Expressing kinetic energy as sum of vector and tensor terms

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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!
 
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Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
 

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