I have trouble understanding this definition:

Let $Q$ be some manifold and $L: TQ \to \mathbb{R}$ a smooth function. Then for some local coordinates $(q, \dot{q})$ on $TQ$ the conjugated momentum is defined as $\frac{\partial L}{\partial \dot{q}}$, which is an element of the co-tangential bundle $T^{*}Q$.

How is the expression $\frac{\partial L}{\partial \dot{q}}$ to be interpreted? If one simply expresses $L$ in local coordinates by $$L \circ(q^{-1}, \dot{q}^{-1}): \mathbb{R}^{2n} \to \mathbb{R}$$ and differentiates it with respect to the second variable one gets a function $\mathbb{R}^n \to \mathbb{R}$ and not an element of the co-tangential bundle $T^{*}Q$. Is the correct expression $$\partial_2 ( L \circ(q^{-1}, \dot{q}^{-1}))\circ (q, \dot{q}) \in T^{*}Q\ ?$$

  • $\begingroup$ yes, the issue is that the derivatives you get away from $0 \in \Bbb R^n$ are not actually independent of the choice of local coordinates in any reasonable sense. The only invariant data is what you're writing down along the 0-section. $\endgroup$ – user98602 Dec 31 '18 at 0:39

The notation $\partial L/\partial\dot{q}$ is no more than a symbol that denotes the Frechet derivative of $L$ with respect to $\dot{q}$. More precisely, let $$ L:Q\times TQ\to\mathbb{R}. $$ Then for all $x\in Q$, $T_xQ$ is a vector space equipped with the norm induced by the metric on $Q$. In this sense, one may define a bounded linear operator $A:TQ\to\mathbb{R}$, such that $$ \lim_{Y\to 0}\frac{\left\|L(x,X+Y)-L(x,X)-A(Y)\right\|_{\mathbb{R}}}{\left\|Y\right\|_{T_xQ}}=0 $$ for a given $x\in Q$ and a given $X\in T_xQ$. This linear operator $A$, if it exists, is called the partial derivative of $L$ with respect to $X$, also denoted by $\partial L/\partial X$ for intuition. Further, if one considers a trajectory $q:\mathbb{R}\to Q$, we have $\dot{q}:\mathbb{R}\to T_{q}Q$. Hence when considering the special form $L(q,\dot{q})$, we also take $\partial L/\partial\dot{q}$ instead of $\partial L/\partial X$, again, for intuition.

Finally, whatever the notation is assigned to the operator defined above, either $A$ or $\partial L/\partial X$ or $\partial L/\partial\dot{q}$, it is (1) linear and is (2) from $TQ$ to $\mathbb{R}$. On the other hand, the collection of all linear operators from $TQ$ to $\mathbb{R}$ is definitely $T^*Q$. Therefore, it follows that $$ A\in T^*Q,\quad\text{or}\quad\frac{\partial L}{\partial X}\in T^*Q,\quad\text{or}\quad\frac{\partial L}{\partial\dot{q}}\in T^*Q. $$

Hope this could be helpful for you.


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