Classical mechanics of particles and fields

Classical mechanics of particles and fields D.V. Skryabin Department of Physics, University of Bath PACS numbers: The concise and transparent exposition of many topics covered in this unit can be found in: L. Landau and E. Lifshitz, Mechanics. Course of theoretical physics vol. 1. (library has many copies) 1. Introduction to variational calculus Main idea of the Lagrangian mechanics is that the true trajectory from the poin in space-time to the point 2 gives the minimal value of the integral A = A = (kin.energy pot.energy) ( ) mẋ 2 U(x) 2 It means: if we calculate A for some x(t) we always will find a number, which is greater than A calculated for the true, i.e. physically realizable, x(t). A is called action. To formulate this idea more rigorously we need to learn how to find functions which give stationary points (e.g. maxima or minima) for this type of integrals. We need to learn Variational calculus Functional is a mathematical operation, which takes a function x(t) and produces a number: A = A[x(t)] FIG. 1: Possible trajectories. The true one gives minimum of the action integral.

2 Typical example of a functional is a definite integral: A[x(t)] = L(x(t), t) Here L is a GIVEN function of its arguments (x and t) and x is itself a function of t. A is a number which obviously depends on our choice of function x(t) for a given L. One shell remember that our functional operates on an infinite set of functions x(t). Useful analogy is a function of a single independent variable operating on the entire set of real numbers from to +. To find extrema of a functional we, first, take a small deviation δx from a chosen (out of infinite number of possibilities) function x(t). δx is called variation of x. Then we calculate A[x + δx] A[x] upto the first order in δx and the result is called variation (or first variation) of A Let us consider A[x(t)] = t 2 using the outlined method. L(x(t), t) and calculate its variation and variational derivative A[x(t) + δx(t)] = L(x(t) + δx(t), t) but L is a usual function so we can expand into Taylor series. and Thus A[x(t) + δx(t)] A[x(t)] = L(x(t) + δx(t), t) = L(x(t), t) + L x δx(t) +... [ L(x(t), t) + L ] x δx(t) +... L(x(t), t) L δa = x δx. Function between the integral sign and δx is called variational derivative of a functional A: δa/δx. The above definition of the variational derivative is analogous to the following fact known from the usual calculus: if F (t) = x(t), then df/ = x(t). δa δx = L x Function x 0 (t) for which our functional reaches its stationary value must be such that δa = 0. x 0 (t) is then called an extremum of the functional A

3 FIG. 2: Lagrange (left): 1736-1813 and Leonard Euler (right): 1707-1783 Because δx is arbitrary, it means that δa is zero only if δa δx = L x = 0. The trivial results δa = L is valid only if L = L(x(t), t), but not if L = L(x, ẋ, t) δx x Let s work out variation of A[x(t)] if L = L(x, ẋ, t) using the same methodology... δa = { δx(t) [ ] [ ]} L L + δẋ x ẋ Using integration by parts we manipulate the second part of the integral above δẋ [ ] L dδx = ẋ [ ] [ ] t2 L L = ẋ ẋ δx t2 δx d t 1 We consider variations with the fixed ends, i.e. δx( ) = δx(t 2 ) = 0. {[ ]} L ẋ This gives δa = [ L x d ] L δx ẋ Thus if x 0 (t) is such that it solves the equation δa δx = L x d L ẋ L x d L ẋ = 0

4 FIG. 3: Snell s law then our functional A[x(t)] reaches its stationary value A 0 for x = x 0 (t) and x 0 (t) is called extremum of A Equation above is called Euler-Lagrange s or simply Lagrange s equation. Thus we have demonstrated that the problem of finding stationary values for the functional can be reduced to differential equations. 1.1 Variational formulation of the geometrical optics: Fermat s principle A ray of light follows a path such that the total path time is stationary (minimal) Mathematically it means that along the true path δ = 0. l = dl v = n(l)dl c where n(l) is refractive index varying along the path. Hence δ n(l)dl = 0. l Thus the equivalent formulation of the Fermat principle is that a ray follows a trajectory such that the total optical path (length refractive index) is stationary. In cartesian coordinates we reformulate the problem, as finding such y(x), that gives a stationary value to the functional x2 F = n(x, y) 1 + (dy/dx) 2 dx x 1 Let s consider an example when n is a function of y only and does not explicitly contain x. Then δf = 1 x2 dxδy d [ ] n(y) = 0 2 dx 1 + ẏ 2 Thus x 1 n(y) 1 + ẏ 2 = k = const

5 FIG. 4: William Hamilton: 1805-1865 k is the constant determined by initial conditions. Take a simple example of an interface (at y = 0) between two homogeneous media with different refractive indices n 1 and n 2. ẏ = tanφ hence Hence which is the Snell s law n cos φ = const = n sin θ n 1 sin θ 1 = n 2 sin θ 2 2. Hamilton s principle of mechanics or principle of stationary action For conservative systems, i.e. when forces can be derived from potentials the following is true Any mechanical system follows a path that yields the stationary value of the action functional: A = t 2 L. L is called Lagrangian and it is the difference between the kinetic and potential energies of the system. δa = 0 for the true path. It is an axiomatic law, which is proved by the comparison with experiment. In many, but NOT ALL, practical cases A has a minimum value, when calculated along the true trajectory. Therefore the above principle is often called principle of least action. However, it can be a maximum sometimes, or it can be a saddle point, i.e. second variations along different coordinates can have different signs. To understand whether a given stationary trajectory corresponds to a minimum or maximum one has to calculate second variations (we are not doing it in this course). Lagrangian of a particle in a potential We do not have time to verify our theory by experiments, therefore we will verify it by comparison with the Newton equations. mẍ = F or equivalently dp = du dx,

where U is the potential energy and p = mẋ is the momentum. Thus one can see that the unusually looking Lagrange equation d L ẋ = L x is surprisingly similar to the 2nd law of Newton and even equivalent to it providing that Using that L = m L ẋ p p = L ẋ, L x = du dx and integrating the first of the above equations we find that L = p2 2m + ψ(x), where ψ is an unknown function of x. Integrating the 2nd of the above equations we find that L = U(x) + φ(p), where φ is an unknown function of φ. Requiring compatibility of the 2 expressions for L we find that ψ(x) = U(x) and that φ(p) = p 2 /(2m). Thus Lagrangian of a particle moving in the potential U(x) is L = p2 U(x) = T U, 2m where T is the kinetic energy and V is the potential one. We have demonstrated that the Lagrangian is the kinetic energy minus potential energy. Principle of Stationary-Action also can be formulated as: The path of a particle obeys the Euler-Lagrange equation [ ] d L = L ẋ x One might like to remember it in the form d [ ] L = L v x In 3D the principle of stationary action gives system of 3 Lagrange equations [ ] d L = L [ ] ẋ x, d L = L [ ] ẏ y, d L = L ż z Here L = m 2 (ẋ2 + ẏ 2 + ż 2 ) U(x, y, z) For shortness the 3D Lagrange equations can be written as [ ] d L r = L r where the vector derivatives should be understood as gradients and then corresponding gradients components in the left-hand side should be made equal to the ones in the right-hand side. Examples Ball dropped under the angle. L = mẋ 2 /2 + mẏ 2 /2 + mgx and Lagrange equation is ẍ = g ÿ = 0 (Galileo equation). Mass on a horizontal spring. T = mẋ 2 /2, U = kx 2 /2 and the Lagrange equation is mẍ = kx. 6

7 3. Universality of Lagrange equations Generalized coordinates, momenta and forces Let us now consider a system of particles, which interact one with another, but not with other bodies or external forces. Lagrangian of this system is a sum of Lagrangians of the free particles, minus the interaction energy. L = 1 N m a v 2 a U( r 1, r 2,...) 2 a=1 a numbers the particles. The fact that the potential energy depends only on positions of particles, but not on their velocities, shows, that interactions are instantaneous. If the propagation of interactions is not instantaneous, then that velocity is different in different inertial reference frames, which contradicts to the relativity principle. Equations of motion are d L = L dx a, m a ẋ a x a = U x a If L is written in terms of cartesian coordinates x, y, z, than the Euler-Lagrange equations are equivalent to Newtons F = ma equations. But what about the cases where we use polar, spherical, or some other coordinates or dynamical variables to describe motion? Many particle systems is a particularly obvious example, when the coordinates different from cartesian ones can be very convenient If we try to describe a system of N particles, we need in general 3N coordinates to specify the position of its components. If external constraints are imposed on the system (as it often happens in practice), the number of degrees of freedom may be less. If there are m constraints applied, the number of degrees of freedom will be s = 3N m (give some example (pendulum, molecule)). The coordinates do not need to be the coordinates of a coordinate system, but can be any set of quantities that completely specifies the state of the system. These quantities are called generalized coordinates: q n. The coordinate space based on the generalized coordinates is called configuration space. The instantaneous state of the system is fully specified by a point in the s = 3N m dimensional configuration space. The time evolution of the system can be described by a path in the configuration space. One of the great advantages of using Lagrangians is that Euler-Lagrange equation look the same in all coordinate systems. You can use any convenient degrees of freedom and both inertial and non-inertial frames. In any case d L q i = L q i. If we assume x a = f a (q 1, q 2,... q s ), ẋ a = k f a q k q k and substitute these expressions into the Lagrangian L = 1 2 Na=1 m a v a 2 U( r 1, r 2,...), we find L = i,k 1 2 a ik( q) q i q k U( q) The kinetic energy in the generalized coordinates is still a quadratic function of the velocities, but may depend on the coordinates as well

8 Whatever is the choice of the coordinates we have q k = q k (x a, y a, z a, t) d [ ] L q k = L q k For example consider 2D motion and choose (r, φ) as the generalized coordinates: x = r cos φ, y = r sin φ. Transformation of the coordinate system for the 2nd law of Newton is not a simple task, but using Lagrangian approach it is straightforward to write T = mẋ2 2 + mẏ2 2 = m 2 [r2 φ2 + ṙ 2 ]. Let us assume that the force is f = Ar α 1 /α = r U, then the potential energy U = Ar α. Thus Lagrangian is The corresponding equations of motion are Definitions: p k = L q k is the generalized momentum F k = L q k is the generalized force Thus EL equations are L = m 2 (ṙ2 + r 2 φ2 ) Ar α m( r r φ 2 ) = A α rα 1, m d (r2 φ) = 0 dp k = F k In polar coordinates L/ φ = mr 2 φ = l is the angular momentum and L/ ṙ is the momentum of radial motion. Generalized angular force L/ φ is the torque: dl/ = τ 3.1 Lagrange s equation and motion in a non-inertial frame of reference The power of the Lagrange s approach is that the Lagrange eq. d L L = r r is valid in any coordinate system including the ones, which are not inertial. Inertial frame of reference is the frame, where free motion of a particle, i.e. motion without forces, happens with a constant velocity. It is known that the standard form of the Newton s 2nd law applies only in the inertial frames. Let s assume that in an inertial frame of reference K 0 a particle has the Lagrangian L 0 = 1 2 m v2 0 U

9 Naturally the corresponding equation of motion coincides with the 2nd law of Newton m d v 0 = U r 0 First we consider a translational motion with acceleration and then rotational. We consider a non-inertial frame K 1 moving along a straight line with the velocity V relative to K 0 v 0 = v 1 + V (t) We do NOT assume that V is time independent, hence the system K 1 can accelerate. The Lagrange equation in the new system is d L 1 r = L 1 1 r 1 Now we take L 0 and replace v 0 with v 1 to get L 1 L 1 = 1 2 m( v 1) 2 + m v 1 V + 1 2 m V 2 U Before making further transformations we need to note, that if we consider two Lagrangians L a(x, ẋ, t) and L b (x, ẋ, t), such that L b = L a + df/, where f = f(x, t) (does not depend on ẋ!). Then one can easily show that the action A b = A a + f(x 2, t 2) f(x 1, ). Hence δa a = δa b (δx() = δx(t 2) = 0). Hence the resulting equations of motion are the same in the a and b cases. Thus the Lagrangian is defined up to an additive total time derivative of an arbitrary function of co-ordinates and time (but not velocity). Omitting V 2 term as not giving any contribution into the Lagrange equation and replacing v 1 = d r 1 / we find v 1 V = d( V r 1 )/ r 1 d V / Omitting the total time derivative (see above) from the Lagrangian we finally have L 1 = 1 2 m( v 1) 2 m W r 1 U, W = d V / W is the acceleration of the K 1 The equation of motion one can derive is m d v 1 = U r 1 m W Thus an accelerated translational motion of a reference frame is equivalent to an additional force m W. This type of forces resulting from the motion of a reference frame are called inertial forces. Now we want to study effects of rotation and we assume that there is a 3rd reference frame K. Its origin coincides with the origin of K 1, but it rotates with respect to it with the variable angular velocity ω(t). Thus K executes both translational and rotational motions relative to the inertial frame K 0. v 0, r 0 is the velocity and position in K 0, v 1, r 1 in K 1 and v, r in K. Note that r 1 = r, since the origin of these two systems is the same and the vectors go to the same point. Thus we have and v 1 = v + ω r L = 1 2 m( v)2 + m v ( ω r) + 1 2 m( ω r)2 m W r U

[note (see the intro to the relativistic chapter), that the rotation of the frame leads to the term depending on the direction of v in the Lagrangian.] Calculation of derivatives needed for the Lagrange eq. is difficult and we first carefully write dl, where L( r, v, t) dl = m v d v + md v ω r + m v ω d r+ 10 +m( ω r) ( ω d r) m W d r ( U/ r) d r = If we organize the equation above in a way that d r and d v enter via scalar products only, then it is easy to get partial derivatives = m v d v + md v ω r + md r v ω+ +m( ω r) ω d r m W d r ( U/ r) d r Thus The Lagrange equation L v = m v + m ω r L r = m v ω + m ω r ω m W ( U/ r) md v/ = ( U/ r) m W + +m r ω + 2m v ω + m ω r ω Thus there are 3 inertial forces due to rotation of a reference frame. The first term is non-zero only if the rotation is non-uniform. The 2nd term is the Coriolis force. The Coriolis force is linear in ω and is the only one which depends on v. The last term is the centrifugal force. Note that acceleration of the frame is equivalent to destruction of the isotropy and homogeneity of space, through the violation of the symmetries of the Lagrangian. 4. Symmetries and conservation laws Noether s theorem (flexible formulation) All fundamental conservation laws of nature can be derived from symmetries. In particular, from the symmetries of space and time, expressed as symmetries of a Lagrangian. 4.1 Conservation of momentum Elementary case. Let s consider 1D Lagrangian L = mẋ 2 /2 + U, d L ẋ = L x

11 FIG. 5: Emmy Noether: 1882-1935 p = L/ ẋ = p x is the momentum. If U does not depend on x, then dp x / = 0, p x = const which means that momentum is conserved. The generalization of this observation is that if Lagrangian L(q i, q i, t) does not depend explicitly a generalized coordinate q n, so that L/ q n = 0, then the associated momentum L/ q n = p n, is a conserved quantity or an integral of motion so that p n = const. If homogeneity of space is not destroyed by presence of a position dependent potential, then all three components of momentum of a mechanical system are conserved. Homogeneity of space and homogeneity of Lagrangian are the same things. Mathematically it is expressed as: if we take the Lagrangian L( r, r, t) and shift the coordinate system by an arbitrary vector ɛ, i.e. replace r with r + ɛ, ( r r + ɛ), then the Lagrangian has exactly the same appearance in the new coordinate system. Example L = m/2(ẋ 2 + ẏ 2 ). Introducing new coordinates x 0 = x + a, y 0 = y + b ( ɛ = (a, b)) we find that Lagrangian L 0 = m/2(ẋ 2 0 + ẏ0) 2 is the same. If L = m/2(ẋ 2 + ẏ 2 ) mgy, then clearly new Lagrangian will be different. This is because the homogeneity of space along y direction is destroyed by the field of gravity. Proof for the space homogeneous in all 3 directions: Let s consider a small change of the Lagrangian dl caused by a small (time independent) differential change of the radius vector d r (using vector notations and that d r = 0) dl( r, r, t) = L r d r + L r d r = ( i x L + j y L + k z L) d r If Lagrangian is known to be invariant (symmetric) with respect to an arbitrary shift of the reference frame, it means that change of the Lagrangian caused by the shift is zero, i.e., dl = 0, which is possible only if x L = y L = z L = 0 = L r This, through the Lagrange equation, implies, that all 3 components of the momentum are conserved. In a less trivial situation, there can be only a specific direction in space shifts along which preserve the Lagrangian. Let s associate the vector r 0 with this direction, and calculate change of the Lagrangian caused by the shift ɛ r 0, where ɛ 1 dl = L r r 0ɛ

Since in this case only the scalar ɛ is arbitrary, we have that L r r 0 = 0, which through the Lagrange equations is equivalent to hence d [ ] L r r 0 = 0 p r 0 = const The above equation implies that the projection of p on r 0 is an integral of motion. Directing a coordinate axis along r 0 will obviously make the momentum component associated with this coordinate to be a conserved quantity. Thus conservation of momentum follows from the homogeneity of space or invariance of a Lagrangian with respect to the coordinate shifts. Other words translational invariance, implies conservation of momentum. All above is equally valid for the many particle case, when we need to sum the momenta of the individual particles, and for example the last equation will be replaced with p a r 0 = const. a Also all above is valid for any generalized coordinates q i different from the Cartesian ones. A generalized coordinate q i, such that L/ q i = 0, is called a cyclic coordinate. Momenta associated with cyclic coordinates do not change with time. Below is the brief summary of the lecture presentation, which is fine, but in future it will be replaced with the more transparent and detailed explanations presented above. Corresponding change in velocities is Q i Q i + ɛk i(q 1, Q 2, Q 3) V i V i + ɛ d Ki(Q1, Q2, Q3) Symmetry implies, that when we make the above substitution in the Lagrangian, we should have that Conservation law is then Worked example: i dl dɛ = 0 dl ( ) L dɛ = Q i Q i ɛ + L V i = V i ɛ i ( ) L K i + L dk i = d L K i = 0 Q i V i V i i L V i K i = const i 12

13 6. Lagrangian of a mechanical system is given by Are x and y cyclic coordinates?? Verify that x x + ɛ, y y + 2ɛ is a symmetry. It is obvious by a direct substitution. Work out a conserved momentum: ṗ = 0 L = m 2 (5ẋ2 2ẋẏ + 2ẏ 2 ) + C(2x y) It can be done by couple of equivalent methods. First, lets calculate a change of L, i.e. dl, caused by the nonzero ɛ in the above substitution. It is clear that symmetry implies that dl = 0, simply because the original Lagrangian and Lagrangian with x replaced by x + ɛ and y replaced by y + 2ɛ are the same. On the other hand dl = L(x + ɛ, y + 2ɛ, ẋ, ẏ) L(x, y, ẋ, ẏ) = ɛ( xl + 2 yl). However Lagrange s eqs. are dpx = xl, dp y = yl. Hence dl = ɛ d (px + 2py) = 0. Thus px + 2py = const Another method is to apply r 0 p = const directly. In this example r 0 = (1, 2) and p = (p x, p y). 4.2 Time invariant Lagrangians and conservation of energy If Lagrangian does not explicitly depend on time, i.e. it varies in time only by means of coordinates and velocities, then L L = L(q i, v i ), t = 0, dl 0 Thus in this case the differential change of the Lagrangian dl caused by is given by dl = dl = [ L dq i i q i + L ] dv i = d L v i v i i v i Thus [ ] [ ] d L v i L = 0, H = p i v i L i v i i H is called Hamiltonian or generalized energy. Hamiltonian is a conserved quantity only if Lagrangian does not change after the substitution: t t + ɛ In general H T + U, but in some cases it is. For example, in Cartesian basis i p i v i = 2T and hence H = T + U. 4.3 Conservation of angular momentum Isotropy of space means that mechanical properties of a closed system do not vary, when it is rotated as a whole in space. Formally it means that the Lagrangian is rotationally invariant. If a particle with a radius vector r undergoes rotation by dφ around the vertical axis (specifying direction of d φ and making angle θ with r), then the resulting change of the radius vector is The resulting velocity change is d r = d φ r d v = d φ v

14 FIG. 6: The Lagrangian is unchanged on rotation hence dl resulting from d r and d v is 0. dl = a ( L r a d r a + L v a d v a ) = 0 From the Euler-Lagrange equations d L v x = L we have x ( ṗ a d r a + p a dv a ) = ( ṗ a δφ r a + p a dφ v a ) = 0 a a Using permutation of the mixed product A ( B C) = B ( C A) = C ( A B), we find d φ a ( r a ṗ a + v a p a ) = d φ d r a p a = 0 a dφ is arbitrary and the conservation of the total angular momentum around the chosen axis of symmetry is obvious r a p a = const a 5. 1D motion in a potential Consider Lagrangian L = 1 2 mẋ2 U(x) then hamiltonian is H = 1 2 mẋ2 + U(x) ẋ 2 = 2 (H U(x)) m

15 FIG. 7: Potential of the harmonic oscillator. Phase plane of the harmonic oscillator. E = H(t = 0) is a constant which can be found from initial conditions. It has meaning of the energy stored in the system at all moments of time including the initial time t = 0. E U(x) for any x. This is simply because there is also kinetic energy If U(x) = E then the corresponding x is a turning point, i.e. the point where velocity of a particle changes its sign going through zero (kinetic energy=0). ± dx 2 = (E U(x)) m t t 0 = t t 0 = ± x 2 x 0 dx m (E U(x )) If there are 2 turning points, then period of oscillations is twice of the time required to go from one turning point to the other T = x2 2m x 1 dx E U(x) the simplest example is the harmonic oscillator with U(x) = kx 2 /2. Classically forbidden regions: E < U(x) Allowed regions: E U(x) 5.1 Pendulum More sophisticated example is the pendulum x = l sin φ, y = l cos φ (we count y from the point of support), U = mgl cos φ, T = ml 2 φ2 /2 EL equation is l φ = g sin φ

16 FIG. 8: Small amplitude of oscillations sin φ φ 1 3! φ3 +... Then problem is reduced to the standard harmonic oscillator. l φ gφ. Hence ω 2 g/l and period 2π/ω 2π l/g. H = ml φ 2 /2 mgl cos φ 0 < ml 2 φ2 /2 = E + mgl cos φ = E U(φ) Kinetic energy is zero at the turning points φ = φ 0, i.e. E = U(φ 0 ) Therefore, at the turning points, we have E = mgl 2 cos φ 0 Since energy is conserved along the trajectory, we derive ml 2 φ2 /2 = mgl(cos φ cos φ 0 ) If we supply enough energy to the pendulum then oscillatory motion changes to rotation. In the critical case φ 0 = π and E cr = U(π) = mgl. In this case the exact integration of equations of motion is possible. General trigonometric expression tells us that cos φ + 1 = 2 cos 2 φ. Therefore for φ 2 0 = π our equations of motion are reduced to φ 2 = 4g cos 2 φ g l 2, φ = ±2 l cos φ 2 ± corresponds to clockwise or anti-clockwise motion. The ode in question is now 1st order and separable. For convenience we take ψ = φ/2 dψ g cos ψ = l t New integration variable u = tan ψ 1 u2, cos ψ = 2 1 + u, du 2 dψ = 1 2 (1 + u2 )

17 FIG. 9: this gives us dψ = 2du/(1 + u 2 ) and 2 g du g 1 u = t 2 l, arctanh(u) = t l (t 2, u = tanh 1 ) g 2 l φ = 4arctan tanh t g l 2 t = 0, then φ = 0, t =, then φ = π It means if we push pendulum with just the correct energy, it will move towards its upright position over the infinite time. If we just below the required energy it will be an oscillatory motion with very long period, if we are just above it will be rotational motion with very long revolution time. 5.2 Stability of equilibria Dynamical system is said to be in an equilibrium if time derivatives of all generalized coordinates and velocities are equal to zero. Equilibrium is stable if small deviations do not drive generalized coordinates far from their values at equilibrium. (these definitions are not very rigorous but sufficient for our purposes) Consider a pendulum l φ = g sin φ There are two equilibrium values of φ: φ 1 = 0, φ 2 = π. φ 1 is stable and φ 2 is unstable. Let s prove it mathematically. Consider a small deviations x from φ 1, i.e. make substitution φ = φ 1 + x lẍ = g sin(φ 1 + x) g[sin φ 1 + (cos φ 1 )x +...] lẍ = gx x = C cos ωt + D sin ωt = Ae iωt + A e iωt = 2 A cos(ωt + ψ 0 ), lω 2 = g

18 x is bounded and φ stays close to φ 1 providing C, D are small. Consider small deviations x from φ 2, i.e. make substitution φ = φ 2 + x lẍ = g sin(φ 2 + x) g[sin φ 2 + (cos φ 2 )x +...] lẍ = gx x = Ae λt + Be λt, lλ 2 = g x is increasing with time to very large values and φ does not stay close to φ 2 even if A, B are small. Pendulum falls down towards its stable equilibrium. Using small deviations from equilibria as new generalized coordinates we can work out kinetic and potential energies upto terms which are quadratic in x and ẋ. Write down a quadratic Lagrangian and quadratic Hamiltonian. Derive EL equation which will coincide with above equations. If Lagrangian is quadratic in generalized coordinates and velocities, it means that EL equations derived from this Lagrangian will be linear T = ml2 2 ( φ) 2 ml2 2 (ẋ)2 Taylor expansion for cos, cos φ = cos φ 0 (sin φ 0 )(φ φ 0 ) 1/2(cos φ 0 )(φ φ 0 ) 2 +... U = mgl cos φ mgl(cos φ 1,2 1 2 cos φ 1,2x 2 ) L(x, ẋ) = ml2 2 ẋ2 mgl 2 cos φ 1,2x 2 generalized coordinate here is x Important Note: terms in the Lagrangian which do not depend on the generalized coordinates and velocities can be disregarded. This is because they do not make any contribution into the EL equation Quadratic Hamiltonian is H(x, ẋ) = ml2 2 ẋ2 + mgl 2 cos φ 1,2x 2 Using this Hamiltonian we can calculate energy of oscillations near the stable equilibrium φ 1 = 0. In this case we have H = ml2 2 ẋ2 + mgl 2 x2 and known solution to the equations of motion is x = 2 A cos(ωt + ψ 0 ). Thus H = 2 A 2 mgl One can ask what is happening with H-quadratic calculated for unstable solution. Can H depend on time for this solution?? Breaking conservation of energy?? and making our approximation not

physical?? No! Everything is Ok. H near unstable point φ 2 = π is H = ml2 2 ẋ2 mgl 2 x2 substituting in there our solution x = Ae λt + Be λt we find that H = 2mgl AB Note if initial conditions are such that either A = 0 or B = 0, then H = 0. 19 Problems: One-star problems are for everyone. 2- and 3-star problems are only for enthusiasts. 1. Consider functional A[x(t)] = t 2 L(x, ẋ, ẍ, t). Assume that δx(,2 ) = δẋ(,2 ) = 0 and work out δa δx =?? 2. Find extremum of A[x(t)] = 1 0 (x2 + ẋ 2 + 2xe t ) for x(0) = 0, x(1) = e Answer is x 0 (t) = e 2 2(e 2 1) (et e t ) + tet 2 3. Mass on a vertical spring. Consider mass m oscillating on a vertical spring in the field of gravity. Sketch plot showing your coordinate system. Write down expressions for kinetic energy, potential energy, Lagrangian and Euler-Lagrange equations. Make sure that the EL equations coincide with the corresponding 2nd law of Newton. 4. Lagrangian of a particle is L = m 2 (ẋ2 +ẏ 2 +ż 2 )+mgy. Specify cyclic coordinates and conserved momenta 5. Work out 3D Lagrangian of a free particle in polar coordinates (r, φ, z), x = r cos φ, y = r sin φ. Specify cyclic coordinates and write down expression for the corresponding conserved momenta. 6. Lagrangian of a mechanical system is given by L = m 2 (5ẋ2 2ẋẏ + 2ẏ 2 ) + C(2x y) Are x and y cyclic coordinates?? Work out two generalized momenta: p x and p y. Are they conserved?? Verify that x x + ɛ, y y + 2ɛ is a symmetry. Workout a conserved momentum and verify its conservation by a direct calculation: ṗ = 0 7. A particle of mass m moves in one dimension under the influence of a force F : F = k x 2 e t/τ where k and τ are positive constants. Compute the Lagrangian and Hamiltonian functions. Compare the Hamiltonian and the total energy and discuss the conservation of energy for the system.

8. Consider particle with kinetic energy T = ẋ 2 and potential energy U = x 2 x 4. Specify intervals of E values which allow oscillatory motion of the particle. E is the total (initial) energy. Specify stable and unstable equilibria. 9. Work out an approximate expression for dependence of the pendulum frequency on amplitude of oscillations. Plan: Start from l φ = g sin φ and assume that φ is small. Replace sin φ with its Taylor expansion, but retain the first term nonlinear in φ. To solve the resulting equation assume φ = a(t)e iω0t + a(t) e iω0t, where ω 0 is the frequency of the pendulum found in the approximation sin φ φ. Assume that a is a slow function of time: ä ω 0 ȧ. 2nd part. Higher order harmonics. Consider substitution x = a(t)e iω0t + a(t) e iω0t + b(t)e i3ω0t + b(t) e i3ω0t and using the same approximations as above derive the system of first order differential equations for a and b. Using amplitude-phase substitutions a = r 1 e iθ and b = r 3 e i3θ find solutions of this system. 20 10. Write Lagrangian for the pendulum if its hanging point oscillates horizontally as f cos Ωt. Solve the corresponding Lagrange equation in the limit of small oscillations. 11. Write Lagrangian for the pendulum if its hanging point oscillates vertically as f cos Ωt. Write the corresponding Lagrange equation in the limit of small oscillations around φ = 0 and φ = π.