This is related to a question about a simple model of a polymer chain that I have asked yesterday. I have a Hamiltonian that is given as:

$H = \sum\limits_{i=1}^N \frac{p_{\alpha_i}^2}{2m} + \frac{1}{2}\sum\limits_{i=1}^{N-1} m \omega^2(\alpha_i - \alpha_{i+1})^2 $

where $\alpha_i$ are generalized coordinates and the $p_{\alpha_i}$ are the corresponding conjugate momenta. I want to find the equations of motion. From Hamilton's equations I get

$\frac{\partial H}{\partial p_{\alpha_i}} = \dot{\alpha_i} = \frac{p_{\alpha_i}}{m} \tag{1}$

$- \frac{\partial H}{\partial {\alpha_i}} = \dot{p_{\alpha_i}} = -m \omega^2 (\alpha_i - \alpha_{i+1} ) \tag{2}$

, for $i = 2,...,N-1$. Comparing this to my book, (1) is correct, but (2) is wrong. (2) should really be

$- \frac{\partial H}{\partial {\alpha_i}} = \dot{p_{\alpha_i}} = -m \omega^2 (2\alpha_i - \alpha_{i+1} - \alpha_{i-1}) \tag{$2_{correct}$}$

Clearly, I am doing something wrong. I suspect that I'm not chain-ruling correctly. But I also don't get, where the $\alpha_{i-1}$ is coming from. Can anybody clarify?

  • $\begingroup$ btw, how do I number equations in the latex environment of stackexchange? this looks kinda ugly. $\endgroup$
    – seb
    Commented Mar 23, 2013 at 10:16
  • $\begingroup$ You can use \tag{your_label} following the equation. $\endgroup$ Commented Mar 23, 2013 at 10:25

1 Answer 1


Instead of using the chain rule (although it of course gives the same answer) expand the square of the $i^\mathrm{th}$ term in the sum parentheses to obtain $$ \alpha_i^2 - 2\alpha_i\alpha_{i+1}+\alpha_{i+1}^2 $$ differentiating this with respect to $\alpha_i$ gives $$ 2\alpha_1 - 2\alpha_{i+1} $$ Now, from the $(i-1)^\mathrm{th}$ term $$ \alpha_{i-1}^2 - 2\alpha_{i-1}\alpha_i + \alpha_i^2 $$ you get an additional $$ -2\alpha_{i-1} + 2\alpha_i $$ when you take the $\alpha_i$ derivative. Putting these results together gives the answer in the book.

  • $\begingroup$ ok, so when I would write this entire sum out and then I differentiate w.r.t. $\alpha_i$, of course I have this contribution from the $(i-1)^{th}$ term! brilliant. thanks for claryfing. I totally missed that. $\endgroup$
    – seb
    Commented Mar 23, 2013 at 10:26
  • $\begingroup$ Sure thing sugar bean. $\endgroup$ Commented Mar 23, 2013 at 10:35
  • $\begingroup$ Yeah, the confusing part with the way you do it is that you use the index $ i$ for the summing and for differentiating. A better way to do it to write $ \frac{\partial}{\partial \alpha_{k}} $ so that you don't confuse the indices. Both indices are dummy variables. So that can get confusing. Then once you get that, you can relabel your $k's$ into $i's$. $\endgroup$
    – Ben S
    Commented Jan 20, 2017 at 17:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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