Physics 5153 Classical Mechanics. Solution by Quadrature-1
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1 October 14, :47:49 1 Introduction Physics 5153 Classical Mechanics Solution by Quadrature In the previous lectures, we have reduced the number o eective degrees o reedom that are needed to solve the equations o motion. This was done by inding integrals (constants) o the motion. The irst step was to determine which variables are cyclic, since these lead to conserved momenta. Once the initial conditions are speciied or the problem, the momenta are known or all time. I the system is conservative, we ound that the Jacobi integral (T T 0 + V = h) is also an integral (constant) o the motion. In cases where all the variables are cyclic except or one, the Jacobi integral can be directly integrated q = (q) dt = dq (q) and the time dependence o the remaining generalized coordinate determined. In this case we say that the solution is ound by quadratures. That is, the solution is given as an indeinite integral o a single variable. Our goal in this lecture is to determine when an equation o motion is separable, and can be put into a number o equations o the orm o Eq. 1, that is we would like to know when an equation o motion can be solved by quadratures. 1.1 Liouville System First recall that a system having n degrees o reedom requires n integrals (constants) o the motion or the complete solution o the equations o motion. These n integrals o motion are eectively the initial conditions or the second order dierential equation o motion. As stated in the introduction, is n 1 variables are cyclic, then the variables can be removed by use o the Routhian method. This leaves us with an equation o motion in one variable. Furthermore, is the system is conservative, then the equation is solved directly by quadratures. Let s assume that we have a conservative holonomic system. In addition, assume that there are not enough cyclic variables to guarantee separability as given above. It may still be separable, i it is an orthogonal system, that is, it is a natural system and there are no cross terms ( q i q j i j) in the kinetic energy. Let s consider the ollowing general example, which we will show is separable, that is, solvable by quadratures. Assume that the kinetic and potential energies are given in the ollowing orms T = 1 q i V = 1 v i (q i ) () where = (1) i (q i ) > 0 (3) Let s consider the Lagrange equations in the ollowing orm ( ) d T T + V = 0 (4) dt q i q i q i Solution by Quadrature-1
2 October 14, :47:49 Substituting the kinetic and potential energies rom Eq. into Lagrange s equation gives the ollowing equation d dt ( q i) 1 i q j + 1 v i V i = 0 (5) q i q i q i Since this is a natural system, the Jacobi (energy) integral is given by T + V = 1 j=1 q j + V = h 1 j=1 Next substitute Eq 6 into Eq. 5 and simpliy Now multiply this equation by q i and obtain j=1 q j = 1 (h V ) (6) d dt ( q i) h i + 1 v i = 0 (7) q i q i d dt ( q i ) h i q i q i + v i q i q i = 0 d dt ( q i ) = d dt (h i v i ) (8) where the right hand side o the second term is a total derivative since each term only depends on q i. Integrating, the result is q i = [h i v i + c i ] (9) Next sum over i, where we ind the ollowing c i = 0 (10) which can be shown using Eqs., 6 and 9. Hence the c i and h together orm n independent constants o the motion. (Note that there are only n 1 independent c i, otherwise all the c i would have to be zero.) The remaining n constants o the motion, can be derived using Eq. 9 in the ollowing orm dq i dt = (hi v i + c i ) (11) which implies dq 1 (h1 v 1 + c 1 ) = dq (h v + c ) = = dq n (hn v n + c n ) = dt dτ (1) where τ is a time-like parameter. Each dierential expression is a unction o a single q i, so the problem is reduced to quadratures. Integrating these equations produces the remaining constants o the motion. This system can be generalized to become a Liouville system, replacing dq i with M i (q i )dq i to obtain T = 1 M i (q i ) q i (13) Solution by Quadrature-
3 October 14, :47:49 where we assume that M i (q i ) > 0. The potential energy (V ) remains as beore. A natural system having the kinetic and potential energies in this orm is called a Liouville system. By making the transormation q i M i (q i ) q i, the solutions or a Liouville system are as in Eq. 1 given by where dq 1 φ1 (q 1 ) = dq φ (q ) = = dq n φn (q n ) = dt = dτ (14) φ i (q i ) = M i (h i v i + c i ) (i = 1,,..., n) (15) Using Eq. 3, multiplying Eq. 14 by i, and summing over i, we obtain i dq i φi (q i ) = dt i dq i φi (q i ) = t β 1 (16) Similarly, taking dierences o the indeinite integrals o Eq. 14, we have dq 1 φ1 (q 1 ) dq j φj (q j ) = β j (j = 1,,..., n) (17) where the irst integral is chosen arbitrarily as a reerence. Thus, these two sets o equations provide n independent constant the β 1, β,..., β n, which, with the previous n 1 c i and h, constitute the required n independent constants o the motion. In evaluating the integrals o Eqs. 16 and 17, a question arises concerning the sign o φ i (q i ). Since is positive, we rom Eq. 14 ind that φ i (q i ) has the same sign as dq i. This is important in the study o libration motions, that is, motions in which one or more q s oscillate between ixed limiting values. 1. Example As an example, let s consider the spherical pendulum shown in Fig. 1. Reduce the problem to quadratures and obtain the integrals o the motion. Beore attempting the solution, we determine the kinetic and potential energies using an appropriate set o generalized coordinates. We start by writing the Lagrangian in Cartesian coordinates L = 1 m(ẋ + ẏ + ż ) mgz (18) Beore speciying the point transormation to an appropriate set o generalized coordinates, we must determine the number o degrees o reedom. The Lagrangian as given above has 3 coordinates, but there is a constraint among the coordinates (x +y +z l = 0). Hence, there are degrees o reedom. The point transormations that are required to express the Lagrangian in the generalized coordinates (, φ) given in the Fig. 1 are x = l sin cos φ y = l sin sin φ z = l cos ẋ = l cos cos φ φl sin sin φ ẏ = l cos sin φ + φl sin cos φ ż = l sin (19) Solution by Quadrature-3
4 October 14, :47:49 z PSrag replacements x φ l y Figure 1: Spherical pendulum along with coordinate system. where the constraint has been taken into account. Hence, the Lagrangian is transormed to the ollowing orm We can now solve the problem using two dierent methods Method 1 L = 1 ml ( + φ sin ) mgl cos (0) In this method, the problem will be solved by removing the cyclic variables rom the Lagrangian by using the Routhian method. We use the Lagrangian given in Eq. 0 L = 1 ml ( + φ sin ) mgl cos (1) Note that the Lagrangian has a single cyclic coordinate φ, with the associated generalized momentum given by L φ = ml φ sin = α φ () where α φ is the constant momentum in the φ direction. Next we calculate the Routhian R = L α φ φ = 1 ml αφ ml sin mgl cos (3) where φ has been eliminated as expected. Notice, that the Routhian can be written as a natural system T = 1 ml V αφ R = T V (4) = ml sin + mgl cos Hence, we can immediately write down the (Jacobi) energy integral h = T + V = 1 ml + αφ ml sin + mgl cos (5) Solution by Quadrature-4
5 October 14, :47:49 This can be solved or the velocity = ml (h mgl cos α φ /(ml ) sin ) (6) which can be written as ollows This equation can be integrated to give ml sin d dt = ml sin (h mgl cos ) αφ t t 0 = 0 ml sin d ml sin (h mgl cos ) α φ The sign o the square root should be the same as d stated earlier, since any change in d causes a positive change in dt. Next we calculate φ(). Start with the generalized momentum Next substitute or dt integrating this equation gives dφ = dt ml sin αφ dφ = d sin ml sin (h mgl cos ) αφ φ φ 0 = 0 α φ d sin ml sin (h mgl cos ) αφ Thus, we have obtained the 4 constants o the motion α φ, h, t 0, and φ Method We will express the system as a Liouville system. The kinetic and potential energies in the generalized coordinates shown in Fig. 1 are (7) (8) (9) (30) (31) T = 1 ml ( + φ sin ) V = mgl cos (3) Notice that the system is orthogonal, since the kinetic energy depends only on the square o the generalized velocities and there are no cross terms. The standard orm o the Liouville problem, T = 1 M i (q i ) q i V = 1 v i (q i ) with = i (q i ) > 0 (33) Writing the kinetic and potential energies in this orm we get T = 1 ( ml sin ) { } 1 sin + φ 1 { V = m ml sin gl 3 sin cos } (34) Solution by Quadrature-5
6 October 14, :47:49 Thereore, we ind the coeicients to be () = ml sin M () = 1 sin v () = m ql 3 sin cos (35) φ (φ) = 0 M φ (φ) = 1 v φ (φ) = 0 Next, we give the φ i (q i ) or this problem φ i (q i ) = M i (h i v i + c i ) { φ = sin [ml sin (h mgl cos ) + c ] φ φ = c φ and using Eq. 10, we ind the c i to be Finally, we arrive at the result c φ = c = [ ml φ sin ] = α φ (36) 0 ml sin d φ () = t t 0 and 0 d φ φ () = dφ cφ φ 0 0 α φ d φ () = φ φ 0 (37) which ater substituting or φ i (q i ), we ind the same result as using the other method. Notice that we have arrived at the 4 constant need, c, c φ, φ 0, and t 0, which all given by the initial conditions. Solution by Quadrature-6
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