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I've seen mainly two definitions of generalized momenta, $p_k$, and I wasn't sure which one is always true/ the correct one: $$p_k\equiv\dfrac{\partial\mathcal T}{\partial \dot q_k}\text{ and }p_k\equiv\dfrac{\partial\mathcal L}{\partial \dot q_k}.$$

I've seen mostly written the $2$nd one. Notice that if the potential energy is velocity independent, the $2$nd definiton reduces to $1$st. What about the general case?

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1 Answer 1

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I've seen mainly two definitions of generalized momenta, $p_k$, and I wasn't sure which one is always true/ the correct one: $$p_k\equiv\dfrac{\partial\mathcal T}{\partial \dot q_k}\text{ and }p_k\equiv\dfrac{\partial\mathcal L}{\partial \dot q_k}.$$

I've seen mostly written the $2$nd one. Notice that if the potential energy is velocity independent, the $2$nd definiton reduces to $1$st. What about the general case?

In the general case, we use the latter definition for the "generalized" or "canonical" momentum: $$ p_k \equiv \frac{\partial L}{\partial \dot q_k}\;,\tag{1} $$ where $$ L = T - U\;, $$ where $T$ is the kinetic energy and $U$ is the potential energy.

The term: $$ \pi_k \equiv \frac{\partial T}{\partial \dot q_k}\;,\tag{2} $$ is usually called the "mechanical" momentum, to contrast it with the "generalized" or "canonical" momentum.


A common case where Eq. (1) and Eq. (2) differ is when we use a Lagrangian to describe the dynamics of a particle in an electromagnetic field. The difference is due to the velocity-dependent part of the electromagnetic potential: $e\vec v \cdot \vec A$, where $\vec A$ is the vector potential, $e$ is the charge, and $\vec v$ is the velocity.

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