Newtonian Mechanics - kuweb.math.ku.dk/~solovej/MATFYS/MatFys1.pdfNewtonian Mechanics In these notes classical mechanics will be viewed as a mathematical model for the descrip-tion

Chapter 1

Newtonian Mechanics

In these notes classical mechanics will be viewed as a mathematical model for the descrip-tion of physical systems consisting of a certain (generally finite) number of particles withfixed masses subject to certain interactions and possibly also to external forces. As is wellknown since the 17’th century such models provide accurate descriptions of a vast varietyof mechanical systems. The development in the 20’th century of quantum mechanics andof the theories of relativity, on the other hand, have revealed that such models should beviewed as approximations with limitations on their range of validity.

1.1 Classical space-time

The first basic ingredient of the model we are about to describe is a set M, called theuniverse and whose elements are called events. It is assumed that the events can be labeledby four real numbers, i.e. there exists a bijective map x : M → R4, called a coordinatesystem on M. Composing one coordinate system x with a bijective map ψ : R4 → R4 weobtain a second coordinate system y = ψ x. Thus coordinate systems are far from beingunique. We call ψ the coordinate transformation from x-coordinates to y-coordinates.

Next, we postulate the existence of two functions T : M×M → R and d : D → R,where D = (p, q) ∈ M×M | T (p, q) = 0. The function T is called the time-differencefunction and T (p, q) has the physical interpretation of the lapse of time between the eventp and the event q as measured by standard clocks. Two events p and q are said to besimultaneous, if T (p, q) = 0. Thus D is the set of pairs of simultaneous events, and thephysical interpretation of d(p, q) is the (spatial) distance between the simultaneous eventsp and q. Note that the distance between non-simultaneous events has no physical meaningand is not defined.

In view of the physical interpretation of the functions T and d, these must be subjectto further constraints. This is achieved by postulating the existence of a coordinate systemx = (x1, x2, x3, t) such that

d(p, q) = ‖x(p)− x(q)‖ , p, q ∈ D , (1.1)

1

Chap. 1 Newtonian Mechanics Version of 20.11.08 2

andT (p, q) = t(q)− t(p) , p, q ∈M×M , (1.2)

where we use the notation x = (x1, x2, x3) and ‖ · ‖ denotes the Euclidean norm in R3

defined by

‖v‖ =√v21 + v2

2 + v23 , v = (v1, v2, v3) ∈ R3 .

A coordinate system x fulfilling (1.1) and (1.2) will be called a Galilean coordinate systemand we call x1(p), x2(p), x3(p) the space-coordinates of the event p, and t(p) is called itstime-coordinate.

The relation (1.1) expresses the Euclidean nature of the distance d. A few remarks onthe Euclidean norm and its properties will be useful for the following discussion. We definethe scalar ( or inner) product of two vectors v, w ∈ R3, written 〈v, w〉 or simply v · w, by

v · w = 〈v, w〉 = v1w1 + v2w2 + v3w3 .

Note that‖v‖2 = 〈v, v〉 > 0 for v 6= 0 = (0, 0, 0) ,

and it is clear from the definition that 〈v, w〉 is a linear real-valued function of v ∈ R3,for fixed w, as well as of w for fixed v. Moreover, 〈v, w〉 = 〈w, v〉. We say, that 〈·, ·〉 is asymmetric and positive definite bilinear form on R3.

The angle θ ∈ [0, π] between two non-trivial vectors v and w is defined by

cos θ =〈v, w〉‖v‖‖w‖

. (1.3)

In particular, we say that v and w are orthogonal if their scalar product vanishes:

v and w are orthogonal if and only if 〈v, w〉 = 0 .

Note that the definition (1.3) presupposes that the right hand side belongs to the interval[−1, 1], which is a consequence of the Cauchy-Schwarz inequality (see Exercise 1.1d)).

A mapping S : R3 → R3 is called an isometry if it preserves the Euclidean distance,that is if

‖S(v)− S(w)‖ = ‖v − w‖ , v, w ∈ R3 .

It can be shown (see Exercise 1.2) that such a mapping is of the form

S(v) = a+ S0(v) ,

where a ∈ R3 is a fixed vector and S0 is a linear isometry, also called an orthogonaltransformation. Moreover, any such orthogonal transformation is either a rotation aroundan axis through 0 or it is a rotation around an axis through 0 composed with a reflection in aplane through 0 (see Exercise 1.4). A matrix that represents an orthogonal transformationwith respect to the standard basis for R3 is called an orthogonal matrix, and the set oforthogonal matrices is denoted by O(3). They are discussed further in Exercise 1.3.

The following result gives a characterization of coordinate transformations betweenGalilean coordinate systems.


Proposition 1.1. Let x and y be two Galilean coordinate systems. Then there exists aconstant t0 ∈ R, a function ϕ : R → R3 and a function A : R → O(3), such that y = ψ x,where

ψ(x, t) = (ϕ(t) +A(t)x, t− t0) , x ∈ R3 , t ∈ R . (1.4)

Proof. Setting x = (x, t) and y = (y, s) and choosing a fixed event q0, we have accordingto (1.2)

t(p)− t(q0) = s(p)− s(q0) , p ∈M ,

and hences(p) = t(p)− t0 , p ∈M , (1.5)

where t0 = t(q0)− s(q0).Since x is a bijective mapping there exists, for each τ ∈ R, a unique pτ ∈M such that

x(pτ ) = (0, 0, 0, τ), i.e. pτ is the event occurring at the origin of x-coordinates at x-time τ .From (1.1) we get

‖y(p)− y(pτ )‖ = ‖x(p)‖ if t(p) = τ, and ‖y(p)− y(q)‖ = ‖x(p)− x(q)‖ if t(p) = t(q).

Setting ϕ(τ) = y(pτ ) and

Sτ (v) = y(x−1(v, τ))− ϕ(τ) , v ∈ R3 ,

this can be restated as

‖Sτ (v)‖ = ‖v‖ and ‖Sτ (v)− Sτ (w)‖ = ‖v − w‖ w, v ∈ R3 .

It follows from the remark above (see Exercises 1.2-3) that, for each τ , there exists a(unique) orthogonal matrix A(τ) such that Sτ (v) = A(τ)v, considering v here (as well asx in (1.4)) as a column vector. Together with (1.5) this implies

y(p) = (ϕ(t(p)) +A(t(p))x(p), t(p)− t0) , p ∈M ,

which is equivalent to the statement of the Proposition. 2

We see from (1.4) that the time coordinates are related by a constant shift, a propertythat is commonly expressed by saying that, in classical physics, time is absolute, since itis uniquely determined once a zero-point has been chosen. Consequently, we shall in thefollowing usually restrict attention to systems with a common time coordinate.

The spatial content of eq.(1.4) can be described geometrically as follows. As alreadynoted in the proof, the function ϕ describes the motion of the origin of x-coordinates withrespect to the coordinate system y. More precisely, setting x = 0 in (1.4), we see that ϕ(t)is the y-coordinate set of the origin of x-coordinates at x-time t. Similarly, denoting byε1 = (1, 0, 0), ε2 = (0, 1, 0), ε3 = (0, 0, 1) the standard basis vectors of the vector space R3

and choosing two simultaneous events p, q at x-time t such that x(p) − x(q) = εi we seethat

y(p)− y(q) = A(t)εi = Ai(t) ,

where Ai(t) denotes the i’th column of A(t). In other words, the columns of A(t) are they-coordinates of the standard basis vectors in x-coordinate space at x-time t. As previously


mentioned (see also Exercise 1.4) the orthogonal matrix A(t) represents a rotation aroundan axis through the origin, possibly composed with a reflection in a plane. We call A(t)orientation preserving if the reflection is absent, and ψ is by definition orientation preserv-ing if this is the case for all t. This allows us to characterize an orientation preservingGalilean coordinate transformation as being the composition of a time-dependent rotationaround a (time-dependent) axis, represented by A(t), composed with a time-dependentspatial translation, represented by ϕ.

1.2 The Galilean principle of relativity and inertial systems

We shall restrict our attention to ordinary differential equations as candidates for the fun-damental laws of mechanics. It is a well known fact that the laws governing the behavior ofmechanical systems are not invariant under arbitrary Galilean coordinate transformations.A falling body behaves differently in a coordinate system at rest at the surface of the earthand in a freely falling system. In fact, the Galilean transformations are sufficiently generalthat there are no interesting differential equations preserving their form under all suchtransformations.

The Galilean principle of relativity is an invariance principle restricting the possibleforms of the basic equations of motion for a mechanical system under a restricted set ofcoordinate transformations. It can we formulated as follows.

The Galilean principle of relativity. There exist particular Galilean coordinatesystems, called inertial systems, in which the equations of motion for mechanical systemsassume identical form. Any coordinate system in uniform motion relative to an inertialsystem is likewise an inertial system.

We shall discuss the invariance issue in more detail in the next section. The second partof the statement can, in view of the results of the preceding section, be stated as follows:

If x is an inertial system of coordinates, and ψ : R4 → R4 is given by

ψ(x, t) = (a+ vt+Ax, t+ s) , x ∈ R3 , t ∈ R , (1.6)

then y = ψx is also an inertial coordinate system. Here ϕ(t) = a+vt describes the motionof the origin of x-coordinates on a straight line through a with constant velocity v in y-coordinate space, and the matrix A represents a constant rotation of the x-coordinate axesaround a fixed axis through the origin of x-coordinates, possibly composed with reflectionin a fixed plane. The parameter s is a time translation. We call ψ as given by (1.6) aninertial coordinate transformation and denote it by ψ[a, s, v, A].

It is important to note, that the set of inertial coordinate transformations forms a groupwith respect to composition of mappings.

Definition 1.2 (Group). A group G is a set G with an element e ∈ G and maps

G×G 3 (g, h) 7→ gh ∈ G, G 3 g 7→ g−1 ∈ G (1.7)

such thatg1(g2g3) = (g1g2)g3, ge = eg = g, gg−1 = g−1g = e

for all g, g1, g2, g3 ∈ G. The first map in (1.7) is called the multiplication, the second mapthe inverse.


The set O(3) of orthogonal matrices is a group with matrix multiplication and matrixinverse (see Exercise 1.3).

Thus saying that the inertial coordinate transformations form a group means that com-posing two such transformations yields a third inertial transformation, that the inverse ofan inertial coordinate transformation is an inertial transformation and that the identitymapping of R4 is an inertial coordinate transformation. The last property is obvious, sincethe identity mapping is obtained when a = s = v = 0 and A = I, the 3×3 identity matrix.The other two properties follow from the formula

ψ[a1, s1, v1, A1] ψ[a2, s2, v2, A2] = ψ[a1 + v1s2 +A1a2, s1 + s2, v1 +A1v2, A1A2] , (1.8)

which is obtained by straightforward computation (see Exercise 1.5). In particular, onegets −s, added in ver-

sion of Nov. 12ψ[a, s, v, A]−1 = ψ[−A−1(a− vs), −s, −A−1v, A−1] , (1.9)

and every inertial coordinate transformation can be decomposed into four basic types oftransformations:

a space translation Ta = ψ[a, 0, 0, I]

a time translation T s = ψ[0, s, 0, I]

a boost Bv = ψ[0, 0, v, I]

an O(3)-transformation RA = ψ[0, 0, 0, A] ,

according toψ[a, s, v, A] = T s Bv Ta RA. (1.10)

The group of inertial coordinate transformations is often called the Galilei group.The Galilean principle of relativity was generalized by Einstein in 1905 to include all

physical phenomena, and not only mechanical ones. Since its content is otherwise thesame, except that the definition of an inertial coordinate system is different, we shall in thefollowing refer to the Galilean principle of relativity simply as the principle of relativity.

1.3 The Newtonian equations of motion

An ordinary differential equation of order n, describing the motion of a point x(t) in Rk asa function of the variable t, is an equation of the form

dnx

dtn= f(x,

dx

dt, . . . ,

dn−1x

dtn−1, t) , (1.11)

where f is a function with values in Rk defined on a subset Ω ⊆ Rkn × R. We will mostlyassume that Ω is the whole space Rkn × R and that f is a C∞-function, i.e. all its partialderivatives exist and are continuous at all points of Ω. A solution to (1.11) is a functionx : I → Rk, where I is an interval, such that (x(t), x′(t), . . . , x(n−1)(t), t) ∈ Ω for t ∈ I,and such that

x(n)(t) = f(x(t), x′(t), . . . , x(n−1)(t), t) t ∈ I .


As usual, the i’th derivative x(i) of the vector function x = (x1, . . . , xk) is defined bydifferentiating the coordinates:

x(i)(t) = (x(i)1 (t), . . . , x(i)

k (t)) .

The variable t will in the following be chosen to be the time coordinate in some inertialsystem x. For a system consisting of a single (point-)particle we have k = 3 and x(t) is thespatial coordinate set of the particle at time t. For a system consisting of N particles wecollect their coordinate sets x1(t), . . . , xN (t) into one set x(t) = (x1(t), . . . , xN (t)) ∈ R3N ,such that k = 3N in this case.

Let us first consider the case of a first order equation for a single particle

dx

dt= f(x, t) .

Under an inertial coordinate transformation with A = I and s = 0, we have y = x+ a+ vtand hence dy

dt = dxdt + v. The equation can therefore be rewritten in the form

dy

dt= f(y − a− vt, t) + v .

Thus, the two equations have identical form if and only if the function f fulfills

f(x, t) = f(x− a− vt, t) + v

for arbitrary x, a, v ∈ R3 and t ∈ R. Since the left hand side is independent of a it followsthat f is independent of x. Hence f(x, t) = f(t) fulfills f(t) = f(t) − v, which clearly isimpossible. We conclude that an equation of first order is incompatible with the principleof relativity.

Next, consider an equation of second order for a single particle

d2x

dt2= f(x,

dx

dt, t) .

Under an inertial transformation with A = I and s = 0 we have d2xdt2

= d2ydt2

and the equationcan be rewritten as

d2y

dt2= f(y − a− vt,

dy

dt− v, t) .

The two equations have the same form if and only if f satisfies

f(x, z, t) = f(x− a− vt, z − v, t)

for all x, z, a, v ∈ R3 and t ∈ R. As above this implies that f(x, z, t) = f(t) is independentof x, z. Next, applying an inertial transformation with a = s = v = 0, we have y = Ax

and hence dydt = Adx

dt and d2ydt2

= Ad2xdt2

, since A is constant. Therefore, we can rewrite theequation as

d2y

dt2= Af(t) .

Invariance of the equation requires

f(t) = Af(t) ,


which implies f(t) = 0, since 0 is the only vector invariant under arbitrary rotations. Weconclude that the only second order equation for a single particle that is compatible withthe principle of relativity is

d2x

dt2= 0 .

The solutions to this equation are of the form

x(t) = a+ vt ,

where a, v ∈ R3 are constants. This means that the possible motions of the particleare along straight lines with constant velocity v, which we recognize as the contents ofNewton’s first law. We have shown that it is a consequence of the principle of relativityand the assumption that the equation of motion is a second order differential equation.

We promote this latter assumption to a general principle for systems with an arbitrarynumber of particles, i.e. we postulate

Newton’s second law In an inertial system the equation of motion for a system ofN particles is a second order differential equation for the coordinates of the particles asfunctions of time.

This means that the equation of motion is of the form

x = f(x, x, t) (1.12)

where x = (x1, . . . , xN ) and we have introduced the notation x and x for the first andsecond time derivatives of x, respectively. As in the case of a single particle, the principleof relativity implies constraints on the possible forms of the right hand side of (1.12).One finds (see Exercise 1.6) that, as a consequence of invariance under translations andboosts, f can depend only on the coordinate differences xi−xj and the velocity differencesxi − xj , 1 ≤ i < j ≤ N . Furthermore, f is independent of time t because of invarianceunder time shifts T s. Writing f = (f1, . . . , fN ) we see that invariance of (1.12) underO(3)-transformations implies (see Exercise 1.6) that the fi’s must obey the identity x1, . . . , xN added

in version of Nov.12

fi(Ax1, . . . , AxN , Ax1, . . . , AxN ) = Afi(x1, . . . , xn, x1, . . . , xN ) . (1.13)

This requirement is, however, far from determining the form of the fi’s uniquely (seeExercise 1.7). One of the major achievements of 17’th century physics, largely due toNewton, was the discovery of the universal law of gravitation which is the statement thatthe gravitational interaction between N particles is determined by (1.12) with

fi(x) =∑j 6=i

kj(xj − xi)‖xj − xi‖3

, (1.14)

where k1, . . . , kN are constants characteristic of the individual particles, i.e. kj does notdepend on which other particles the j’th one is brought into interaction with. By definitionkj = Gmj , where G is Newton’s constant and mi is the mass of particle i. Note that fi

is independent of the particle velocities xi and clearly fulfills (1.13). Note also that f issingular if two or more particle positions coincide. By measuring the accelerations xi fora suitable set of particle configurations one can determine the masses mi from (1.12) and


(1.14) and once this has been done eq.(1.12) provides in principle a complete descriptionof the motion of the particles, as discussed in more detail below.

It is an empirical fact that gravity is but one out of several forces of nature and thatthese forces can counter balance each other. This provides a means for measuring strengthand direction of forces by comparison with, say, the elastic force of an ideal spring. Forthe gravitational interaction between two particles, 1 and 2, one finds that the elastic forceacting on particle 1 needed in order that its acceleration vanishes, equals

Kk1k2x1 − x2

‖x1 − x2‖3,

where K is a universal constant, depending only on the units of measurement chosen, butindependent of the types of particles considered.1 We interpret this quantity as minus theforce F21 exerted on particle 1 by particle 2 and note that it has the same form as theindividual terms on the right hand side of (1.14) up to multiplicative constants. Hence, bychoosing standard units for the measurement of force, we can write (1.12) in the form

mixi = Fi , (1.15)

where Fi is the total force acting on the particle i given by

Fi =∑j 6=i

Fji (1.16)

andFij = −Gmimj

xi − xj

‖xi − xj‖3. (1.17)

To summarize the preceding discussion in mathematical terms, we assume that theconcept of force acting on a particle with coordinates (x, t) is represented by a vectorF (x, t) in R3 and that Newton’s second law

mx = F (x, t) (1.18)

holds in any inertial system of coordinates, where m is the mass of the particle. This isthe most commonly stated form of Newton’s second law. Note that, although we haveexpressed F explicitly only as a function of the coordinates of the particle on which theforce acts, it may depend on other variables as well. F.ex. eqs.(1.16) and (1.17) express thetotal force acting on particle i as the vector sum of the forces exerted by each of the otherparticles, which depend also on the positions of those particles. In addition, F may alsodepend on the velocity x, which is the case for e.g. the Lorentz force on a charged particlemoving a magnetic field, but not on higher derivatives of x (otherwise, the equation wouldnot be of second order). Thus, in general, to determine the motion of a system of particlesit is not possible to solve the equation (1.12) for each individual particle, since F alsodepends on the motion of the other particles. Instead, one must solve simultaneously thewhole system of coupled differential equations for the particles. This is discussed furtherbelow.

1This universality of K is called the equivalence principle and forms a basic ingredient in the generaltheory of relativity.


Note that, although the principle of relativity is assumed to hold for mechanical systemsonly, consistency of the assumption about the vectorial character of force, or alternativelyof eq.(1.18), implies that any force F (x, t) must transform according to eq.(1.13) under aninertial coordinate transformation ψ[a, s, v, A] from x = (x, t) to y = (y, t′). That is, theforce F ′(y, t) in the y-system is given by

F ′(y, t′) = AF (x, t) . (1.19)

It is worth while stressing that Newton’s second law should not be considered as adefinition of force, which would render it empty of physical content. Rather it receives itscontent from the fact that force is a physical quantity that can be measured.

1.4 Configuration space, phase space and state space

That the equation of motion for a particle is of second order is intimately connected to theempirical fact that one can in general specify arbitrarily the location and the velocity of aparticle at a given instant of time, and once this has been done the motion of the particleis uniquely determined by the acting forces, and similarly for systems of particles. Thefollowing theorem on the existence and uniqueness of solutions to a second order equationis crucial in this connection. the → then in ve-

sion of Nov. 16

Theorem 1.3. Let f : Ω → Rk be a C∞-function defined on an open subset of R2k+1 andlet x0, x0 ∈ Rk and t0 ∈ R be given such that (x0, x0, t0) ∈ Ω. Then there exists a uniquesolution x : I → Rk to the equation

x = f(x, x, t) (1.20)

defined on a maximal open interval I such that t0 ∈ I and the initial conditions x(t0) = x0

and x(t0) = x0 hold.

We shall not prove this theorem here. A proof can be found in C. Berg: MetriskeRum, §7. More detailed results can be found in e.g. E. A. Coddington and N. Levinson:Ordinary Differential Equations. The regularity assumption made for f is unnecessarilystrong, but, on the other hand, some regularity condition stronger that continuity is needed(see Exercise 1.8).

In order to apply the theorem to the equations of motion for a system of N particles,

mixi = Fi , i = 1, . . . , N , (1.21)

we set as before k = 3N , x = (x1, . . . , xN ) , and f = (m−11 F1, . . . ,m

−1N FN ) in terms

of which the equations take the form (1.20). Thus, assuming that each Fi is a C∞-function of the positions and velocities of the particles and of time, we conclude thatthere exists a unique maximal solution with prescribed initial values (x1(t0), . . . , xN (t0))and (x1(t0), . . . , xN (t0)).

A result analogous to Theorem 1.3 is valid for equations of arbitrary order n, in whichcase the values of x and all derivatives up to order n − 1 at some instant t0 need to bespecified in order to ensure existence and uniqueness of a solution. Thus, for a third orderequation one would need to specify location and velocity as well as the acceleration at a


given instant of time. Since this would be in conflict with experience such an equation isnot a candidate equation of motion.

Note that the theorem does not in general provide solutions defined on the full timeaxis, but only on a subinterval, i.e. solutions may not be globally defined. For linearequations, that is when F is a linear function of x and x, all solutions are globally defined,but for non-linear F solutions may grow arbitrarily large in finite time or develop othertypes of singularities (see Exercise 1.9). For the sake of simplicity, we will assume in thefollowing that all solutions are globally defined.

It is convenient to describe the solutions to (1.21) as motions in the phase space R6N

of the system. Here we think of phase space as the space of initial data, and a motionin phase space is by definition a function z : I → R6N , where I is an interval (see alsoDefinition 1.4). Theorem 1.3 asserts that specifying a point z0 = (x0, x0) in phase spaceand a time t0 there is a unique motion z : t→ (x(t), x(t)) determined by (1.21) which passesthrough the given point at time t0, i.e. such that z(t0) = z0. Assuming that all solutionsare globally defined we obtain in this way for each fixed t a mapping Ψt,t0 : R6N → R6N

defined byΨt,t0(z0) = z(t) for z(t0) = z0 ∈ R6N .

Since this holds for all t0, t ∈ R it follows from Theorem 1.3 (see Exercise 1.11) that Ψt0,t Exercise nr. fixedin version ofNov. 16is bijective and fulfills

Ψt2,t1 Ψt1,t0 = Ψt2,t0 and (Ψt1,t0)−1 = Ψt0,t1 (1.22)

for arbitrary t0, t1, t2 ∈ R. The family of maps Ψt,t0 , t, t0 ∈ R, is called the phase flow ofthe equations (1.21).

It is also useful to describe the solutions to (1.21) as motions in configuration spaceR3N , considered as the space of initial configurations in space of the system. Since theinitial positions alone do not determine the temporal development of the system there isno analogue of the phase flow in configuration space. The dimension of configuration spaceis half the dimension of phase space and is called the number of degrees of freedom of thesystem.

Finally, let us mention that a point in phase space of a system of N particles is oftencalled a state of the system. Thus, given a state of the system at a given instant of time, itsstate is determined at all times, provided the acting forces are known. This notion of stateis, however, too restrictive for some purposes. Thus, for typical thermodynamic systems,the number of particles is sufficiently large that their individual positions and velocitiescannot be determined in practice. A more convenient description of the system is obtainedby a probability distribution on phase space. More precisely, such a thermodynamic stateof the system at a given time t0 is given by a non-negative function p on phase space,representing the probability density of states, fulfilling∫

R6N

p(x, v)d3Nxd3Nv = 1 .

The probability density pt at time t is then given by p(Ψ−1t,t0

(x, v))

added in versionof Nov. 13

pt(x, v) = p(Ψ−1t,t0

(x, v)) = p(Ψt0,t(x, v)) .


Note that Ψt0,t is the “backwards flow” from t to t0. The fact, that this is still a probabilitydistribution is known as Liouville’s Theorem. We will prove this in Chapter 3. Thermody-namic equilibrium states are defined as states that are invariant under the phase flow, i.e.such that pt = p for all t.

In general a thermodynamic equilibrium state can be extremely complicated as it de-pends on a large number of variables—the 3N variables of the microscopic particles. Inthermodynamics one is mostly concerned with special states that are determined from onlya small number of macroscopic variables, e.g., the temperature.

As an example we may consider N particles moving freely in a box [0, L]3 ⊆ R3 exceptthat the velocities are reflected when a particle meets the boundary of the box. The phaseflow then preserves the lengths of the individual velocities v1, . . . , vN . For such a systemthere are the special thermodynamic equilibrium states called the thermal Gibbs statedetermined only by the temperature parameter T > 0. They are given by the Maxwell-Boltzmann distribution (see Exercise 1.12)

p(x1, . . . , xN , v1, . . . , vN ) = L−3NN∏

i=1

[(mi

2πkBT

)3/2

exp

(−

12miv

2i

kBT

)], (1.23)

where the constant kB is Boltzmann’s constants (1.38 × 10−23 Joule/Kelvin). Since theprobability distribution depends only on the lengths of the velocities it is a thermodynamicequilibrium state.

We have used that the kinetic energy of a particle of mass mi and velocity vi is Ekin =12miv

2i (for an explanation of this see the next section). In this thermal state the expected

energy isN∑

i=1

∫12miv

2i p(x1, . . . , xN , v1, . . . , vN )d3Nxd3Nv =

32NkBT. (1.24)

(See Exercise 1.13). This gives the relation U = 32NkBT between the internal energy U

and the temperature of an ideal gas of N particles. In Exercise 1.14 we show how to arriveat the ideal gas equation

PV = NkBT (1.25)

relating the volume V = L3, pressure P and temperature T for this system.

1.5 Conservative Force Fields“in mind” addedin version of Nov.12We will in this section consider k-dimensional spaces, for physical space, of course, we will

have the situation k = 3 in mind. For many purposes it is, however, interesting to considerthe general case. We could either have other degrees of freedom or, as we have seen, severalparticles.

In this section we shall discuss in more details time independent force fields and theconcept of work and energy .

Definition 1.4 (Parametrized curve). A parametrized curve or motion on Rk is a contin-uous map γ : I → Rk defined on an interval I ⊆ R. We say that γ is piecewise-C1 if it hasa continuous derivative except at finitely many points, where the left and right derivativesmake sense, but may be different. The derivative γ(t) is called the velocity at time t.


Definition 1.5 (Curve). We say that two parameterized curves γ : I → Rk and γ : J → Rk

are reparametrizations of each other (or equivalent) if there exists a strictly monotonecontinuous function φ : J → I, with φ(J) = I (surjectivity) such that γ = γ φ. In thiscase we say that the two parametrized curves represent the same curve (or more preciselythat a curve is an equivalence class of parametrized curves). If we only identify parametrizedcurves by monotone increasing reparametrizations φ we say that the curve is oriented. Acurve is called smooth if it can be represented by a C1-parametrization γ : I → Rk withnon-vanishing velocity, i.e., γ(t) 6= 0 for all t ∈ I.

Definition 1.6 (Work Integral). Let F : O → Rk be a continuous force field on an openset O ⊆ Rk and γ : I → O be a piecewise C1 motion. We define the line or work integralof the force F along γ to be

Wγ(F ) =∫

IF (γ(t)) · γ(t)dt.

An often used notation for the work integral is

Wγ(F ) =∫

γF · dr.

An important property of the work integral is that it depends only on the oriented curverepresented by γ and not on the details of the motion. We formulate this more precisely.

Lemma 1.7. If φ : J → I is an increasing2 C1-map between intervals I, J ⊆ R, then

Wγ(F ) = Wγφ(F ).

Proof. This is a simple calculation using integration by substitution

Wγφ(F ) =∫

JF (γ(φ(s)) · γ(φ(s))φ′(s)ds =

∫IF (γ(t)) · γ(t)dt = Wγ(F ).

The notion of work is closely related to the notion of energy. The kinetic energy of aparticle with mass m and velocity v is Ekin = 1

2mv2. We may rephrase Newton’s second

law (1.18) in terms of work and kinetic energy.

Theorem 1.8 (Energy and work). If a particle of mass m performs a C2-motion γ : I →O subject to Newton’s second law (1.18) under the influence of a continuous force fieldF : O → Rk then

12mγ(t2)2 −

12mγ(t1)2 =

∫ t2

t1

F (γ(t)) · γ(t)dt

for all t1, t2 ∈ I. The left side above is the change in kinetic energy from t1 to t2. Theright side is the work integral of F along the motion γ from t1 to t2 (assuming here thatt1 ≤ t2).

2It is not necessary for the map φ to be monotone, it only has to map the left endpoint to the leftendpoint and the right endpoint to the right endpoint.


Proof. By differentiation and using Newton’s second law we arrive at

d

dt

(12mγ(t)2

)= mγ(t) · γ(t) = F (γ(t)) · γ(t),

which proves the above identity by the fundamental theorem of calculus.

We explained above that the work integral depends only on the curve and not on thedetails of the motion representing it. We now consider a very important class of force fieldsfor which the work integral depends only on the end points of the motion. An equivalentway of saying this is that the work along any closed path (a path that ends and begins atthe same point) is zero (see Exercise 1.15).

Definition 1.9 (Conservative Force Field). A continuous force field F : O → Rk is saidto be conservative on the open set O ⊆ Rk if for all piecewise C1 motions γ : [a, b] → Owith γ(a) = γ(b) we have

Wγ(F ) =∫ b

aF (γ(t)) · γ(t)dt = 0.

As we shall see below the gravitational force field 1.17 is conservative on the set whereit is defined (i.e., when no two points are on top of each other).

Examples of force fields which are not conservative are friction forces, characterizedby their work integral always being negative. Friction forces are characterized by beingdirected opposite to the velocity. Since they depend on velocity these forces do not fallinto the general discussion above. One can however still define their work integral.

Conservative force fields may be defined by a potential function.

Theorem 1.10. A continuous force field F : O → Rk on an open set O ⊆ Rk is conserva-tive if and only if there exists a C1-function V : O → R such that F = −∇V .

Proof. If there exists a function V as stated in the theorem we have for all piecewise C1

paths γ : [a, b] → O V (γ(b)) −V (γ(a)) →V (γ(a))−V (γ(b))in version ofNov. 16

∫ b

aF (γ(t)) · γ(t)dt = −

∫ b

a∇V (γ(t)) · γ(t)dt = −

∫ b

a

d

dtV (γ(t))dt = V (γ(a))− V (γ(b))

which vanishes if γ(a) = γ(b) and thus F is conservative.On the other hand if F is conservative we will now show how to define a function V

such that −∇V = F . For simplicity let us assume that that there exists a point x0 ∈ Osuch that for all points x ∈ O we may find a piecewise C1 path γ : [a, b] → O such thatγ(a) = x0 and γ(b) = x. A set with this property is said to be pathwise connected. Thisassumption is only for simplicity. Any open subset of Rk may be written as a disjointunion of open pathwise connected sets. This follows since the relation that two points areconnected by a piecewise C1 path is an equivalence relation and the equivalence classes areopen.

We now simply define

V (x) = −∫ b

aF (γ(t)) · γ(t)dt


where γ : [a, b] → O is any piecewise C1 path with γ(a) = x0 and γ(b) = x. Since F isconservative the value of V above does not depend on the choice of path γ from x0 to x(see Exercise 1.15).

In order to differentiate V at x = (x1, . . . , xk) we use that O is open and choosea ball B ⊆ O centered at x. If we vary the first coordinate arriving at another pointx′ = (x′1, x2, . . . , xk) in the ball B, we may connect x to x′ by the path going along thefirst coordinate axis. We may then connect x0 to x′ by first connecting x0 to x followed bythe coordinate path from x to x′. From the definition of V we see that

V (x′)− V (x) = −∫ x′1

x1

F1(t, x2, . . . , xk)dt,

where F1 is the first coordinate of F . It now follows from the fundamental theorem ofcalculus that

V (x′)− V (x)x′1 − x1

→ −F1(x)

as x′1 → x1. Since this is true for all coordinates we have ∇V (x) = −F (x). Since F iscontinuous it follows that V is C1.

The above theorem allows us to formulate the theorem of conservation of energy.C1 → C2 in ver-sion of Nov. 16Theorem 1.11 (Conservation of Energy). If a particle of mass m performs a piecewise C2

motion γ : I → Rk according to Newton’s second law under the influence of a conservativeforce field F = −∇V then the total energy

E(t) =12mγ(t)2 + V (γ(t))

is conserved. We call the term V (γ(t)) the potential energy.

Proof. By differentiation we have from Newton’s second law

E′(t) = mγ(t) · γ(t) +∇V (γ(t)) · γ(t) = (mγ(t)− F (γ(t))) · γ(t) = 0.

Theorem 1.12. All rotationally invariant force fields F (x) = f(‖x‖2)x, where f : R+ → Rare conservative

Proof. For a piecewise C1-motion γ : [a, b] → Rk we have

Wγ(F ) =∫ b

af(‖γ(t)‖2)γ(t) · γ(t)dt

=12

∫ b

af(‖γ(t)‖2)

d‖γ(t)‖2

dtdt =

12

∫ ‖γ(b)‖2

‖γ(a)‖2f(s)ds,

which explicitly depends only on the endpoints γ(a) and γ(b).


If we compare with how the potential V was defined in the proof of Theorem 1.10 wesee from the above proof that the potential of a spherically symmetric force field may bechosen spherically symmetric, i.e., to depend only on ‖x‖.

This shows in particularly that the gravitational force is conservative. Below we shallwrite the potential for the gravitational field explicitly.

For N particles moving in R3 we say that they are subject to conservative force fieldsif the force field is conservative as a field on R3N . This means that there is a functionC1-function V on R3N such that the force on particle i is

Fi = −∇xiV (x1, . . . , xN ).

If the N particles have masses m1, . . . ,mN and perform a C2 motion (γ1(t), . . . , γN (t))according to Newton’s second law then the total energy

E(t) =∑

i

12miγi(t)2 + V (γ1(t), . . . , γN (t))

is conserved. Indeed,

E′(t) =∑

i

miγi(t) · γi(t) +∑

i

∇xiV (γ1(t), . . . , γN (t)) · γi(t) = 0.

The gravitational force between N particles of masses m1, . . . ,mN located at pointsx1, . . . , xN in R3 is conservative. In fact, if we define

V (x1, . . . , xN ) = −∑

1≤i<j≤N

Gmimj

‖xi − xj‖(1.26)

then−∇xiV (x1, . . . , xN ) = −

∑j 6=i

Gmimjxi − xj

‖xi − xj‖3(1.27)

in accordance with (1.17) (see Exercise 1.16). Equation numbercorrected in ver-sion of Nov. 20

Theorem 1.13. Given a system of N particles in R3 subject to a conservative force fieldFi(x1, . . . , xN ) = −∇xiV (x1, . . . , xN ), i = 1, . . . , N , which is invariant under translations.Then there exists a function W (y2, . . . , yN ) of N−1 variables and a constant vector F0 ∈ R3

such that

V (x1, . . . , xN ) = W (x2 − x1, . . . , xN − x1) + F0 · (x1 + . . .+ xN ).

The second term is the potential of a constant force F0. One refers to the first term, whichdepends on the distances between the particles only, as the interior potential and the lastterm as an exterior constant force. (The fact that the first term above is written with x1

as the reference point is not important, see Exercise 1.17.)

Proof. Translation invariance means that F (x1, . . . , xN ) = F (x1 − a, . . . , xN − a), i.e.,

∇xiV (x1, . . . , xN ) = ∇xiV (x1 − a, . . . , xN − a)


for all a ∈ Rk and all i = 1, . . . , N . This implies that

V (x1, . . . , xN ) = V (x1 − a, . . . , xN − a) + g(a),

for all a ∈ Rk and some function g : Rk → R with g(0) = 0. Differentiating this equationwrt. a gives by the chain rule

∇g(a) =N∑

i=1

∇xiV (x1 − a, . . . , xN − a) =N∑

i=1

∇xiV (x1, . . . , xN ).

In other words both sides of this equation are constant. Hence g(a) = b · a+ c for a vectorb and a constant c. Since g(0) = 0 we have c = 0. Hence g(a) = b · a.

Choosing a = x1 gives

V (x1, . . . , xN ) = V (0, x2 − x1, . . . , xN − x1) + b · x1

= V (0, x2 − x1, . . . , xN − x1)−1Nb · ((x2 − x1) + . . .+ (xN − x1))

+1Nb · (x1 + . . .+ xN )

= W (x2 − x1, . . . , xN − x1) + F0 · (x1 + . . .+ xN ),

whereW (y2, . . . , yN ) = V (0, y2, . . . , yN )− 1

Nb · (y2 + . . .+ yN )

and F0 = 1N b.

If we have two particles then the interior potential is of the form W (x2−x1) and hencethe force with which particle 1 acts on particle 2 is −∇W (x2 − x1) which is equal andopposite to ∇W (x2 − x1), the force with which particle 2 acts on particle 1 . This is theaction-reaction principle or Newton’s third law. If we also assume that the force field isinvariant under rotations we get the strong action-reaction principle that the equal andopposite forces are along the line joining the particles (see Exercise 1.7).

1.6 Exact and closed differentials

In the previous section we discussed conservative forces and we characterized them as fieldsF , which could be written as the gradient of a function. details → detail in

version of Nov. 16In this section we want to discuss this type of problem in more detail and give otherapplications in physics. modeled → mod-

elled in version ofNov. 16In the previous sections we discussed functions and vector fields on the physical space

modelled by Rk. In this case it makes sense to think of the gradient of a function as a vectorfield in space. It is well known that the gradient makes sense as a vector independently ofthe orthogonal coordinate system chosen to calculate it (see Exercise 1.18). It can indeedbe characterized as pointing in the direction in which the function has the fastest increaseand its size is the derivative in this direction.

In general, however, we may in physics consider functions f of several variables, wherethe variables are fundamentally different. In thermodynamics for instance one may consider


the pressure P (U, V ) of a system as a function of the total energy U and the volume Vof the system. In this case it is not too meaningful to think of the gradient of P as avector. In fact, the two components ∂P

∂U and ∂P∂V have different units, hence talking about

the lengths of such a vector or the angle between such vectors is meaningless.Instead one introduces the notion of the differential

dP =∂P

∂UdU +

∂P

∂VdV.

In physics dP, dU, dV are often interpreted as infinitesimal quantities.We will here give a rigorous definition of the differential of a function and give a precise

meaning to the expression above.

Definition 1.14 (Differential of a function). If f : O → R is a C1- function on an open setO ⊆ Rk we define the differential of f at the point x ∈ O to be the linear map dfx : Rk → Rgiven by

dfx(X) =k∑

i=1

∂f

∂xi(x)Xi, X ∈ Rk

In other words dfx(X) is the directional derivative of f at the point x in the direction givenby the vector X.

As usual, we identify the coordinates xi, i = 1, . . . , k with the coordinate functions “Its differential is”→ “The differen-tial of xi is “ inversion of Nov. 16

x = (x1, . . . , xk) 7→ xi. The differential of xi is

dxi(X) = Xi.

Note that dxi is the same map at all points. Hence we may write

dfx =∂f

∂x1(x)dx1 + . . .

∂f

∂xk(x)dxk,

or simply

df =∂f

∂x1dx1 + . . .

∂f

∂xkdxk,

where we have omitted the subscript referring to the point in which the differential iscalculated.

Definition 1.15 (General differential). A general differential or 1-form ω on an opensubset of O ⊆ Rk is a continuous map from O to the linear functions from Rk to R of theform3

ωx =k∑

i=1

ωi(x)dxi,

where the maps ωi : O → R, i = 1, . . . , k are continuous. Thus for a vector X ∈ Rk wehave

ωx(X) =k∑

i=1

ωi(x)Xi

3Any map from O to the linear functions from Rk to R can be written in this form. The extra informationis that the coordinate functions ωi are continuous.


If ω is the differential of a function, i.e., if there is a C1 function f : O → R such thatω = df we say that ω is an exact differential.

If F : O → Rk is a vector field we have the corresponding differential ωF = F1dx1 +. . .+ Fkdxk, i.e., for any vector X ∈ Rk

ωF (X) = F ·X = 〈F,X〉.

Thus the correspondence between vector fields and differentials requires the scalar productto make sense, i.e., that it makes sense to calculate lengths and angles.

For vector fields we had the notion of line or work integral. We have a correspondingnotion for differentials.

Definition 1.16. If ω =∑k

i=1 ωidxi is a differential on an open set O ⊆ Rk and γ : I → Ois a piecewise C1 motion we define the line integral of ω along γ∫

γω =

∫Iωγ(t)(γ(t))dt =

k∑i=1

∫Iωi(γ(t))γi(t)dt.

Theorem 1.10 for vector fields may be translated to a statement about differentials.

Theorem 1.17 (Characterization of exact differentials). A differential ω defined on anopen set O ⊆ Rk is exact if and only if

∫γ ω = 0 for any closed parametrized curve γ :

[a, b] → O, i.e., a curve for which γ(a) = γ(b).

For a C2 function f : O → Rk we have the well-known property that

∂2f

∂xi∂xj=

∂2f

∂xj∂xi.

Thus a C1-differential ω =∑k

i=1 ωidxi (where C1 means that the coordinate functions ωi

are C1) can only be exact if∂ωi

∂xj=∂ωj

∂xi, (1.28)

for all i, j = 1, . . . , k.

Definition 1.18 (Closed differentials). A C1 differential ω is said to be closed if it satisfiesthe relation (1.28) for all i, j = 1, . . . , k.

We have seen that exact differentials are necessarily closed. The interesting questionis whether closed differentials are necessarily exact. It turns out that this depends on thedomain on which the differential is defined, more precisely it depends on the topology ofthe domain. The set should have a property referred to as being simply connected4.

Theorem 1.19. If the open set O ⊆ Rk has the property that each of its connected com-ponents are simply connected then a differential form ω defined on O is exact if and onlyif it is closed.

4An open set is simply connected if any closed curve in the set can be continuously deformed to a pointentirely within the set.


We will not prove this theorem here. It requires the study of surface integrals and inparticular Stoke’s integral formula (or in 2 dimensions Green’s integral formula).

The set R2 \ 0 is not simply connected. In Exercise 1.19 an example is given of adifferential which is closed but not exact on this set.

Example 1.20 (Magnetic fields in 2-dimensions). A charged particle moving in an opensubset O of the 2-dimensional plane R2 may be subject to a magnetic field perpendicularto the plane. The magnetic field is described by a function B : O → R which is the thirdcomponent of the 3-dimensional magnetic field vector.

If we are given a differential α = α1dx + α2dy on O we say that α is the magneticvector potential corresponding to the magnetic field B = ∂α2

∂x − ∂α1∂y . Hence if α is closed

the corresponding magnetic field vanishes. Strangely, this does not mean that the vectorpotential has no effect. This would only be the case if the differential is exact.

In quantum mechanics if one considers a charged particle moving in the set R2 \ 0under the influence of the magnetic vector potential given by the closed differential inExercise 1.19 then the motion of the particle is not the same as that of a free particle. Thisvector potential is called the Bohm-Aharonov vector potential and the effect is seen onlyin quantum mechanics.

Definition 1.21 (Integrating factor). If ω is a differential on an open set O ⊆ Rk, we saythat a function τ : O → R with τ not identically equal to zero, is an integrating factor forω if the differential τω is exact.

The following theorem is a consequence of a more general version of the existence anduniqueness for ordinary differential equations stated in Theorem 1.3.

Theorem 1.22 (Integrating factors in 2 dimensions). In 2-dimensions any differentiallocally has an integrating factor.

Example 1.23 (Integrating factors, temperature and entropy). As a last example wediscuss a 3-dimensional thermodynamical system described by the internal energy U andthe volume V 5. Such a system has a pressure P (U, V ). The pressure is the force per areaexerted by the system on the boundary of its domain.

An important differential studied in thermodynamics is the heat differential

dU + P (U, V )dV

The importance of this differential can heuristically be described as follows. If the systemreceives an infinitesimal energy this will be either stored as an infinitesimal internal energydU or used for the system to do work, i.e., to change its volume infinitesimally. The workdone by the system in changing its volume by dV is PdV (why?). Thus the total energyreceived by the system equals dU + PdV .

The heat differential is not exact, but it has an integrating factor, which is the reciprocaltemperature (this is indeed a way to introduce the temperature). Thus there exists afunction S(U, V ) such that

dS = T (U, V )−1(dU + PdV ).5In thermodynamics only the volume of the region in which the system lives plays a role. The geometry

of the region is unimportant


The function S, which is determined in this way, up to an overall additive constant iscalled the entropy of the system. The fact that the heat differential has an integratingfactor follows at least locally from Theorem 1.22, since the system depends only on 2variables U and V .

There are however also systems that depend on more than two variables. An exampleis if we have several, say M , thermodynamic systems as above, which are in thermalcontact, this means that they can transfer their internal energies. Such a combined systemis described by the total internal energy U and the volumes V1, V2, . . . , VM of the individualsubsystems.

In this case it is an important result which follows from the Second Law of Thermody-namics (one of the fundamental principles of thermodynamics) that the heat differential

dU + P1dV1 + . . .+ PMdVM ,

where P1, . . . , PM are the pressures in each subsystem, has the common temperature of thecombined system as integrating factor, i.e., we again have

TdS = dU + P1dV1 + . . .+ PMdVM ,

where S is the entropy. Here P1, P2,. . . ,PM , S, and T are functions of U, V1, . . . , VM .

Exercises

Exercise 1.1. Prove the following statements relating to the scalar product 〈·, ·〉 on R3.

a) If the vectors e1, e2, e3 ∈ R3 fulfill

〈ei, ej〉 =

1 if i = j

0 if i 6= j,

then (e1, e2, e3) is a basis for R3. It is called an orthonormal basis for R3.

b) Let (e1, e2, e3) be an orthonormal basis for R3 and let v ∈ R3. Use the linearityproperties of the scalar product to show that the coordinate set (x1, x2, x3) of v withrespect to the basis (e1, e2, e3) is given by

xi = 〈v, ei〉 , i = 1, 2, 3 .

c) Use the linearity properties of the scalar product to verify the identities

〈v, w〉 =12‖v‖2 +

12‖w‖2 − 1

2‖v − w‖2

〈v, w〉 =14‖v + w‖2 − 1

4‖v − w‖2


d) Derive the Cauchy-Schwarz inequality

|〈v, w〉| ≤ ‖v‖‖w‖ , v, w ∈ R3 .

Hint. One possible way to proceed is to consider the polynomial f(x) = ‖v + xw‖2

and use that, since it is nonnegative, its discriminant must be less than or equal tozero.

Exercise 1.2. Let S : R3 → R3 be an isometry. Prove that

S(v) = a+ S0(v) ,

where a ∈ R3 is a fixed vector and S0 : R3 → R3 is a linear isometry, in the following twosteps.

a) Seta = S(0)

and define S0 byS0(v) = S(v)− a , v ∈ R3 .

Verify that S0(0) = 0 and that S0 is an isometry and use the first identity in Exercise1c) to show

〈S0(v), S0(w)〉 = 〈v, w〉 , v, w ∈ R3 .

b) Show, using a), that

‖S0(λv)− λS0(v)‖2 = 0‖S0(v + w)− S0(v)− S0(w)‖2 = 0 ,

and conclude that S0 is linear.

Exercise 1.3. Assume S0 : R3 → R3 is an orthogonal transformation and let A be the3× 3-matrix representing S0 with respect to the standard basis (ε1, ε2, ε3) for R3, that is

S0(v) = Av , v ∈ R3 .

a) Show that the columns of A constitute an orthonormal basis for R3 and that this isequivalent to

AtA = I ,

where At denotes the transpose of A (defined by substituting the columns of A bythe corresponding rows), and I is the 3× 3 identity matrix.

Conclude that A is invertible and that

A−1 = At . (1.29)

Matrices fulfilling (1.29) are called orthogonal matrices and the set of orthogonal3× 3-matrices is denoted by O(3).

b) Show that A is an orthogonal matrix if and only if the corresponding orthogonaltransformation S0 maps an orthonormal basis to an orthonormal basis.


c) Show that I is an orthogonal matrix and that, if A,B are orthogonal matrices, thenAB and A−1 are also orthogonal matrices.

Exercise 1.4. In this exercise A denotes an orthogonal 3 × 3-matrix and S0 the corre-sponding linear map, S0v = Av , v ∈ R3.

a) Let λ ∈ R. Show that the equation

Av = λv

has a solution v 6= 0 if and only if

det(A− λI) = 0 . (1.30)

Show that the left hand side of (1.30) is a third order polynomial in λ and that ithas at least one root λ1 ∈ R and conclude from ‖Av‖ = ‖v‖ that

λ1 = ±1.

b) Explain that there exists a vector v1 ∈ R3 such that

Av1 = λ1v1 and ‖v1‖ = 1 ,

and that, given such a vector v1, there exist vectors v2, v3 ∈ R3 such that (v1, v2, v3)is an orthonormal basis for R3.

Show that the matrix representing S0 with respect to the basis (v1, v2, v3) is of theform λ1 0 0

0 a b0 c d

,

where the columns (0, a, c) and (0, b, d) are orthogonal and of norm 1. Show that itcan be written in one of the following two forms:

B1 =

λ1 0 00 cos θ − sin θ0 sin θ cos θ

B2 =

λ1 0 00 cos θ sin θ0 sin θ − cos θ

.

for some θ ∈ [0, 2π[.

c) i) If λ1 = 1, show that B1 represents a rotation through an angle θ around thev1-axis, i.e. show that vectors parallel to v1 are left invariant by S0 and that if v isorthogonal to v1 then S0(v) is also orthogonal to v1 with same norm and forming anangle θ with v. λ → λ1 3 times in

version of Nov. 13

ii) If λ1 = 1, show that B2 represents a rotation through an angle θ around thev1-axis composed with reflection in the (v1, v2)-plane.

iii) If λ1 = −1, show that B1 represents a rotation through an angle θ around thev1-axis composed with reflection in the (v2, v3)-plane.

iv) If λ1 = −1, show that B2 represents a rotation through an angle π around theaxis obtained by rotating the v2-axis through an angle θ/2 in the (v2, v3)-plane.


Exercise 1.5. Verify formulas (1.8), (1.9) and (1.10).[a, s, v, A] →[0, 0, 0, A] inversion of Nov. 16

Exercise 1.6. Show that, for a system of N particles, the principle of relativity impliesthat the force acting on a particle can depend only on the spatial coordinate differencesand the velocity differences in an inertial system of coordinates, and that it transformsaccording to eq.(1.13) under an inertial coordinate transformation ψ[0, 0, 0, A].

Exercise 1.7. Show, for the case of two interacting particles, that if the forces Fi, i = 1, 2,are of the form

F1 = g1(‖x1 − x2‖)(x2 − x1) F2 = g2(‖x1 − x2‖)(x1 − x2) ,

where g1 and g2 are C∞-functions on the positive real axis, then the equations of motion(1.12) are invariant under inertial coordinate transformations.

Show also, that if the forces are assumed to be independent of velocities then invarianceof the equations of motion under inertial coordinate transformations implies that the forcesmust be of the form given above (for x1 6= x2).

Exercise 1.8. Consider the first order differential equation

dx

dt= f(x, t) ,

where the continuous function f : R2 → R is defined by

f(x, t) =

√x if x ≥ 0

0 if x < 0 .

Find two different solutions x1 and x2 such that x1(0) = x2(0) = 0.

Exercise 1.9. (a) Show that the force field F1(x, y) = (x2 + y, y2 + x) is conservative onR2.

(b) Find the work integral of F1 along a motion going in a straight line from (0, 0) to (1, 0)and then in a straight line from (1, 0) to (1, 1).

(c) Find the work integral for the force field F2(x, y) = (y2, x2) along the same motion asabove.

(d) Is the force field F2 above conservative?

Exercise 1.10. Find all maximal solutions to the first order differential equation

dx

dt= 1 + x2 , (x, t) ∈ R2 ,

and verify that they are not globally defined.

Exercise 1.11. Prove the relations (1.22) for the phase flow.

Exercise 1.12. Use the fact that∫∞−∞ e−x2

dx =√π to show that the Maxwell-Boltzmann

distribution (1.23) is a probability distribution, i.e., that∫p(x1, . . . , xN , v1, . . . , vN )d3Nxd3Nv = 1


Exercise 1.13. Use the fact that ddα

∫∞−∞ e−αx2

dx = −∫∞−∞ x2e−αx2

dx (can you provethis?) and

∫∞−∞ e−x2

dx =√π to show the identity (1.24). Show that the corresponding

relation in k space dimensions would be U = k2NkBT . This is often referred to as the

equipartition of energy that each degree of freedom contributes an equal amount 12NkBT

to the total energy. If the particles have internal degrees of freedom they will also contributeto the total energy.

Exercise 1.14 (Ideal gas equation). (Difficult). We study again the system of N particlesin the box [0, L]3 given by the Maxwell-Boltzmann distribution (1.23). When a particleof velocity v = (v(1), v(2), v(3)) (where the superscripts refer to the 3 coordinates) hits the sub → super in

version of Nov. 20wall x(1) = 0 its velocity changes to (−v(1), v(2), v(3)) (notice that we must have v(1) < 0).If it has mass m its momentum has therefore changed by −2mv(1) in the first coordinatedirection. The lost momentum, i.e., 2mv(1) is said to be transferred to the wall. The totaltransfer of momentum in absolute value per unit time and unit area is called the pressureP of the system.

Argue that the the expected transfer of momentum to the wall per unit time and unitarea is given by

L−2N∑

i=1

∫0<x

(1)i <−v

(1)i

|2miv(1)i |p(x1, . . . , xN , v1, . . . , vN )d3Nxd3Nv.

Use the result of the previous exercise to show that this integral is L−3NkBT . Since thetransfer of momentum per unit time and unit area is the pressure P on the wall we havearrived at the ideal gas equation (1.25).

Exercise 1.15. Show that a force field is conservative if and only if any work integral ofthe field depends only on the endpoints of the motion.

Exercise 1.16. Prove the relation (1.27) for the gravitational force.

Exercise 1.17. Given a function W (y2, . . . , yN ) of N − 1 variables in Rk show that thereexists a function W ′ such that

W (x2 − x1, x3 − x1, . . . , xN − x1) = W ′(x1 − x2, x3 − x2 . . . , xN − x2).

Exercise 1.18. (a) Let f : Rk → R be a C1-function. For any k × k matrix let gA(x) =f(Ax). Show that ∇gA(x) = At∇f(Ax).

(b) We want to investigate how the gradient changes under a coordinate transformationgiven by an invertible k × k matrix A. Thus introduce new coordinates y = Ax.In the new coordinates a C1-function f : R3 → R becomes g(y) = f(A−1y). Show,using the result of the previous question, that if and only if the matrix A representsan orthogonal transformation (see Exercise 1.3 for the case k = 3) does the gradienttransform like a vector, i.e., ∇g(y) = A∇f(A−1y), for all functions f .

Exercise 1.19. Consider the differential

α(x,y) = − y

x2 + y2dx+

x

x2 + y2dy

defined on the set R2 \ 0.


(a) Show that α is a closed differential.

(b) Show that if γ : [0, 2π] → R2 is the motion γ(t) = (cos(t), sin(t)) then∫γ α 6= 0 and

conclude that α is not exact.

Exercise 1.20 (Entropy of an ideal gas). Consider an ideal gas of N particles. We haveseen that the pressure P (U, V ) and temperature T (U, V ) are given as functions of theenergy U and volume V from the relations

V P (U, V ) = NkBT (U, V ), U =32NkBT (U, V ).

Show that T (U, V )−1 is an integrating factor for the heat differential

dU + P (U, V )dV

and that the entropy must be given by

S(U, V ) = NkB ln((V/N)(U/N)3/2

)+ CN

for some constant C independent of U and V . Show that C is independent of N if weassume that S, V , and U are proportional to N .

Exercise 1.21. Let F : Ω → R3 be a C1 vector field and γ : [a, b] → Ω a piecewise C1 R3 → Ω in vesionof Nov. 3parametrized curve and ϕ : [c, d] → [a, b] a strictly monotone C1-function with ϕ(c) = b

and ϕ(d) = a. Let γ = γ ϕ.Show that dx → dr in ver-

sion of Nov. 20∫γF · dr = −

∫γF · dr

Exercise 1.22. Show that the vector fields

F (x, y) = (−y, x) , (x, y) ∈ R2 ,

and

G(x, y) =

(−y√x2 + y2

,x√

x2 + y2

), (x, y) ∈ R2 \ (0, 0) ,

are not conservative.Hint. Evaluate the integrals of F and G along a suitable closed curve.

Exercise 1.23. Consider a particle in one dimension with potential energy U(x) = −x4,i.e. its equation of motion is

mx = −U ′(x) = 4x3 .

a) Show that the energy x removed in ver-sion of Nov. 20

E =12mx2 − x4

is conserved.


b) Use a) to show that for a solution x(t), t ∈ [t1, t2], the time t2 − t1 it takes to movefrom a point x1 to a point x2 is given by

t2 − t1 =∫ x2

x1

dx√2m(E + x4)

,

assuming that x1 < x2 and that x 6= 0 on [t1, t2].

c) Use b) to show that, for E > 0, the particle escapes to infinity in finite time, i.e. thesolution is not globally defined.

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Newtonian Mechanics - kuweb.math.ku.dk/~solovej/MATFYS/MatFys1.pdfNewtonian Mechanics In these notes classical mechanics will be viewed as a mathematical model for the descrip-tion