Calculation step in deriving Hamiltonian from Lagrangian

Question

Suppose we have the following equation :

$$L=\frac{1}{2}\sum_{ij}\dot{q_i}\space T_{ij}\space\dot{q_j}$$

Here $T_{ij}$ is a symmetric matrix, that depends on $q_i,t$. Suppose, I want to evaluate $\frac{\partial L}{\partial\dot{q_j}}$. So now we have,

$$\frac{\partial L}{\partial\dot{q_k}}=\frac{1}{2}\frac{\partial }{\partial\dot{q_k}}\sum_{ij}\dot{q_i}\space T_{ij}\space\dot{q_j}$$

In one of the lecture notes, my professor wrote that this is equivalent to,

$$\sum_i T_{ik}\space\dot{q_k}$$

However, this doesn't seem correct to me. According to me,

$$\frac{\partial L}{\partial\dot{q_k}}=\frac{1}{2}\frac{\partial }{\partial\dot{q_k}}\sum_{ij}\dot{q_i}\space T_{ij}\space\dot{q_j}=\sum_i T_{ik}\space\dot{q_i}$$

This should be the correct equation ( note that I've replaced $\dot{q_k}$ by $\dot{q_i}$, which is what we are summing over. )

Suppose both $i,j$ range from $1$ to $3$. Then according to my teacher,

$$\frac{\partial L}{\partial \dot{q_3}}=T_{13}\dot{q_3}+T_{23}\dot{q_3}+T_{33}\dot{q_3}$$

According to me however,

$$\frac{\partial L}{\partial \dot{q_3}}=T_{13}\dot{q_1}+T_{23}\dot{q_2}+T_{33}\dot{q_3}$$

Can someone tell me who is correct ? Was there a simple printing error in my teacher's notes, or am I making an obvious mistake in here ?

Answer 1

We have,

$$ L = \frac12 \sum \dot{q}_i T_{ij} \dot{q}_j$$

Taking the derivative of both side with $p$th coordinate and interchanging derivative with sum:

$$ \partial_{\dot{q}_p} L = \frac12 \sum \partial_{\dot{q}_p} (\dot{q}_i T_{ij} \dot{q}_j)$$

Tricky term is $\partial_{\dot{q}_p} (\dot{q}_i T_{ij} \dot{q}_j)$, we will work that out seperately:

$$ \partial_{\dot{q}_p} (\dot{q}_i T_{ij} \dot{q}_j)= \delta_{ip} T_{ij} \dot{q}_j + \dot{q}_i T_{ij} \delta_{pj} $$

Now, let's put this back in the summation:

$$ \partial_{\dot{q}_p} L = \frac12 \sum_{i,j} \delta_{ip} T_{ij} \dot{q}_j + \dot{q}_i T_{ij} \delta_{pj} + \dot{q}_i \dot{q}_j\partial_{\dot{q}_p} T_{ij} = \frac12 \left[\sum_{i,j} \delta_{ip} T_{ij} \dot{q}_j + \sum_{i,j} \dot{q}_i T_{ij} \delta_{pj} \right]$$

Now suppose we consider the second summation $ \sum_{i,j} \dot{q}_i T_{ij} \delta_{pj}$, since $i,j$ are dummy and further more the matrix is symmetric we can swap $i <--> j$, this gives the same thing as first term giving us:

$$ \partial_{\dot{q}_p} L= \sum_{i,j} \dot{q}_i T_{ij} \delta_{pj}$$

At this point , we can use the property of the $\delta$ symbol. Since the it is only equal $1$ when $p=j$, we can replace all our $j$ with $p$:

$$ \partial_{\dot{q}_p} L= \sum_{i,p} \dot{q}_i T_{ip} $$

But, $p$ is just one number (derivative index, so we have:

$$ \partial_{\dot{q}_p} L= \sum_{i} \dot{q}_i T_{ip} = \dot{q}_1 T_{1p} + \dot{q}_2 T_{2p} + \dot{q}_3 T_{3p} $$

*:Note that I am abusing tensor notation a bit, but I am doing it so that it is smthn OP is familiar with. Many physics book don't get the right index placement so who cares anyways.

Answer 2

Just use Einstein summation to clear things up. $$L=\frac{1}{2}T_{ij}\dot{q_i}\dot{q_j}$$ $$\frac{\partial L}{\partial \dot{q_k}}=\frac{1}{2}T_{ij}\left[\dot{q_i}\frac{\partial \dot{q_j}}{\partial \dot{q_k}}+\dot{q_j}\frac{\partial \dot{q_i}}{\partial \dot{q_k}}\right]=\frac{1}{2}T_{ij}[\dot{q_i}\delta_{jk}+\dot{q_j}\delta_{ik}]=\frac{1}{2}T_{ij}\dot{q_i}\delta_{jk}+\frac{1}{2}T_{ij}\dot{q_j}\delta_{ik}=\frac{1}{2}\dot{q_i}T_{ij}\delta_{jk}+\frac{1}{2}\dot{q_j}T_{ij}\delta_{ik}=\frac{1}{2}\dot{q_i}T_{ik}+\frac{1}{2}\dot{q_j}T_{jk}$$ where we have used $T_{ij}=T_{ji}$ for evaluation of the second term in the sum. Since the summation is over $i,j$ these two are really the same quanity, hence we have $$\frac{\partial L}{\partial \dot{q_k}}=\dot{q_i}T_{ik}=\sum_i\dot{q_i}T_{ik}$$ So$$\frac{\partial L}{\partial \dot{q_3}}= \dot{q_1}T_{13}+\dot{q_2}T_{23}+\dot{q_3}T_{33}$$ It would seem that you are correct, not the professor.