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15. Lagrangian dynamics
LAST TIMES


PRINCIPLE OF LEAST ACTION
(or Hamilton's principle)


We now reformulate mechanics by requiring that every mechanical system is characterized by a Lagrangian which is a function of the coordinates and velocities of the particles constituting the system. The actual trajectories are then those that minimize the action

\begin{displaymath}J=\int_{t_1}^{t_2}Ldt\end{displaymath}

subject to boundary conditions at the endpoints. The integration is with respect to time.

Logically we consider this to be a postulate replacing Newton's laws as the foundation of mechanics. It 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 e equation is linear in the Lagrangian. We can multiply 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

L=L(x,v,t)

the equation of motion is

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

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

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

If we put 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, any coordinate that suffices to describe all possible configurations of the system will do.

Example: THE PENDULUM
\begin{figure}
\epsfysize=160pt
\epsffile{npend.eps}
\end{figure}
Instead of using the Cartesian coordinates of the mass it is convenient describe the motion by the angle $\theta$. We used to call the length of the pendulum L. To avoid confusion with the Lagrangian we rename it r.

The kinetic and potential energies are

\begin{displaymath}T=\frac{mr^2\dot{\theta}^2}{2}; V(\theta)=-mgr\cos\theta\end{displaymath}

With L=T-V the equation of motion becomes

\begin{displaymath}\frac{\partial L}{\partial\theta}-\frac{d}{dt}(\frac{\partial...
...tial\dot{\theta}})=
-mgr\sin\theta-mr^2\frac{d\dot\theta}{dt}=0\end{displaymath}

giving the expected result

\begin{displaymath}\ddot{\theta}+\frac{g}{r}\sin\theta\end{displaymath}



The constraint on the motion

r2=x2+y2=const.

depends on the coordinates only (not on the velocities). Such constraint are called holonomic. In both Lagrangian and Newtonian mechanics such constraints can be handled by simple substitution.

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 J=\delta\int_{t_1}^{t_2}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 L}{\partial q_i}-\frac{d}{dt}(\frac{\partial 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

L=T-V


\begin{displaymath}=\frac{M}{2}\dot{X}^2+\frac{m}{2}[\dot{X}^2+r^2\dot{\theta}^2
+2\dot{X}\dot{\theta}r\cos\theta]+mgr\cos\theta\end{displaymath}

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

and V(x) is velocity independent, the equation of motion can be written

\begin{displaymath}\frac{\partial L}{\partial x}=f=\frac{d}{dt}\frac{\partial 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:
$p_i=\frac{\partial L}{\partial\dot{q}_i}=$ generalized momentum
$f_i=\frac{\partial L}{\partial q}=$ 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}L=\frac{mr^2\dot{\theta}^2}{2}+mgr\cos\theta\end{displaymath}

The generalized force is

\begin{displaymath}f_\theta=\frac{\partial 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 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

L=T-V


\begin{displaymath}=\frac{M}{2}\dot{X}^2+\frac{m}{2}[\dot{X}^2+r^2\dot{\theta}^2
+2\dot{X}\dot{\theta}r\cos\theta]+mgr\cos\theta\end{displaymath}

This Lagrangian doesn't depend explicitly on X hence

\begin{displaymath}p_X=\frac{\partial 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.
The systematic approach makes it much easier to avoid errors!.

SUMMARY
We have

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Birger Bergersen
1999-02-21