Expressing kinetic energy as sum of vector and tensor terms

[CAF123]

Homework Statement

A system of N particles described by the vector coordinates $\mathbf{r}_k, k = 1,2, \dots, N$ subject to 3N - f constraints can be expressed in terms of generalised coordinates $q_i, i=1,2, \dots, f$ by $\mathbf{r}_k = \mathbf{r}_k(q_1, q_2, \dots, q_f, t)$

a) Prove the 'cancellation of dots' rule
b) Show that the kinetic energy of the system can be written as $$T = \frac{1}{2}\sum_{k=1}^{N} m_k \mathbf{r}_k^2 = M_o + \sum_{i=1}^f M_i \dot{q_i} + \frac{1}{2}\sum_{i=1}^f \sum_{j=1}^f M_{ij} \dot{q_i} \dot{q_j} $$ expressing $M_o, M_{i}, M_{ij}$ in terms of $\mathbf{r}_k$ and $t$.

Homework Equations

[/B]
$$\mathbf{\dot{r}}_k = \sum_{i=1}^f \frac{\partial \mathbf{r}_k}{\partial q_i} \dot{q_i} + \frac{\partial \mathbf{r}_k}{\partial t}$$

The Attempt at a Solution

a) and b) are both fine I think. I put my work for b) below anyway to show what I have.

The term is $$\frac{1}{2} \sum_{k=1}^N m_k \left( \frac{\partial \mathbf{r}_k}{\partial q_i} \dot{q_i} + \frac{\partial \mathbf{r}_k}{\partial t}\right)^2,$$ using the Einstein summation convention for the $i$ sum. Then I can write this like $$\frac{1}{2} \sum_{k=1}^k m_k \left( \frac{\partial \mathbf{r}_k}{\partial q_i} \frac{\partial \mathbf{r}_k}{\partial q_j} \dot{q_i}\dot{q_j} + \left( \frac{\partial \mathbf{r}_k}{\partial t}\right)^2 + 2 \frac{\partial \mathbf{r}_k}{\partial q_i} \dot{q_i} \frac{\partial \mathbf{r}_k}{\partial t}\right)$$ from which I can read $$M_o = \frac{1}{2} \sum_{k=1}^N m_k \left(\frac{\partial \mathbf{r}_k}{\partial t}\right)^2 \,\,\,\,\, M_i = \sum_{k=1}^N m_k \frac{\partial \mathbf{r}_k}{\partial q_i} \frac{\partial \mathbf{r}_k}{\partial t} \,\,\,\,\,\, M_{ij} = \sum_{k=1}^N m_k \frac{\partial \mathbf{r}_k}{\partial q_i} \frac{\partial \mathbf{r}_k}{\partial q_j} $$

I have edited my previous version of the OP in which I asked for some guidance on how to progress. Really I am confident in the above result now, so if anyone could clarify that would be great, otherwise if the moderators wish to delete this then that would also be fine. Thanks!

 

[mark726]

The term is $$\frac{1}{2} \sum_{k=1}^N m_k \left( \frac{\partial \mathbf{r}_k}{\partial q_i} \dot{q_i} + \frac{\partial \mathbf{r}_k}{\partial t}\right)^2,$$ using the Einstein summation convention for the $i$ sum. Then I can write this like $$\frac{1}{2} \sum_{k=1}^k m_k \left( \frac{\partial \mathbf{r}_k}{\partial q_i} \frac{\partial \mathbf{r}_k}{\partial q_j} \dot{q_i}\dot{q_j} + \left( \frac{\partial \mathbf{r}_k}{\partial t}\right)^2 + 2 \frac{\partial \mathbf{r}_k}{\partial q_i} \dot{q_i} \frac{\partial \mathbf{r}_k}{\partial t}\right)$$ from which I can read $$M_o = \frac{1}{2} \sum_{k=1}^N m_k \left(\frac{\partial \mathbf{r}_k}{\partial t}\right)^2 \,\,\,\,\, M_i = \sum_{k=1}^N m_k \frac{\partial \mathbf{r}_k}{\partial q_i} \frac{\partial \mathbf{r}_k}{\partial t} \,\,\,\,\,\, M_{ij} = \sum_{k=1}^N m_k \frac{\partial \mathbf{r}_k}{\partial q_i} \frac{\partial \mathbf{r}_k}{\partial q_j} $$

Your result for $M_o$ is correct. However, your results for $M_i$ and $M_{ij}$ are incorrect. The correct expressions are: $$M_i = \sum_{k=1}^N m_k \frac{\partial \mathbf{r}_k}{\partial q_i} \cdot \frac{\partial \mathbf{r}_k}{\partial t}$$ and $$M_{ij} = \sum_{k=1}^N m_k \frac{\partial \mathbf{r}_k}{\