generalized quantity

• (Jacobi’s) Generalized Mometum

  $$ \frac{\partial L}{\partial\dot{q}_j}\equiv p_j $$

• (Jacobi’s) Generalized Energy | Jacobi’s energy integral → Hamitonian function

  $$ h(\mathbf{q},\mathbf{\dot{q}};t)\equiv\sum_j\left(\dot{q}_j\frac{\partial L}{\partial\dot{q}_j}\right)-L(\mathbf{q},\mathbf{\dot{q}},t) $$

  ⇒generalized energy theorem

  $$ \frac{dH\left(\mathbf{q},\mathbf{p},t\right)}{dt}=\frac{dh(\mathbf{q},\mathbf{\dot{q}},t)}{dt}=\sum_j\dot{q}j\left[Q_j^{EXC}+\sum{k=1}^m\lambda_k\frac{\partial g_k}{\partial q_j}(\mathbf{q},t)\right]-\frac{\partial L(\mathbf{q},\mathbf{\dot{q}},t)}{\partial t} $$

symmetries & Invariance

Hamiltonian Invariance 哈密顿不变性

由generalized energy theorem， 非保守力为零的情况下，哈密顿是否显性依赖时间等同于拉格朗日量。

If the following two requirements are satisfied

1. the kinetic energy has a homogeneous 单一 quadratic 平方 dependence on the generalized velocities, that is, the transformation to generalized coordinates is independent of time, $\frac{\partial x_{\alpha,i}}{\partial t}=0.$
2. the potential energy is not velovity dependent, thus the terms $\frac{\partial U}{\partial\dot{q}_i}=0.$

Then

$$ H=T+U=E $$