<p>First, let's clarify that there is nothing inherently &quot;real&quot; about the angle <span class="math-container">$\phi$</span>. In fact, the expression you cite is a <em>definition</em> of the angle <span class="math-container">$\phi$</span>. You can write the shear <span class="math-container">$\gamma = |\gamma| \, e^{i\phi}$</span>, of course, in which case <span class="math-container">$\phi$</span> is merely the argument of the complex number.</p>
<p>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.</p>
<p>Firstly, mathematically: If you rotate an observed image by an angle, say <span class="math-container">$\alpha$</span>, the shear <span class="math-container">$\gamma'$</span> you measure from the rotated image is <span class="math-container">$\gamma' = \gamma \, e^{2i\alpha}$</span>. In other words, shear has spin weight 2, which follows from the transformation of the Hessian matrix under a rotation. So by defining <span class="math-container">$\phi$</span> with the factor of 2 (i.e. spin weight), we get the simple behaviour <span class="math-container">$\phi \to \phi' = \phi + \alpha$</span> under rotations.</p>
<p>The second, related reason comes from looking at images. The image of a small circular source under shear looks like this:</p>
<p><a href="https://i.sstatic.net/CV2kj.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/CV2kj.png" alt="Shear vs ellipticity" /></a>
<em>(Figure credit: <a href="https://arxiv.org/abs/astro-ph/0509252" rel="nofollow noreferrer">arXiv:astro-ph/0509252</a>)</em></p>
<p>The x-axis is <span class="math-container">$\gamma_1$</span>, the y-axis is <span class="math-container">$\gamma_2$</span>. If you look e.g. at the outermost ring of figures, where <span class="math-container">$|\gamma| \approx 1$</span>, you see that the ellipse with shape <code>–</code> (i.e. orientation 0°) is made by <span class="math-container">$\gamma_1 = 1$</span>, <span class="math-container">$\gamma_2 = 0$</span>. Similarly, the ellipse with shape <code>|</code> (90°) is made by <span class="math-container">$\gamma_1 = -1$</span>, <span class="math-container">$\gamma_2 = 0$</span>; for shape <code>/</code> (45°) it's <span class="math-container">$\gamma_1 = 0$</span>, <span class="math-container">$\gamma_2 = 1$</span>; and for shape <code>\</code> (135°) it's <span class="math-container">$\gamma_1 = 0$</span>, <span class="math-container">$\gamma_2 = -1$</span>. Therefore, by defining <span class="math-container">$\gamma = \gamma_1 + i \, \gamma_2 = |\gamma| \, e^{2i\phi}$</span>, you make <span class="math-container">$\phi$</span> agree with the observed shape of a small circular source.</p>

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2022-01-26T08:11:46.037Z
Oh.. I perfectly understood. Thank you for your awesome explanation.
BAO
CC BY-SA 4.0
2022-01-25T23:19:17.833Z
2022-01-25T23:19:17.833Z
ntessore
<p><img src="https://i.sstatic.net/xLcBx.png" alt="1" />From the review paper Bartelmann and schneider, 1999) [Weak Gravitational Lensing], page 48.</p>
<p>Like above, <span class="math-container">$\gamma_1 \equiv \left| \gamma \right| \cos(2\phi), \gamma_2 \equiv \left| \gamma \right|\sin(2\phi)$</span>.</p>
<p>we have to definite like this to match the definition of <span class="math-container">$\gamma_1 \text{ and } \gamma_2$</span> with real angle, <span class="math-container">$\phi$</span>.</p>
<p>This point is right what I want to know.</p>
<p>Why does the defining with <span class="math-container">$2\phi$</span> reflect &quot;real angle&quot;?</p>

<astrophysics><gravitational-lensing><weak-lensing>
In shear component $\gamma_1, \gamma_2$ of weak lensing, Why does the defining with $2\phi$ reflect "real angle"?
BAO
# In shear component $\gamma_1, \gamma_2$ of weak lensing, Why does the defining with $2\phi$ reflect "real angle"?

(Asked by: username_0 on 2022-01-25)

![1](https://i.sstatic.net/xLcBx.png)From 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"?

**Tags:** astrophysics, gravitational-lensing, weak-lensing

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## 1 answers

### Answer by username_1 on 2022-01-25. Score: 2 | Accepted answer

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](https://i.sstatic.net/CV2kj.png)](https://i.sstatic.net/CV2kj.png)
*(Figure credit: [arXiv:astro-ph/0509252](https://arxiv.org/abs/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.

> **Comments:**
> **username_0**: Oh.. I perfectly understood. Thank you for your awesome explanation.  

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