# Lagrangians and the principle of least action

References: Tom Lancaster and Stephen J. Blundell, Quantum Field Theory for the Gifted Amateur, (Oxford University Press, 2014) – Section 1.4.

A fundamental application of functional derivatives is in the derivation of the principle of least action and the Euler-Lagrange equation. A particle with mass ${m}$ following some trajectory ${x\left(t\right)}$ will have a potential energy ${V\left(x\right)}$ and a kinetic energy ${T\left(x\right)}$ at each point on the trajectory. If the particle follows the trajectory between times ${t=0}$ and ${t=\tau}$, then these energies have averages given as functionals

 $\displaystyle \bar{V}\left[x\right]$ $\displaystyle =$ $\displaystyle \frac{1}{\tau}\int_{0}^{\tau}V\left(x\left(t\right)\right)dt\ \ \ \ \ (1)$ $\displaystyle \bar{T}\left[x\right]$ $\displaystyle =$ $\displaystyle \frac{1}{\tau}\int_{0}^{\tau}T\left(x\left(t\right)\right)dt\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{m}{2\tau}\int_{0}^{\tau}\dot{x}^{2}dt \ \ \ \ \ (3)$

From example 1 in the earlier post, we have

$\displaystyle \frac{\delta\bar{V}\left[x\right]}{\delta x\left(t\right)}=\frac{1}{\tau}\frac{\partial V}{\partial x} \ \ \ \ \ (4)$

From example 2 in the same post, we have

 $\displaystyle \frac{\delta\bar{T}\left[x\right]}{\delta x\left(t\right)}$ $\displaystyle =$ $\displaystyle \frac{1}{\tau}\left(\frac{\partial T}{\partial x}-\frac{d}{dt}\left[\frac{\partial T}{\partial\dot{x}}\right]\right)\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\tau}\left(0-m\ddot{x}\right)\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{m\ddot{x}}{\tau} \ \ \ \ \ (7)$

In classical physics, all forces are ultimately represented by conservative forces (gravity and electromagnetism), so a force can be represented by the gradient of a potential: ${F=-\frac{\partial V}{\partial x}}$. Newton’s law ${F=ma}$ then says

$\displaystyle m\ddot{x}=-\frac{\partial V}{\partial x} \ \ \ \ \ (8)$

so in terms of functional derivatives

$\displaystyle \frac{\delta\bar{T}\left[x\right]}{\delta x\left(t\right)}=\frac{\delta\bar{V}\left[x\right]}{\delta x\left(t\right)} \ \ \ \ \ (9)$

That is, the variation in the average kinetic energy over the path is equal to the variation in the average potential energy. This suggests that the quantity

$\displaystyle L\equiv T-V \ \ \ \ \ (10)$

known as the Lagrangian, is in some sense fundamental, since the variation in its integral over a path is zero. To give a name to this integral, we define the action ${S}$ as

$\displaystyle S\equiv\int_{0}^{\tau}Ldt \ \ \ \ \ (11)$

giving rise to the condition

$\displaystyle \frac{\delta S\left[x\right]}{\delta x\left(t\right)}=0 \ \ \ \ \ (12)$

Technically, this means that the action is stationary when the trajectory ${x\left(t\right)}$ is the actual trajectory followed by a particle. In pracitice, the action nearly always turns out to be a minimum, so this is known as the principle of least action. Varying the trajectory slightly causes the action to increase and gives a path that is not followed by the particle.

If we know the Lagrangian for a particle (or more generally, for a system of particles), we can work out the equation(s) of motion by applying the principle of least action. To do this, we consider the Lagrangian as a function of a generalized coordinate ${x\left(t\right)}$ and its first derivative ${\dot{x}\left(t\right)}$:

$\displaystyle L=L\left(x,\dot{x}\right) \ \ \ \ \ (13)$

We can then apply the principle of least action:

$\displaystyle \frac{\delta S\left[x\right]}{\delta x\left(t\right)}=\frac{\delta}{\delta x\left(t\right)}\left[\int_{0}^{\tau}L\left(x,\dot{x}\right)dt\right]=0 \ \ \ \ \ (14)$

This has the same form as example 2 from the earlier post that we used above, so we get

$\displaystyle \frac{\delta}{\delta x\left(t\right)}\left[\int_{0}^{\tau}L\left(x,\dot{x}\right)dt\right]=\frac{\partial L}{\partial x}-\frac{d}{dt}\left[\frac{\partial L}{\partial\dot{x}}\right]=0 \ \ \ \ \ (15)$

Thus we get the Euler-Lagrange equation for a single particle moving in one dimension

$\displaystyle \boxed{\frac{\partial L}{\partial x}-\frac{d}{dt}\frac{\partial L}{\partial\dot{x}}=0} \ \ \ \ \ (16)$

Example As a simple example, suppose we have a particle whose potential and kinetic energies are given by

 $\displaystyle V\left(x,\dot{x}\right)$ $\displaystyle =$ $\displaystyle \frac{1}{2}kx^{2}\ \ \ \ \ (17)$ $\displaystyle T\left(x,\dot{x}\right)$ $\displaystyle =$ $\displaystyle \frac{1}{2}m\dot{x}^{2} \ \ \ \ \ (18)$

Then
the equation of motion is given by

 $\displaystyle L\left(x,\dot{x}\right)$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(m\dot{x}^{2}-kx^{2}\right)\ \ \ \ \ (19)$ $\displaystyle \frac{\partial L}{\partial x}-\frac{d}{dt}\frac{\partial L}{\partial\dot{x}}$ $\displaystyle =$ $\displaystyle -kx-m\ddot{x}=0\ \ \ \ \ (20)$ $\displaystyle m\ddot{x}$ $\displaystyle =$ $\displaystyle -kx \ \ \ \ \ (21)$

This is just the equation of motion for a mass on a spring with spring constant ${k}$.