about the same axis like a , eqn (1.52) generalizes to

2 I = mj Rj , (1.53) j X th where mj is the mass of the j entity and Rj is its (perpendicular) distance from the axis of rotation. In the continuum limit of a solid object, the summa- tion is replaced by a corresponding three-dimensional integral over the mass . The important thing to note is that I depends on the distribution of the mass relative to the axis of rotation. Spinning skaters, for example, change their moment of inertia by stretching out or drawing in their arms and leg; con- sequently, their angular speed increases or decreases in accordance with the conservation of angular momentum (Iω = constant). We should point out that the behaviour of a rigid body can always be anal- ysed in terms of the motion of its centre of mass and a rotation about an axis through it. For a symmetric diatomic molecule, such as hydrogen or oxygen, therefore, the relevant moment of inertia is calculated from an axis that passes through the mid-point of the bond. For rotation in the plane of the molecule, d 2 d 2 m I = m + m = d 2 2 2 2       where m is the atomic mass and d is the bond length. For rotation about an axis collinear with the bond, I 0. ≈ 1.7.5 A comparison with

Exercise 1.5 Given that the H2 bond length Let us conclude this section with a tabular summary that emphasizes the simi- is 0.075 nm, evaluate its moment of inertia. larity between the basic notions and formulae involved in rotational and linear If the average rotational kinetic energy of H2 at a temperature of T = 300 K is k T , where k is the , determine the Table 1.1 The correspondence between the formulae of linear and rotational motion. average angular momentum of the molecule. Hence estimate the rotational frequency of a Linear Rotational hydrogen molecule at room temperature. displacement x angle θ

dx dθ velocity v = angular velocity ω = dt dt

dv d2x dω d2θ acceleration a = = = dt dt2 dt dt2

inertia (mass) m moment of inertia I

momentum p = mv angular momentum L = Iω

dp dv dL dω force F = = m torque G = = I dt dt dt dt

1 p2 1 L2 kinetic energy (KE) mv2 = rotational KE Iω2 = 2 2m 2 2I 1.8 Limitations of classical mechanics 21 motion. The results are given in Table 1.1. To simplify the comparison, the lin- ear motion has been restricted to lie along the x-axis and only the magnitudes of vector quantities have been considered.

1.8 Limitations of classical mechanics

While classical mechanics is immensely powerful in helping us to understand everyday phenomena, it has been found to breakdown in the limit of light-like speeds, the vicinity of ultra-heavy objects and on atomic length scales. It is then superseded by the of relativity and . Even in these circumstances, however, many of the elementary notions of classical mechanics still play a central role.

1.9 Answers to exercises

1.1 4.0kms−1. 1.2 Yes, because its direction (and, hence, velocity) is continually changing. 1.3 180m. 1.4 1.5 µm. Conservation of non-zero momentum 6.0 µm. ⇒ 1.5 I =4.7 10−48 kgm2. Root-mean-square L =2.0 10−34 kgm2 s−1. Therefore,× ω 4.2 1013 rads−1 or f 6.7 GHz.× ≈ × ≈

Quantity Value (SI units)

Planck constant (h) 6.626 10−34 Js × Boltzmann constant 1.380 10−23 JK−1 × Avogadro constant 6.022 1023 mol−1 × −12 −1 Permittivity of free space (ǫo) 8.854 10 F m × Elementary charge (e) 1.602 10−19 C × Mass of an electron 9.109 10−31 kg × Mass of a proton 1.673 10−27 kg × Mass of a neutron 1.675 10−27 kg ×