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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.

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2 Answers 2

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$\newcommand{\td}[0]{\dot{\theta}}$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 $L_\theta - \hat{n} \cdot \vec{r}_i \times \partial_{\dot{\vec{r}}_i}U$.

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  • $\begingroup$ I'm editing it right now now, it is a minus sign. Searching a little bit i saw someone trying to solve this and made the following statement, that $$\frac{\partial r_{i}}{\partial \theta}$$ I'm having trouble understanding this but also might be useful for making a substitution on what I had. I like to solve a problem by various ways as a way to avoid getting bloicked on a test. $\endgroup$ Commented Oct 16, 2013 at 1:32
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The solution to this classical exercise from Goldstein's "Classical Mechanics" book was started correctly, and the relation with the cross product can be proved with the reasoning I will expose with details in what follows.

But before that, we should remember that the Cartesian coordinates $\mathbf{r}_{i}$ do not depend explicitly on the $\dot{q}_{j}$'s, but only on the $q_{j}$ and $t$. Therefore, $\partial \mathbf{r}_{i}/ \partial \dot{q}_{j}=0$ for all $i$'s and $j$'s.

So we may continue from the last equation written in the question body and write:

$$ p_{\theta} = \frac{\partial L}{\partial \dot{\theta}} = \frac {\partial T}{\partial \dot{\theta}} - \sum_{i} (\nabla_{v}U)\cdot\frac{\partial \mathbf{\dot{r}}_{i}}{\partial \dot{\theta}} $$

Observe that I used the gradient in the velocity space to group the three Cartesian coordinates of each particle $i$. Now we just write explicitly the kinetic equation term as

$$ T = \frac{1}{2} \sum_{i} m_{i} \mathbf{\dot{r}}_{i}^2 $$ and take the partial derivative of it with respect to $\dot{\theta}$, obtaining

$$ p_{\theta} = \sum_{i} m_{i} \mathbf{\dot{r}}_{i} \cdot \frac{\partial \mathbf{\dot{r}}_{i}}{\partial \dot{\theta}} - \sum_{i} (\nabla_{v}U)\cdot\frac{\partial \mathbf{\dot{r}}_{i}}{\partial \dot{\theta}} $$

Now we use the following relation, which may be straightforwardly verified, $$ \frac{\partial \mathbf{\dot{r}}_{i}}{\partial \dot{\theta}} = \frac{\partial \mathbf{r}_{i}}{\partial \theta} $$

arriving at the following expression $$ p_{\theta} = \sum_{i} m_{i} \mathbf{r}_{i} \cdot \frac{\partial \mathbf{r}_{i}}{\partial \theta} - \sum_{i} (\nabla_{v}U)\cdot\frac{\partial \mathbf{r}_{i}}{\partial \theta} $$

The $\theta$ coordinate is such that $d\theta$ corresponds to a rotation of the system, keeping the magnitude of the vector $\mathbf{r}_{i}$ constant. We may assume without loss of generality that the $z$ is the axis of rotation rotation, so that a rotated vector $\mathbf{r'} = (r'_{x}, r'_{y}, r'_{z})$ may be written as
$$ \mathbf{r'} = (r_{x}\cos\theta-r_{y}\sin\theta, r_x \sin\theta + r_{y} \cos \theta , r_{z}) $$

We may now easily derive the rotated vector as a function of $\theta$, and obtain $$ \frac{\partial \mathbf{r'}_{i}}{\partial \theta} = (-r_{x}\sin\theta-r_{y}\cos\theta, r_x \cos\theta - r_{y} \sin \theta , 0) $$ This expression can be easily identified with the following cross product $$ \frac{\partial \mathbf{r'}}{\partial \theta} = \mathbf{n} \times \mathbf{r'} \,\, , $$ where $\mathbf{n}$ is just the direction of rotation.

Making the substitution on the generalized momentum equation we obtain $$ p_{\theta} = \sum_{i} m_{i} \mathbf{r}_{i} \cdot \mathbf{n} \times \mathbf{r}_{i} - \sum_{i} (\nabla_{v}U)\cdot\mathbf{n} \times \mathbf{r}_{i} $$ Using the cyclic relation of the $\bf A \cdot B \times C$ product, we may write $$ p_{\theta} = \mathbf{n} \cdot \sum_{i} \mathbf{r}_{i} \times (m_{i} \mathbf{r}_{i}) - \sum_{i} \mathbf{n} \cdot \mathbf{r}_{i} \times (\nabla_{v}U) $$
which is the expression for the momentum $p_{\theta}$ that we wanted to prove.

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