Chapter 7. The Principle of Least Action 7.1 Force Methos vs. Energy Methos We have so far stuie two istinct ways of analyzing physics problems: force methos, basically consisting of the application of Newton s Laws, an energy methos, consisting of the application of the principle of conservation of energy (the conservations of linear an angular momenta can also be consiere as part of this). Both have their avantages an isavantages. More precisely, energy methos often involve scalar quantities (e.g., work, kinetic energy, an potential energy) an are thus easier to hanle than forces, which are vectorial in nature. However, forces tell us more. The simple example of a particle subjecte to the earth s gravitation fiel will clearly illustrate this. We know that when a particle moves from position y 0 to y in the earth s gravitational fiel, the conservation of energy tells us that 1 2 m v2 2 v 0 ΔK + ΔU grav = 0 + mg y y 0 = 0, (7.1) which implies that the final spee of the particle, of mass m, is v = v 2 0 + 2g( y 0 y). (7.2) Although we know the final spee v of the particle, given its initial spee v 0 an the initial an final positions y 0 an y, we o not know how its position an velocity (an its spee) evolve with time. On the other han, if we apply Newton s Secon Law, then we write This equation is easily manipulate to yiel mge y = mae y = m v e y. (7.3) v = t t 0 v t = gτ + v 0 t 0 + v 0, = g t t 0 (7.4) - 138 -
where v 0 is the initial velocity (it appears in equation (7.4) as a constant of integration). If we conveniently choose t 0 0, then it follows that (note that v can be negative) v = v 0 gt, (7.5) while we note that we chose the y-axis to increase upwar (as coul also be guesse from equation (7.3)). We can integrate once more to get y = t 0 vτ t = ( v 0 gτ )τ 0 (7.6) = y 0 + v 0 t 1 2 g. Equations (7.5) an (7.6) allow us to track the position an velocity of the particle at any time after t = 0, something impossible to glean from the conservation of energy alone. We can further manipulate these two equations by taking the square of equation (7.5) v 2 = v 0 2 2v 0 gt + g 2 = v 2 0 2g v 0 t 1 2 g = v 2 0 2g( y y 0 ), (7.7) where equation (7.6) was inserte in the last step. Equation (7.7) is the same as equation (7.2) obtaine through the principle of conservation of energy. We conclue that, although more ifficult to apply, the force metho is nonetheless significantly more powerful. One is left to woner whether there exists a metho that woul combine the ease of application of the energy methos but yiel the power of the force methos. This is the topic we stuy for the rest of this chapter. 7.2 The Principle of Least Action an Newton s Secon Law Minimizing principles have been known an use for a long time in physics to solve problems of all kin. For example, the propagation of a beam of light between two points can be etermine by minimizing the time of travel (the Principle of Least Time), or it is often foun that a system will settle to rest in a configuration that minimizes it energy (e.g., a ball in a parabolic well). It is therefore not surprising that physicists have sought an overarching minimization principle from which all of physics coul be erive from (incluing, for example, Newton s Laws). As it turns out, there exists such a principle: The Principle of Least Action, which can be state as follows - 139 -
The motion of a system from time to is such that the line integral (calle the action) I = L, (7.8) where L = K U (with K an U the kinetic an potential energies, respectively) is minimize for the actual path of the motion. The quantity L is calle the Lagrangian, after the French-Italian scientist Joseph-Louis Lagrange (1736 1813) who is responsible for establishing the theory that rests on the minimization of the action. Although we will soon show how equation (7.8) can be minimize, mathematically, to erive the so-calle Lagrange equations, it is more instructive at first to see how we can use the a priori knowlege of Newton s Secon Law to guess how the Lagrangian is such a funamental physical quantity. Let us then return to our earlier problem of the particle subjecte to the earth s gravity, keeping in min that we wish to attack it using scalar quantities, such as the kinetic an gravitational potential energies. The question is as follows: Given that Newton s Secon Law specifies that (see equation (7.3)) m 2 y + mg = 0, (7.9) 2 how coul we use the kinetic an gravitational potential energies K an U to arrive at the same result? The first part to the answer rests with the relation we establishe in Chapter 2 between the gravitational force an the potential energy, i.e., or in this case F = U, (7.10) F = U y. = mgy y = mg. (7.11) Next we consier the inertial term m 2 y 2 in equation (7.9) an rewrite it as m 2 y 2 = m v. (7.12) - 140 -
Since we note that equation (7.12) is a function of the spee v it is quite clear that we cannot, once again, use the potential energy for this term as it only epens on the position y. On the other han, the kinetic energy can be avantageously consiere because K = 1 2 mv2, (7.13) while an K v = mv (7.14) K v = m v = m 2 y 2. (7.15) It follows from equations (7.11) an (7.15) that we can write K v + U y = m 2 y + mg 2 = 0. (7.16) We thus recover equation (7.3), which is what we sought to accomplish. We now introuce a small change in notation by putting a ot above a variable to enote its time erivative. That is, v = y y. (7.17) Finally, we also note that K is solely a function of the velocity y an inepenent of the position y, an that conversely, as was mentione earlier, U is only a function of the position y an not y, such that K y = U y We therefore introuce a new function (i.e., the Lagrangian) = 0. (7.18) - 141 -
an write the equation of motion (7.16) as L( y, y ) K ( y ) U ( y) (7.19) L y L = 0. (7.20) y This is the general form of the equation of motion we were seeking. 1 We can now verify that this applies equally well to other physical situations in general (an not only a particle in a gravitational fiel) by testing it on well-known problems. 7.2.1 Exercises 1. The Harmonic Oscillator. A particle of mass m is attache to a massless spring of force constant k, an allowe to slie without friction on a horizontal plane. Fin the mass equation of motion. Solution. The kinetic energy of the system is K = 1 2 m x2, (7.21) while the potential energy is that store in the spring U = 1 2 kx2. (7.22) The Lagragian is therefore L = K U = 1 2 m x2 1 2 kx2, (7.23) while 1 In equation (7.20) we use the partial erivatives L y an L y instea of total erivative L y an L y to express the fact that the Lagrangian L = L y, y is a function of both the position an the velocity. This is convention an notation use in mathematics to make this situation explicit. - 142 -
L x = k 2 = kx ( x 2 ) x (7.24) an x L = m x 2 2 = m ( 2 x ) 2 = m x. (7.25) The equation of motion is then L L = m x + kx x = 0, (7.26) as coul easily be verifie using Newton s Secon Law. This equation of motion can be integrate to yiel = Acos x t k m t +φ, (7.27) with A an φ some constants that epen on the initial conitions (i.e., position an spee) when the mass was release an allowe to move. 2. The Sliing Disk. Fin the equation of motion for a isk of mass m an raius R that is sliing own a frictionless ramp, incline by an angle α, from rest at a height h. Solution. The kinetic energy of the isk is K = 1 2 m y2, (7.28) with y its position along the incline ( y = 0 at the top an increasing ownwar). The potential energy is U = mg h + R ysin( α ), (7.29) - 143 -
an the Lagrangian L = K U = 1 2 m y2 mg h + R ysin( α ). (7.30) We now calculate L y = mg h + R y = mgsin( α ), y y sin( α ) (7.31) an y L y = m y 2 2 = m ( 2 y ) 2 = m y. (7.32) The equation of motion for the isk s centre of mass is therefore L y L y = m y mgsin ( α ) = 0, (7.33) as coul easily be verifie using Newton s Secon Law. This equation of motion can be integrate to yiel y( t) = h + R + 1 2 gsin( α ). (7.34) Note that we i not make use of any forces (e.g., the normal force that the incline applies to the isk) to solve the problem. 3. The Rolling Disk. Fin the equation of motion for the position of the centre of mass of a isk of mass m an raius R that is rolling without slipping own a ramp, incline by an angle α, from rest at a height h (see Figure 1). Solution. The kinetic energy of the isk is the sum of translational an rotational components - 144 -
Figure 1 - A isk rolling on an incline plane without slipping. K = 1 2 m y2 + 1 2 I θ 2 = 1 2 m y2 + 1 4 mr2 θ 2, (7.35) where I = mr 2 2 is the moment of inertia of the isk about its axis of rotation an y is the velocity of its centre of mass. But we can also see from the figure that y = Rθ (7.36) an therefore y = Rθ. (7.37) Inserting equation (7.37) in equation (7.35) we get K = 1 2 m y2 + 1 4 m y2 = 3 4 m y2. (7.38) The potential energy is as calculate in equation (7.29) of Prob. 2 an the Lagrangian U = mg h + R ysin( α ), (7.39) - 145 -
L = K U = 3 4 m y2 mg h + R ysin( α ). (7.40) We now calculate L y = mg h + R y = mgsin( α ), y y sin( α ) (7.41) an y L y = 3 4 m y 2 = 3 4 m ( 2 y ) = 3 2 m y. (7.42) The equation of motion for the crate is therefore L y L y = 3 2 = 0. m y mgsin ( α ) (7.43) This equation of motion can be integrate to yiel y( t) = h + R + 1 3 gsin( α ). (7.44) We see that, when compare with equation (7.34) for the sliing isk of Prob. 2, the rotation of the isk brings an aitional factor of 2 3 in the position of the isk. That is, the isk goes slower own the incline by a factor of 2 3. 4. The Penulum. Fin the equation of motion for the angular position θ of the penulum mae of a light string of length an a mass m shown in Figure 2. Solution. The kinetic energy of the system is - 146 -
Figure 2 - A simple penulum of mass an length. K = 1 2 I θ 2 = 1 2 m 2 θ 2, (7.45) while the potential energy is given by The Lagrangian is therefore U = mg cos( θ ). (7.46) L = K U = 1 θ 2 m 2 2 + mg cos( θ ), (7.47) with the neee erivatives L θ = mg sin θ θ L θ = 1 θ 2 2 m 2 = m 2 θ. (7.48) The equation of motion is thus L θ L θ = m 2 θ + mg sin θ = 0. (7.49) - 147 -
For cases of small oscillations sin( θ ) θ, an the equation of motion becomes θ + g θ 0, (7.50) with for solution θ t Acos g t +φ. (7.51) The quantities A an φ are some constants that epen on the initial conitions (i.e., position an velocity) when the mass was release an allowe to oscillate. 5. The Couple Oscillator. The Lagrangian formalism is also very powerful to erive the equations of motion for systems that have more than one egree of freeom. That is, in the preceing problems we ha to solve for only one variable (or egree of freeom; e.g., the position of the mass attache to a spring, the centre of mass position of a isk coming own an incline, or the angular isplacement of a penulum), but it is not uncommon that more than one variables are involve in the escription of a physical system. For example, let us consier the couple oscillator compose of three massless springs (two of force constant κ an one of constant κ 12 ) an two particles of similar mass M, as shown in Figure 3. We will use the Lagrangian formalism to simultaneously etermine the equations of motion for the positions x 1 an x 2 of the two masses. Solution. The kinetic energy of the system is mae of the sum of the corresponing energies of the iniviual particles K = 1 2 M x 2 2 ( 1 + x 2 ), (7.52) while the potential energy of the system must inclue a term, for the mile spring of force constant κ 12, that epens on the ifference in the positions of the two particles The Lagrangian is then U = 1 2 κ x 2 2 ( 1 + x 2 ) + 1 2 κ 12 ( x 2 x 1 ) 2. (7.53) with the neee erivatives 1 2 κ x 2 2 ( 1 + x 2 ) 1 2 κ 12 x 2 x 1 L = 1 2 M x 2 2 1 + x 2 2, (7.54) - 148 -
Figure 3 - Two masses connecte to each other by a spring an to fixe points by other springs. L = κ x 1 +κ 12 ( x 2 x 1 ) x 1 L = κ x 2 κ 12 ( x 2 x 1 ) x 2 (7.55) an L x 1 = M x 1 L x 2 = M x 2. (7.56) We thus have two equations of motion, one for x 1 an another for x 2, L x 1 L = M x 1 + ( κ +κ 12 )x 1 κ 12 x 2 = 0 x 1 L x 2 L = M x 2 + ( κ +κ 12 )x 2 κ 12 x 1 = 0. x 2 (7.57) Determining the solution for such a set of couple ifferential equations is well beyon the scope of of our iscussion, but it can be shown to be x 1 x 2 ( t) = A 1 cos( ω 1 t +φ 1 ) + A 2 cos( ω 2 t +φ 2 ) ( t) = A 1 cos( ω 1 t +φ 1 ) + A 2 cos ω 2 t +φ 2, (7.58) with - 149 -
ω 1 = κ + 2κ 12 M, ω 2 = κ M, (7.59) an where A 1, A 2, φ 1, an φ 2 are constants that epen on the initial conitions. 7.2.2 Minimization of the Action 2 Although we have successfully applie Lagrange s equation to several physical systems, we have yet to prove that it stems from an overarching principle involving the minimization of the action integral efine in equation (7.8). We thus en this chapter by accomplishing this task. In general, for a function f ( x) to exhibit a minimum at some point x 0 it must be that its slope vanishes at that point. 3 That is, the variation δ f in the function shoul verify that δ f = f x δ x = 0 (7.60), then both for a minimum to occur about x 0. If the function epens on more than one variable, e.g., f = f x, y erivatives f x an f y must vanish about a point x 0, y 0 there. An we fin that, upon using the so-calle chain rule of calculus, δ f = f x δ x + f y δ y = 0 (7.61) about ( x 0, y 0 ). We will apply this technique to fin the minimum of the action to the simpler case of a system with only one egree of freeom x (a generalization to several egrees of freeom is straightforwar), by first nothing that it is a function of both x an x. That is, = δ L x, x δ I x, x t L = x δ x + L 2 x δ x = 0. (7.62) 2 This section contains avance mathematical concepts on which you will not be teste; it is solely provie for completeness. 3 A maximum or an inflection point in f x will also have vanishing erivatives; to ensure that a minimum (maximum) occurs we shoul verify that the secon erivative 2 f x 2 is positive (negative). We will assume in our analysis that a minimum occurs, although this shoul, in principle, be verifie. - 150 -
In equation (7.62) we applie the chain rule, as in equation (7.61), but consiering x an x as the inepenent variables (like x an y were in equation (7.61)). We now transform the quantity δ x as follows δ x = δ x x( t) x t + Δt = δ lim Δt 0 Δt δ x t + Δt = lim Δt 0 Δt = ( δ x). δ x( t) (7.63) The secon term in the integran of equation (7.62) can therefore be written as L t L δ x = x x δ x 2. (7.64) We nee to further transform this integral. But before we o so, let us consier the following erivative L x δ x = L δ x + L x δ x, (7.65) since we know that the erivative of a prouct of two functions follows the rule Inserting equation (7.65) in equation (7.64) we have ( uv) = u v + u v. (7.66) L δ x = x L x δ x t = L 2 x δ x = L x δ x t1 L δ x L δ x L δ x. (7.67) We now note, however, that the Principle of Least Action consiers variations from the true path of motion between two fixe points ( x 1, ) an ( x 2, ) at which the variations δ x = 0. That is, - 151 -
x x 2 δx x 1 t Figure 4 Variations about the true path of motion (soli line) between to fixe points an at which. = δ x( ) δ x = 0. (7.68) This is shown in Figure 4. The action is evaluate between these fixe points for all curves by integrating the corresponing Lagrangians; the true path of motion is the one for which the action is minimize. Inserting equation (7.68) in the last of equations (7.67) we fin L δ x x L = δ x, (7.69) an the variation of the action δ I becomes (from equation (7.62)) δ I ( x, x ) = t L x 2 = 0. L δ x (7.70) Since equation (7.70) is vali for any variation δ x, it must be that L L = 0. (7.71) x We recognize the Lagrange Equation we erive earlier from Newton s Secon Law (see equation (7.20)). This equation amazingly emerges from the Principle of Least Action. It is to be note that, although all the examples we previously consiere assume that the kinetic an potential energies were respectively only functions of the velocity an the position, it is not necessarily the case in general (for example, the potential energy for a charge particle subjecte to an electromagnetic fiel oes epen on its velocity). - 152 -