Classical Mechanics 3: Hamilton's action principle.

We saw before that Newton's 2nd law can be written in a more general form as
\[ \frac{d}{dt} \frac{\partial L}{\partial v}(x, \dot{x}) = \frac{\partial L}{\partial x}(x, \dot{x}), \]
known as the Euler-Lagrange equations. Hamilton discovered a principle that explains the origin of these equations. Consider some path of the system given by a curve \(\gamma\), i.e.
\[ x(t) = \gamma(t) \]
\[ \dot{x}(t) = \frac{d}{dt}\gamma(t) \]
Then we may define a quantity associated with the path \(\gamma\):
\[ S = \int L(\gamma, \dot{\gamma})dt \]
called the action. Hamilton discovered the following.

Theorem The path taken by a mechanical system is one which extremizes the action.

To prove this, suppose we perturb the path a small amount, while leaving the endpoints fixed, i.e. \(\gamma \mapsto \gamma + \epsilon (\delta\gamma)\) with \(\epsilon > 0\) small and \(\delta\gamma\) a path that is \(0\) at its endpoints. Then
\[ L(\gamma + \epsilon\delta\gamma, \dot\gamma + \epsilon\delta\dot{\gamma}) = L(\gamma, \dot{\gamma}) + \epsilon \frac{\partial L}{\partial x}\delta\gamma + \epsilon \frac{\partial L}{\partial v} \delta\dot\gamma + o(\epsilon^2) \]
Thus
\[ S[\gamma + \epsilon\delta\gamma] = S[\gamma] + \epsilon \int \frac{\partial L}{\partial x} \delta \gamma dt + \epsilon \int \frac{\partial L}{\partial v} \delta \dot\gamma dt + o(\epsilon^2) \]
Integrating by parts, and using the fact that \(\delta\gamma\) is \(0\) on the endpoints, we have
\[ \int \frac{\partial L}{\partial v}\delta\dot\gamma dt = -\int \frac{d}{dt} \frac{\partial L}{\partial v} \delta\gamma dt \]

Combining the above, we have
\[ \frac{\delta S}{\delta \gamma}(\delta \gamma) = \int \left(\frac{\partial L}{\partial x} - \frac{d}{dt}\frac{\partial L}{\partial v} \right) \delta\gamma dt \]
Thus the variational derivative of \(S\) is
\[ \frac{\delta S}{\delta \gamma} = \frac{\partial L}{\partial x} - \frac{d}{dt} \frac{\partial L}{\partial v} \]
So a path \(\gamma\) is a critical point of \(S\) (i.e. it extremizes \(S\)) if and only if the Euler-Lagrange equations are satisfied.