A Space and time

B Galileo’s principle of relativity

C Newton’s principle of determinacy

A Notation

Affine n-dimensional space Aⁿ is distinguished from ℝⁿ in that there is “no fixed origin.” The group ℝⁿ acts on Aⁿ as the group of parallel displacements (Figure 1):

[Thus the sum of two points of Aⁿ is not defined, but their difference is defined and is a vector in ℝⁿ.]

Figure 1

Parallel displacement

A euclidean structure on the vector space ℝⁿ is a positive definite symmetric bilinear form called a scalar product. The scalar product enables one to define the distance

between points of the corresponding affine space Aⁿ. An affine space with this distance function is called a euclidean space and is denoted by Eⁿ.

B Galilean structure

The galilean space-time structure consists of the following three elements:

1.

The universe—a four-dimensional affine⁴ space A⁴. The points of A⁴ are called world points or events. The parallel displacements of the universe A⁴ constitute a vector space ℝ⁴.

2.

Time—a linear mapping t : ℝ⁴ → ℝ from the vector space of parallel displacements of the universe to the real “time axis.” The time interval from event a ∈ A⁴ to event b ∈ A⁴ is the number t(b − a) (Figure 2). If t(b − a) = 0, then the events a and b are called simultaneous.

Figure 2

Interval of time t

The set of events simultaneous with a given event forms a three-dimensional affine subspace in A⁴. It is called a space of simultaneous events A³.

The kernel of the mapping t consists of those parallel displacements of A⁴ which take some (and therefore every) event into an event simultaneous with it. This kernel is a three-dimensional linear subspace ℝ³ of the vector space ℝ⁴.

The galilean structure includes one further element.

3.

The distance between simultaneous events

is given by a scalar product on the space ℝ³. This distance makes every space of simultaneous events into a three-dimensional euclidean space E³.

We mention three examples of galilean transformations of this space. First, uniform motion with velocity v:

Next, translation of the origin:

Finally, rotation of the coordinate axes:

where G : ℝ³ → ℝ³ is an orthogonal transformation.

Let M be a set. A one-to-one correspondence φ₁ : M → ℝ × ℝ³ is called a galilean coordinate system on the set M. A coordinate system φ₂ moves uniformly with respect to φ₁ if is a galilean transformation. The galilean coordinate systems φ₁ and φ₂ give M the same galilean structure.

C Motion, velocity, acceleration

The derivative

is called the velocity vector at the point t₀ ∈ I.

The second derivative

is called the acceleration vector at the point t_o.

Problem. Is it possible for the trajectory of a differentiable motion on the plane to have the shape drawn in Figure 3? Is it possible for the acceleration vector to have the value shown?

Figure 3

Trajectory of motion of a point

Answer. Yes. No.

A curve in galilean space which appears in some (and therefore every) galilean coordinate system as the graph of a motion, is called a world line (Figure 4).

Figure 4

World lines

The direct product of n copies of ℝ³ is called the configuration space of the system of n points. Our n mappings x_i: ℝ → ℝ³ define one mapping

of the time axis into the configuration space. Such a mapping is also called a motion of a system of n points in the galilean coordinate system on ℝ × ℝ³.

D Newton’s equations

In particular, the initial positions and velocities determine the acceleration. In other words, there is a function F: ℝ^N × ℝ^N × ℝ→ ℝ^N such that

(1)

Newton used Equation (1) as the basis of mechanics. It is called Newton’s equation.

E Constraints imposed by the principle of relativity

If we subject the world lines of all the points of any mechanical system⁹ to one and the same galilean transformation, we obtain world lines of the same system (with new initial conditions) (Figure 5).

Figure 5

Galileo’s principle of relativity

From this it follows that the right-hand side of Equation (1) in an inertial coordinate system does not depend on the time:

From invariance under passage to a uniformly moving coordinate system (which does not change ẍ_i or x_j − x_k, but adds to each ẋ_j a fixed vector v) it follows that the right-hand side of Equation (1) in an inertial system of coordinates can depend only on the relative velocities

Thus, if φ_i: ℝ→ ℝ³(i = 1,..., n) is a motion of a system of points satisfying (1), and G: ℝ³ → ℝ³ is an orthogonal transformation, then the motion Gφ_i: ℝ→ ℝ³(i,..., n) also satisfies (1). In other words.

where Gx denotes (Gx₁,...,Gx_n), x_i ∈ ℝ³.

A Example 1: A stone falling to the earth

Experiments show that

* (2)

where x is the height of a stone above the surface of the earth.

If we introduce the “potential energy” U = ɡx, then Equation (2) can be written in the form

Experiments show that the radius vector of the stone with respect to some point 0 on the earth satisfies the equation

(3)

The vector in the right-hand side is directed towards the earth. It is called the gravitational acceleration vector g. (Figure 6.)

Figure 6

A stone falling to the earth

B Example 2: Falling from great height

Like all experimental facts, the law of motion (2) has a restricted domain of application. According to a more precise law of falling bodies, discovered by Newton, acceleration is inversely proportional to the square of the distance from the center of the earth:

where r = r₀ + x (Figure 7).

Figure 7

The earth’s gravitational field

This equation can also be written in the form (3), if we introduce the potential energy

(3)

inversely proportional to the distance to the center of the earth.

C Example 3: Motion of a weight along a line under the action of a spring

Experiments show that under small extensions of the spring the equation of motion of the weight will be (Figure 8)

Figure 8

Weight on a spring

This equation can also be written in the form (3) if we introduce the potential energy

It is experimentally established that for any two bodies the ratio of the accelerations ẍ₁/ẍ₂ under the same extension of a spring is fixed (does not depend on the extent of extension of the spring or on its characteristics, but only on the bodies themselves). The value inverse to this ratio is by definition the ratio of masses:

D Example 4: Conservative systems

Definition. The motion of n points, of masses m₁,..., m_n, in the potential field with potential energy U is given by the system of differential equations

(4)

A Definitions

A system with one degree of freedom is a system described by one differential equation

(1)

The kinetic energy is the quadratic form*

The potential energy is the function

The sign in this formula is taken so that the potential energy of a stone is larger if the stone is higher off the ground.

The total energy is the sum

In general, the total energy is a function, , of x and ẋ.

Theorem (The law of conservation of energy). The total energy of points moving according to the equation (1) is conserved: is independent of t.

Proof.

□

B Phase flow

Equation (1) is equivalent to the system of two equations:

(2)

The image of φ is called the phase curve. Thus the phase curve is given by the parametric equations

C Examples

Example 1. The basic equation of the theory of oscillations is

In this case (Figure 9) we have:

The energy level sets are the concentric circles and the origin. The phase velocity vector at the phase point (x, y) has components (y, − x). It is perpendicular to the radius vector and equal to it in magnitude. Therefore, the motion of the phase point in the phase plane is a uniform motion around 0: x = r₀ cos(φ₀ − t), y = r₀ sin(φ₀ − t). Each energy level set is a phase curve.

Figure 9

Phase plane of the equation

Example 2. Suppose that a potential energy is given by the graph in Figure 10. We will draw the energy level sets . For this, the following facts are helpful.

Figure 10

Potential energy and phase curves

1.

Any equilibrium position of (2) must lie on the x axis of the phase plane. The point x = ξ, y = 0 is an equilibrium position if ξ is a critical point of the potential energy, i.e., if (∂U/∂x) | _{x = ξ} = 0.

2.

Each level set is a smooth curve in a neighborhood of each of its points which is not an equilibrium position (this follows from the implicit function theorem). In particular, if the number E is not a critical value of the potential energy (i.e., is not the value of the potential energy at one of its critical points), then the level set on which the energy is equal to E is a smooth curve.

Problem. Show that the time it takes to go from x₁ to x₂ (in one direction) is equal to

Problem. Draw the phase curves, given the potential energy graphs in Figure 11.

Figure 11

Potential energy

Answer. Figure 12.

Figure 12

Phase curves

Problem. Draw the phase curves for the “equation of an ideal planar pendulum”: .

Problem. Draw the phase curves for the “equation of a pendulum on a rotating axis”: .

Problem. Find the tangent lines to the branches of the critical level corresponding to maximal potential energy E = U(ξ) (Figure 13).

Figure 13

Critical energy level lines

Answer. .

Problem. Let S(E) be the area enclosed by the closed phase curve corresponding to the energy level E. Show that the period of motion along this curve is equal to

Answer. .

D Phase flow

Let M be a point in the phase plane. We look at the solution to system (2) whose initial conditions at t = 0 are represented by the point M. We assume that any solution of the system can be extended to the whole time axis. The value of our solution at any value of t depends on M. We denote the resulting phase point (Figure 14) by

Figure 14

Phase flow

Example. The phase flow given by the equation is the group g^t of rotations of the phase plane through angle t around the origin.

Problem. Draw the image of the circle under the action of a transformation of the phase flow for the equations (a) of the “inverse pendulum,” and (b) of the “nonlinear pendulum,” .

Answer. Figure 15.

Figure 15

Action of the phase flow on a circle

A Definitions

By a system with two degrees of freedom we will mean a system defined by the differential equations

(1)

where f is a vector field on the plane.

A system is said to be conservative if there exists a function U: E² → ℝsuch that f = −∂U/∂x. The equation of motion of a conservative system then has the form¹⁴ .

B The law of conservation of energy

Theorem. The total energy of a conservative system is conserved, i.e.,

Proof. by the equation of motion.

Remark. In a system with one degree of freedom it is always possible to introduce the potential energy

Problem. Find an example of a system of the form , which is not conservative.

C Phase space

The equation of motion (1) can be written as the system:

(2)

The system (2) defines the phase velocity vector field in four space as well as¹⁵ the phase flow of the system (a one-parameter group of diffeomorphisms of four-dimensional phase space). The phase curves of (2) are subsets of four-dimensional phase space. All of phase space is partitioned into phase curves. Projecting the phase curves from four space to the x₁, x₂ plane gives the trajectories of our moving point in the x₁, x₂ plane. These trajectories are also called orbits. Orbits can have points of intersection even when the phase curves do not intersect one another. The equation of the law of conservation of energy

defines a three-dimensional hypersurface in four space: E(x₁, x₂, y₁, y₂) = E₀; this surface, π_E0, remains invariant under the phase flow: g^tπ_E0 = π_E0. One could say that the phase flow flows along the energy level hypersurfaces. The phase velocity vector field is tangent at every point to π_E0. Therefore, π_E0 is entirely composed of phase curves (Figure 16).

Figure 16

Energy level surface and phase curves

Example 1 (“small oscillations of a spherical pendulum”). Let . The level sets of the potential energy in the x₁, x₂ plane will be concentric circles (Figure 17).

Figure 17

Potential energy level curves for a spherical pendulum

The equations of motion, , , are equivalent to the system

A solution has the form

It follows from the law of conservation of energy that

i.e., the level surface π_E0 is a sphere in four space.

Example 2 (“Lissajous figures”). We look at one more example of a planar motion (“small oscillations with two degrees of freedom”):

The potential energy is

From the law of conservation of energy it follows that, if at the initial moment of time the total energy is

then all motions will take place inside the ellipse U(x₁, x₂) ≤ E.

Our system consists of two independent one-dimensional systems. Therefore, the law of conservation of energy is satisfied for each of them separately, i.e., the following quantities are preserved

Consequently, the variable x₁ is bounded by the region , and x₂ oscillates within the region |x₂| ≤ A₂. The intersection of these two regions defines a rectangle which contains the orbits (Figure 18).

Figure 18

The regions U ≤ E, U₁ ≤ E and U₂ ≤ E

Consider the following method of describing an orbit in the x₁, x₂ plane. We look at a cylinder with base 2A₁ and a band of width 2A₂. We draw on the band a sine wave with period 2πA₁/ω and amplitude A₂ and wind the band onto the cylinder (Figure 19). The orthogonal projection of the sinusoid wound around the cylinder onto the x₁, x₂ plane gives the desired orbit, called a Lissajous figure.

Figure 19

Construction of a Lissajous figure

The form of a Lissajous figure very strongly depends on the frequency ω. If ω = 1 (the spherical pendulum of Example 1), then the curve on the cylinder is an ellipse. The projection of this ellipse onto the x₁, x₂ plane depends on the difference φ₂ − φ₁ between the phases. For φ₁ = φ₂ we get a segment of the diagonal of the rectangle; for small φ₂ − φ₁ we get an ellipse close to the diagonal and inscribed in the rectangle. For φ₂ − φ₁ = π/2 we get an ellipse with major axes x₁, x₂; as φ₂ − φ₁ increases from π/2 to π the ellipse collapses onto the second diagonal; as φ₂ − φ₁ increases further the whole process is repeated from the beginning (Figure 20).

Figure 20

Series of Lissajous figures with ω = 1

Now let the frequencies be only approximately equal: ω ≈ 1. The segment of the curve corresponding to 0 ≤ t ≤ 2π is very close to an ellipse. The next loop also reminds one of an ellipse, but here the phase shift φ₂ − φ₁ is greater than in the original by 2π(ω − 1). Therefore, the Lissajous curve with ω ≈ 1 is a distorted ellipse, slowly progressing through all phases from collapsed onto one diagonal to collapsed onto the other (Figure 21).

Figure 21

Lissajous figure with ω ≈ 1

If one of the frequencies is twice the other (ω = 2), then for some particular phase shift the Lissajous figure becomes a doubly traversed arc (Figure 22).

Figure 22

Lissajous figure with ω = 2

Problem. Show that this curve is a parabola. By increasing the phase shift φ₂ − φ₁ we get in turn the curves in Fig. 23.

Figure 23

Series of Lissajous figures with ω = 2

In general, if one of the frequencies is n times bigger than the other (ω = n), then among the graphs of the corresponding Lissajous figures there is the graph of a polynomial of degree n (Figure 24); this polynomial is called a Chebyshev polynomial.

Figure 24

Chebyshev polynomials

A Work of a force field along a path

Recall the definition of the work by a force F on a path S. The work of the constant force F (for example, the force with which we lift up a load) on the path is, by definition, the scalar product (Figure 25)

Figure 25

Work of the constant force F along the straight path S

Suppose we are given a vector field F and a curve l of finite length. We approximate the curve l by a polygonal line with components ΔS_i and denote by F_i the value of the force at some particular point of ΔS_i; then the work of the field F on the path l is by definition (Figure 26)

In analysis courses it is proved that if the field is continuous and the path rectifiable, then the limit exists. It is denoted by ∫_l (F, dS).

Figure 26

Work of the force field F along the path l

B Conditions for a field to be conservative

Proof. Suppose that the work of a field F does not depend on the path. Then

is well defined as a function of the point M. It is easy to verify that

i.e., the field is conservative and U is its potential energy. Of course, the potential energy is defined only up to the additive constant U(M₀), which can be chosen arbitrarily.

Conversely, suppose that the field F is conservative and that U is its potential energy. Then it is easily verified that

i.e., the work does not depend on the shape of the path.

Problem. Show that the vector field F₁ = x₂, F₂ = −x₁ is not conservative (Figure 27).

Figure 27

A non-potential field

Problem. Is the field in the plane minus the origin given by , conservative? Show that a field is conservative if and only if its work along any closed contour is equal to zero.

C Central fields

Proof. According to the previous problem, we may set F(r) = Φ(r)e_r, where r is the radius vector with respect to 0, r is its length and the unit vector e_r = r/|r| its direction. Then

and this integral is obviously independent of the path. □

A The law of conservation of angular momentum

Lemma. Let a and b be two vectors changing with time in the oriented euclidean space ℝ³. Then

Proof. By definition . By the lemma, . Since the field is central it is apparent from the equations of motion that the vectors and r are collinear. Therefore . □

B Kepler’s law

We introduce polar coordinates r, φ on our plane with pole at the center of the field 0. We consider, at the point r with coordinates (|r| = r, φ), two unit vectors: e_r directed along the radius vector so that

and e_φ, perpendicular to it in the direction of increasing φ. We express the velocity vector in terms of the basis e_r, e_φ (Figure 29).

Figure 29

Decomposition of the vector ṙ in terms of the basis e_r, e_φ

Lemma. We have the relation

Proof. Clearly, the vectors e_r and e_φ rotate with angular velocity , i.e.,

Differentiating the equality r = re_r gives us

Consequently, the angular momentum is

Thus, the quantity is preserved. This quantity has a simple geometric meaning.

Kepler called the rate of change of the area S(t) swept out by the radius vector the sectorial velocity C (Figure 30):

The law discovered by Kepler through observation of the motion of the planets says: in equal times the radius vector sweeps out equal areas, so that the sectorial velocity is constant, dS/dt = const. This is one formulation of the law of conservation of angular momentum. Since

this means that the sectorial velocity

is half the angular momentum of our point of mass 1, and therefore constant.

Figure 30

Sectorial velocity

Example. Some satellites have very elongated orbits. By Kepler’s law such a satellite spends most of its time in the distant part of its orbit, where the magnitude of is small.

A Reduction to a one-dimensional problem

We look at the motion of a point (of mass 1) in a central field on the plane:

It is natural to use polar coordinates r, φ.

By the law of conservation of angular momentum the quantity is constant (independent of t).

Theorem. For the motion of a material point of unit mass in a central field the distance from the center of the field varies in the same way as r varies in the one-dimensional problem with potential energy

Proof. Differentiating the relation shown in Section 7 , we find

Since the field is central,

Therefore the equation of motion in polar coordinates takes the form

But, by the law of conservation of angular momentum,

where M is a constant independent of t, determined by the initial conditions. Therefore,

Remark. The total energy in the derived one-dimensional problem

is the same as the total energy in the original problem

since

B Integration of the equation of motion

The total energy in the derived one-dimensional problem is conserved. Consequently, the dependence of r on t is defined by the quadrature

Since , , and the equation of the orbit in polar coordinates is found by quadrature,

C Investigation of the orbit

We fix the value of the angular momentum at M. The variation of r with time is easy to visualize, if one draws the graph of the effective potential energy V(r) (Figure 31).

Figure 31

Graph of the effective potential energy

Let E be the value of the total energy. All orbits corresponding to the given E and M lie in the region V(r) ≤ E. On the boundary of this region, V = E, i.e., ṙ = 0. Therefore, the velocity of the moving point, in general, is not equal to zero since for M ≠ 0.

The inequality V(r) ≤ E gives one or several annular regions in the plane:

If 0 ≤ r_min < r_max < ∞, then the motion is bounded and takes place inside the ring between the circles of radius r_min and r_max.

The shape of an orbit is shown in Figure 32. The angle φ varies monotonically while r oscillates periodically between r_min and r_max. The points where r = r_min are called pericentral, and where r = r_max, apocentral (if the center is the earth—perigee and apogee; if it is the sun—perihelion and aphelion; if it is the moon—perilune and apolune).

Figure 32

Orbit of a point in a central field

In general, the orbit is not closed: the angle between the successive pericenters and apocenters is given by the integral

The angle between two successive pericenters is twice as big.

It can be shown that if the angle Φ is not commensurable with 2π, then the orbit is everywhere dense in the annulus (Figure 33).

Figure 33

Orbit dense in an annulus

We now look at the case r_max = ∞. If lim_r→∞ U(r) = lim_r→∞ V(r) = U_∞ < ∞, then it is possible for orbits to go off to infinity. If the initial energy E is larger than U, then the point goes to infinity with finite velocity . We notice that if U(r) approaches its limit slower than r⁻², then the effective potential V will be attracting at infinity (here we assume that the potential U is attracting at infinity).

D Central fields in which all bounded orbits are closed

It follows from the following sequence of problems that there are only two cases in which all the bounded orbits in a central field are closed, namely,

and

Problem 1. Show that the angle Φ between the pericenter and apocenter is equal to the semiperiod of an oscillation in the one-dimensional system with potential energy W(x) = U(M/x) + (x²/2).

Hint. The substitution x = M/r gives

Answer.

It follows that (the logarithmic case corresponds to α = 0)). For example, for α = 2 we have Φ_cir = π/2, and for α = − 1 we have Φ_cir = π.

Hint. The substitution x = yx_max reduces Φ to the form

Answer. . Note that Φ₀ does not depend on M.

Solution. If all bounded orbits are closed, then, in particular, Φ_cir = 2π(m/n) = const. According to Problem 3, U = ar^α(α ≥ − 2), or U = b ln r (α = 0). In both cases . If α > 0, then according to Problem 4, lim_E→∞Φ(E, M) = π/2. Therefore, Φ_cir = π/2, α = 2. If α < 0, then according to Problem 5, lim_{E→ − ∞} Φ(E, M) = π/(2 + α). Therefore, , α = −1. In the case α = 0 we find , which is not commensurable with 2π. Therefore, all bounded orbits can be closed only in fields where U = ar² or U = −k/r. In the field U = ar², a > 0, all the orbits are closed (these are ellipses with center at 0, cf. Example 1, Section 5). In the field U = −k/r all bounded orbits are also closed and also elliptical, as we will now show.

E Kepler’s problem

This problem concerns motion in a central field with potential U = −k/r and therefore V(r) = −(k/r) + (M²/2r²) (Figure 34).

Figure 34

Effective potential of the Kepler problem

By the general formula

Integrating, we get

To this expression we should have added an arbitrary constant. We will assume it equal to zero; this is equivalent to the choice of an origin of reference for the angle φ at the pericenter. We introduce the following notation:

Now we get φ = arc cos((p/r) − 1)/e, i.e.,

This is the so-called focal equation of a conic section. The motion is bounded (Figure 35) for E < 0. Then e < 1, i.e., the conic section is an ellipse. The number p is called the parameter of the ellipse, and e the eccentricity. Kepler’s first law, which he discovered by observing the motion of Mars, consists of the fact that the planets describe ellipses, with the sun at one focus.

Figure 35

Keplerian ellipse

The parameter and eccentricity are related with the semi-axes by the formulas

i.e.,

, where c = ae is the distance from the center to the focus (cf. Figure 35).

Remark. An ellipse with small eccentricity is very close to a circle.¹⁷ If the distance from the focus to the center is small of first order, then the difference between the semi-axes is of second order: . For example, in the ellipse with major semi-axes of 10 cm and eccentricity 0.1, the difference of the semi-axes is 0.5 mm, and the distance between the focus and the center is 1 cm.

Proof. We denote by T the period of revolution and by S the area swept out by the radius vector in time T. 2S = MT, since M/2 is the sectorial velocity. But the area of the ellipse, S, is equal to πab, so T = 2πab/M. Since

(from a = p/(1 − e²)), and

then ; but 2|E| = k/a, so T = 2πa^3/2k^−1/2. □

Hint. The orbit differs from a circle only to second order, and we can disregard this difference. The radius has the intended value since the initial energy has the intended value. Therefore, we get the true orbit (Figure 36) by twisting the intended orbit through 1°.

Figure 36

An orbit which is close to circular

Problem. The first cosmic velocity is the velocity of motion on a circular orbit of radius close to the radius of the earth. Find the magnitude of the first cosmic velocity v₁ and show that (cf. Section 3B).

Hint. We take as our unit of length the radius of the space ship’s circular orbit, and we choose a unit of time so that the period of revolution around this orbit is 2π. We must study solutions to Newton’s equation

close to the circular solution with r₀ = 1, φ₀ = t. We seek those solutions in the form

By substituting the expressions for r and φ in Newton’s equation, we get, after simple computation, the variational equations in the form

After solving these equations for the given initial conditions , , we get the answer given above.

A Conservative fields

We consider a motion in the conservative field

where U = U(r), r ∈ E³.

The law of conservation of energy holds:

B Central fields

For motion in a central field the vector does not change: dM/dt = 0.

Every central field is conservative (this is proved as in the two-dimensional case), and

since , and the vector ∂U/∂r is collinear with r since the field is central.

Proof. ; therefore r(t) ⊥ M, and since M = const., all orbits lie in the plane perpendicular to M.²⁰

C Axially symmetric fields

Let z be the axis, oriented by the vector e_z in three-dimensional euclidean space E³; F a vector in the euclidean linear space ℝ³; 0 a point on the z axis; r = x − 0 ∈ ℝ³ the radius vector of the point x ∈ E³ relative to 0 (Figure 37).

Figure 37

Moment of the vector F with respect to an axis

Definition. The moment M_z relative to the z axis of the vector F applied at the point r is the projection onto the z axis of the moment of the vector F relative to some point on this axis:

Proof. . Since , it follows that r and lie in a plane passing through the z axis, and therefore is perpendicular to e_z.

Therefore,

A Internal and external forces

Newton’s equations for the motion of a system of n material points, with masses m_i and radius vectors r_i ∈ E³ are the equations

The vector F_i is called the force acting on the i-th point.

The forces F_i are determined experimentally. We often observe in a system that for two points these forces are equal in magnitude and act in opposite directions along the straight line joining the points (Figure 38).

Figure 38

Forces of interaction

If all forces acting on a point of the system are forces of interaction, then the system is said to be closed. By definition, the force acting on the i-th point of a closed system is

If the system is not closed, then it is often possible to represent the forces acting on it in the form

where F_ij are forces of interaction and F′_i(r_i) is the so-called external force.

Example. (Figure 39) We separate a closed system into two parts, I and II. The force F_i applied to the i-th point of system I is determined by forces of interaction inside system I and forces acting on the i-th point from points of system II, i.e.,

Figure 39

Internal and external forces

B The law of conservation of momentum

Definition. The momentum of a system is the vector

Proof. , since for forces of interaction F_ij = −F_ji.

Definition. The center of mass of a system is the point

In fact, (∑m_i)r = ∑ (m_ir_i), from which it follows that .

Proof. . Therefore, . □

C The law of conservation of angular momentum

Definition. The angular momentum of a material point of mass m relative to the point 0, is the moment of the momentum vector relative to 0:

The angular momentum of a system relative to 0 is the sum of the angular momenta of all the points in the system:

Proof. . The first term is equal to zero, and the second is equal to

by Newton’s equations.

The sum of the moments of two forces of interaction is equal to zero since

Therefore, the sum of the moments of all forces of interaction is equal to zero:

Therefore, . □

We denote the sum of the moments of the external forces by .

D The law of conservation of energy

Definition. The kinetic energy of a point of mass m is

Definition. The kinetic energy of a system of mass points is the sum of the kinetic energies of the points:

where the m_i are the masses of the points and ṙ_i are their velocities.

Proof.

Therefore,

□

Let r = (r₁,..., r_n) be the radius vector of a point in the configuration space, and F = (F₁,..., F_n) the force vector. We can write the theorem above in the form

In other words:

Definition. A system is called conservative if the forces depend only on the location of a point in the system (F = F(r)), and if the work of F along any path depends only on the initial and final points of the path:

Theorem. For a system to be conservative it is necessary and sufficient that there exist a potential energy, i.e., a function U(r) such that

Proof. By what was shown earlier,

□

Let all the forces acting on the points of a system be divided into forces of interaction and external forces:

where F_ij = −F_ji = f_ije_ij.

Proof. If a system consists entirely of two points i and j, then, as is easily seen, the potential energy of the interaction is given by the formula

We then have

Therefore, the potential energy of the interaction of all the points will be

□

If the external forces are also conservative, i.e., , then the system is conservative, and its total potential energy is

For such a system the total mechanical energy

is conserved.

Definition. A decrease in the mechanical energy E(t₀) − E(t₁) is called an increase in the non-mechanical energy E′:

E Example: The two-body problem

Suppose that two points with masses m₁ and m₂ interact with potential U, so that the equations of motion have the form

We now look at the vector r = r₁ − r₂. Multiplying the first of the equations of motion by m₂, the second by m₁, and computing, we find that , where U = U(|r₁ − r₂|) = U(|r|).

In particular, in the case of a Newtonian attraction, the points describe conic sections with foci at their common center of mass (Figure 40).

Figure 40

The two body problem

A Example

Let r(t) satisfy the equation m(d²r/dt²) = −(∂U/∂r). We set t₁ = αt and m₁ = α²m. Then r(t₁) satisfies the equation . In other words:

B A problem

Suppose that the potential energy of a central field is a homogeneous function of degree v:

Answer. γ = βα⁻², .

A Variations

Example. Let γ = {(t, x): x = x(t), t₀ ≤ t ≤ t ₁} be a curve in the (t, x)-plane; ẋ =dx/dt; L = L(a, b, c) a differentiable function of three variables. We define a functional Φ by

In case , we get the length of γ.

Theorem. The functional is differentiable, and its derivative is given by the formula

Proof.

where

Integrating by parts, we find that

B Extremals

Theorem. The curve γ: x = x(t) is an extremal of the functional Φ(γ) = on the space of curves passing through the points x(t₀) = x₀ and x(t₁) = x₁, if and only if

Lemma. If a continuous function f(t), t₀ ≤ t ≤ t₁ satisfies for any continuous²⁷ function h(t) with h(t₀) = h(t₁) = 0, then f(t) ≡ 0.

Proof of the lemma. Let f(t*) > 0 for some t*, t₀ < t* < t₁. Since f is continuous, f(t) > c in some neighborhood Δ of the point t*: t₀ < t* − d < t < t* + d < t₁. Let h(t) be such that h(t) = 0 outside Δ, h(t) > 0 in Δ, and h(t) = 1 in Δ/2 (i.e., for t s.t. . Then, clearly, (Figure 42). This contradiction shows that f(t*) = 0 for all t*, t₀ < t* < t₁. □

Figure 42

Construction of the function h

Proof of the theorem. By the preceding theorem,

The term after the integral is equal to zero since h(t₀) = h(t₁) = 0. If γ is an extremal, then F(h) = 0 for all h with h(t₀) = h(t₁) = 0. Therefore,

where

for all such h. By the lemma, f(t) ≡ 0. Conversely, if f(t) ≡ 0, then clearly F(h) ≡ 0. □

Example. We verify that the extremals of length are straight lines. We have:

C The Euler-Lagrange equation

Definition. The equation

is called the Euler-Lagrange equation for the functional

Theorem. The curve γ is an extremal of the functional on the space of curves joining (t₀, x₀) and (t₁, x₁), if and only if the Euler- Lagrange equation is satisfied along γ.

D An important remark

For example, the same functional—length of a curve—is given in cartesian and polar coordinates by the different formulas

The extremals are the same—straight lines in the plane. The equations of lines in cartesian and polar coordinates are given by different functions: x₁ = x₁(t), x₂ = x₂(t), and r = f(t), φ = φ(t).

However, both these vector functions satisfy the Euler—Lagrange equation

only, in the first case, when x_cart = x₁, x₂ and , and in the second case when x_pol = r, φ and .

A Hamilton’s principle of least action

Theorem. Motions of the mechanical system (1) coincide with extremals of the functional

is the difference between the kinetic and potential energy.

Proof. Since U = U(r) and , we have and ∂L/∂r_i = −∂U/∂r_i.☐

Corollary. Let (q₁,..., q_3n) be any coordinates in the configuration space of a system of n mass points. Then the evolution of q with time is subject to the Euler-Lagrange equations

Definition. In mechanics we use the following terminology: is the Lagrange function or lagrangian, q_i are the generalized coordinates, are generalized velocities, are generalized momenta, ∂L/∂q_i are generalized forces, is the action, −(∂L/∂q_i) = 0 are Lagrange’s equations.

The last theorem is called “Hamilton’s form of the principle of least motion” because in many cases the action q(t) is not only an extremal but is also a minimum value of the action functional .

B The simplest examples

Example 1. For a free mass point in E³,

in cartesian coordinates q_i = r_i we find

Here the generalized velocities are the components of the velocity vector, the generalized momenta are the components of the momentum vector, and Lagrange’s equations coincide with Newton’s equations dp/dt = 0. The extremals are straight lines. It follows from Hamilton’s principle that straight lines are not only shortest (i.e., extremals of the length but also extremals of the action .

Example 2. We consider planar motion in a central field in polar coordinates q₁ = r, q₂ = φ. From the relation we find the kinetic energy and the lagrangian , where U = U(q₁).

The generalized momenta will be , i.e.,

The first Lagrange equation takes the form

We already obtained this equation in Section 8.

Since q₂ = φ does not enter into L, we have ∂L/∂q₂ = 0. Therefore, the second Lagrange equation will be . This is the law of conservation of angular momentum.

In general, when the field is not central (U = U(r, φ)), we find .

A Definition

The Legendre transformation of the function f is a new function g of a new variable p, which is constructed in the following way (Figure 43). We draw the graph of f in the x, y plane. Let p be a given number. Consider the straight line y = px. We take the point x = x(p) at which the curve is farthest from the straight line in the vertical direction: for each p the function px − f(x) = F(p, x) has a maximum with respect to x at the point x(p). Now we define g(p) = F(p, x(p)).

Figure 43

Legendre transformation

B Examples

Examples 1. Let f(x) = x². Then .

Example 4. Let f(x) be a convex polygon. Then g(p) is also a convex polygon, in which the vertices of f(x) correspond to the edges of g(p), and the edges of f(x) to the vertices of g(p). For example, the corner depicted in Figure 44 is transformed to a segment under the Legendre transformation.

Figure 44

Legendre transformation taking an angle to a line segment

C Involutivity

Proof. In order to apply the Legendre transform to g, with variable p, we must by definition look at a new independent variable (which we will call x), construct the function

and find the point p(x) at which G attains its maximum: ∂G/∂p = 0, i.e., g′(p) = x. Then the Legendre transform of g(p) will be the function of x equal to G(x, p(x)).

We will show that G(x, p(x)) = f(x). To this end we notice that G(x, p) = xp − g(p) has a simple geometric interpretation: it is the ordinate of the point with abscissa x on the line tangent to the graph of f(x) with slope p (Figure 45). For fixed p, the function G(x, p) is a linear function of x, with ∂G/∂x = p, and for x = x(p) we have G(x, p) = xp − g(p) = f(x) by the definition of g(p).

Figure 45

Involutivity of the Legendre transformation

D Young’s inequality

By definition of the Legendre transform, F(x, p) = px − f(x) is less than or equal to g(p) for any x and p. From this we have Young’s inequality:

Example 1. If , then and we obtain the well-known inequality for all x and p.

E The case of many variables

Problem. Let f be the quadratic form . Show that its Legendre transform is again a quadratic form , and that the values of both forms at corresponding points coincide (Figure 46):

Figure 46

Legendre transformation of a quadratic form

A Equivalence of Lagrange’s and Hamilton’s equations

We consider the system of Lagrange’s equations , where p = , with a given lagrangian function L: ℝⁿ × ℝⁿ × ℝ → ℝ, which we will assume to be convex³⁰ with respect to the second argument .

Theorem. The system of Lagrange’s equations is equivalent to the system of 2n first-order equations (Hamilton’s equations)

where is the Legendre transform of the lagrangian function viewed as a function of .

Proof. By definition, the Legendre transform of with respect to is the function , in which is expressed in terms of p by the formula , and which depends on the parameters q and t. This function H is called the hamiltonian.

The total differential of the hamiltonian

is equal to the total differential of for :

Both expressions for dH must be the same. Therefore,

Applying Lagrange’s equations , we obtain Hamilton’s equations.

B Hamilton’s function and energy

Example. Suppose now that the equations are mechanical, so that the lagrangian has the usual form L = T − U, where the kinetic energy T is a quadratic form with respect to :

Example. For the form f(x) = x² this is a well-known property of a tangent to a parabola. For the form we have p = mx and g(p) = p²/2m = mx²/2 = f(x).

Proof of the theorem. Reasoning as in the lemma, we find that . □

Example. For one-dimensional motion

In this case and Hamilton’s equations take the form

Proof. We consider the variation in H along the trajectory H(p(t), q(t), t). Then, by Hamilton’s equations,

C Cyclic coordinates

Corollary 2. Let q₁ be a cyclic coordinate. Then p₁ is a first integral. In this case the variation of the remaining coordinates with time is the same as in a system with the n − 1 independent coordinates q₂,..., q_n and with hamiltonian function

depending on the parameter c = p₁.

Proof. We set p′ = (p₂,..., p_n) and q′ = (q₂,..., q_n). Then Hamilton’s equations take the form

The last equation shows that p₁ = const. Therefore, in the system of equations for p′ and q′, the value of p₁ enters only as a parameter in the hamiltonian function. After this system of 2n − 2 equations is solved, the equation for q₁ takes the form

and is easily integrated. □

A The phase flow

Definition. The phase flow is the one-parameter group of transformations of phase space

where p(t) and q(t) are solutions of Hamilton’s system of equations (Figure 47).

Figure 47

Phase flow

B Liouville’s theorem

Theorem 1. The phase flow preserves volume: for any region D we have (Figure 48)

Figure 48

Conservation of volume

Suppose we are given a system of ordinary differential equations ẍ = f(x), x = (x₁,..., x_n), whose solution may be extended to the whole time axis. Let {g^t} be the corresponding group of transformations:

(1)

Let D(0) be a region in x-space and v(0) its volume;

C Proof

Proof. For any t, the formula for changing variables in a multiple integral gives

Calculating ∂g^tx/∂x by formula (1), we find

We will now use a well-known algebraic fact:

Lemma 2. For any matrix A = (a_ij),

where tr is the trace of A (the sum of the diagonal elements).

Using this, we have

But tr . Therefore,

which proves Lemma 1. □

Proof of theorem 2. Since t = t₀ is no worse than t = 0, Lemma 1 can be written in the form

and if divf ≡ 0, dv/dt ≡ 0. □

In particular, for Hamilton’s equations we have

D Poincaré’s recurrence theorem

This theorem applies, for example, to the phase flow g^t of a two-dimensional system whose potential U(x₁, x₂) goes to infinity as (x₁, x₂) → ∞; in this case the invariant bounded region in phase space is given by the condition (Figure 49)

Figure 49

The way a ball will move in an asymmetrical cup is unknown; however Poincaré’s theorem predicts that it will return to a neighborhood of the original position.

Poincaré’s theorem can be strengthened, showing that almost every moving point returns repeatedly to the vicinity of its initial position. This is one of the few general conclusions which can be drawn about the character of motion. The details of motion are not known at all, even in the case

The following prediction is a paradoxical conclusion from the theorems of Poincaré and Liouville: if you open a partition separating a chamber containing gas and a chamber with a vacuum, then after a while the gas molecules will again collect in the first chamber (Figure 50).

Figure 50

Molecules return to the first chamber.

Proof of Poincaré’s theorem. We consider the images of the neighborhood U (Figure 51):

Figure 51

Theorem on returning

All of these have the same volume. If they never intersected, D would have infinite volume. Therefore, for some k ≥ 0 and l ≥ 0, with k > l,

Therefore, g^{k − 1}U ∩ U ≠ ∅. If y is in this intersection, then y = gⁿx, with x ∈ U(n = k − l). Then x ∈ U and gⁿx ∈ U(n = k − l). □

E Applications of Poincaré’s theorem

Example 1. Let D be a circle and g rotation through an angle α. If α = 2π(m/n), then gⁿ is the identity, and the theorem is obvious. If α is not commensurable with 2π, then Poincaré’s theorem gives

(Figure 52)

Figure 52

Dense set on the circle

It easily follows that

Example 2. Let D be the two-dimensional torus and φ₁ and φ₂ angular coordinates on it (longitude and latitude) (Figure 53).

Figure 53

Torus

Consider the system of ordinary differential equations on the torus

Clearly, div f = 0 and the corresponding motion

preserves the volume dφ₁ dφ₂. From Poincaré’s theorem it is easy to deduce

Example 3. Let D be the n-dimensional torus Tⁿ, i.e., the direct product³⁴ of n circles:

A point on the n-dimensional torus is given by n angular coordinates φ = (φ₁,..., φ_n). Let α = (α₁,..., α_n), and let g^t be the volume-preserving transformation

A Example

Let γ be a smooth curve in the plane. If there is a very strong force field in a neighborhood of γ, directed towards the curve, then a moving point will always be close to γ. In the limit case of an infinite force field, the point must remain on the curve γ. In this case we say that a constraint is put on the system (Figure 54).

Figure 54

Constraint as an infinitely strong field

We consider the system with potential energy

depending on the parameter N (which we will let tend to infinity) (Figure 55).

Figure 55

Potential energy U_N

We consider the initial conditions on γ:

Theorem. The following limit exists, as N → ∞:

The limit q₁ = ψ(t) satisfies Lagrange’s equation

where (T is the kinetic energy of motion along γ).

We obtain exactly the same result if we replace the plane by the 3n-dimensional configuration space of n points, consisting of a mechanical system with metric (the m_i are masses), replace the curve γ by a submanifold of the 3n-dimensional space, replace q₁ by some coordinates q₁ on γ, and replace q₂ by some coordinates q₂ in the directions perpendicular to γ. If the potential energy has the form

then as N → ∞, a motion on γ is defined by Lagrange’s equations with the lagrangian function

B Definition of a system with constraints

Definition. Let γ be an m-dimensional surface in the 3n-dimensional configuration space of the points r₁,..., r_n with masses m₁,..., m_n. Let q = (q₁,..., q_m) be some coordinates on γ:r_i = r_i(q). The system described by the equations

is called a system of n points with 3n − m ideal holonomic constraints. The surface γ is called the configuration space of the system with constraints.

A Definition of a differentiable manifold

We assume that if points p and p′ in two charts U and U′ have the same image in M, then p and p′ have neighborhoods V ⊂ U and V′ ⊂ U′ with the same image in M (Figure 56). In this way we get a mapping φ′⁻¹ φ: V → V′.

Figure 56

Compatible charts

B Examples

Example 1. Euclidean space ℝⁿ is a manifold, with an atlas consisting of one chart.

Example 2. The sphere S² = {(x, y, z): x² + y² + z² = 1} has the structure of a manifold, with atlas, for example, consisting of two charts (U_i, φ_i, i = 1, 2) in stereographic projection (Figure 57). An analogous construction applies to the n-sphere

Figure 57

Atlas of a sphere

Example 3. Consider a planar pendulum. Its configuration space—the circle S¹—is a manifold. The usual atlas is furnished by the angular coordinates φ: ℝ¹ → S¹, U₁ = (− π, π), U₂ = (0, 2π) (Figure 58).

Figure 58

Planar, spherical and double planar pendulums

Example 4. The configuration space of the “spherical” mathematical pendulum is the two-dimensional sphere S² (Figure 58).

Example 5. The configuration space of a“planar double pendulum” is the direct product of two circles, i.e., the two-torus T² = S¹ × S¹ (Figure 58).

Example 6. The configuration space of a spherical double pendulum is the direct product of two spheres, S² × S².

Example 7. A rigid line segment in the (q₁, q₂)-plane has for its configuration space the manifold ℝ² × S¹, with coordinates q₁, q₂, q₃ (Figure 59). It is covered by two charts.

Figure 59

Configuration space of a segment in the plane

Example 8. A rigid right triangle O AB moves around the vertex O. The position of the triangle is given by three numbers: the direction OA ∈ S² is given by two numbers, and if OA is given, one can rotate OB ∈ S¹ around the axis OA (Figure 60).

Figure 60

Configuration space of a triangle

Connected with the position of the triangle O AB is an orthogonal right-handed frame, e₁ = OA/|OA|, e₂ = OB/|OB|, e₃ = [e₁, e₂]. The correspondence is one-to-one; therefore the position of the triangle is given by an orthogonal three-by-three matrix with determinant 1.

The set of all three-by-three matrices is the nine-dimensional space ℝ⁹. Six orthogonality conditions select out two three-dimensional connected manifolds of matrices with determinant + 1 and −1. The rotations of three-space (determinant + 1) form a group, which we call SO(3).

Therefore, the configuration space of the triangle OAB is SO(3).

Problem. Show that SO(3) is homeomorphic to three-dimensional real projective space.

Example 9. Consider a system of k rods in a closed chain with hinged joints.

Problem. How many degrees of freedom does this system have?

Example 10. Embedded manifolds. We say that M is an embedded k-dimensional sub-manifold of euclidean space ℝⁿ (Figure 61) if in a neighborhood U of every point x ∈ M there are n − k functions f₁: U → ℝ, f₂: U → ℝ,..., f_{n − k}: U → ℝ such that the intersection of U with M is given by the equations f₁ = 0,..., f_{n − k} = 0, and the vectors grad f₁,..., grad f_{n − k} at x are linearly independent.

Figure 61

Embedded submanifold

It is easy to give M the structure of a manifold, i.e., coordinates in a neighborhood of x(how?).

It can be shown that every manifold can be embedded in some euclidean space. In Example 8, SO(3) is a subset of ℝ⁹.

Problem. Show that SO(3) is embedded in ℝ⁹, and at the same time, that SO(3) is a manifold.

C Tangent space

If M is a k-dimensional manifold embedded in Eⁿ, then at every point x we have a k-dimensional tangent space TM_x. Namely, TM_x is the orthogonal complement to {grad f₁,..., grad f_{n − k}} (Figure 62). The vectors of the tangent space TM_x based at x are called tangent vectors to M at x. We can also define these vectors directly as velocity vectors of curves in M:

Figure 62

Tangent space

D The tangent bundle

The union of the tangent spaces to M at the various points, TM_x, has a natural differentiable manifold structure, the dimension of which is twice the dimension of M.

E Riemannian manifolds

If M is a manifold embedded in euclidean space, then the metric on euclidean space allows us to measure the lengths of curves, angles between vectors, volumes, etc. All of these quantities are expressed by means of the lengths of tangent vectors, that is, by the positive-definite quadratic form given on every tangent space TM_x (Figure 63):

Figure 63

Riemannian metric

For example, the length of a curve on a manifold is expressed using this form as l(γ) = , or, if the curve is given parametrically, γ: [t₀, t₁] → M, t → x(t) ∈ M, then

Remark. Let U be a chart of an atlas for M with coordinates q₁,..., q_n. Then a Riemannian metric is given by the formula

where dq_i are the coordinates of a tangent vector.

F The derivative map

Definition. The derivative of a differentiable mapping f: M → N at a point x ∈ M is the linear map of the tangent spaces

which is given in the following way (Figure 64):

Figure 64

Derivative of a mapping

Let v ∈ TM_x. Consider a curve φ: ℝ → M with φ(0) = x, and velocity vector (dφ/dt)|_{t = 0} = v. Then f_*xv is the velocity vector of the curve f ◦ φ: ℝ → N,

Problem. Show that the vector f_*xv does not depend on the curve φ, but only on the vector v.

Problem. Show that the map f_*x: TM_x → TN_f(x) is linear.

Problem. Let x = (x₁,..., x_m) be coordinates in a neighborhood of x ∈ M, and y = (y₁,..., y_n) be coordinates in a neighborhood of y ∈ N. Let ξ be the set of components of the vector v, and η the set of components of the vector f_*xv. Show that

Taking the union of the mappings f_*x for all x, we get a mapping of the whole tangent bundle

Problem. Show that f_* is a differentiable map.

Problem. Let f: M → N, g: N → K, and h = g ◦ f: M → K. Show that h_* = g_* ◦ f_*.

A Definition of a lagrangian system

Let M be a differentiable manifold, TM its tangent bundle, and L: TM → ℝ a differentiable function. A map γ: ℝ → M is called a motion in the lagrangian system with configuration manifold M and lagrangian function L if γ is an extremal of the functional

where is the velocity vector

Example. Let M be a region in a coordinate space with coordinates q = (q₁,...,q_n). The lagrangian function L: TM → ℝ may be written in the form of a function of the 2n coordinates. As we showed in Section 12, the evolution of coordinates of a point moving with time satisfies Lagrange’s equations.

Theorem. The evolution of the local coordinates q = (q₁,..., q_n) of a point γ(t) under motion in a lagrangian system on a manifold satisfies the Lagrange equations

where is the expression for the function L: TM → ℝ in the coordinates q and on TM.

B Natural systems

Let M be a Riemannian manifold. The quadratic form on each tangent space,

is called the kinetic energy. A differentiable function U: M → ℝ is called a potential energy.

Example. Consider two mass points m₁ and m₂ joined by a line segment of length l in the (x, y)-plane. Then a configuration space of three dimensions

is defined in the four-dimensional configuration space ℝ² × ℝ² of two free points (x₁, y₁) and (x₂, y₂) by the condition (Figure 65).

Figure 65

Segment in the plane

There is a quadratic form on the tangent space to the four-dimensional space (x₁, x₂, y₁, y₂):

Our three-dimensional manifold, as it is embedded in the four-dimensional one, is provided with a Riemannian metric. The holonomic system thus obtained is called in mechanics a line segment of fixed length in the (x, y)-plane. The kinetic energy is given by the formula

C Systems with holonomic constraints

Consider the configuration manifold M of a system with constraints as embedded in the 3n-dimensional configuration space of a system of free points. The metric on the 3n-dimensional space is given by the quadratic form . The embedded Riemannian manifold M with potential energy U coincides with the system defined in Section 17 or with the limiting case of the system with potential , which grows rapidly outside of M.

D Procedure for solving problems with constraints

1.

Determine the configuration manifold and introduce coordinates q₁,..., q_k (in a neighborhood of each of its points).

2.

Express the kinetic energy as a quadratic form in the generalized velocities

3.

Construct the lagrangian function L = T − U(q) and solve Lagrange’s equations.

Example. We consider the motion of a point mass of mass 1 on a surface of revolution in three-dimensional space. It can be shown that the orbits are geodesics on the surface. In cylindrical coordinates r, φ, z the surface is given (locally) in the form r = r(z) or z = z(r). The kinetic energy has the form (Figure 66)

in coordinates φ and z, and

in coordinates r and φ. (We have used the identity .)

Figure 66

Surface of revolution

The lagrangian function L is equal to T. In both coordinate systems φ is a cyclic coordinate. The corresponding momentum is preserved; is nothing other than the z-component of angular momentum. Since the system has two degrees of freedom, knowing the cyclic coordinate φ is sufficient for integrating the problem completely (cf. Corollary 3, Section 15).

We can obtain more easily a clear picture of the orbits by reasoning slightly differently. Denote by α the angle of the orbit with a meridian. We have , where |v| is the magnitude of the velocity vector (Figure 66).

By the law of conservation of energy, H = L = T is preserved. Therefore, |v| = const, so the conservation law for p_φ takes the form

(“Clairaut’s theorem”).

This relationship shows that the motion takes place in the region |sin α| ≤ 1, i.e., r ≥ r₀ sin α₀. Furthermore, the inclination of the orbit from the meridian increases as the radius r decreases. When the radius reaches the smallest possible value, r = r₀ sin α₀, the orbit is reflected and returns to the region with larger r (Figure 67).

Figure 67

Geodesics on a surface of revolution

Problem. Show that the geodesics on a convex surface of revolution are divided into three classes: meridians, closed curves, and geodesics dense in a ring r ≥ c.

Problem. Study the behavior of geodesics on the surface of a torus ((r − R)² + z² = ρ²).

E Non-autonomous systems

A lagrangian non-autonomous system differs from the autonomous systems, which we have been studying until now, by the additional dependence of the lagrangian function on time:

In particular, both the kinetic and potential energies can depend on time in a non-autonomous natural system:

A system of n mass points, constrained by holonomic constraints dependent on time, is defined with the help of a time-dependent submanifold of the configuration space of a free system. Such a manifold is given by a mapping

which, for any fixed t ∈ ℝ, defines an embedding M → E³ⁿ. The formula of section D remains true for non-autonomous systems.

Example. Consider the motion of a bead along a vertical circle of radius r (Figure 68) which rotates with angular velocity ω around the vertical axis passing through the center O of the circle. The manifold M is the circle. Let q be the angular coordinate on the circle, measured from the highest point.

Figure 68

Bead on a rotating circle

Let x, y, and z be cartesian coordinates in E³ with origin O and vertical axis z. Let φ be the angle of the plane of the circle with the plane xOz. By hypothesis, φ = ωt. The mapping i: M × ℝ → E³ is given by the formula

From this formula (or, more simply, from an “infinitesimal right triangle”) we find that

In this case the lagrangian function L = T − U turns out to be independent of t, although the constraint does depend on time. Furthermore, the lagrangian function turns out to be the same as in the one-dimensional system with kinetic energy

and with potential energy

The form of the phase portrait depends on the ratio between A and B. For 2B < A (i.e., for a rotation of the circle slow enough that ω²r < g), the lowest position of the bead (q = π) is stable and the characteristics of the motion are generally the same as in the case of a mathematical pendulum (ω = 0).

For 2B > A, i.e., for sufficiently fast rotation of the circle, the lowest position of the bead becomes unstable; on the other hand, two stable positions of the bead appear on the circle, where cos q = −A/2B = −g/ω²r. The behavior of the bead under all possible initial conditions is clear from the shape of the phase curves in the (Figure 69).

Figure 69

Effective potential energy and phase plane of the bead

A Formulation of the theorem

Definition. A lagrangian system (M, L) admits the mapping h if for any tangent vector v ∈ TM,

Example. Let . The system admits the translation h: (x₁, x₂, x₂) → (x₁ + s, x_1, x₃) along the x₁ axis and does not admit, generally speaking, translations along the x₂ axis.

Noether’s theorem. If the system (M, L) admits the one-parameter group of diffeomorphisms h^s: M → M, s ∈ ℝ, then the lagrangian system of equations corresponding to L has a first integral I: TM → ℝ.

In local coordinates q on M the integral I is written in the form

B Proof

First, let M = ℝⁿ be coordinate space. Let φ: ℝ → M, q = φ(t) be a solution to Lagrange’s equations. Since preserves L, the translation of a solution, h^s ◦ φ: ℝ → M also satisfies Lagrange’s equations for any s.³⁸

We consider the mapping Φ: ℝ × ℝ → ℝⁿ, given by q = Φ(s, t) = h^s(φ(t)) (Figure 70).

Figure 70

Noether’s theorem

We will denote derivatives with respect to t by dots and with respect to s by primes. By hypothesis

(1)

where the partial derivatives of L are taken at the point .

As we stated above, the mapping Φ|_s = _const: ℝ → ℝⁿ for any fixed s satisfies Lagrange’s equation

We introduce the notation and substitute ∂F/∂t for ∂L/∂q in (1).

Writing as dq′/dt, we get

□

Remark. The first integral is defined above using local coordinates q. It turns out that the value of I(v) does not depend on the choice of coordinate system q.

C Examples

Example 1. Consider a system of point masses with masses m_i:

constrained by the conditions f_j(x) = 0. We assume that the system admits translations along the e₁ axis:

In fact, (d/ds)|_{s = 0}h^sx_i = e₁. According to the remark at the end of B, the quantity

is preserved, i.e., the first component P₁ of the momentum vector is preserved. We showed this earlier for a system without constraints.

Example 2. If a system admits rotations around the e₁ axis, then the angular momentum with respect to this axis,

is conserved.

It is easy to verify that if h^s is rotation around the e₁ axis by the angle s, then (d/ds)|_{s = 0}h^sx_i = [e₁, x_i], from which it follows that

Problem 1. Suppose that a particle moves in the field of the uniform helical line x = cos φ, y = sin φ, z = cφ. Find the law of conservation corresponding to this helical symmetry.

Answer. In any system which admits helical motions leaving our helical line fixed, the quantity I = cP₃ + M₃ is conserved.

Problem 2. Suppose that a rigid body is moving under its own inertia. Show that its center of mass moves linearly and uniformly. If the center of mass is at rest, then the angular momentum with respect to it is conserved.

Problem 3. What quantity is conserved under the motion of a heavy rigid body if it is fixed at some point O? What if, in addition, the body is symmetric with respect to an axis passing through O?

Problem 4. Extend Noether’s theorem to non-autonomous lagrangian systems.

Hint. Let M₁ = M × ℝ be the extended configuration space (the direct product of the configuration manifold M with the time axis ℝ).

Define a function L₁: TM₁ by

i.e., in local coordinates q, t on M₁ we define it by the formula

We apply Noether’s theorem to the lagrangian system (M₁, L₁).

If L₁ admits the transformations h^s: M₁ → M₁, we obtain a first integral I₁: TM₁ → ℝ. Since ∫ L dt = ∫L₁ dτ. this reduces to a first integral I: TM × ℝ → ℝ of the original system. If, in local coordinates (q, t) on M₁ we have I₁ = I₁(q, t, dq/dτ, dt/dτ), then .

In particular, if L does not depend on time, L₁ admits translations along time, h^s(q, t) = (q, t + s). The corresponding first integral I is the energy integral.

A Example

Consider the holonomic system (M, L), where M is a surface in three-dimensional space {x}:

Definition. The quantity

is called the constraint force (Figure 71).

Figure 71

Constraint force

If we take the constraint force R(t) into account, Newton’s equations are obviously satisfied:

The physical meaning of the constraint force becomes clear if we consider our system with constraints as the limit of systems with potential energy U + NU₁ as N → ∞, where U₁(x) = ρ²(x, M). For large N the constraint potential NU₁ produces a rapidly changing force F = −N ∂U₁/∂x; when we pass to the limit (N → ∞) the average value of the force F under oscillations of x near M is R. The force F is perpendicular to M. Therefore, the constraint force R is perpendicular to M: (R, ξ) = 0 for every tangent vector ξ.

B Formulation of the D’Alembert-Lagrange principle

In mechanics, tangent vectors to the configuration manifold are called virtual variations. The D’Alembert-Lagrange principle states:

for any virtual variation ξ, or stated differently, the work of the constraint force on any virtual variation is zero.

C The equivalence of the D’Alembert-Lagrange principle and the variational principle

Definition. The curve x is called a conditional extremal of the action functional

if the differential δΦ is equal to zero under the condition that the variation consists of nearby curves³⁹ joining x₀ to x₁ in M.

We will write

(1)

Clearly, Equation (1) is equivalent to the Lagrange equations

in some local coordinate system q on M.

Theorem. A curve x: ℝ → M ⊂ ℝ^N is a conditional extremal of the action (i.e., satisfies Equation (1)) if and only if it satisfies D’Alembert’s equation

(2)

Lemma. Let f: {t: t₀ ≤ t ≤ t₁} → ℝ^N be a continuous vector field. If, for every continuous tangent vector field ξ, tangent to M along x (i.e., ξ(t) ∈ TM_x(t) with ξ(t) = 0 for t = t₀, t₁), we have

then the field f(t) is perpendicular to M at every point x(t) (i.e., (f(t), h) = 0 for every vector h ∈ TM_x(t) (Figure 72).

Figure 72

Lemma about the normal field