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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?

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$$p_i = \frac{\partial L}{\partial \dot{q}^i}$$ 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 $$p_i = \frac{\partial L}{\partial \dot{q}^i} = \frac{\partial T}{\partial \dot{q}^i}\big(q, \dot{q}\big) + c\, A_i\big(q, t\big) \neq \frac{\partial T}{\partial \dot{q}^i}\big(q, \dot{q}\big)$$

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  • $\begingroup$ Apparent Joos and Freeman do expand their definition to accommodate the EM vector potential, etc. But they sure don't advertise that fact. In some cases generalized "momentum" is a momentum-type quantity. Angular momentum, for example. But there are a number of situation where it doesn't look nor act like a momentum of any kind. My question is along the lines of "what would it break" if we used a stricter definition of kinetic energy and momentum. Part of the problem is that there are more ways of defining terms in Lagrangian mechanics than there are authors doing the defining. $\endgroup$ Oct 31 '18 at 21:26
  • $\begingroup$ @StevenHatton If you alter the definition of generalized momentum then what you break is the symplecitc structure of the Hamiltonian evolution of the position-momentum variables (phase space variables). You also break the duality between the Lagrangian picture and the Hamiltonian picture. Keeping the symplecitc nature of the phase space (the space of positions and momenta) allows you to benefit from a lot of structure, like ways to find more general versions of conservation laws, "conservation of information" on some level, etc. $\endgroup$ Nov 2 '18 at 0:23
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  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.)

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  • $\begingroup$ The appealing thing about Joos and Freeman's definition is that, within a limited scope, kinetic energy is expressible as a quadratic in velocity. That limited scope addresses a vast amount of physics. They continue to present the Hamiltonian as a quadratic in the generalized momentum (assuming all position variables are cyclic, and the system is conservative). $\endgroup$ Oct 31 '18 at 22:39

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