Hamiltonian

Definition

The total time derivative of the Lagrangian \(L\) is $$\frac{dL}{dt} = \sum_i\frac{\partial L}{\partial q_i}\frac{dq_i}{dt} + \sum_i\frac{\partial L}{\partial \dot{q}_i}\frac{d\dot{q}_i}{dt} +\frac{\partial L}{\partial t}.$$ Since, by the Lagrange's equation, $$\frac{\partial L}{\partial q_i} = \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_i}\right),$$ we have \begin{align} \frac{d L}{dt} &= \sum_i\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_i}\right)\dot{q}_i + \sum_j\frac{\partial L}{\partial \dot{q}_i}\frac{d\dot{q}_i}{dt} + \frac{\partial L}{\partial t}\\ &= \sum_i \frac{d}{dt}\left(\dot{q}_i\frac{\partial L}{\partial \dot{q}_i}\right) + \frac{\partial L}{\partial t}. \end{align} So, $$\frac{d}{dt}\left(\sum_i\dot{q}_i\frac{\partial L}{\partial \dot{q}_i}-L\right) + \frac{\partial L}{\partial t} = 0.$$ Define the Hamiltonian, or energy function, as $$H(q_1,...,q_n,\dot{q}_1,...,\dot{q}_n,t) = \sum_i\dot{q}_i\frac{\partial L}{\partial \dot{q}_i}-L.$$ Then, we have the equation $$ \bbox[5px,border:2px solid #666] { \frac{dH}{dt} = -\frac{\partial L}{\partial t} } $$

Kinetic Energy

For a holonomic system, \begin{align} \vec{v}_i&=\frac{d\vec{r}_i}{dt} = \sum_j \frac{\partial \vec{r}_i}{\partial q_j}\dot{q}_j+\frac{\partial \vec{r}_i}{\partial t}\\ v^2_i &= \sum_j \frac{\partial \vec{r}_i}{\partial q_j}\dot{q}_j+\frac{\partial \vec{r}_i}{\partial t} \cdot \sum_k \frac{\partial \vec{r}_i}{\partial q_k}\dot{q}_k+\frac{\partial \vec{r}_i}{\partial t}\\ &= \sum_j\sum_k \frac{\partial \vec{r}_i}{\partial q_j}\cdot\frac{\partial \vec{r}_i}{\partial q_k}\dot{q}_j\dot{q}_k + 2\sum_k\frac{\partial \vec{r}_i}{\partial q_k}\cdot\frac{\partial \vec{r}_i}{\partial t}\dot{q}_k + \frac{\partial \vec{r}_i}{\partial t}\cdot\frac{\partial \vec{r}_i}{\partial t} \end{align} Define $$T_0\equiv \sum_i\frac{1}{2}m_i\left(\frac{\partial \vec{r}_i}{\partial t}\right)^2,$$ $$T_1\equiv \sum_j\left(\sum_im_i\frac{\partial \vec{r}_i}{\partial q_j}\cdot\frac{\partial\vec{r}_i}{\partial t}\right)\dot{q}_j$$ and $$T_2\equiv\frac{1}{2}\sum_{j,k}\left(\sum_im_i\frac{\partial\vec{r}_i}{\partial q_j}\cdot\frac{\partial \vec{r}_i}{\partial q_k}\right)\dot{q}_j\dot{q}_k$$ Thus, the total kinetic energy can be divided into three terms $$T = \sum_i\frac{1}{2}m_iv^2_i = T_0 + T_1 + T_2$$ If the constraint is time-independent, then the transformation $$\vec{r}_i = \vec{r}_i(q_1,...,q_N)$$ is also time-independent, then $$T_0=T_1=0$$

Hamiltonian VS total energy

If the potential energy \(V\) is velocity independent $$V=V(q_1,...,q_N,t)$$ and constraint is time-independent, i.e. $$T=T_2,$$ then \begin{align} \frac{\partial L}{\partial \dot{q}_\lambda} &= \frac{\partial}{\partial \dot{q}_\lambda}(T_2-V) \\ &= \frac{\partial}{\partial \dot{q}_\lambda}\left[\frac{1}{2}\sum_{j,k}\left(\sum_im_i\frac{\partial\vec{r}_i}{\partial q_j}\cdot\frac{\partial\vec{r}_i}{\partial q_k}\dot{q}_j\dot{q}_k\right)\right]\\ &= \frac{1}{2}\sum_{j,k}\left(\sum_im_i\frac{\partial \vec{r}_i}{\partial q_j}\cdot\frac{\partial \vec{r}_i}{\partial q_k}\right)(\delta_{\lambda i}\dot{q}_k + \dot{q}_j\delta_{\lambda k})\\ &= \frac{1}{2}\sum_k\left(\sum_im_i\frac{\partial\vec{r}_i}{\partial q_\lambda}\cdot\frac{\partial\vec{r}_i}{\partial q_k}\right)\dot{q}_k + \frac{1}{2}\sum_j\left(\sum_im_i\frac{\partial\vec{r}_i}{\partial q_j}\cdot\frac{\partial\vec{r}_i}{\partial q_\lambda}\right)\dot{q}_j\\ &= \sum_j\left(\sum_im_i\frac{\partial\vec{r}_i}{\partial q_j}\cdot\frac{\partial\vec{r}_i}{\partial q_\lambda}\right)\dot{q}_j. \end{align} The Hamiltonian is then \begin{align} H &\equiv \sum_\lambda \frac{\partial L}{\partial \dot{q}_\lambda}\dot{q}_\lambda - L \\ &= \sum_{\lambda, j}\left(\sum_im_i\frac{\partial \vec{r}_i}{\partial q_j}\cdot\frac{\partial\vec{r}_i}{\partial q_\lambda}\right)\dot{q}_j\dot{q}_\lambda-(T_2-V)\\ &= 2T_2-(T_2-V)\\ &= T_2 + V\\ &= T + V, \end{align} which is the total energy. Therefore, Hamiltonian is equivalent to the total energy on the condition that potnetial energy does not depend on generalized velocity and constraint does not depend on time.