Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

  Nonlinear potential well (1D Class. Mech)

+ 1 like - 0 dislike
547 views

Consider the period $T$ of the bound states of a nonlinear potential well $U(x)$. So $x(t)$ is just bouncing back and forth between $x_0$ and $x_1$.

$T(x_0)\propto\int_{x_0}^{x_1}\frac{\mathrm{d}x}{\sqrt{E-U(x)}}$ where $E=U(x_0)=U(x_1)$.

How do we prove $T$ is differentiable in energy i.e. $\mathrm{d}T/\mathrm{d}x_0$ exists at least for bound states. Leibniz integral rule doesn't really apply here. Do I have to go into the business of action-angle? Thanks.

asked Jul 28, 2015 in General Physics by anonymous [ revision history ]
recategorized Jul 28, 2015

1 Answer

+ 2 like - 0 dislike

Write $f(x_0,E,\delta)$ for the integral, taken between $x_0+\delta$ and $x_1-\delta$, where $\delta>0$ is tiny. Then you may differentiate under the integral with respect to E to get $\frac{df}{dE}(x_0,E,\delta)$. Then you must show that the limit of the result remains well-defined when $\delta\to 0$ to get $\frac{df}{dE}(x_0,E,0)$. Now you may use the chain rule to calculate the derivative of $T(x_0)=f(x_0,U(x_0),0)$.

answered Jul 29, 2015 by Arnold Neumaier (15,787 points) [ revision history ]

So basically you treat $E$ as an independent variable in the first place, evaluate the integral, then plug in $E=U(x_0)$ and differentiate in $x_0$ via chain rule. Sounds brilliant. Thanks.

I was thinking of transforming the integration domain of any $T(x_0)$ to a fixed interval say $(0,\frac{1}{2})$ via change of variable. Then show that the derivative of the new integrand with respect to $x_0$ exists a.e and integrable.

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOver$\varnothing$low
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...