• 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.


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


(propose a free ad)

Site Statistics

198 submissions , 156 unreviewed
4,911 questions , 2,087 unanswered
5,311 answers , 22,546 comments
1,470 users with positive rep
805 active unimported users
More ...

  Obtaining an angular distribution from the Klein-Nishina formula

+ 1 like - 0 dislike

I'm having trouble on how one might obtain an angular distribution from the Klein-Nishina formula for compton scattering of polarised photons, more specifically on how this report is obtaining it. (http://geant4.cern.ch/G4UsersDocuments/UsersGuides/PhysicsReferenceManual/html/node56.html)

The experiment concerns gamma ray compton polarimetry and detects correlations of two scattered back to back entangled photons by having two detectors perpendicular to the beam and rotating one about the azimuth, centered on the scatterers.

They obtain an equation for the scattered intensity relative to incident flux (equation 22) which I am fine with, however the transition from equation 22 to equation 44 and the end result, equation 45 seems ambiguous to me. Could anyone shed some light on how $\psi$ (Detector separation angle) is introduced here?

In short, the report states:

$I = \int_{\phi_{min}}^{\phi_{max}} \int_{\delta_{min}}^{\delta_{max}}\frac{d\sigma}{d\Omega} \,d\delta d\phi$

where $I$ is a beam intensity relative to the incident beam, $\delta$ is the polar scattering angle and $\phi$ is a polarisation angle about the beam axis.

the Klein-Nishina formula for compton scattering of polarised radiation is:

$\frac{d\sigma}{d\Omega} = \frac{1}{2} r_o^2 \frac{\frac{1}{2-\cos{\delta}} + 2 - \cos{\delta} - 2\sin^2{\delta} \cos^2{\phi}}{2 - \cos^2{\delta}}$

and then an equation is obtained in the form $I = a - bcos^2(\psi)$

where $\psi$ is the detector separation angle.

How has this distribution dependent on the detector separation angle $\psi$ resulted from the previous equations?

Sorry if this is obvious, as it seems mysterious to me. The relevant derivations and diagrams are present from appendix A, page 13 of the report.

Thanks for your time.

This post imported from StackExchange Physics at 2016-01-23 21:11 (UTC), posted by SE-user Bolly Vonning
asked Jan 23, 2016 in Experimental Physics by Bolly Vonning (5 points) [ no revision ]
retagged Jan 23, 2016

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:
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).
To avoid this verification in future, please log in or register.

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

Your rights