Defining generalized momentum in terms of kinetic energy versus a Lagrangian

Question

Reputable authors (e.g., Bergmann, Wells, Susskind) define generalized momentum using the Lagrangian $L$ as $$p_{i}\equiv\frac{\partial L}{\partial\dot{q}^{i}}.\tag{1}$$

Joos and Freeman define generalized momentum for holonomous-scleronomous systems using the kinetic energy $T$ as $$p_{i}\equiv\frac{\partial T}{\partial\dot{q}^{i}}.\tag{2}$$

There is no direct contradiction due to the qualification that the system is holonomous-scleronomous. Nonetheless, it begs the question: what would be the consequences of one definition over the other in more general circumstances? Put differently, why choose one over the other?

Answer 1

$$p_i = \frac{\partial L}{\partial \dot{q}^i}$$ is the fundamental definition of generalized momentum. There is a very good reason for choosing it and that reason is the natural, canonical switch from the second derivative Euler-Lagrange equations of motion to the first derivative vector field form of Hamiltonian equations of motion. Indeed, the Euler-Lagrange equations are $$\frac{d}{dt} \, \frac{\partial L}{\partial \dot{q}^i}\big(q,\dot{q}, t\big) = \frac{\partial L}{\partial {q}^i}\big(q,\dot{q}, t\big) $$ After setting $$p_i = \frac{\partial L}{\partial \dot{q}^i}\big(q,\dot{q}, t\big)$$ you can rewrite the Euler-Lagrange system as the double system \begin{align} p_i &= \frac{\partial L}{\partial \dot{q}^i}\big(q,\dot{q}, t\big)\\ \frac{d}{dt} p_i &= \frac{\partial L}{\partial {q}^i}\big(q,\dot{q}, t\big) \end{align} Solve the first set of equations with respect to $\dot{q} = f(q, p, t)$ and you get \begin{align} \dot{q}^i &= f^i(q, p, t)\\ \dot{p}_i &= \frac{\partial L}{\partial {q}^i}\big(q, \, f(q, p, t), \, t\big) \end{align} As it turns out, if you define the function $$H(q, p, t) = \sum_{k}\, p_k \, \dot{q}^k - L(q, \dot{q}, t) = \sum_k\, p_k \, f^k(q,p,t) - L\big(q,\, f(q,p,t),\, t\big)$$ then \begin{align} \frac{\partial H}{\partial q^i}\big(q, p, t\big) &= \sum_k\, p_k \, \frac{\partial f^k}{\partial q^i}\big(q,p,t\big) - \frac{\partial L}{\partial q^i}\big(q,\, f(q,p,t),\, t\big) \\ & \,\,\,\,\,\, - \sum_k \, \frac{\partial L}{\partial \dot{q}^k}\big(q,\, f(q,p,t),\, t\big) \frac{\partial f^k}{\partial q^i}\big(q,\, p,\, t\big) = \\ &= \sum_k\, p_k \, \frac{\partial f^k}{\partial q^i}\big(q,p,t\big) - \frac{\partial L}{\partial q^i}\big(q,\, f(q,p,t),\, t\big) - \sum_k \, p_k \frac{\partial f^k}{\partial q^i}\big(q,\, p,\, t\big) = \\ &= - \, \frac{\partial L}{\partial q^i}\big(q,\, f(q,p,t),\, t\big) \end{align} Hence $$\dot{p}_i = \frac{\partial L}{\partial q^i}\big(q,\, f(q,p,t),\, t\big) = - \, \frac{\partial H}{\partial q^i}\big(q, p, t\big)$$ Analogously, \begin{align} \frac{\partial H}{\partial p_i}\big(q, p, t\big) &= f^i(q, p, t) + \sum_k\, p_k \, \frac{\partial f^k}{\partial p_i}\big(q,p,t\big) - \sum_k \, \frac{\partial L}{\partial \dot{q}^k}\big(q,\, f(q,p,t),\, t\big) \frac{\partial f^k}{\partial p_i}\big(q,\, p,\, t\big) = \\ &= f^i(q, p, t) + \sum_k\, p_k \, \frac{\partial f^k}{\partial p_i}\big(q,p,t\big) - \sum_k \, p_k \frac{\partial f^k}{\partial q^i}\big(q,\, p,\, t\big) = \\ &= - \, \frac{\partial L}{\partial p_i}\big(q,\, f(q,p,t),\, t\big) = \\ &= f^i(q, p, t) \end{align} Hence $$\dot{q}^i = f^i\big(q, p , t\big) = \frac{\partial H}{\partial p_i}\big(q, p, t\big)$$ Consequently the Euler-Lagrange system turns into the Hamiltonian system \begin{align} \dot{q}^i &= \frac{\partial H}{\partial p_i}\big(q, p, t\big)\\ \dot{p}_i &= - \, \frac{\partial H}{\partial q^i}\big(q, p, t\big) \end{align} The Hamiltonian form of the system has far reaching consequences. Only in the case where the Lagrangian is of the form kinetic minus potential energy $$L\big(q, \dot{q}, t\big) = T\big(q, \dot{q}, t\big) \, - \, U\big( q, t\big)$$ with $T$ being the kinetic energy, we obtain $$p_i = \frac{\partial L}{\partial \dot{q}^i} = \frac{\partial T}{\partial \dot{q}^i}$$ This is a special case.

However, in electrodynamics for example, you may encounter the Lagrangian of a charged particle with charge $c$ moving in an electromagnetic field. In this case, the Lagrangian is something like $$L = T(q, \dot{q}) + \sum_{k=1}^3 c\,A_k\big(q, t\big) \, \dot{q}^k \, - \, c\,\phi\big(q, t\big) $$ where $A_k(q,t)$ is the magnetic vector potential and $\phi(q,t)$ is the electric scalar potential. As you can check here, the generalized momenta are $$p_i = \frac{\partial L}{\partial \dot{q}^i} = \frac{\partial T}{\partial \dot{q}^i}\big(q, \dot{q}\big) + c\, A_i\big(q, t\big) \neq \frac{\partial T}{\partial \dot{q}^i}\big(q, \dot{q}\big)$$

Answer 2

1. The canonical/conjugate momentum (1) is the natural/fundamental notion in Lagrangian formalism. (Also recall that there exist velocity-dependent potentials $U(q,\dot{q},t)$.)

2. The kinetic momentum (2) only exists if there is a natural notion of a kinetic term $T$ in the Lagrangian $L$. (The kinetic term $T$ is by the way not always the kinetic energy $K$. Think e.g. of a relativistic point particle, cf. this Phys.SE post.)