Principle of Least Action

Principle (Least Action). Define the Action of a conservative system:

S = \int_{t_1}^{t_2} L(q, \dot{q}, t) dt \tag{1}

in which L = T - V is the Lagrangian. The real trajectory of this system q_c(t) is that which let L(q, \dot{q}, t) be the extremum (usually the least).

Euler-Lagrange Equation

S is a functional of trajectories. We use the variational method to derive the real trajectory q_c(t).

We apply a virtual infinitely small amount of perturbation \delta q(t) to q_c(t):

q(t) = q_c(t) + \delta q(t), \delta q(t) \rightarrow 0 \tag{2}

Apply Equation 2 to Equation 1:

\delta S = \int_{t_1}^{t_2} dt (\frac{\partial L}{\partial q_i} \delta q_i + \frac{\partial L}{ \partial \dot{q_i}} \delta \dot{q_i})

We also have:

\frac{\partial L}{\partial \dot{q_i}}\delta \dot{q_i} = \frac{d}{dt} (\frac{\partial L}{\partial \dot{q_i}}\delta q_i) - \frac{d}{dt}(\frac{\partial L}{\partial \dot{q_i}})\delta q_i

Then, we have:

\delta S = \int_{t_1}^{t_2} dt (\frac{\partial L}{\partial q_i} \delta q_i - \frac{d}{dt}(\frac{\partial L}{\partial \dot{q_i}})\delta q_i) + \left. \frac{\partial L}{\partial \dot{q_i}}\delta q_i \right|_{t_1}^{t_2} = 0

The term outside of the integral is 0, because \delta q(t_1) = \delta q(t_2) = 0. Inside the integral, because this equation is true \forall \delta q_i, and also different q_i s are independent, we have:

Equation (Euler-Lagrange).

\frac{\partial L}{\partial q_i} - \frac{d}{dt}(\frac{\partial L}{\partial \dot{q_i}}) = 0, i = 1, 2, ..., f \tag{3}

Non-uniqueness of L

In classic systems, we can actually define a new equivalent Lagrangian of the system

L^{\prime} (q, \dot{q}, t) = L(q, \dot{q}, t) + \frac{d}{dt}f(q, t) \tag{4}

Proof. Substituting Equation 4 into Equation 3, we have:

\frac{\partial}{\partial q_i}\frac{df}{dt}=\frac{d}{dt}(\frac{\partial}{\partial \dot{q}_i}\frac{df}{dt})

We also have:

\frac{df}{dt}=\frac{\partial f}{\partial q_j}\dot{q}_j+\frac{\partial f}{\partial t}

Another substitution:

\begin{aligned} L.H.S. &= \frac{\partial}{\partial q_i}\left( \frac{\partial f}{\partial q_j}\dot{q}_j+\frac{\partial f}{\partial t} \right) \\ &= \frac{\partial ^2f}{\partial q_i\partial q_j}\dot{q}_j+\frac{\partial ^2f}{\partial q_i\partial t}+\frac{\partial f}{\partial q_j}\frac{\partial \dot{q}_j}{\partial q_i} \end{aligned}

We also have {\partial \dot{q}_j} / {\partial q_i}=0 because q and \dot{q} are independent. So:

L.H.S.=\frac{\partial ^2f}{\partial q_i\partial q_j}\dot{q}_j+\frac{\partial ^2f}{\partial q_i\partial t}

Similarly,

\begin{aligned} R.H.S.&=\frac{d}{dt}(\frac{\partial}{\partial \dot{q}_i}\left( \frac{\partial f}{\partial q_j}\dot{q}_j+\frac{\partial f}{\partial t} \right) ) \\ &=\frac{d}{dt}(\frac{\partial \dot{q}_j}{\partial \dot{q}_i}\frac{\partial f}{\partial q_j})=\frac{d}{dt}(\delta _{ij}\frac{\partial f}{\partial q_j}) \\ &=\frac{d}{dt}(\frac{\partial f}{\partial q_i}) =\frac{\partial ^2f}{\partial q_k\partial q_i}\dot{q}_k+\frac{\partial ^2f}{\partial q_i\partial t} \\ &=L.H.S. \end{aligned}

That’s the end of our proof.