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!

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

ReplyDeleteYour 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 !