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}.