Energy Function and the Conservation of Energy
- Classical Mechanics 1.6
Energy Function
Let’s think of the total derivative of the Lagrangian with respect to time:
\[ \begin{align*} \odv{L}{t} &= \sum_i \pdv{L}{q_i} \odv{q_i}{t} + \sum_i \pdv{L}{\dot{q}_i} \odv{\dot{q}_i}{t} + \pdv{L}{t} \nl &= \sum_i \odv{}{t} \left( \pdv{L}{\dot{q}_i} \right) \dot{q}_i + \sum_i \pdv{L}{\dot{q}_i} \odv{\dot{q}_i}{t} + \pdv{L}{t} \nl &= \sum_i \odv{}{t} \left( \dot{q}_i \pdv{L}{\dot{q}_i} \right) + \pdv{L}{t} \end{align*} \]
Rearranging gives us
\[ \odv{}{t} \left( \sum_i \dot{q}_i \pdv{L}{\dot{q}_i} - L \right) + \pdv{L}{t} = 0 \]
We define the energy function $h$ (which is equivalent to the Hamiltonian $H$) as:
\[ \begin{align*} h(\b{q}, \dot{\b{q}}, t) &= \sum_i \dot{q}_i \pdv{L}{\dot{q}_i} - L \nl &= \dot{\b{q}} \cdot \pdv{L}{\dot{\b{q}}} - L \nl &= \dot{\b{q}} \cdot \b{p} - L \end{align*} \]
Thus, we have
\[ \odv{h}{t} = -\pdv{L}{t} \]
If the Lagrangian does not explicitly depend on time, then $h$ is conserved.
Degree Analysis
We saw that the total kinetic energy $T$ can always be written as:
\[ T = T_0 + T_1 + T_2 \]
where $T_0$ is independent of the generalized velocities $\dot{q}_i$, $T_1$ is linear in $\dot{q}_i$, and $T_2$ is quadratic in $\dot{q}_i$. For most systems, the potential energy $V$ can be also split into three parts. Therefore, the Lagrangian can be expressed as:
\[ L = L_0 + L_1 + L_2 \]
Now, recall that Euler’s theorem for homogeneous functions states that if a function $f(x_1, x_2, \ldots, x_n)$ is homogeneous of degree $k$, then:
\[ \sum_{i=1}^{n} x_i \pdv{f}{x_i} = k f \]
Applying this to the energy function $h$, we have:
\[ h = 2L_2 + L_1 - L = L_2 - L_0 \]
If the transformation equation defining the generalized coordinates does not depend on time, then $T=T_2$. If, further, the potential does not depend on velocities, then $L_2=T$ and $L_0=-V$. Thus, in this common case, the energy function is simply the total mechanical energy:
\[ h = T + V = E \]
Thus, if the Lagrangian does not explicitly depend on time, then the total mechanical energy is conserved.
Non-conservative Systems
Finally, for the case where non-conservative forces are present, and they are the frictional forces derivable from a Rayleigh dissipation function $\mathcal{F}$, we have:
\[ \odv{h}{t} = -\pdv{L}{t} - \sum_i \pdv{\mathcal{F}}{\dot{q}_i} \dot{q}_i \]
Applying Euler’s theorem again, we get:
\[ \odv{h}{t} = -\pdv{L}{t} - 2\mathcal{F} \]
If the Lagrangian does not explicitly depend on time, for the system where $h$ is the total mechanical energy $E$, we have:
\[ \odv{E}{t} = -2\mathcal{F} \]
which matches the result we’ve obtained earlier.
