[NOTES/CM-03003] Conservation of Energy

Time Translations

It is known  that invariance of Lagrangian  under space translations, rotations leads to conservation laws for momentum and angular momentum respectively. What about conservation of energy? Is it also related to a symmetry of the Lagrangian?  The  symmetry transformation corresponding to conservation of energy is is time translation \( t \to t+ \epsilon\). Thus ff the Lagrangian does not depend on time explicitly, there is a conservation law which is reduces to the familiar energy conservation for mechanical systems, for example systems with many particles.

Conservation of Energy

This  conserved quantity associated with time translations is known as the Hamiltonian of the system and  reduces to the total energy   $(=KE+PE)$   for a mechanical system. For other system also, the Hamiltonian qualifies to be identified with energy.

If the Lagrangian does not contain $t$ explicitly, we have \(\frac{\partial{L}}{\partial{t}}=0 \) and hence \begin{equation} \frac{dL}{dt}=\sum_k \frac{\partial{L}}{\partial{ q_k}}\dot q_k+\sum_k\frac{\partial{L}}{\partial{\dot q_k}}(\ddot q_k). \end{equation} Using  Euler Lagrange EOM we get

\begin{eqnarray}
\frac{dL}{dt}&=& \sum_k\Big[\frac {d}{dt}\Big(\frac{\partial{L}}{\partial{\dot
q_k}}\Big)\dot q_k + \frac{\partial{L}}{\partial{\dot q_k}}\ddot q_k\Big]
=\frac{d}{dt}\sum_{k=1}\frac{\partial{L}}{\partial{\dot q_k}}\dot q_k,
\end{eqnarray}

or

\begin{equation} \frac{d}{dt}\Big(\sum_k\frac{\partial{L}}{\partial{\dot q_k}}\dot{q}_k-L\Big )=0. \end{equation}

Hence $H$, defined by, \begin{equation} H\stackrel{\rm def}{\equiv}\sum_{k=1}\frac {\partial{L}}{\partial{\dot q_k}}\dot q_k-L, \end{equation} is a constant of motion. We can also write \begin{equation} H= \sum_{k=1}^{N}p_k \dot{q}_k -L, \end{equation} where $p_k=\frac{\partial{L}}{\partial{\dot q_k}}$ is called the {\bf canonical momentum conjugate} to the coordinate $\dot q_k$ and $H$ will be called {\bf Hamiltonian} of the system

In an alternate form of dynamics, the canonical momenta take over  the role played by velocities and Hamiltonian becomes central quantity  which governs the dynamics. The EOM can be written in an alternate form called the Hamiltonian EOM.

Example:

Let us consider a single particle moving in force field described by potential energy $V(\vec{r})$.Then \begin{equation} L=\frac{1}{2}m\dot{\vec {r}}^{2} - V(\vec{r}); \qquad \vec {r}=(x,y,z), \end{equation} the canonical momenta are \begin{equation} p_x=\frac{\partial{L}}{\partial{\dot x}} = m\dot x;\qquad p_y=\frac{\partial{L}}{\partial{\dot y}}=m \dot y;\qquad p_z=\frac{\partial{L}}{\partial{\dot z}}=m\dot z. \end{equation} and the Hamiltonian is given by

\begin{eqnarray}
H&=&\sum p_k\dot q_k-L \\
&=&\frac{1}{2}m\dot{\vec {r}}^{\,\,2}+V(\vec{r}).\qquad\qquad [\text{Verify This}]
\end{eqnarray}

Thus the canonical momenta, in this example, coincide with components of momentum $m\dot{\vec r}$ and Hamiltonian is equal to the energy.

However, it must be remembered that the canonical momenta are not always equal to `ordinary' momenta and the Hamiltonian need not be a sum of kinetic and potential energy when the system is described by a velocity dependent generalized potential. An example is that of a charged particle in magnetic field.