# PHY5210 W08

## Chapter 7: Lagrange's Equations

In Taylor, read sections 7.7 to 7.8 for today, and 10.1 to 10.2 for Friday.

### Recall

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

### Examples of Lagrange's Equations

#### A Bead on a Spinning Wire Hoop

A bead of mass m is constrained to move on a wire formed into a hoop of radius R. The hoop sits in a vertical plane and is rotated about the vertical diameter at a constant angular velocity ω. Specify the bead's position by the angle θ measured from the downward vertical direction. Write down the Lagrangian, determine the equation of motion, and find any equilibrium positions that may exist. Determine the stability of any equilibrium positions.

We can use spherical coordinates, with the origin at the center of the hoop, to describe the bead's position, (r, θ, φ) = (R, θ, ωt). In spherical coordinates, ds² = dr² + r²dθ² + r²sin²θdφ², and v² = ds²/dt² = rdot² + r²θdot² + r²sin²θφdot². For the constrained coordinates of the bead the kinetic energy is T = ½m(R²θdot² + R²sin²θω²). The potential energy is all gravitational, U = -mgR&cosθ. The Lagrangian is

L = T - U = ½m(R²θdot² + R²sin²θω²) + mgR&cosθ

The generalized coordinate is θ. The Lagrange equations for θ yield

mR²sinθcosθω² - mgRsinθ = (d/dt)(mR²θdot) = mR²θddot.

An equilibrium position is a value of θ where both θdot and θddot are zero. Set θddot = 0 and solve for any values of θ that satisfy the equation.

(cosθω² - g/R)sinθ = θddot = 0.

This is satisfied for θ = 0, π, or ±arccos(g/Rω²), where the last one(s) exists only if |g/Rω²| ≤ 1. To evaluate the stability of these equilibria, we need to consider what will happen if the bead is moved slightly from the equilibrium location, say by an amount Δθ, with a very small velocity Δθdot. If the equilibrium is stable, the bead will try to return to its starting location, meaning it will accelerate opposite to the motion such that θddot/Δθ < 0.

Near the θ = 0 equilibrium, a small displacement produces an acceleration of

θddot ≈ (ω² - g/R)Δθ

So θddot/Δθ ≈ (ω² - g/R) is less then zero when g/Rω² > 1. The equilibrium at θ = 0 is stable if g/Rω² > 1, and unstable otherwise.

Near the θ = π equilibrium, a small displacment produces an acceleration of

θddot ≈ -(-ω² - g/R)Δθ

So θddot/Δθ ≈ (ω² + g/R) which is always positive. The equilibrium at θ = π is unstable.

For convenience, call the other equilibrium angles θ0 = arccos(g/Rω²). We will need to use the relations cos(θ0 + Δθ) = cosθ0cosΔθ - sinθ0sinΔθ ≈ (g/Rω²) - sqrt(1 - (g/Rω²)²)Δθ and sin(θ0 + Δθ) = cosθ0sinΔθ + sinθ0cosΔθ ≈ (g/Rω²)Δθ + sqrt(1 - (g/Rω²)²). Near the θ = θ0 equilibrium, a small displacement produces an acceleration of

θddot ≈ -(1 - (g/Rω²)²)ω²Δθ.

This gives θddot/Δθ ≈ ((g/Rω²)² - 1)ω² < 0 when g/Rω² < 1 as required for this to be an equilibrium point. Therefore, when this equilibrium point exists, it is stable.

#### Oscillations of the Bead near Equilibrium

What is the frequency of small oscillations of the bead near the points of stable equilibrium?

We have almost all the information needed now to answer this question. If we expand the expression for θddot to first order in Δtheta;, as done above, then the frequency for small oscillations, Ω, appears as the constant in the relation θddot = -Ω²Δθ.

For the stable equilibrium near θ = 0 the expression is

θddot ≈ -(g/R - ω²)Δθ

Therefore the frequency is Ω = sqrt(g/R - ω²).

For the stable equilibria near θ = ±arccos(g/Rω²)

θddot ≈ -(1 - (g/Rω²)²)ω²Δθ.

Therefore the frequency is Ω = ωsqrt(1 - (g/Rω²)²) = sqrt(ω² - (g/Rω)²).

### Generalized Momenta and Ignorable Coordinates

Recall that we introduced the terminology that ∂L/∂qi = Fi is a generalized force and ∂L/∂qidot = pi is a generalized momentum. And with this notation Lagrange's equations read

Fi = dpi/dt.

It is sometimes the case that the Lagrangian for a system won't depend a coordinate qk. In this situation, the corresponding generalized force will be zero, Fk = 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).

Recall that we introduced the terminology that ∂L/∂qi = Fi is a generalized force and ∂L/∂qidot = pi is a generalized momentum. And with this notation Lagrange's equations read

Fi = dpi/dt.

It is sometimes the case that the Lagrangian for a system won't depend a coordinate qk. In this situation, the corresponding generalized force will be zero, Fk = 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).

### Conclusion

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

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(r1, ..., rn) is the Lagrangian in the original reference frame, then in the translated frame the Lagrangian is L'(r1 + εxhat, ..., rn + εxhat). 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/∂x1) + ... + ε(∂L/∂xn) = ε ∑ ∂L/∂xi = 0

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

L/∂xi = (d/dt)(∂L/∂xidot) = (d/dt)pix

Therefore, the above relation becomes

∑ ∂L/∂xi = ∑ (d/dt)pix = (d/dt) ∑pix = (d/dt)Px = 0

So we see that conservation of momentum is related to the uniformity of space.

#### Conservation of Energy

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 qi, their time derivatives, qidot, and possibly t,

L = L(q1, ..., qn, q1dot, ..., qndot, t)

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

dL/dt = (∂L/∂q1)q1dot + ... + (∂L/∂qn)qndot + (∂L/∂q1dot)q1ddot + ... + (∂L/∂qndot)qnddot + ∂L/∂t

This looks a little nicer when written with the sum notation

dL/dt = ∑[(∂L/∂qi)qidot + (∂L/∂qidot)qiddot] + ∂L/∂t

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

L/∂qi = (d/dt) (∂L/∂qidot) = (d/dt)pi = pidot.

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

dL/dt = ∑[pidotqidot + piqiddot] + ∂L/∂t = (d/dt)[∑(piqidot)] + ∂L/∂t.

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

(d/dt)[∑(piqidot) - L] = 0.

or that the quantity ∑(piqidot) - L is a constant. This quantity is special, and is called the Hamiltonian

H = ∑(piqidot) - L.

The Hamiltonian is important for a number of reasons: there's a formulation of the laws of mechanics based on the Hamiltonian, similar to the Lagrangian formulation we are using now; the Hamiltonian plays an important role in quantum mechanics; and, under certain restrictions, the Hamiltonian is the total energy of the system. We shall prove this last point here.

The condition that is needed for the Hamiltonian to be the total energy of the system is that the relation between the generalized coordinates and Cartesian coordinates be time independent,

rα = rα ( q1, ..., qn)

Generalized coordinates that satisfy this condition are called natural.

To show that the Hamiltonian is the total energy for natural coordinates, begin by writing the kinetic energy in terms of the generalized coordinates, starting with the velocities

dtrα = ∑j (∂rα/∂qj) dtqj

and inserting this into the expression for kinetic energy

T = ½ ∑α mα (dtrα)² = ½ ∑α,j,k mα (∂rα/∂qj) (∂rα/∂qk) dtqj dtqk

Since the cartesian coordinates are assumed to be time independent, their partial derivatives are time independent. These can be combined with the mass and summed over \$alpha; to form a quantity we'll call Ajk,

Ajk = ∑α mα (∂rα/∂qj) (∂rα/∂qk)

With this definition, the kinetic energy becomes

T = ½ ∑j,k Ajk dtqj dtqk

Now, to prove that the Hamiltonian is the total energy, we need to evaluate the quantity ∑(piqidot). Begin by evaluating the generalized momentum

pi = ∂L/∂qidot = ∂T/∂qidot = ∑j Aij dtqj

Using this we have

i(piqidot) = ∑i,j Aij dtqi dtqj = 2T.

This gives us the result we want, since inserting this in the definition of the Hamiltonian yields

H = ∑(piqidot) - L = 2T - (T - U) = T + U = E.

Recall that this entire argument began by consider how the Lagrangian would change by a translation in time. We've shown that if the Lagrangian is unaffected by a translation in time then energy is conserved. This complements the earlier proof that if the Lagrangian is unaffected by a translation in space then momentum is conserved. Likewise, one can show that if the Lagrangian is unchanged by a rotation along a particular axis, then angular momentum along that axis is conserved. Thus we arrive at an interesting connection between the fundamental conservation laws of energy, momentum, and angular momentum, and the uniformity of the laws of physics in time, space, and angle of rotation.

### Lagrange's Equations for Magnetic Forces*

Interesting, but does involve the use of the vector potential, something beyond the present level. I recommend that students refer back to this when doing E&M at the appropriate level. This is an example of how topics in physics tie together; our divisions into standalone subjects is somewhat arbitrary.