Classical Mechanics – Differentiation with Index in Euler-Lagrange Equation

classical-mechanicseuler-lagrange-equationindex-notation

When differentiating some additive term contained within the Lagrangian of a system, I am having trouble with feeling comfortable with my answer. Being new to index notation might be the issue .

So
$m_{ij},b_{ij}$ are both constants (in the question we assume $m_{ij}$ is symmetric)

$\ L= \frac{1}{2}m_{ij}\dot q^i\dot q^j +b_{ij}\dot q^iq^j $

So I want to calculate $\frac{\partial L}{\partial q^i}$ , so I can calculate the EOM

(I am adding extra steps that I don't usually write out to demonstrate my thought process , just in case its incorrect and I have miss understood certain concepts)

  1. I differentiate with respect to $q^k$ as I was of the understanding that differentiating with respect to an index that is a free index doesn't work (i.e is not defined ??? $\frac{\partial b_{ij}}{\partial q^i}$ )
    \begin{equation}
    \frac{\partial L}{\partial q^i} = \ 0 + \frac{\partial b_{ij}}{\partial q^k} {\dot q^iq^j} +b_{ij}\frac{\partial\dot q^i}{\partial q^k}{q^j} +b_{ij}\dot q^i\frac{\partial q^j}{\partial q^k}
    \end{equation}

first two additive terms (is that the correct verbiage??) are zero. I am left with:

( assumed $\frac{\partial q^j}{\partial q^i}=\frac{\partial q^i}{\partial q^j}=\delta _{ij} =\delta _{ji}$ )
\begin{equation}
\frac{\partial L}{\partial q^i} = b_{ij}\dot q^i\delta _{jk}
\end{equation}
so k is the free index and i,j are dummy variables (using Einstein summation convention , we can summed over repeated indices)

\begin{equation}
\frac{\partial L}{\partial q^i} = b_{ik}\dot q^i
\end{equation}
so j->k (the k delta goes to 1)

Now where I am stuck is determining which answer is correct and why or are these answer the same thing . My thoughts is there is nothing that would lead me to assume $b_{ij}$ is symmetric . The question says $m_{ij}$ and $b_{ij}$ are constant and only $m{ij}$ is symmetric. I would lean towards the first answer as i->k since I am differentiating with respect to $q^i$??? otherwise I do not know .
\begin{equation}
\frac{\partial L}{\partial q^i} = \ b_{ji} \dot q^j{} or \frac{\partial L}{\partial q^i} = \ b_{ij} \dot q^i{}
\end{equation}

Thank you

Best Answer

I think the initial derivative index might be the cause of some of your worry, I would rewrite

$$\frac{\partial L}{\partial q^k} = b_{ij}\dot{q}^i\delta_{jk} = b_{ik}\dot{q}^i$$

Also note that the repeated indices are dummy indices since they are summed over using Einstein's summation convention, and can be relabeled as you wish.

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