Friday, 19 September 2008

Lagrangian Mechanics: From the Euler-Lagrange Equation to Newton's Laws

Last week I wrote about the Euler-Lagrange Equation, and how it can be obtained from the Principle of Least Action. Today, I'd like to show that this equation is consistent with Newtonian Mechanics.

For reference, the Euler-Lagrange equation for some arbitrary coordinate q is:

\frac{\partial L}{\partial q} - \frac{d}{dt}\frac{\partial L}{\partial \dot{q}} = 0


L, the Lagrangian, is the kinetic energy of a system minus the potential.

Let us now consider the case of a free particle of mass m, moving in a potential V(x). The particle's instantaneous speed is given by
v = \frac{dx}{dt}=\dot{x}
. The kinetic energy is the familiar
T=\frac{1}{2}m\dot{x}^2
.

Applying the Euler-Lagrange Equation, we have:

\begin{align*}
\frac{\partial L}{\partial x} = -\frac{\partial V(x)}{\partial x} \\
\mathrm{and}~~\frac{\partial L}{\partial \dot{x}} = m\dot{x} \\
\Rightarrow \frac{d}{dt}\frac{\partial L}{\partial \dot{x}} = m\ddot{x}
\end{align*}


Putting these together, we have:

\begin{align*}
-\frac{\partial V(x)}{\partial x} - m\ddot{x} = 0 \\
\Rightarrow m\ddot{x} = -\frac{\partial V}{\partial x}
\end{align*}


The right-hand side is the gradient of a potential energy (in 1D). Force can be defined in terms of the gradient of a potential V:

F = -\nabla V

And since
\ddot{x} = \frac{d^2 x}{dt^2}
is acceleration, a, the result of applying the Euler-Lagrange equation to a classical-mechanical Lagrangian is the familiar form of Newton's Second Law:

F = ma


In other words, applying the Euler-Lagrange equation to a suitable Lagrangian provides an equation of motion!

1 comment:

  1. Very good article ! (this one and the previous one about the principle of least action).

    Your calculus are very clear, however I still try to understand where do the remaining hypothesis (ie. not a lot of things: you only minimize the time-integral of L, defined as T-V) come from.

    How can the principle can be interpreted ? Is it the idea that the amount of "lost" energy must be minimal, thus Lagrangian is defined as the difference between kinetic and potentiel forms of energy?


    I would be very happy if you could give me your correction &| your interpretation about this.

    Thanks !

    ReplyDelete