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

S = ∫_{1}^{2} L(q_{i}, q_{i}dot, t) dt.

∂L/∂q_{i} - (d/dt)(∂L/∂q_{i}dot) = 0

L = T - U

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 = ∑ ½m_{a}v_{a}² - U(**r**_{1}, **r**_{2}, ..., t).

Of course, this function is the potential energy of the system (in Cartesian coordinates).
To see this, put the Lagrangian into Lagrange's equations for x_{a}, y_{a}, z_{a}:

(d/dt)(∂L/∂v_{xa}) = ∂L/∂x_{a}

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

m_{a}(dv_{xa}/dt) = -∂U/∂x_{a}

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

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.

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** = r**rhat**, and its velocity is **r**dot = rdot**rhat** + rφdot**φhat**.
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.

Lagrange's equation for the r coordinate is

∂L/∂r = (d/dt)(∂L/∂rdot

or

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

We can rewrite this, calling -∂U/∂r = F_{r} to get

F_{r} = m(rddot - rφdot²) = ma_{r}

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

Lagrange's equation for the φ coordinate is

∂L/∂φ = (d/dt)(∂L/∂φdot

or

-∂U/∂φ = (d/dt)(mr²φdot)

The left side of this equation is the torque, Γ, in two dimensions (F_{φ} = -(1/r)∂U/∂φ, so Γ = rF_{φ} = -∂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.

The derivatives of the Lagrangian are known as a generalized force (∂L/∂q_{i} = F_{i}) and a generalized momentum (∂L/∂q_{i}dot = p_{i}).
When thought of this way, Lagrange's equations can be expressed in the form

dp_{i}/dt = F_{i}

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.

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, q_{i}, and 3N Lagrange equations.

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.

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).

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.

A double pendulum is made by suspending a mass m_{1} from a fixed point by a massless rod of length l_{1}, and suspending a mass m_{2} from m_{1} by a massless rod of length l_{2}.
Both rods are free to pivot in the x-y plane.
Find the Lagrangian for the system.

We can express the positions of the two masses in terms of just two coordinates, φ_{1} and φ_{2}, the angles of the rods l_{1} and l_{2} with respect to the vertical.
Measuring y vertically downward from the point of attachement, the position of m_{1} is (x_{1},y_{1}) = l_{1}(sinφ_{1}, cosφ_{1}) and the position of m_{2} is (x_{2},y_{2}) = l_{1}(sinφ_{1}, cosφ_{1}) + l_{2}(sinφ_{2}, cosφ_{2}).
The kinetic energy of m_{1} is T_{1} = ½m_{1}v² = ½ m_{1}(x_{1}dot² + y_{1}dot²) = ½m_{1}l_{1}²φ_{1}dot².
The kinetic energy of m_{2} is T_{2} = ½m_{2}(x_{2}dot² + y_{2}dot²) = ½m_{2}l_{1}²φ_{1}dot² + ½m_{2}l_{2}²φ_{2}dot².

The potential energy of m_{1} is U_{1} = -m_{1}gy_{1} = -m_{1}gl_{1}cosφ_{1}, and the potential energy of m_{2} is U_{2} = -m_{2}gy_{2} = -m_{2}g(l_{1}cosφ_{1} + l_{2}cosφ_{2}).

The Lagrangian is

L = T_{1} + T_{2} - U_{1} - U_{2} = ½m_{1}l_{1}²φ_{1}dot² + ½m_{2}l_{1}²φ_{1}dot² + ½m_{2}l_{2}²φ_{2}dot² + m_{1}gl_{1}cosφ_{1} + m_{2}g(l_{1}cosφ_{1} + l_{2}cosφ_{2})

or, after rearranging terms

L = ½(m_{1} + m_{2})l_{1}²φ_{1}dot² + (m_{1} + m_{2})gl_{1}cosφ_{1} + ½m_{2}l_{2}²φ_{2}dot² + m_{2}gl_{2}cosφ_{2}