Advanced Classical Mechanics/Constraints and Lagrange's Equations

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

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. These restrictions are either kinematical or geometrical in nature.

$f({\overrightarrow{r}}_1,{\overrightarrow{r}}_2,{\overrightarrow{r}}_3,\cdots ,t)=0$

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

${({\overrightarrow{r}}_i-{\overrightarrow{r}}_j)}^2-c_{ij}^2=0$

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

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

$f({\overrightarrow{r}}_1,{\overrightarrow{r}}_2,{\overrightarrow{r}}_3,\cdots ,{\overrightarrow{v}}_1,{\overrightarrow{v}}_2,{\overrightarrow{v}}_3,\cdots ,t)=0.$

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

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

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 ($\theta ,\varphi$).

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 $\delta {\overrightarrow{r}}_i$. The virtual displacement must be consistent with the constraints.

For each particle we have

${\overrightarrow{F}}_i={\dot{\overrightarrow{p}}}_i\mathrm{s}\mathrm{o}\overrightarrow{\mathrm{F}}\mathrm{-}{\dot{\overrightarrow{\mathrm{p}}}}_{\mathrm{i}}\mathrm{=}\mathrm{0}$

and summing over the particles we have

$\sum _i({\overrightarrow{F}}_i-{\dot{\overrightarrow{p}}}_i)\cdot \delta {\overrightarrow{r}}_i=0.$

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

${\overrightarrow{F}}_i={\overrightarrow{F}}_i^{(a)}+{\overrightarrow{f}}_i$

so we have

$\sum _i({\overrightarrow{F}}_i-{\dot{\overrightarrow{p}}}_i)\cdot \delta {\overrightarrow{r}}_i=\sum _i({\overrightarrow{F}}_i^{(a)}-{\dot{\overrightarrow{p}}}_i)\cdot \delta {\overrightarrow{r}}_i+\sum _i{\overrightarrow{f}}_i\cdot \delta {\overrightarrow{r}}_i=0.$

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

$\sum _i({\overrightarrow{F}}_i^{(a)}-{\dot{\overrightarrow{p}}}_i)\cdot \delta {\overrightarrow{r}}_i=0,$

D'Alembert's principle. The forces of constraints 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.

Generalized Coordinates

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

${\overrightarrow{r}}_i={\overrightarrow{r}}_i(q_1,q_2,q_3,\cdots ,t)$

and so

${\overrightarrow{v}}_i=\frac{d{\overrightarrow{r}}_i}{dt}=\sum _k\frac{\partial {\overrightarrow{r}}_i}{\partial q_k}{\dot{q}}_k+\frac{\partial {\overrightarrow{r}}_i}{\partial t}$

and a virtual displacement is

$\delta {\overrightarrow{r}}_i=\sum _j\frac{\partial {\overrightarrow{r}}_i}{\partial q_j}\delta q_j.$

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

$\sum _i{\overrightarrow{F}}_i\cdot \delta {\overrightarrow{r}}_i=\sum _i{\overrightarrow{F}}_i\cdot \left(\sum _j\frac{\partial {\overrightarrow{r}}_i}{\partial q_j}\delta q_j\right)=\sum _j(\sum _i{\overrightarrow{F}}_i\cdot \frac{\partial {\overrightarrow{r}}_i}{\partial q_j})\delta q_j=\sum _j{Q}_j\delta q_j$

where $Q_j$ is called a generalized force.

The second term takes a bit more work. We have

$\sum _i{\dot{\overrightarrow{p}}}_i\cdot \delta {\overrightarrow{r}}_i=\sum _i{m}_i{\ddot{\overrightarrow{r}}}_i\cdot \delta {\overrightarrow{r}}_i=\sum _i{m}_i{\ddot{\overrightarrow{r}}}_i\left(\sum _j\frac{\partial {\overrightarrow{r}}_i}{\partial q_j}\delta q_j\right).$

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

$\sum _i{m}_i{\ddot{\overrightarrow{r}}}_i\cdot \frac{\partial {\overrightarrow{r}}_i}{\partial q_j}=\sum _i[\frac{d}{dt}(m_i{\dot{\overrightarrow{r}}}_i\cdot \frac{\partial {\overrightarrow{r}}_i}{\partial q_j})-m_i{\dot{\overrightarrow{r}}}_i\cdot \frac{d}{dt}\left(\frac{\partial {\overrightarrow{r}}_i}{\partial q_j}\right)]$

                              $=\sum _i[\frac{d}{dt}(m_i{\overrightarrow{v}}_i\cdot \frac{\partial {\overrightarrow{r}}_i}{\partial q_j})-m_i{\dot{\overrightarrow{r}}}_i\cdot \frac{\partial {\overrightarrow{v}}_i}{\partial q_j}]$

We can use the fact that

$\frac{\partial {\overrightarrow{v}}_i}{\partial {\dot{q}}_j}=\frac{\partial {\overrightarrow{r}}_i}{\partial q_k}$

from the defintion of ${\overrightarrow{v}}_i$ to get

                              $=\sum _i[\frac{d}{dt}(m_i{\overrightarrow{v}}_i\cdot \frac{\partial {\overrightarrow{v}}_i}{\partial {\dot{q}}_j})-m_i{\dot{\overrightarrow{r}}}_i\cdot \frac{\partial {\overrightarrow{v}}_i}{\partial q_j}]$

                              $=\frac{d}{dt}\left[\frac{\partial }{\partial {\dot{q}}_j}\sum _i\left(\frac{1}{2}{m}_i{v}_i^2\right)\right]-\frac{\partial }{\partial q_j}\left[\sum _i\left(\frac{1}{2}{m}_i{v}_i^2\right)\right].$

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

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

Lagrange's Equations

If the constraints are holonomic, we can pick the $q_j$ to be independent so the various $\delta q_j$ are completely arbitrary and the quantity in braces must vanish to yield

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

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

${\overrightarrow{F}}_i=-{\overrightarrow{\mathrm{\nabla }}}_{i}V$ so

$Q_j=\sum _i{\overrightarrow{F}}_i\cdot \frac{\partial {\overrightarrow{r}}_i}{\partial q_j}=-\sum _i{\overrightarrow{\mathrm{\nabla }}}_{i}V\cdot \frac{\partial {\overrightarrow{r}}_i}{\partial q_j}=-\frac{\partial V}{\partial q_j}$

In this case we can rearrange the equation to give

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

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

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

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 $q_i$. 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 $q_i$, 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,

$\frac{dL}{dt}=\sum _i[\frac{\partial L}{\partial q_i}{\dot{q}}_i+\frac{\partial L}{\partial {\dot{q}}_i}{\ddot{q}}_i]+\frac{\partial L}{\partial t}.$

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

$\frac{dL}{dt}=\sum _i[\frac{d{p}_i}{dt}{\dot{q}}_i+p_i{\ddot{q}}_i]+\frac{\partial L}{\partial t}=\sum _i\frac{d(p_i{\dot{q}}_i)}{dt}+\frac{\partial L}{\partial t}$

and rearranging

$\frac{\partial L}{\partial t}=\frac{d}{dt}(L-\sum _i{p}_i{\dot{q}}_i).$

So if

$\partial L/\partial t=0$

then the Hamiltonian,

$H=L-\sum _i{p}_i{\dot{q}}_i,$

is conserved.

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.

$\overrightarrow{F}=q(\overrightarrow{E}+\overrightarrow{v}\times \overrightarrow{B}).$

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

$\overrightarrow{E}=-\overrightarrow{\mathrm{\nabla }}\varphi -\frac{\partial \overrightarrow{A}}{\partial t}$

where $\overrightarrow{A}$ is the vector potential. The magnetic field may be written as

$\overrightarrow{B}=\overrightarrow{\mathrm{\nabla }}\times \overrightarrow{A}.$

Let's consider the function

$V=q\varphi (\overrightarrow{r},t)-q\dot{\overrightarrow{r}}\cdot \overrightarrow{A}(\overrightarrow{r},t)$

and calculate

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

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

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

Notice that the terms proportional to $\dot{x}$ cancel leaving

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

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 $\overrightarrow{v}\times \overrightarrow{B}$, so the following Lagrangian will yield the equations of motion

$L=\frac{1}{2}m{\dot{r}}^2+q\overrightarrow{\dot{r}}\cdot \overrightarrow{A}(\overrightarrow{r},t)-q\varphi (\overrightarrow{r},t).$

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

$H=\frac{\partial L}{\partial \dot{\overrightarrow{r}}}\cdot \dot{\overrightarrow{r}}-L=\frac{1}{2}m{\dot{r}}^2+q\varphi (\overrightarrow{r},t)$

is conserved. The force is not conservative but the system is. Template:Subpage navbar