and that its potential energy is V = 10 2 d.x. Then the Lagrangian is L- T-V - / 3 (2%)'- 30 (82)'] ax- fedx. where L is the Lagrangian density, C = Ho (4) The action is therefore (5) In this action, the "path" is the function y(x, t). (c) To find the equations of motion, first we must vary the action as we vary y(x, t) y(x, t) + dy(x, t). Substitute this into the action (5), and keeping only first order terms in dy, show that 8s = " at [" ax 10 21 91 - To ox (6) (d) Proceeding as we did for the point particle, integrate by parts in (6) to obtain three terms for of. Hamilton's principle asserts that the correct equation of motion for y(x, t) follows from oS = 0. Since we have three terms, each must vanish indepedently. Argue that the first term vanishes if we specify the initial and final configurations of the string at times t; and t. Argue that the second term vanishes if we specify boundary conditions at the endpoints of the string - recall that there are two choices, Neumann and Dirichlet, so you must consider both. Finally, argue that in order for the third term to vanish, the function y(x, t) must satisfy the wave equation, with wave speed vo = VTo/ Ho. We can actually derive the wave equation for the string using Hamilton's principle in a slightly more general way. Start with (7 ) Define the conjugate momenta by Oc Oc P= ay' where y' = dy/O.x.(e) Vary the action (7) with respect to dy, to first order, and in terms of the conjugate momenta, show that the wave equation for the string can be written as o 0P Ot dx Do this in general, that is, without using an explicit form for L. 0. (8) (f) Explicitly compute P' and P* for our Lagrangian density (4) and show that (8) is the wave equation for a string you obtained in part (d). Physically, then, what is P*? Problem 1 The string Lagrangian Nearly a century after Newton's mechanics was developed, an alternate approach to mechan- ics was developed by Joseph-Louis Lagrange and William Rowan Hamilton, now referred to as Lagrangian mechanics and Hamiltonian mechanics. or sometimes as analytical mechanics. Although no new fundamental physics was introduced. these methods provided a new and more powerful view on mechanics based on variational techniques. Much later. it was found that these methods can be generalized to quantum physics. which is not readily done for Newtonian mechanics. In this problem, we will learn how to derive the wave equation for a one-dimensional (nonrelativistic) string using Lagrangian mechanics. The Lagrangian L, of a (classical) system is defined by L=T-YV, where T 1s the kinetic energy of the system and V' is the potential energy of the system. For a point particle of mass m moving in one dimension along the r-axis under the influence of a potential V' (r), the Lagrangian takes the form _ s L=T-V =-mi*V(x), 2 where #(t) = du/dt. This Lagrangian is an implicit function of time, but it has no explicit time dependence. The action is defined as S = /F L(t)dt. where P is a path, x(#), between an initial position ; at initial time ;, and a final position rp at a final time f; > ;. The action is a functional: it takes a function as an input and gives a number as output. To indicate that S is a functional, we write S[z]. Then, for any path x(t), the action is T S[z] = / Sn(i(0)? =V (a(t)) | dt. S o f : It is very important to emphasize that S can be computed for any path, not just the physical path that the particle takes. Hamilton's principle states that the path P which a system actually takes is one for which the action S 1s stationary. That 1s. if this path P is varied infinitesimally, the action does not change to first order in the variation. Consider the physical path a point particle of mass m takes to be x(t). Then the perturbed path takes the form x()+dx(), where dx(#) represents the deviation or variation from the origin path at time 7. as illustrated in the figure below: x(1) + ex(t) ax (t) We restrict to variations for which the initial and final endpoint are left unchanged; that is, we require that or(t;) = 6x(t;) = 0. (a) Substitute the varied path x(t) + or(t) into the action (1) and expand to first order in 6.r(t). Show that the variation in the action is (to first order) S[a + or] = S[x] + 85. where oS = where V'(x) = dV/dx. (b) To proceed, we need to rewrite the variation in the action in the form of = for(t) [. . . ] dt. that is, so that there are no derivatives acting on the or. Integrate by parts in (2) to move the derivative off or, and then use the fact that or(t,) = or(t;) = 0 to show that (2) becomes 6S = ox(t) [-mi(t) - V'(x(t))] dt. (3) Hamilton's principle says that the action must be stationary for the physical path, which means that of = 0. This must be true for every variation or(t), and therefore, it must be that the integrand in (3) is itself zero. That is, it must be that mix(t) = -V'(x(t)). But this is just Newton's 2nd law for our particle moving in the potential V(x)! We have thus "derived" Newton's 2nd law from Hamilton's principle. More accurately, we have de- termined the correct equations of motion that the physical path of the particle must satisfy through a variational principle. Let us now build the action for a (nonrelativistic) one-dimensional string of constant mass density Ho, under constant tension To. and with endpoints at a = 0 and a = a (not necessarily fixed - just a string of finite length a), and by applying Hamilton's principle, derive the wave equation. We will consider only transverse displacements; let the transverse direction be denoted y so that the transverse motion of the string is described by y(x, t). First, recall that the kinetic energy of the string is T = " Mode) (34 )