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
5,082 questions , 2,232 unanswered
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

  Integrating over level sets of a function

+ 4 like - 0 dislike
1129 views

I have a background in math, but I think the following construction appears often in physics.

Is there a term for a functional which maps one smooth function $f: R^n \rightarrow R$ to another $L(f):R \rightarrow R$ which represents the integral along each level set? That is, $L(f)(a) = \int_{f^{-1}(a)} d\mu $ where the measure $d\mu$ might be the induced measure from the metric on $R^n$, or it might also depend on the original $f$ in some intelligent way.

For instance, if the initial function is $f:R^2 \rightarrow R$ is defined $f(x)=x^2+y^2$ then $L(f)(a) = 2 \pi \sqrt{a}$.

If the original function is $f:R \rightarrow R$ is defined $f(x)=x^2$, then we'd have $L(f)(a)=0$ for $a<0$, $1$ for $a=0$, and $2$ for $a>0$. In both cases it might be more natural to integrate with a term derived from the value of $f'$ normal to the level set.

I suppose one could generalize to any smooth map $f:M \rightarrow M'$ from a Riemannian manifold to a another smooth manifold and get a new function that represents a computed volume of the preimage for each point: $L(f):M' \rightarrow R$.

My question: Is there a good way to compute $L(f)$ for some class of function $f$?

Any thoughts would be appreciated. It is reminiscent of (but I hope simpler than) stationary phase approximations and path integral calculations.

http://en.wikipedia.org/wiki/Stationary_phase_approximation

This post imported from StackExchange Mathematics at 2014-06-02 20:26 (UCT), posted by SE-user Obidiah
asked Dec 28, 2010 in Mathematics by Obidiah (20 points) [ no revision ]

2 Answers

+ 2 like - 0 dislike

If I remember correctly, what you are describing is close to the concept of a distribution function (it is the name that I am slightly unsure about).

Given a measure space $(X,\mathcal{E},\mu)$, and a measurable function $f:X\to \mathbb{R}$, a concept often used in classical and harmonic analysis is the notion of the distribution function $\lambda_f:\mathbb{R}_+\to\mathbb{R}$ defined as

$$ \lambda_f(a) = \mu\left( \{|f| < a\} \right) $$

that is, $\lambda_f$ is returns the measure of the set on which $|f|$ is bounded by $a$. One can reformulate the $L^p$ norms of $f$ in terms of the distribution function

$$ \|f\|_{L^p(X,\mu)} = p \int_0^\infty t^{p-1}\lambda_f(t) dt $$

a device that is very useful for analytical estimates. Now, formally speaking the functional $L(f)$ you wrote down is very similar to the derivative $\lambda_f'$.


Another possibility that you may want to look into is the notion of pull-back distributions. If $f:M\to\mathbb{R}$ is a smooth map with non-vanishing gradient, then given any distribution (generalized function) $k$ on $\mathbb{R}$ you can define the pull-back distribution $f_*k$ on $M$. The definition of $f_*k$ depends closely on the notion of $L(f)$ you wrote down (see e.g. the introduction to distributions by Friedlander and Joshi), and is a generalization of the co-area formula.


I am not exactly sure how to answer your second question of "is there a good way to compute". If you are asking whether given an analytic expression for the function $f$ whether you can have a "closed form" expression for $L(f)$, the answer is surely no, since the expression also depends on the chosen Riemannian metric/volume form/measure on $M$. If you are asking whether there are any convenient expressions to capture $L(f)$, I think you need to say a bit more about why you care about $L(f)$. You can certainly just write $L(f) = f_*\delta_a$ as the pull-back of the Dirac delta distribution on $\mathbb{R}$; whether that expression is a useful one for your purpose, I don't know.

This post imported from StackExchange Mathematics at 2014-06-02 20:26 (UCT), posted by SE-user Willie Wong
answered Dec 30, 2010 by Willie Wong (580 points) [ no revision ]
other way: $\lambda_{f}(a)=\mu(x:|f|>a)$ (Folland)

This post imported from StackExchange Mathematics at 2014-06-02 20:26 (UCT), posted by SE-user TKM
+ 1 like - 0 dislike

Using the Fourier transform representation of the Dirac measure on the level set, the volume of the level set (which is the required integral according to the examples given in the question) is given by (using the same notation as in the question):

$L(f)(a) = \frac{1}{2\pi} \int_{-\infty}^\infty \mathrm{d}k \,exp(-ika)\, \int\,|\nabla f|\,exp(ikf)\,\mathrm{d}\mu$

On a Riemannian manifold, the gradient is given by:

$\nabla f = \sum_{ij}g^{ij}\frac{\partial f}{\partial x^i}\frac{\partial}{\partial x^j}$

The two examples given in the question can be worked out according to this formula (the second example very easily, the first one involves integrals of trigonometric functions of a squared angle, but it can also be solved analytically).

This post imported from StackExchange Mathematics at 2014-06-02 20:26 (UCT), posted by SE-user David Bar Moshe
answered Dec 28, 2010 by David Bar Moshe (4,355 points) [ no revision ]

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$\hbar$ysics$\varnothing$verflow
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
...