PHY6200 W07

Chapter 6: Calculus of Variations


In Taylor, read sections 7.1 to 7.2 for today, and 7.2 to 7.3 for Monday.


(∂f/∂qi) - d/dt(∂f/∂q'i) = 0

for every qi. This is the generalized form of the Euler-Lagrange equations for an arbitrary number of variables.

Shortest path between two points in spherical coordinates

Solve the problem for the shortest path between two points in 3 dimensions, using spherical coordinates. For convenience, and without loss of generality, we can place the origin at the starting point such that r(t1 = 0) = (r=0, θ=0, φ=0), and r(t2) = (r=r2, θ=θ2, φ=φ2).

From the integral expression for L written above, we see that the function to put into the Euler-Lagrange equations is

f = √[r'(t)² + r(t)² θ'(t)² + r(t)²sin²θ(t) φ'(t)²]

The problem has three equations to solve for: r(t), θ(t), and &phi(t). Each has its own Euler-Lagrange equation:

(∂f/∂r) - d/dt(∂f/∂r') = (r θ'² + r sin²θ φ'² - dr'/dt)/f = 0
(∂f/∂θ) - d/dt(∂f/∂θ') = [r² sinθcosθ φ'² - (d/dt)(r² θ')]/f = 0
(∂f/∂φ) - d/dt(∂f/∂φ') = -[(d/dt)(r²sin²θ φ')]/f = 0

The last equation says that r²sin²θ φ' = constant, and since this is true for all the points, it is true for the starting point, (0, θ1, φ1). This can be satisfied only if φ' = 0 implying that φ = constant = φ1 = φ2. Insert φ' = 0 in the first and second equations to yield

r θ'² = dr'/dt
(d/dt)(r² θ') = 0

The second of these two says that r² θ' = constant. This must be true for all points on the path, in particular, the two endpoints. At the origin, r=0, so the constant must be zero, then use of the other endpoint tells us that &theta'=0. Therefore we find that θ = constant = θ1 = θ2. Inserting θ' = 0 into the first of the two equations yields dr'/dt = 0, therefore r' = constant = c, and r(t) = ct + r0. Since r=0 at t=0, the constant of integration r0 = 0, and using r=r2 at t = t2, we find c = r2/t2. Putting this all together we have

r(t) = r2 t/t2   θ(t) = θ2   and   φ(t) = φ2.

This is the parametric equation for a straight line passing through the origin in spherical coordinates.

Chapter 7: Lagrange's Equations

Now we are ready to begin Lagrangian mechanics. Lagrangian mechanics can be derived in a number of ways. It can be shown to be equivalent to Newtonian mechanics, and it can be shown to be equivalent to Hamiltonian mechanics. Our interest is to learn what it is and how to use it, with only passing interest in how it can be derived and demonstrating its equivalence to alternative formulations of mechanics. We'll begin with a definition of the "action", and show how Lagrangian, show its equivalence to Newtonian mechanics for some simple examples, and begin applying it to problems.

Lagrange's Equations for Unconstrained Motion

The Principle of Least Action

The principle of least action, also known as Hamilton's principle, states that a system will follow a path such that the action is an extremum. The action S is defined as the integral over the path of a function of the coordinates, velocities, and time called the Lagrangian

S = ∫12 L(qi, qidot, t) dt.

This is a variational problem of the sort examined in chapter 6, and results in a set of differential equations, the Euler-Langrange equations, but in the context of Lagrangian mechanics they are known as Lagrange's equations:

L/∂qi - (d/dt)(∂L/∂qidot) = 0

The question now is what should be the form of the Lagrangian L? The answer is that L = T - U, kinetic minus potential energy expressed in coordinates relative to an inertial frame. I won't derive that result, but let's motivate it from some simple arguments.

Lagrangian for a Free Particle

First, notice that the variation of S is unchanged by adding a total time derivative of a function of position and time to the Lagrangian:

S = ∫12 [L(qi, qidot, t) + (df(qi,t)/dt)] dt.

Upon integration, the time derivative gives the difference of the function at the endpoints of the path:

S = ∫12 L(qi, qidot, t) dt + f(qi2,t2) - f(qi1,t1).

Since this derivative term changes the action by a constant, it doesn't change where the action is stationary (minimum, maximum or inflection point). This result will be used in the argument below.

We know that in an inertial frame, a free particle will move with constant velocity, both direction and magnitude. And the uniformity of space means there is no preferred direction of the velocity, so the Lagrangian can depend on the magnitude, or square of the velocity, L = L(v²). Now suppose that we observe the particle from another inertial frame moving with a small relative velocity to the first, δu, so that the particle is observed to move with constant velocity v' = v + δu according to the principle of Galilean relativity. This second frame is still an inertial frame, so the Lagrangian must have the same form in this frame as the first, that is L' = L(v'²) = L(v² + 2v &sdot δu + δu²). Expand this expression to first order in δu:

L(v'²) = L(v²) + (∂L/∂v²) 2v &sdot δu

This change of frames shouldn't change the path that minimizes the action. Therefore, L(v'²) must differ from L(v²) by, at most, a total time derivative of a function of the coordinates and time, that is, the remaining term must be a total time derivative of some function of the position and time. Such a time derivative is, at most, linear in the velocity [df(q,t)/dt = ∂f/∂t + ∂f/∂q (dq/dt)]. Therefore, since (∂L/∂v²) 2v &sdot δu contains one term of velocity explicitly, we can infer that ∂L/∂v² is independent of the velocity. Since L is a function of v², it must contain at most a term proportional to v² which we can write as ½mv².

By straightforward extension, we conclude that for n free and non-interacting particles, the lagrangian is L = ∑ ½mava². This is, of course, the kinetic energy of the particles, which we denote by T.

Lagrangian for Interacting Particles

If the particles are interacting, then the Lagrangian is modified by a function U that can depend on the coordinates and time, but not the velocities

L = ∑ ½mava² - U(r1, r2, ..., t).

Of course, this function is the potential energy of the system. To see this, put the Lagrangian into Lagrange's equations:

(d/dt)(∂L/∂vxa = ∂L/∂xa

and likewise for y and z components of each particle, and take the appropriate derivatives to find the equations of motion:

ma(dvxa/dt) = -∂U/∂xa

and likewise for y and z. This is Newton's second law, where -∂U/∂xa is the x component of the force on particle a, Fxa.

This simple argument justifies the form of the Lagrangian, L = T - U, at least for cartesian coordinates. The full derivation is more complex, because the Lagrangian can be expressed in a wide variety of coordinates. Notice also that this example requires conservative forces, that is, forces derivable from a potential energy. As you may guess from the form of the Lagrangian, that is generally true. The Lagrangian formalism can be adapted to include non-conservative forces, but if these are important, perhaps Newtonian mechanics is a better approach to solve the problem.

© 2007 Robert Harr