# [Physics] Canonical momentum Velocity dependent Lagrangian

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I have a homework problem wich I think I'm on the verge of solving but need help with some relations:

Show that if the potential $U$ in the Lagrangian contains velocity-dependent terms, the canonical momentum corresponding to a coordinate of rotation $\theta$ of the entire system is no longer the mechanical angular momentum $L_{\theta}$ but is given by
$$p_{\theta}=L_{\theta}-\sum_{i}\mathbf{n}.\mathbf{r_{i}}\times\nabla_{v_{i}}U.$$

This is what I have so far:
I know that the position vectors $\mathbf{r_{i}}$ are functions of the $q_i$ generalized coordinates. On the other hand $U=U(\mathbf{r_{i}},\mathbf{\dot{r_{i}}})$

The canonical momentum with respect to a coordinate of rotation $\theta$ is given by:
$$p_{\theta}=\frac{\partial{L}}{\partial\dot{\theta}}=\frac{\partial T}{\partial{ \dot{\theta}}}-\sum_{i}\bigg(\frac{\partial u}{\partial{{r_{i}}}}\frac{\partial r_{i}}{\partial \dot{\theta}}+\frac{\partial U}{\partial \dot{r_{i}}}\frac{\partial \dot{r_{i}}}{\partial \dot{\theta}}\bigg)$$

Using:

$$\frac{\partial \dot{r_{i}}}{\partial \dot{\theta}}=\frac{\partial r_{i}}{\partial \theta}$$

We have:

$$p_{\theta}=\frac{\partial{L}}{\partial\dot{\theta}}=\frac{\partial T}{\partial{ \dot{\theta}}}-\sum_{i}\bigg(\frac{\partial u}{\partial{{r_{i}}}}\frac{\partial r_{i}}{\partial \dot{\theta}}+\frac{\partial U}{\partial \dot{r_{i}}}\frac{\partial r_{i}}{\partial \theta}\bigg)$$

I know that $$\frac{\partial U}{\partial \dot{r_{i}}}$$ can be rewritten as the scalar product of gradient of U and the unit velocity but I don't see how a cross product can be made to appear.

$\newcommand{\td}[0]{\dot{\theta}}$If you change $\td$ by an amount $\Delta \td$, then the velocity of the $i$th particle will change by an amount $\Delta \td \hat{n} \times \vec{r}_i$. The positions stay the same. Thus the resulting change in the lagragian will be $\partial_{\dot{\vec{r}}_i}L \cdot \Delta \td \hat{n} \times \vec{r}_i = \partial_{\dot{\vec{r}}_i}(T-U) \cdot \Delta \td \hat{n} \times \vec{r}_i$ Now if we distribute across $T-U$, the term with $T$ gives the mechanical angular momentum $L_\theta$. The term with $U$ gives $-\partial_{\dot{\vec{r}}_i}U \cdot \Delta \td \hat{n} \times \vec{r}_i = -\hat{n} \cdot \vec{r}_i \times \partial_{\dot{\vec{r}}_i}U \Delta \td$. Therefore the derivate of $L$ with respect to $\dot{\theta}$ is $L_\theta - \hat{n} \cdot \vec{r}_i \times \partial_{\dot{\vec{r}}_i}U$.