1From the review paper Bartelmann and schneider, 1999) [Weak Gravitational Lensing], page 48.

Like above, $\gamma_1 \equiv \left| \gamma \right| \cos(2\phi), \gamma_2 \equiv \left| \gamma \right|\sin(2\phi)$.

we have to definite like this to match the definition of $\gamma_1 \text{ and } \gamma_2$ with real angle, $\phi$.

This point is right what I want to know.

Why does the defining with $2\phi$ reflect "real angle"?


1 Answer 1


First, let's clarify that there is nothing inherently "real" about the angle $\phi$. In fact, the expression you cite is a definition of the angle $\phi$. You can write the shear $\gamma = |\gamma| \, e^{i\phi}$, of course, in which case $\phi$ is merely the argument of the complex number.

So let's talk then about why it makes sense to define the shear angle in this way. There are two ways to understand this choice.

Firstly, mathematically: If you rotate an observed image by an angle, say $\alpha$, the shear $\gamma'$ you measure from the rotated image is $\gamma' = \gamma \, e^{2i\alpha}$. In other words, shear has spin weight 2, which follows from the transformation of the Hessian matrix under a rotation. So by defining $\phi$ with the factor of 2 (i.e. spin weight), we get the simple behaviour $\phi \to \phi' = \phi + \alpha$ under rotations.

The second, related reason comes from looking at images. The image of a small circular source under shear looks like this:

Shear vs ellipticity (Figure credit: arXiv:astro-ph/0509252)

The x-axis is $\gamma_1$, the y-axis is $\gamma_2$. If you look e.g. at the outermost ring of figures, where $|\gamma| \approx 1$, you see that the ellipse with shape (i.e. orientation 0°) is made by $\gamma_1 = 1$, $\gamma_2 = 0$. Similarly, the ellipse with shape | (90°) is made by $\gamma_1 = -1$, $\gamma_2 = 0$; for shape / (45°) it's $\gamma_1 = 0$, $\gamma_2 = 1$; and for shape \ (135°) it's $\gamma_1 = 0$, $\gamma_2 = -1$. Therefore, by defining $\gamma = \gamma_1 + i \, \gamma_2 = |\gamma| \, e^{2i\phi}$, you make $\phi$ agree with the observed shape of a small circular source.

  • $\begingroup$ Oh.. I perfectly understood. Thank you for your awesome explanation. $\endgroup$
    – BAO
    Jan 26, 2022 at 8:11

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