pts.simulation.healpix Namespace Reference

Handling of HEALPix grids. More...

Classes
class	HEALPixGrid
	This class represents a HEALPix grid. More...

Functions
def	getDistanceOnSky (thetaA, phiA, thetaB, phiB)
	Get the shortest distance (in radians) between two points (A, B) on the celestial sphere with the given zenith and azimuth. More...
def	getEmptyHEALPixGrid (order, numValuesPerPixel=1)
	Construct a new, empty HEALPixGrid of the given order and with the given number of values per pixel. More...

Detailed Description

Handling of HEALPix grids.

This module offers functionality to work with HEALPix grids output by the HEALPixSkyInstrument.

Function Documentation

getDistanceOnSky()

def pts.simulation.healpix.getDistanceOnSky	(		thetaA,
			phiA,
			thetaB,
			phiB
	)

Get the shortest distance (in radians) between two points (A, B) on the celestial sphere with the given zenith and azimuth.

The distance can easily be derived by realizing that the angle between the unit vectors pointing to A and B is given by

$$\delta =\arccos {\overrightarrow{r}}_A\dot{}{\overrightarrow{r}}_B.$$

In spherical coordinates, the dot product of the two unit vectors with sky coordinates $(\theta {}_A,\varphi {}_A)$ and $(\theta {}_B,\varphi {}_B)$ yields

$${\overrightarrow{r}}_A\dot{}{\overrightarrow{r}}_B=\sin \theta {}_A\sin \theta {}_B(\cos \varphi {}_A\cos \varphi {}_B+\sin \varphi {}_A\sin \varphi {}_B)+\cos \theta {}_A\cos \theta {}_B=\sin \theta {}_A\sin \theta {}_B\cos (\varphi {}_A-\varphi {}_B)+\cos \theta {}_A\cos \theta {}_B=\frac{1}{2}[\cos (\theta {}_A-\theta {}_B)-\cos (\theta {}_A+\theta {}_B)]\cos (\varphi {}_A-\varphi {}_B)+\frac{1}{2}[\cos (\theta {}_A-\theta {}_B)+\cos (\theta {}_A+\theta {}_B)].$$

getEmptyHEALPixGrid()

def pts.simulation.healpix.getEmptyHEALPixGrid	(		order,
			numValuesPerPixel = 1
	)

Construct a new, empty HEALPixGrid of the given order and with the given number of values per pixel.

The order $k$ relates to the HEALPix parameter $N_{side}=2^k$, which gives the number of pixels along one axis of the 12 HEALPix base pixels. The total number of pixels is then given by $12N_{side}^2$. Each pixel has the exact same size, $\frac{\pi }{3N_{side}^2}$. Assuming square pixels (which is approximately true), the angular resolution $R$ can be computed from

$$R\approx \sqrt{\frac{\pi }{3N_{side}^2}}=\frac{1}{N_{side}}{\left(\frac{10800}{\sqrt{3\pi }}\right)}^{\prime }\approx {\left(\frac{3520}{N_{side}}\right)}^{\prime }$$

For example, most Planck maps are published with an order $k=10$, which corresponds to a resolution of $\approx {3.4}^{\prime }$ (nominally $5^{\prime }$). $k=6$ would correspond to a resolution of $\approx 1^{\circ }$.