May 13, 2021 - Lensing through deflections

Intro

After some hiatus we come back to the project with the hopes of lensing the CIB map "properly", by deflecting the sources and then getting the lensed CIB maps from these.

Jason says that the kappa maps are in

/scratch/r/rbond/jasonlee/cib_lensing2020/kappa_maps/total/

and his code is

/scratch/r/rbond/jasonlee/cib_lensing2020/gal_catalog/mapmaker/project_galaxies_by_redshift_0545.py

Conventions

We will be using the conventions from the paper "Lensed CMB simulation and parameter estimation" https://arxiv.org/pdf/astro-ph/0502469.pdf

There the lensed field [math]\displaystyle{ \tilde X(\hat n) }[/math] is equal to the unlensed field [math]\displaystyle{ X(\hat n') = X(\hat n + \nabla \psi) }[/math].

The relation connecting convergence and gravitational potential is

[math]\displaystyle{ \psi_{\ell m} = \frac{2\kappa_{\ell m}}{\ell(\ell+1)} }[/math]

but one must be careful to also know whether this is used to go from unlesed fields to the lensed ones, or vice versa. The implicit plus sign makes this consistent with

[math]\displaystyle{ \tilde X(\hat n) = X(\hat n + \nabla \psi) }[/math]

Simple analytic example

Let's play around with

[math]\displaystyle{ \kappa = A \cos \theta }[/math]

for some small positive A. This contains only [math]\displaystyle{ \ell = 1 }[/math] multipole, so using our convention from previous section we have

[math]\displaystyle{ \psi = A \cos \theta }[/math]

The vector gradient is

[math]\displaystyle{ \nabla \psi = \left(\frac{\partial \psi}{\partial \theta}, \frac{1}{\sin \theta}\frac{\partial \psi}{\partial \phi}\right) = \left(-A \sin \theta, 0\right) = A \sin \theta \left(\cos \pi, \sin \pi\right) }[/math]

which means [math]\displaystyle{ d = A \sin \theta }[/math]

[math]\displaystyle{ \alpha = \pi }[/math]

Using equations A15, A16 from the paper, this relates the unlensed coordinates [math]\displaystyle{ \theta', \phi' }[/math] to the lensed ones through

[math]\displaystyle{ \cos \theta' = \cos d \cos \theta + \sin d \cos \theta = \cos(\theta - d) \Rightarrow \theta' = \theta - d }[/math]

and

[math]\displaystyle{ \phi' = \phi }[/math].

Let's see if it makes sense. Around the north pole [math]\displaystyle{ \theta = 0 }[/math] we are sitting on maximum convergence. Things are magnified and lensing is pushing everything away from the pole. This is consistent with what we are finding, given for small [math]\displaystyle{ \theta }[/math]

[math]\displaystyle{ \theta' = \theta - A \sin \theta \approx \theta (1-A) }[/math]

[math]\displaystyle{ \theta = \theta'(1 + A) }[/math]

The lensed point is further away from the pole, as expected. It is also by a proper amount, because convergence at the pole is A, so image magnification should be 1 + 2A, which corresponds to each linear dimension to increase by a factor of (1+A).

The fact that we are moving along lines of constant [math]\displaystyle{ \phi }[/math] is also correct.

Bit more complicated analytic example

Let's play around with

[math]\displaystyle{ \kappa = \frac{A}{\sqrt{2}}\left( \cos \theta + \sin \theta \cos \phi \right) }[/math]

again for some small positive A. It turns out this is exactly the same field as before, only with the north pole rotated to [math]\displaystyle{ \theta = \pi/4, \phi = 0 }[/math].

Again this contains only [math]\displaystyle{ \ell = 1 }[/math] multipole, so using our convention from previous section we have

[math]\displaystyle{ \psi = \frac{A}{\sqrt{2}}\left( \cos \theta + \sin \theta \cos \phi \right) }[/math]

The vector gradient is

[math]\displaystyle{ \nabla \psi = \left(\frac{\partial \psi}{\partial \theta}, \frac{1}{\sin \theta}\frac{\partial \psi}{\partial \phi}\right) = \frac{A}{\sqrt{2}}\left(-\sin \theta + \cos \theta \cos \phi, \sin\phi\right) }[/math]

which means [math]\displaystyle{ d = \frac{A}{\sqrt{2}} \sqrt{(-\sin \theta + \cos \theta \cos \phi)^2+ \sin^2 \phi} }[/math]

and [math]\displaystyle{ \alpha }[/math] is given by the relative ratios of the two coordinates.

Notice that had we not included [math]\displaystyle{ 1/\sin \theta }[/math] when calculating the gradient, there would not exist a limit for [math]\displaystyle{ d }[/math] around the poles. Another sanity check - at [math]\displaystyle{ \theta = \pi/4, \phi = 0 }[/math] we see no deflection, at [math]\displaystyle{ \theta = 3\pi/4, \phi = 0 }[/math] we see maximal deflection.

Above we found that the displacement length was [math]\displaystyle{ A \sin \theta }[/math]. In the rotated coordinates it should mean that the displacement length (scalar) is for [math]\displaystyle{ A = 1 }[/math] equal to the sinus of the angle between the location vector and the point of maximal convergence. Cosine of this angle is

[math]\displaystyle{ c = \frac{1}{\sqrt{2}} \left(\cos \theta + \cos \phi \sin \theta\right) }[/math]

It is clear that

[math]\displaystyle{ c^2 + d^2/A^2 = 1 }[/math]

so we got the same result as before, only rotate. Good!

Finally, close to the point of maximal magnification, we have

[math]\displaystyle{ \nabla \psi = (-\Delta \theta, -\Delta \phi/\sqrt{2}) }[/math]

This is pointing towards the maximum of magnification, which is what we want: Recall [math]\displaystyle{ \nabla \psi }[/math] goes from the lensed point (further from the maximum) to the unlensed point (closer to the maximum).