In Taylor, read sections 7.5 to 7.6 for today, and review on Monday.

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

∂L/∂q_{i} - (d/dt)(∂L/∂q_{i}dot) = 0

L = T - U

Recall that we introduced the terminology that ∂L/∂q_{i} = F_{i} is a generalized force and ∂L/∂q_{i}dot = p_{i} is a generalized momentum.
And with this notation Lagrange's equations read

F_{i} = dp_{i}/dt.

It is sometimes the case that the Lagrangian for a system won't depend a coordinate q_{k}.
In this situation, the corresponding generalized force will be zero, F_{k} = 0 and the generalized momentum will be conserved.
Such a coordinate is said to by cyclic or ignorable since the generalized momenta are just constants of the motion.

This situation is a simple of example of a broader theorem about conserved quantities in a system called Noether's Theorem (named for Emmy Noether, a late 19th / early 20th century female mathematician).

As we've seen in some of the example problems, conservation of momentum and energy is built into the Lagrangian approach. It is insightful to see this explicitly.

Conservation of momentum is related to the uniformity of space.
We expect the laws of physics to hold equally well in any inertial frame.
So if we translate our reference frame by a small amount, say in the x direction, then we expect the motion of a system to be unchanged (except for the small change in x coordinate).
Explicitly, if L(**r**_{1}, ..., **r**_{n}) is the Lagrangian in the original reference frame, then in the translated frame the Lagrangian is L'(**r**_{1} + ε**x**hat, ..., **r**_{n} + ε**x**hat).
Note that the first derivatives of the coordinates are unchanged.
In order that the motion is unchanged, we require that δL = 0.
Using the notation for differentials shown in a previous lecture, we can express δL in terms of derivatives of coordinates and their derivatives.
For simplicity, let's assume that the generalized coordinates are cartesian coordinates, xi, yi, zi.
The differential becomes

δL = ε(∂L/∂x_{1}) + ... + ε(∂L/∂x_{n}) = ε ∑ ∂L/∂x_{i} = 0

Using Lagrange's equations the partial derivatives can be equated to the time derivative of the x component of the momentum:

∂L/∂x_{i} = (d/dt)(∂L/∂x_{i}dot) = (d/dt)p_{ix}

Therefore, the above relation becomes

∑ ∂L/∂x_{i} = ∑ (d/dt)p_{ix} = (d/dt) ∑p_{ix} = (d/dt)P_{x} = 0

Conservation of energy rests on the Lagrangian being independent of time; it can depend on qi and qidot, but must lack explicit t dependence.
To see this, we need to evaluate the total time derivative of the Lagrangian.
Recall that the Lagrangian depends on generalized coordinates q_{i}, their time derivatives, qidot, and possibly t,

L = L(q_{1}, ..., q_{n}, q_{1}dot, ..., q_{n}dot, t)

We need to use the chain rule to evaluate the total derivative:

dL/dt = (∂L/∂q_{1})q_{1}dot + ... + (∂L/∂q_{n})q_{n}dot + (∂L/∂q_{1}dot)q_{1}ddot + ... + (∂L/∂q_{n}dot)q_{n}ddot + ∂L/∂t

This looks a little nicer when written with the sum notation

dL/dt = ∑[(∂L/∂q_{i})q_{i}dot + (∂L/∂q_{i}dot)q_{i}ddot] + ∂L/∂t

Using Lagrange's equation and our definition of generalized momentum, we can replace the first term in the sum by

∂L/∂q_{i} = (d/dt)
(∂L/∂q_{i}dot) = (d/dt)p_{i} = p_{i}dot.

the derivative of the generalized momentum. The derivative of the Lagrangian in the second term is just the generalized momentum, so

dL/dt = ∑[p_{i}dotq_{i}dot + p_{i}q_{i}ddot] + ∂L/∂t = (d/dt)[∑(p_{i}q_{i}dot)] + ∂L/∂t.

When the Lagrangian isn't explicitly dependent on time, then the final partial derivative is zero and we can write

(d/dt)[∑(p_{i}q_{i}dot) - L] = 0.

or that the quantity ∑(p_{i}q_{i}dot) - L is a constant.
This quantity is special, and is called the Hamiltonian

H = ∑(p_{i}q_{i}dot) - L.

© 2007 Robert Harr