0
$\begingroup$

enter image description here

The question above includes a Lagrangian whose equation of motion is required. The equation of motion can be worked out by $$\frac{d}{dt}\frac{\partial L}{\partial \dot{x_n}} - \frac{\partial L}{\partial x_n} = 0;$$ however, I do not get the solution shown above, especially the part where an extra $x_{n-1}$ term comes in from.

$\endgroup$
1

1 Answer 1

1
$\begingroup$

Did you perchance write something like $$\frac{\partial L}{\partial x_n} = \frac{\partial}{\partial x_n}\left(-\frac{1}{2}\mu x_n^2-\frac{1}{2}\nu(x_{n+1}-x_n)^2\right) =-\mu x_n + \nu(x_{n+1}-x_n)$$ in your calculations? The first equality here is incorrect. Try using a different name than $n$ for the index, so that you don't confuse it with the summation index, and really think about the terms in the sum. I.e. $$\frac{\partial L}{\partial x_k} = \frac{\partial}{\partial x_k} \left(\sum_{n=0}^{N-1}\dots\right),$$ and maybe think it through for a specific index first, say $k=1$.

Note also that there is a small mistake in the given answer. It is missing a factor of $m$ on the left hand side and should actually read $$m\ddot x_n = -\mu x_n + \nu(x_{n+1} + x_{n-1} - 2x_n).$$

$\endgroup$
6
  • $\begingroup$ Yes, I performed the first equality. If I were to take your advice, then I suppose I would be have to perform a chain rule on the term in hand but I still do not understand where the change in index from $n$ to $n-1$ comes from. $\endgroup$ Commented May 9, 2021 at 0:18
  • $\begingroup$ @GitGoodCodes Hmm, I do not know what you mean about having to use the chain rule. Try this: write out the first two terms of the sum in $L$ explicitly. Now, how many $x_1$-terms do you see? $\endgroup$
    – ummg
    Commented May 9, 2021 at 0:22
  • $\begingroup$ I see! So the summation for n = 1 includes the $x_{n-1}$ term as well as an additional $x_n$ term, is that why? $\endgroup$ Commented May 9, 2021 at 0:35
  • 1
    $\begingroup$ @GitGoodCodes Well, not quite, but I think you've got the right idea. Say we are taking the derivative $\partial L/\partial x_k$, so we are looking for terms in $L$ containing $x_k$. Then, of course, we note that the $n=k$ term in the sum contains both $-(1/2)\mu x_k^2$ and $-(1/2)\nu(x_{k+1}-x_k)$. These are the terms you found and included in your calculations. But, additionally, the $n=k-1$ term contains $-(1/2)\nu(x_{(k-1)+1}-x_{k-1})=-(1/2)\nu(x_k-x_{k-1})$, and this is the one you missed! $\endgroup$
    – ummg
    Commented May 9, 2021 at 0:44
  • $\begingroup$ Ahh I see, your previous hints lead me close to the correct logic. This makes a lot of sense. Many thanks! $\endgroup$ Commented May 9, 2021 at 0:47

Not the answer you're looking for? Browse other questions tagged or ask your own question.