Generalized momentum

Given a Lagrangian classical system with $(\mathcal C, L)$ with coordinates $q_i$, the generalized momentum or conjugate momentum is

$$ p_i:=\frac{\partial L}{\partial \dot{q}_i} $$

Motivation: If we compare the Euler-Lagrange equations

$$ \frac{\partial L}{\partial q}-\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\partial \dot{q}}=0 $$

with Newton equation

$$ \frac{d}{dt}p=F $$

we get the feeling that

on the one hand, $F=\frac{\partial L}{\partial q}=\frac{\partial }{\partial q}(T-V)=-\frac{\partial V}{\partial q}$

on the other, $\frac{\partial L}{\partial \dot{q}}$ must play the role of $p$.

It turns out that generalized momentum is a covector. Roughly speaking, it can be understood in the following way. The Lagrangian $L$ is a modification of the kinetic energy $T$, which is a "kind of" squared length of the velocity $\dot{q}_i$. In a vector space with an inner product $g(-,-)$, a length is computed in the following way:

we take the vector $v$

we obtain the dual vector $g(v,-)$.

we compute $Length=g(v,v)$

Therefore, loosely speaking

$$ \frac{\partial Length}{\partial v}=\frac{\partial g(v,v)}{\partial v}=g(v,-) $$

which is a covector.

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Author of the notes: Antonio J. Pan-Collantes

antonio.pan@uca.es

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