next up previous
Next: About this document ...

11. Hamilton's principle
LAST TIMES


PRINCIPLE OF LEAST ACTION
(or Hamilton's principle)


The Euler-Lagrange equation for the variational principle and the Lagrangian equation of motion derived in lecture 8 for a conservative system are of the same form. To see this let the independent variable x represent time and the dependent variable z be a generalized coordinate. This suggests that Newtonian mechanics can be derived from a variational principle. We now reformulate the mechanics of conservative systems by requiring that they are characterized by a Lagrangian which is a function of the generalized coordinates and velocities of the particles constituting the system. The actual trajectory is then the one for which the action

\begin{displaymath}{\cal S}=\int_{t_1}^{t_2}{\cal L}dt\end{displaymath}

is extremal subject to boundary conditions at the endpoints. The integration is with respect to time. Usually the action will be a minimum hence the name principle of least action.

Logically we consider this to be a postulate taking the place of Newton's laws as the foundation of mechanics. It then remains to construct the Lagrangian -something which, of course, depends on the problem at hand. Also, we are not saying that Newton was wrong- we want the Lagrangian formulation to reproduce Newton's law when applicable.

Caveat: I mention in passing that the actual path is not always a minimum for the entire path, but only for sufficiently short segments. This is of no problem in practice. In deriving the equation of motion we only use the extremum condition.

The Euler-Lagrange equation is linear in the Lagrangian. We can multiply ${\cal L}$ by a constant without changing the equation of motion. We choose the Lagrangian to have dimension of energy.

Example:
Particle in one dimension subject to velocity independent force:
We put v=dx/dt. If the Lagrangian is

\begin{displaymath}{\cal L=L}(x,v,t)\end{displaymath}

the equation of motion is

\begin{displaymath}\frac{\partial{\cal L}}{\partial x}-\frac{d}{dt}(\frac{\partial{\cal L}}{\partial v})=0\end{displaymath}

We write ${\cal T}=\frac{1}{2}mv^2$ for the kinetic energy

\begin{displaymath}\frac{\partial{\cal T}}{\partial x}=0;\;\frac{d}{dt}(\frac{\partial{\cal T}}{\partial v})=m\frac{dv}{dt}=ma\end{displaymath}

where a is the acceleration. Similarly if V(x) is the potential energy

\begin{displaymath}\frac{\partial{\cal V}}{\partial x}=-f;\;\frac{\partial{\cal T}}{\partial v}=0\end{displaymath}

If we put ${\cal L=T-V}$ we see that equation of motion becomes the familiar

f=ma



GENERALIZED COORDINATES
In our previous example x was the Cartesian coordinate of the particle. It need not be, as discussed lecture 7 we may use any set of generalized coordinates which involves imposing holonomic constraints on Cartesian coordinates.

MANY DEGREES OF FREEDOM
Most often we are dealing with systems requiring a number of generalized coordinates to describe the motion.

Suppose N coordinates are required to specify the motion (after we have substituted for the holonomic constraints). We say that the system has N degrees of freedom.
The variational principle is now

\begin{displaymath}\delta {\cal S}=\delta\int_{t_1}^{t_2}{\cal L}(q_1,q_2\cdots q_N,\dot{q}_1,\dot{q}_2,\cdots\dot{q}_N)dt=0\end{displaymath}

We can carry out the variation independently for each of the coordinates and obtain a set of N Euler-Lagrange equations

\begin{displaymath}\frac{\partial{\cal L}}{\partial q_i}-\frac{d}{dt}(\frac{\partial{\cal L}}{\partial\dot{q_i}})
=0\end{displaymath}

i.e. one equation for each coordinate.

Example
PENDULUM WITH MOVABLE SUPPORT

\begin{figure}
\epsfysize=210pt
\epsffile{mpend.eps}
\end{figure}
A pendulum of length r, mass m. Its support has mass M and it can slide without friction horizontally (coordinate X). The horizontal and vertical components of the pendulum mass are

\begin{displaymath}x=X+r\sin\theta;\;y=-r\cos\theta\end{displaymath}

The velocity components are

\begin{displaymath}\dot{x}=\dot{X}+r\cos\theta\;\dot{\theta};\;\dot{y}=r\sin\theta\;\dot{\theta}\end{displaymath}

Hence the Lagrangian is

\begin{displaymath}{\cal L=T-V}\end{displaymath}\begin{displaymath}=\frac{M}{2}\...
...\dot{\theta}^2
+2\dot{X}\dot{\theta}r\cos\theta]+mgr\cos\theta\end{displaymath} (1)

We will come back to the equations of motion for this system later.

GENERALIZED FORCES AND MOMENTA.
When the kinetic energy is on the form

\begin{displaymath}{\cal T}=\frac{1}{2}m\dot{x}^2\end{displaymath}

and ${\cal V}(x)$ is velocity independent, the equation of motion can be written

\begin{displaymath}\frac{\partial{\cal L}}{\partial x}=f=\frac{d}{dt}\frac{\partial{\cal L}}{\partial\dot{x}}
=m\frac{dv}{dt}=\dot{p}\end{displaymath}

where p is the momentum and f the force

In the general case:

$p_i=\frac{\partial{\cal L}}{\partial\dot{q}_i}=$ generalized momentum
In lecture 7 we defined the generalized conservative force as the partial derivative of the potential energy with respect to the generalized coordinate. We now modify the definition so that


$f_i=\frac{\partial{\cal L}}{\partial q}=$ generalized force


The Lagrangian equations of motion can thus be written:

\begin{displaymath}f_i=\frac{dp_i}{dt}\end{displaymath}



If the Lagrangian does not depend explicitly on one of the coordinates the corresponding generalized force is zero and the corresponding generalized momentum is conserved!

Example THE PENDULUM

\begin{displaymath}{\cal L}=\frac{mr^2\dot{\theta}^2}{2}+mgr\cos\theta\end{displaymath}

The generalized force is

\begin{displaymath}f_\theta=\frac{\partial{\cal L}}{\partial\theta}=-mgr\sin\theta\end{displaymath}

Physically the generalized force associated with the angle $\theta $ is the torque!
The generalized momentum is

\begin{displaymath}p_\theta=\frac{\partial{\cal L}}{\partial\dot\theta}=mr^2\dot{\theta}\end{displaymath}

which we recognize as the angular momentum. The Lagrangian equation of motion is thus just

Rate of change of angular momentum=torque

Example
PENDULUM WITH MOVABLE SUPPORT
The Lagrangian (1) doesn't depend explicitly on X hence

\begin{displaymath}p_X=\frac{\partial{\cal L}}{\partial\dot{X}}=(M+m)\dot{X}+m\dot{\theta}r\cos\theta\end{displaymath}

is conserved. A bit of reflection will convince you that this is just the equation for the conservation of linear momentum in the x-direction.

So there is nothing new!
We could have obtained the above results without resorting to Lagrangians.
However, if the system is complicated the Lagrangian approach offers the possibility of proceeding in a systematic fashion, without having to worry about free body diagrams, normal forces, or pseudo forces due to acceleration of coordinate system.
The systematic, algebraic, approach makes it much easier to avoid errors!

SUMMARY
We have

Example problems: Problems 1 and 3 of 2000 midterm , with solution.

Return to title page.

 
next up previous
Next: About this document ...
Birger Bergersen
2001-01-31