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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?

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  • $\begingroup$ btw, how do I number equations in the latex environment of stackexchange? this looks kinda ugly. $\endgroup$ – seb Mar 23 '13 at 10:16
  • $\begingroup$ You can use \tag{your_label} following the equation. $\endgroup$ – joshphysics Mar 23 '13 at 10:25
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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.

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  • $\begingroup$ 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 Mar 23 '13 at 10:26
  • $\begingroup$ Sure thing sugar bean. $\endgroup$ – joshphysics Mar 23 '13 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 Jan 20 '17 at 17:06

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