-1
$\begingroup$

I can't seem to understand what this question is asking. I am not sure really what the second and third terms actually represent in the equation.

Near the equilibrium position $r_0$, the interatomic potential energy curve can be written as $$U(x) = -U_0+\frac{1}{2!}\cdot(x-r_0)^2\cdot\left(\frac{d^2U}{dx^2}\right)_{r_0}+\frac{1}{3!}\cdot(x-r_0)^3\cdot\left(\frac{d^3U}{dx^3}\right)_{r_0}+...$$ How can we determine $\left(\frac{d^2U}{dx^2}\right)_{r_0}$ and $\left(\frac{d^3U}{dx^3}\right)_{r_0}$ experimentally? Briefly describe possible experiments, and what they can tell us about the behaviour of the potential energy curve near the equilibrium position.

$\endgroup$

closed as off-topic by Emilio Pisanty, peterh, Yashas, Kyle Kanos, Jon Custer Jun 8 '17 at 12:43

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Emilio Pisanty, peterh, Yashas, Kyle Kanos, Jon Custer
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ I don't understand the downvotes. Jason is not asking the solution to the problem, he simply didn't understand the question. It's not trivial if you haven't heard of Taylor series before. $\endgroup$ – Renan Nobuyuki Hirayama Jun 7 '17 at 20:16
  • $\begingroup$ @RenanNobuyukiHirayama On my first glance, it's not to clear what he's asking. He also has most of this in an uncropped screenshot of a PDF on his browser, so you have to zoom in to read it on some screens. The PDF part was fixed as I wrote this; but I assume those were the reasons. $\endgroup$ – JMac Jun 7 '17 at 20:19
  • $\begingroup$ In the future, Jason, please type out the text in images (use MathJax when necessary) - not only does it make it easier to read, it also helps search engines (and therefore people who may answer or read your question) find your question. $\endgroup$ – heather Jun 7 '17 at 20:20
1
$\begingroup$

Have you studied Taylor expansion yet? Basically, if you have a smooth (continuously differentiable) function $f:\mathbb{R}\rightarrow\mathbb{R}$, you can expand it in terms of its derivatives computed in a certain point $x_0$: $$ f(x)=f(x_0)+ f'(x_0)(x-x_0)+\frac{1}{2!}f''(x_0)(x-x_0)^2+\ ... $$ where the prime denotes a derivative.

Then in the question, since $x_0$ is an equilibrium point, you know $U'(r_0) = 0$. Therefore the potential energy can be expressed as $$ U(x)=-U_0 +\frac{1}{2!}U''(r_0)(x-r_0)^2+\frac{1}{3!}U'''(r_0)(x-r_0)^3 $$

where $U(r_0)\equiv-U_0$.

$\endgroup$

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