PHY6200 W07

Chapter 7: Lagrange's Equations


In Taylor, read sections 7.2 to 7.3 for today, and 7.3 to 7.5 for Wednesday.


S = ∫12 L(qi, qidot, t) dt.
L/∂qi - (d/dt)(∂L/∂qidot) = 0
L = T - U

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.

Example: One Particle in Two Dimensions; Polar Coordinates

Write down the Lagrangian for a particle moving in two dimensions under the influence of a central force, in polar coordinates. Find the equations of motion and relate them to known results.

In polar coordinates, the position of the particle is r = rrhat, and its velocity is rdot = rdotrhat + rφdotφhat. The Lagrangian is

L = T - U = ½m(rdot² + r²φdot²) - U(r).

This Lagrangian has two independent coordinates, r and φ, so there will be two equations.

The r Equation

Lagrange's equation for the r coordinate is

L/∂r = (d/dt)(∂L/∂rdot


mrφdot² - ∂U/∂r = (d/dt)(mrdot) = mrddot

We can rewrite this, calling -∂U/∂r = Fr to get

Fr = m(rddot - rφdot²) = mar

where the last step relies on noting that the radial component of the acceleration is just the term in parentheses.

The φ Equation

Lagrange's equation for the φ coordinate is

L/∂φ = (d/dt)(∂L/∂φdot


-∂U/∂φ = (d/dt)(mr²φdot)

The left side of this equation is the torque, Γ, in two dimensions (Fφ = -(1/r)∂U/∂φ, so Γ = rFφ = -∂U/∂φ). The right side of the equation is the time derivative of the angular momentum, L. Therefore, this is equivalent to the familiar expression Γ = dL/dt.

Generalized momenta and generalized forces

The derivatives of the Lagrangian are known as a generalized force (∂L/∂qi = Fi) and a generalized momentum (∂L/∂qidot = pi). When thought of this way, Lagrange's equations can be expressed in the form

dpi/dt = Fi

an expression that looks deceivingly like Newton's second law. But notice that the generalized momentum and force can be completely different from a traditional linear momentum and force in Newtonian mechanics. They are just the appropriate derivatives of the Lagrangian. For example the Lagrange equation in φ yielded a torque for the generalized force, and an angular momentum for the generalized momentum.

Several Unconstrained Particles

These ideas are easily extended to a system with many paricles. If there are N particles moving freely in 3-dimensions then there are 3N coordinates, qi, and 3N Lagrange equations.

Constrained Systems; an Example

One of the strengths of Lagrangian mechanics is the ease with which constrained systems are handled. By constrained systems, I mean systems whose motion is restricted, not simply restricted to a line or a plane, but to any arbitrary one dimensional path or two dimensional surface. For instance, a bead that is free to slide along a wire bent to an arbitrary shape, or a mass confined to move on an arbitrary surface. One need simply express the kinetic and potential energy in terms of appropriate variables; one variable in the case of one-dimensional motion, and two variables in the case of two-dimensional motion.

Example: the Simple Pendulum

As a simple example, write down the Lagrangian for a simple pendulum of mass m and length l and solve for the motion.

The position of the pendulum bob can be described by it's (x,y) position, but since the length l is constant, the bob's motion is restricted to a one-dimensional path, and we can express its location in terms of the angle φ with x = lsinφ and y = l(1-cosφ). In terms of φ the kinetic energy is T = ½ml²φdot², and the gravitational potential energy is U = mgl(1-cosφ). The tension doesn't appear in the Lagrangian approach; it is a constraint force that keeps the bob on its proper path, but doesn't contribute to the kinetic or potential energies. The Lagrangian for the pendulum is

L = T - U = ½ml²φdot² - mgl(1-cosφ).

This Lagrangian has only one generalized coordinate, φ, so there is one equation:

L/∂φ = (d/dt)(∂L/∂φdot).

This evaluates to

-mgl sinφ = (d/dt)(ml²φdot) = ml²φddot.

This is the familiar result for a pendulum. If we make the small angle approximation, sinφ≈φ, the equation predicts simple harmonic motion with angular frequency ω = √(g/l).

Constrained Systems in General

Generalized Coordinates

One of the great advantages of Lagrangian mechanics is the ability to use virtually anything as a coordinate for describing the motion. In the example of the simple pendulum, we used the angle φ of the pendulum to the vertical as the coordinate. Let's look at a more complex example, a double pendulum.

The Double Pendulum

A double pendulum is made by suspending a mass m1 from a fixed point by a massless rod of length l1, and suspending a mass m2 from m1 by a massless rod of length l2. Both rods are free to pivot in the x-y plane.

We can express the positions of the two masses in terms of just two coordinates, φ1 and φ2, the angles of the rods l1 and l2 with respect to the vertical. Measuring y vertically downward fromt he point of attachement, the position of m1 is (x1,y1) = l1(sinφ1, cosφ1) and the position of m2 is (x2,y2) = l1(sinφ1, cos&phi1;) + l2(sinφ2, cosφ2). The kinetic energy of m1 is T1 = ½m1v² = ½ m1(x1dot² + y1dot²) = ½m1l1²φ1dot². The kineric energy of m2 is T2 = ½m2(x2dot² + y2dot²) = ½m2l1²φ1dot² + ½m2l2²φ2dot².

© 2007 Robert Harr