Advanced Classical Mechanics/Constraints and Lagrange's Equations

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There is more to classical mechanics than F = ma. Often the motion of a system is constrained in some way.

Contents

[edit] What are constraints?

  • The particles could be restricted to travel along a curve or surface. Specifically one could have some function of the coordinates of each particle and time vanish.

f\left (\vec r_1,\vec r_2, \vec r_3, \cdots , t \right ) = 0

This is called a holonomic constraint. For example we could have

\left ( \vec r_i - \vec r_j \right )^2 - c_{ij}^2 = 0

which expresses that the distances between two particles that make up a rigid body are fixed.

  • There are non-holonomic constraints. For example, one could have

r^2 - a^2 \geq 0 for a particle travelling outside the surface of a sphere or constraints that depend on velocities as well,

f\left (\vec r_1,\vec r_2, \vec r_3, \cdots , \vec v_1,\vec v_2, \vec v_3, \cdots , t \right ) = 0 .

A familar example of the latter is a ball rolling on a surface.

We will be dealing exclusively with holonomic constraints in this course.

[edit] What are the consequences?

  • The coordinates are no longer independent. They are related through the equations of constraint.
  • The force of constraint are not given so they must be determined from the solution (if you actually want them at all).
  • If the constraints are holonomic, the equations of constraint can be used to eliminate some of the coordinates to get a set of generalized independent coordinates.

These generalized coordinates usually will not fall into pairs or triples that transform as vectors, e.g. the natural coordinates for motion restricted to a sphere are spherical coordinates (θ,φ).

[edit] D'Alembert's Principle

D'Alembert's principle relies on the concept of virtual displacements. The idea is that you can imagine freezing the system in time and jiggling each of the particles in a way consistent with the various constraints at that particular time and determine the work (virtual work) needed to perform these virtual displacements.

We will denote the virtual displacement of a particle as \delta \vec r_i. The virtual displacement must be consistent with the constraints.

For each particle we have

\vec F_i = \dot \vec p_i ~\rm{so}~ \vec F - \dot \vec p_i= 0

and summing over the particles we have

\sum_i \left ( \vec F_i - \dot \vec p_i \right ) \cdot \delta \vec r_i = 0.

Let's divide the force on each particle into applied forces and constraints

\vec F_i = \vec F_i^{(a)} + \vec f_i

so we have

\sum_i \left ( \vec F_i - \dot \vec p_i \right ) \cdot \delta \vec r_i = \sum_i \left ( \vec F_i^{(a)} - \dot \vec p_i \right ) \cdot \delta \vec r_i + \sum_i \vec f_i \cdot \delta \vec r_i= 0.

If we assume that the forces of constraint do no virtual work, then the last term vanishes. This is a reasonable assumption because the force of constraint to restrict a particle to a surface is normal to that surface but a displacement consistent with the constraints is tangent to the surface so the dot product will vanish and the force of constraint performs no virtual work. If the constraints are a function of time, the forces of constraint can perform work on the particles but the whole idea of D'Alembert's principle is that we have frozen time.

This leaves us with

\sum_i \left ( \vec F_i^{(a)} - \dot \vec p_i \right ) \cdot \delta \vec r_i = 0,

D'Alembert's principle. The forces of constrain are gone! However, the coordinates are not independent, so the equation is only a statement about the sum of the various forces, momenta and virtual displacements.

[edit] Generalized Coordinates

We can try to find a set of independent coordinates given the constraints. Let's write

\vec r_i = \vec r_i \left ( q_1, q_2, q_3, \cdots, t \right )

and so

\vec v_i = \frac{d \vec r_i}{d t} = \sum_k \frac{\partial \vec r_i}{\partial q_k} \dot q_k + \frac{\partial \vec r_i}{\partial t}

and a virtual displacement is

\delta \vec r_i = \sum_j \frac{\partial \vec r_i}{\partial q_j} \delta q_j.

Notice that there is no dt in the virtual displacement because time is held fixed. Now let's look at the first term in D'Alembert's equation

\sum_i \vec F_i \cdot \delta \vec r_i = \sum_i \vec F_i \cdot \left ( \sum_j \frac{\partial \vec r_i}{\partial q_j} \delta q_j \right ) = \sum_j \left ( \sum_i \vec F_i \cdot \frac{\partial \vec r_i}{\partial q_j} \right ) \delta q_j = \sum_j Q_j \delta q_j

where Qj is called a generalized force.

The second term takes a bit more work. We have

\sum_i \dot \vec p_i \cdot \delta \vec r_i = \sum_i m_i \ddot \vec r_i \cdot \delta \vec r_i = \sum_i m_i \ddot \vec r_i \left ( \sum_j \frac{\partial \vec r_i}{\partial q_j} \delta q_j \right ).

Let's focus on a particular one of the generalized coordinates j. We have

\sum_i m_i \ddot \vec r_i \cdot \frac{\partial \vec r_i}{\partial q_j} = \sum_i \left [ \frac{d}{dt} \left ( m_i \dot \vec r_i \cdot \frac{\partial \vec r_i}{\partial q_j} \right ) - m_i \dot \vec r_i \cdot \frac{d}{dt} \left ( \frac{\partial \vec r_i}{\partial q_j} \right ) \right ]

                              = \sum_i \left [ \frac{d}{dt} \left ( m_i \vec v_i \cdot \frac{\partial \vec r_i}{\partial q_j} \right ) - m_i \dot \vec r_i \cdot \frac{\partial \vec v_i}{\partial q_j} \right ]

We can use the fact that

\frac{\partial \vec v_i}{\partial \dot q_j} = \frac{\partial \vec r_i}{\partial q_k}

from the defintion of \vec v_i to get

                              = \sum_i \left [ \frac{d}{dt} \left ( m_i \vec v_i \cdot \frac{\partial \vec v_i}{\partial \dot q_j} \right ) - m_i \dot \vec r_i \cdot \frac{\partial \vec v_i}{\partial q_j} \right ]

                              = \frac{d}{dt} \left [ \frac{\partial}{\partial \dot q_j} \sum_i \left ( \frac{1}{2} m_i v_i^2 \right ) \right ] - \frac{\partial}{\partial q_j} \left [ \sum_i \left ( \frac{1}{2} m_i v_i^2 \right ) \right ].

Notice that the quantity in the innermost parenthesis is just the total kinetic energy of the system T so we have

\sum_j \left \{ \left [ \frac{d}{dt} \left ( \frac{\partial T}{\partial \dot q_j} \right ) - \frac{\partial T}{\partial q_j} \right ] - Q_j \right \} \delta q_j = 0 .

[edit] Lagrange's Equations

If the constraints are holonomic, we can pick the qj to be independent so the various δqj are completely arbitrary and the quantity in braces must vanish to yield

\left [ \frac{d}{dt} \left ( \frac{\partial T}{\partial \dot q_j} \right ) - \frac{\partial T}{\partial q_j} \right ] = Q_j .

These expressions are sometimes called Lagrange's equations, but the term Lagrange's equations is often reserved for the case of a conservative force. In this case we have

\vec F_i = - \vec \nabla_i V so

Q_j = \sum_i \vec F_i \cdot \frac{\partial \vec r_i}{\partial q_j} = - \sum_i \vec \nabla_i V\cdot \frac{\partial \vec r_i}{\partial q_j} = - \frac{\partial V}{\partial q_j}

In this case we can rearrange the equation to give

\left [ \frac{d}{dt} \left ( \frac{\partial T}{\partial \dot q_j} \right ) - \frac{\partial (T-V)}{\partial q_j} \right ] = 0 .

Furthermore, if the forces do not depend explicitly on the velocities we can define, the Lagrangian to be L = T − V

and write Lagrange's equations in their traditional form

\left [ \frac{d}{dt} \left ( \frac{\partial L}{\partial \dot q_j} \right ) - \frac{\partial L}{\partial q_j} \right ] = 0 .

[edit] Conserved Quantities

For each coordinate we can define a conjugate momentum to be

p_i = \frac{\partial L}{\partial {\dot q}_i}.

This is called the generalized momentum conjugate to the coordinate qi. Let's look at Lagrange's equations with this definition

\frac{d p_i}{dt} = \frac{\partial L}{\partial q_i}

so if the Lagrangian does not depend on a particular coordinate qi, then the momentum conjugate to the coordinate does not change with time; it is conserved. This conserved momentum is called a first integral. The coordinate that doesn't affect the Lagrangian is called a cyclic coordinate.

In general the Lagrangian will depend on the coordinates, velocities and time; what happens if \partial L/\partial t vanishes? Is there a conserved quantity similar to the momenta?

Let calculate the total derivative of the Lagrangian with respect to time,

\frac{d L}{d t} = \sum_i \left [ \frac{\partial L}{\partial q_i} {\dot q}_i + \frac{\partial L}{\partial {\dot q}_i} {\ddot q}_i \right ] + \frac{\partial L}{\partial t} .

Now let's use the definition of the momenta and the Lagrange's equation to simplify things a bit,

\frac{d L}{d t} = \sum_i \left [ \frac{d p_i}{d t} {\dot q}_i + p_i {\ddot q}_i \right ] + \frac{\partial L}{\partial t} = \sum_i \frac{d ( p_i {\dot q}_i )}{d t} + \frac{\partial L}{\partial t}

and rearranging

\frac{\partial L}{\partial t} = \frac{d}{dt} \left ( L - \sum_i p_i {\dot q}_i \right ).

So if

\partial L/\partial t = 0

then the Hamiltonian,

H = L - \sum_i p_i {\dot q}_i,

is conserved.

[edit] Lagrangians with Non-Conservative Forces

From the analysis so far it would appear that one can only construct a Lagrangian when the forces that act on a particle are conservative (they can be derived from a potential V). It is indeed the case that truly dissipative forces such as friction cannot be directly included in a Lagrangian formulation, but forces that can be written in the form d/dt (\partial L/\partial {\dot q}) may be included in the Lagrangian. Although this seems very restrictive, an important force of this class is the magnetic force on a charged particle.

\vec F = q \left( \vec E + \vec v \times \vec B \right ).

In electrostatics the electric field is simply the gradient of the electrostatic potential, but for more general fields we have

\vec E = -\vec \nabla \phi - \frac{\partial \vec A}{\partial t}

where \vec A is the vector potential. The magnetic field may be written as

\vec B = \vec \nabla \times \vec A.

Let's consider the function

V = q \phi \left (\vec r, t\right )- q {\dot \vec r} \cdot \vec A \left (\vec r, t \right )

and calculate

\frac{d}{dt} \left ( \frac{\partial V}{\partial {\dot x}} \right ) - \frac{\partial V}{\partial x} = - q \frac{d A_x}{d t} - q \frac{\partial \phi}{\partial x} + q \left ( \dot x \frac{\partial A_x}{\partial x} + \dot y \frac{\partial A_y}{\partial x} + \dot z \frac{\partial A_z}{\partial x} \right ).

The total time derivative of the vector potential generates several terms (due to the chain rule) to yield

\frac{d}{dt} \left ( \frac{\partial V}{\partial {\dot x}} \right ) - \frac{\partial V}{\partial x} = - q \left ( \frac{\partial A_x}{\partial t} + \dot x \frac{\partial A_x}{\partial x} + \dot y \frac{\partial A_x}{\partial y} + \dot z \frac{\partial A_x}{\partial z} \right ) - q \frac{\partial \phi}{\partial x} + q \left ( \dot x \frac{\partial A_x}{\partial x} + \dot y \frac{\partial A_y}{\partial x} + \dot z \frac{\partial A_z}{\partial x} \right ).

Notice that the terms proportional to \dot x cancel leaving

\frac{d}{dt} \left ( \frac{\partial V}{\partial {\dot x}} \right ) - \frac{\partial V}{\partial x} = - q \left ( \frac{\partial \phi}{\partial x} + \frac{\partial A_x}{\partial t} \right ) + q \left [ \dot y \left ( \frac{\partial A_x}{\partial y} - \frac{\partial A_y}{\partial x} \right ) + \dot z \left ( \frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x} \right ) \right ].

The first term is simply the charge of the particle times the x-component of the electric field. The second term is the charge of the particles times the x-component of \vec v \times \vec B, so the following Lagrangian will yield the equations of motion

L = \frac{1}{2} m {\dot r}^2 + q \vec \dot r \cdot \vec A \left (\vec r, t\right ) - q \phi \left ( \vec r, t \right ) .

If the vector potential and the scalar potential do not depend on time, then \partial L/\partial t = 0 and

H = \frac{\partial L}{\partial \dot \vec r} \cdot \dot \vec r - L = \frac{1}{2} m {\dot r}^2 + q \phi \left ( \vec r, t \right )

is conserved. The force is not conservative but the system is.

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