### Time dependence

Systems in general are always implicitly dependent on time, since generalised coordinates themselves are functions of time.

But, we could imagine many examples where a trajectory of a system is *explicitly* time-dependent.

For example, we could have a system close to a large moving mass which we don't include in our description directly. Then, the net forces acting on parts of the system will change differentially in a way that explicitly depend on time.

If, however, we insist that that a system is not explicitly time-dependent, then, by definition, it has time-translational symmetry.

### Time-translational symmetry

Suppose then that a given solution of the Euler-Lagrange equations is not explicitly dependent on time, hence has time-translational symmetry. That is, if we adjust the time when the system starts running, then the trajectory will behave exactly as before, but just at a later time.

So consider the transformation $$t \to t + \epsilon$$ We need to convert this to an expression involving the change in the coordinates themselves. Notice from the (highly exaggerated) diagram in figure 4.4, that, at least for the section of the trajectory between the times $[t_B, t_C]$, that a time translation can also be considered as a coordinate translation $$q_i(t) \to q_i(t - \epsilon) = q_i(t) - \dot{q_i}\epsilon$$ so the change in $q_i$ is $$\delta q_i = -\dot{q_i}\epsilon$$

Thus, the change in action over this period is $$\begin{align*}\delta A_{BC} &= \int_{t_B}^{t_C} L(q_i, \dot{q_i}) \textrm{d}t\\ &= \left [ \sum_{i} \frac{\partial L}{\partial \dot{q_i}} \delta q_i \right ]_{t_B}^{t_C}\\ &= \epsilon \left [ - \sum_{i} \pi_i \; \dot{q_i} \right ]_{t_B}^{t_C}\end{align*}$$ since the trajectory is a solution of the Euler-Lagrange equations.

Under this variation, though, the portion of the curve on the time interval $[t_A, t_B]$ is also translated, but this isn't part of the new trajectory. This means that the action from this portion needs to be subtracted. Similarly, the portion of the trajectory from $[t_C, t_D]$ needs to be *added*.

Since we considering an infinitesimal time displacement, we can simply write $$\begin{align*}A_{AB} &= \int_{t_A}^{t_B} L(q_i, \dot{q_i}) \textrm{d}t = L(t_B) \epsilon\\ A_{C'D} &= \int_{t_C}^{t_D} L \textrm{d}t = L(t_C) \epsilon\end{align*}$$

Thus, the equation for the total change in the action is $$0 = \delta A = \epsilon \left [- \sum_{i} \pi_i \; \dot{q_i} \right ]_{t_B}^{t_C} - L(t_B) \epsilon + L(t_C) \epsilon$$

### The Hamiltonian

Dividing by $\epsilon$ and re-arranging, we get $$\left. \sum_{i} \pi_i \; \dot{q_i} \right|_{t_B} - L(t_B) = \left. \sum_{i} \pi_i \; \dot{q_i} \right|_{t_C} - L(t_C) $$

Thus, the conserved quantity associated with time-translation symmetry is $$H = \sum_{i} \pi_i \; \dot{q_i} - L$$ which is called the **Hamiltonian**, or energy, of the system.

We can consider this as the definition of energy - the *quantity conserved under time-translation*. We shall see that we can build a whole new formalism of mechanics based on the Hamiltonian of a system.

Note that, where we can define the Lagrangian as $$L = T - U$$ for some potential, $U$, and kinetic energy, $T$, then the Hamiltonian is given by $$H = T + U$$

### Particle in free space

We know that the Lagrangian for a free particle in space is $$L = \tfrac{1}{2} m \dot{x}^2 - U(x)$$ for some potential $U$.

Then, $$\begin{align*}H &= p_x \dot{x} - L\\ &= m \dot{x}^2 - \left( \tfrac{1}{2} m \dot{x}^2 - U(x) \right)\\ &= \tfrac{1}{2} m \dot{x}^2 + U(x)\\ &= T + U\end{align*}$$ which is just the normal expression for energy.