Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Center of Image - Brightness Distribution

  1. Nov 1, 2017 #1
    Hello,

    I am reading a review on weak gravitational lensing (https://arxiv.org/pdf/astro-ph/0509252.pdf) and they define the center of an image as follows:

    $$\vec \theta_c = \frac{\int d^2 \theta I(\vec \theta) q_I [I(\vec \theta)] \vec \theta}{\int d^2 \theta I(\theta) q_I[I(\vec \theta)]}$$

    where ##I(\vec \theta)## is the brightness distribution of an image isolated in the sky and ##q_I(I)## is some weight function.

    I am having some trouble seeing that this does indeed define the center of an image and was hoping someone could help me see it.
     
  2. jcsd
  3. Nov 3, 2017 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    It looks like a standard weighted average formula, ##\bar x=\frac{\int f(x)x.dx}{\int f(x).dx}##, where f(x) is the weight function. Just the same as finding the centroid of a lamina.
    If your question is how they get that weighting function then I am hampered by not knowing how to interpret d2θ.
     
  4. Nov 3, 2017 #3
    Ah ha! Thank you, I was missing that.

    ##\vec \theta## is an angular position on the flat approximation of the sky, i've attached an image to illustrate it. A common example in the literature is to use the heaviside step function as the weight function, which I think just defines a sharp cutoff of the image along an isophote.
     

    Attached Files:

  5. Nov 3, 2017 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Ok, but can you shed any light on the d2θ? I would have understood dθ as just the usual integration notation. I suspect it has something to do with the fact that we really want to integrate over the annulus rdrdφ, where r is proportional to θ and φ runs from 0 to 2π, but that would give something like θdθ, not d2θ.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Center of Image - Brightness Distribution
  1. Really bright (Replies: 6)

Loading...