How is the generalised momentum a covector?

Question

I know that this has been a question asked here before, but I'm still struggling to understand the concept. V.I. Arnold states that for "the configuration manifold $V$, the 'generalised momentum '$\textbf{p}=\dfrac{\partial L}{\partial \dot{\textbf{q}}}$ is a cotangent vector. I'm struggling to see how this works, a co-vector, $dx_i$, at a point $p\in M$ is defined as

$$dx_i:T_PM\rightarrow \mathbb{R}$$ $$dx_i(\textbf{v})=v_i$$ Where $\textbf{v}\in T_pM$. So a co-vector should map $\textbf{v}$ to the $i$'th component of $\textbf{v}$. How would this work for generalised momentum, since it would mean $$\textbf{p}(\textbf{v})=\dfrac{\partial L}{\partial \dot{\textbf{q}}}(\textbf{v})=v_i$$ But I don't see how this is the case.

Answer 1

You need to decide if your position coordinates are $x$ or $q$, you are using both. Per the last formula, keep using $q$.

As a 1-form $\mathbf p$ is $\sum_i p_i\,dq^i$, so that the evaluation $\mathbf p(\mathbf v)$ has the result $\sum_i p_iv^i$.

Note that as directional derivative $$\frac{∂L}{∂\dot{\mathbf q}}(\mathbf v)=\frac{d}{ds}L(\mathbf q,\dot{\mathbf q}+s\mathbf v)\Big|_{s=0}$$ is indeed a linear functional in $\mathbf v$.

Answer 2

This can be understood by looking at the transform properties of $p_i$. Under a change of general coordinates $q_i \mapsto q'_l(q_i)$, one has $\dot{q}'_l= \frac{\partial q'_l}{\partial q_i}\dot{q}_i$, which implies $\frac{\partial \dot{q}'_l}{\partial \dot{q}_i}= \frac{\partial q'_l}{\partial q_i}$. So we have \begin{eqnarray} p_i &\equiv& \frac{\partial L}{\partial \dot{q}_i}\\ &=& \frac{\partial \dot{q}'_l}{\partial \dot{q}_i}\frac{\partial L}{\partial \dot{q}'_l}\\ &=& \frac{\partial {q}'_l}{\partial {q}_i} p'_l, \end{eqnarray} which shows explicitly that $p_i$ transforms as a covector under the change of general coordinate.

Please Ref. Theodore Frankel "The Geometry of Physics" Section 2.3.