<p>I think the problem here may be that this is a 2D situation.</p>
<p>You say you used
<span class="math-container">$$ \nabla^2\frac{\vec\theta-\vec\theta'}{\mid\vec\theta-\vec\theta'\mid^2}=4\pi\delta^3(\vec{r})$$</span>
(shouldn't there be a minus sign if <span class="math-container">$\vec{r} = \vec\theta-\vec\theta'$</span> ?)
but this is a relationship true only in 3D geometry and <span class="math-container">$\delta^3$</span> is the 3D delta function, while <span class="math-container">$\vec{\theta}$</span> is defined as a &quot;two dimensional vector, which could be an angular position on the sky&quot; in the review paper.</p>
<p>The equivalent in 2D geometry is
<span class="math-container">$$ \nabla^2 \ln \mid\vec\theta-\vec\theta'\mid = 2\pi\delta^2(\vec\theta-\vec\theta')\ ,$$</span>
where <span class="math-container">$\delta^2$</span> is the 2D delta function.</p>
<p>From there, we see
<span class="math-container">\begin{eqnarray}
\nabla^2 \psi(\vec\theta) &amp; = &amp; \frac{1}{\pi}  \int_{\mathbb{R}^2} d^2\theta'\ \kappa(\vec\theta')\nabla^2 \ln \mid\vec\theta-\vec\theta'\mid \\
 &amp; = &amp; \frac{1}{\pi}  \int_{\mathbb{R}^2} d^2\theta'\ \kappa(\vec\theta') 2\pi\delta^2(\vec\theta-\vec\theta') \\
 &amp; = &amp; 2\kappa(\vec\theta)\ .
\end{eqnarray}</span></p>
<p>As for why your solution is incorrect, I suspect it is the second line.
I don't think the definition of divergence you have used is correct. Why would you equate it with the gradient of the vector? That is not generally true.
I am sure the&quot;proof&quot; is as I have presented it.</p>

CC BY-SA 4.0
2022-01-23T15:03:51.503Z
There should be a minus sign, sorry.
I understood most of them. If you answer a few questions, I think my question will be solved.

First, How do you derive the two-dimensional delta function? I derived the three dimensional delta function from 'Coulomb's law'.
(I saw many site related with this like "https://math.stackexchange.com/questions/1932451/how-to-prove-this-relation-between-the-laplacian-of-the-logarithm-and-the-dirac", but I cannot understand)
BAO
CC BY-SA 4.0
2022-01-23T15:16:57.133Z
Second, If even $\vec\theta$ means <$\theta_1$, $\theta_2$>, Cartesian Coordinate, is it right adding the factor '1/'? I understood the $\vec\theta$ is not theta in polar coordinate.
BAO
CC BY-SA 4.0
2022-01-23T18:24:26.977Z
How do you derive the 2D Delta function is a separate question (for Maths SE). Second question. I have no idea, you put no definitions of coordinates or symbols in your question, so I am guessing that the definition of divergence is incorrect, or your switch from a 2D to a 3D integral. My derivation is correct. @BAO
ProfRob
CC BY-SA 4.0
2022-01-24T00:38:07.130Z
Thank you. I fully understood, so I edit in my question with what I realized. Could I ask one more question? Why did you say "$\vec\nabla\cdot\vec\alpha(\vec\theta)=\frac{1}{\pi}\int_{\mathbb{R}^2} \kappa(\vec\theta')\frac{1}{\mid\vec\theta-\vec\theta'\mid} \nabla \frac{\vec\theta-\vec\theta'}{\mid\vec\theta-\vec\theta'\mid^2}\ d\theta'^2$"? is the two dimensional divergence's form like $\frac{1}{\mid\vec\theta-\vec\theta'\mid}\nabla$, not just $ \nabla$ ?
BAO
CC BY-SA 4.0
2022-01-24T08:12:35.613Z
@BAO I am not sure it is correct (other than it seems to lead to the correct solution) and have deleted that part of my answer. However, I don't see why your original version is correct at all. Divergence is not a gradient.
ProfRob
CC BY-SA 4.0
2022-01-24T10:36:19.873Z
Okay, your answer is really helpful!
BAO
CC BY-SA 4.0
2022-01-23T10:58:22.893Z
2022-01-23T19:31:23.957Z
2022-01-23T19:31:23.957Z
ProfRob
<p><a href="https://i.sstatic.net/QU08R.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/QU08R.png" alt="enter image description here" /></a></p>
<p>From the review paper Bartelmann and schneider, 1999) <a href="https://arxiv.org/abs/astro-ph/9912508" rel="nofollow noreferrer">Weak Gravitational Lensing</a>, page 48.</p>
<hr />
<p>Like above, the Laplacian of the deflection function is convergence, <span class="math-container">$\kappa(\vec{\theta})$</span>.</p>
<p>I tried to derive this below, but I obtain 4<span class="math-container">$\kappa(\vec{\theta})$</span>, not 2<span class="math-container">$\kappa(\vec{\theta})$</span>.</p>
<p>Why am I wrong?</p>
<p>--Solution process--</p>
<p>Using <span class="math-container">$ \nabla^2\frac{\vec\theta-\vec\theta'}{\mid\vec\theta-\vec\theta'\mid^2}=4\pi\delta(\vec{r})$</span>,</p>
<p><span class="math-container">$\nabla^2\psi(\vec\theta)=\vec\nabla\cdot\vec\alpha(\vec\theta)=\frac{1}{\pi}\int_{\mathbb{R}^2}d^2\theta'\kappa(\vec\theta')\vec\nabla\frac{\vec\theta-\vec\theta'}{\mid\vec\theta-\vec\theta'\mid^2}=\frac{1}{\pi}\int_V[\kappa(\vec\theta')\nabla^2\frac{\vec\theta-\vec\theta'}{\mid\vec\theta-\vec\theta'\mid^2}]dV$</span></p>
<p><span class="math-container">$=\frac{1}{\pi}\int_V[\kappa(\vec\theta')4\pi\delta(\vec\theta-\vec\theta')]dV=\frac{1}{\pi}\kappa(\vec\theta)4\pi=4\kappa(\vec\theta)$</span></p>
<p>(<span class="math-container">$\nabla^2 = \frac{\partial^2}{\partial^2 x} +\frac{\partial^2}{\partial^2 y}+\frac{\partial^2}{\partial^2 z} $</span> , because it's Cartesian coordinate).</p>
<hr />
<p>I realized something, so I write what I realized.</p>
<p>It seems that applying the divergence theorem was wrong.</p>
<p>Because <span class="math-container">$\mathbb{R}^2$</span> is open surface, not closed surface(because it is projected image on celestial sphere).</p>

CC BY-SA 4.0
2022-01-21T22:19:37.027Z
Might this be better suited to [Mathematics Stack Exchange](https://math.stackexchange.com)?
GrapefruitIsAwesome
CC BY-SA 4.0
2022-01-23T01:51:42.270Z
Yes, I will do. Thank you.
BAO
CC BY-SA 4.0
2022-01-23T06:20:51.813Z
@GrapefruitIsAwesome astrophysicists know about and use math as well
uhoh
CC BY-SA 4.0
2022-01-23T06:23:43.630Z
@BAO I've added a bounty, but "Why is the laplacian of the deflection function convergence?" is not exactly clear, and in the body of the question you only ask "but I obtain 4$\kappa(\vec{\theta})$, not 2$\kappa(\vec{\theta})$.

Why am I wrong?" Do you think you can restate your question more clearly so it's 100% clear exactly what you are expecting as an answer? *Thanks!*
uhoh
CC BY-SA 4.0
2022-01-23T09:49:10.517Z
@uhoh Thank you for the comment. I will do it.
BAO
CC BY-SA 4.0
2022-01-23T18:32:02.230Z
Please update your question giving a full definition of the symbols you have used and the coordinate system.
ProfRob
CC BY-SA 4.0
2022-01-24T00:33:38.447Z
Yes I did it. Thank you.
BAO
CC BY-SA 4.0
2022-01-26T00:35:44.977Z
There is a connection to the divergence, if you will. The gradient of $\ln(|x|)$ is the outward normal. By the divergence theorem, the Laplacian (i.e. div grad) of $\ln(|x|)$ in $n$ dimensions is hence the surface of the $n$-sphere times $\delta(0)$. This tells you that it must be $2\pi$ instead of $4\pi$ in your identity for 2D.
ntessore
CC BY-SA 4.0
2022-01-26T07:57:31.150Z
@ntessore Okay, I will think more about your comment. Thank you! And.. Can I ask some trivial question? In Stack exchange, Shouldn't I write "thank you"?
BAO
CC BY-SA 4.0
2022-01-26T08:23:29.773Z
@BAO Sorry, I wasn't careful when writing the above. It's not the gradient of $\ln(|x|)$ in $n$ dimensions, it's a function $f(x)$ such that $\nabla f(x) = x/|x|^n$. In two dimensions, that's $\ln(|x|)$, while in higher dimensions, it's $|x|^{2-n}/(2-n)$. Then you can show the Green's function property with the divergence theorem.
ntessore
<astrophysics><mathematics><gravitational-lensing><weak-lensing>
Finding the Laplacian of the deflection potential: I obtain 4(⃗ ), not 2(⃗ ). Why am I wrong?
BAO
# Finding the Laplacian of the deflection potential: I obtain 4(⃗ ), not 2(⃗ ). Why am I wrong?

(Asked by: username_0 on 2022-01-20)

[![enter image description here](https://i.sstatic.net/QU08R.png)](https://i.sstatic.net/QU08R.png)

From the review paper Bartelmann and schneider, 1999) [Weak Gravitational Lensing](https://arxiv.org/abs/astro-ph/9912508), page 48.

---

Like above, the Laplacian of the deflection function is convergence, $\kappa(\vec{\theta})$.

I tried to derive this below, but I obtain 4$\kappa(\vec{\theta})$, not 2$\kappa(\vec{\theta})$.

Why am I wrong?

--Solution process--

Using $ \nabla^2\frac{\vec\theta-\vec\theta'}{\mid\vec\theta-\vec\theta'\mid^2}=4\pi\delta(\vec{r})$,

$\nabla^2\psi(\vec\theta)=\vec\nabla\cdot\vec\alpha(\vec\theta)=\frac{1}{\pi}\int\_{\mathbb{R}^2}d^2\theta'\kappa(\vec\theta')\vec\nabla\frac{\vec\theta-\vec\theta'}{\mid\vec\theta-\vec\theta'\mid^2}=\frac{1}{\pi}\int\_V[\kappa(\vec\theta')\nabla^2\frac{\vec\theta-\vec\theta'}{\mid\vec\theta-\vec\theta'\mid^2}]dV$

$=\frac{1}{\pi}\int\_V[\kappa(\vec\theta')4\pi\delta(\vec\theta-\vec\theta')]dV=\frac{1}{\pi}\kappa(\vec\theta)4\pi=4\kappa(\vec\theta)$

($\nabla^2 = \frac{\partial^2}{\partial^2 x} +\frac{\partial^2}{\partial^2 y}+\frac{\partial^2}{\partial^2 z} $ , because it's Cartesian coordinate).

---

I realized something, so I write what I realized.

It seems that applying the divergence theorem was wrong.

Because $\mathbb{R}^2$ is open surface, not closed surface(because it is projected image on celestial sphere).

**Tags:** astrophysics, mathematics, gravitational-lensing, weak-lensing
> **Comments:**
> **username_1**: Might this be better suited to [Mathematics Stack Exchange](https://math.stackexchange.com)?  
> **username_0**: Yes, I will do. Thank you.  
> **username_2**: @GrapefruitIsAwesome astrophysicists know about and use math as well  
> **username_2**: @BAO I've added a bounty, but "Why is the laplacian of the deflection function convergence?" is not exactly clear, and in the body of the question you only ask "but I obtain 4$\kappa(\vec{\theta})$, not 2$\kappa(\vec{\theta})$.  Why am I wrong?" Do you think you can restate your question more clearly so it's 100% clear exactly what you are expecting as an answer? *Thanks!*  
> **username_0**: @uhoh Thank you for the comment. I will do it.  
> **username_3**: Please update your question giving a full definition of the symbols you have used and the coordinate system.  
> **username_0**: Yes I did it. Thank you.  
> **username_4**: There is a connection to the divergence, if you will. The gradient of $\ln(|x|)$ is the outward normal. By the divergence theorem, the Laplacian (i.e. div grad) of $\ln(|x|)$ in $n$ dimensions is hence the surface of the $n$-sphere times $\delta(0)$. This tells you that it must be $2\pi$ instead of $4\pi$ in your identity for 2D.  
> **username_0**: @ntessore Okay, I will think more about your comment. Thank you! And.. Can I ask some trivial question? In Stack exchange, Shouldn't I write "thank you"?  
> **username_4**: @BAO Sorry, I wasn't careful when writing the above. It's not the gradient of $\ln(|x|)$ in $n$ dimensions, it's a function $f(x)$ such that $\nabla f(x) = x/|x|^n$. In two dimensions, that's $\ln(|x|)$, while in higher dimensions, it's $|x|^{2-n}/(2-n)$. Then you can show the Green's function property with the divergence theorem.  

---

## 1 answers

### Answer by username_3 on 2022-01-23. Score: 4 | Accepted answer

I think the problem here may be that this is a 2D situation.

You say you used
$$ \nabla^2\frac{\vec\theta-\vec\theta'}{\mid\vec\theta-\vec\theta'\mid^2}=4\pi\delta^3(\vec{r})$$
(shouldn't there be a minus sign if $\vec{r} = \vec\theta-\vec\theta'$ ?)
but this is a relationship true only in 3D geometry and $\delta^3$ is the 3D delta function, while $\vec{\theta}$ is defined as a "two dimensional vector, which could be an angular position on the sky" in the review paper.

The equivalent in 2D geometry is
$$ \nabla^2 \ln \mid\vec\theta-\vec\theta'\mid = 2\pi\delta^2(\vec\theta-\vec\theta')\ ,$$
where $\delta^2$ is the 2D delta function.

From there, we see
\begin{eqnarray}
\nabla^2 \psi(\vec\theta) & = & \frac{1}{\pi} \int\_{\mathbb{R}^2} d^2\theta'\ \kappa(\vec\theta')\nabla^2 \ln \mid\vec\theta-\vec\theta'\mid \\
& = & \frac{1}{\pi} \int\_{\mathbb{R}^2} d^2\theta'\ \kappa(\vec\theta') 2\pi\delta^2(\vec\theta-\vec\theta') \\
& = & 2\kappa(\vec\theta)\ .
\end{eqnarray}

As for why your solution is incorrect, I suspect it is the second line.
I don't think the definition of divergence you have used is correct. Why would you equate it with the gradient of the vector? That is not generally true.
I am sure the"proof" is as I have presented it.

> **Comments:**
> **username_0**: There should be a minus sign, sorry. I understood most of them. If you answer a few questions, I think my question will be solved.  First, How do you derive the two-dimensional delta function? I derived the three dimensional delta function from 'Coulomb's law'. (I saw many site related with this like "https://math.stackexchange.com/questions/1932451/how-to-prove-this-relation-between-the-laplacian-of-the-logarithm-and-the-dirac", but I cannot understand)  
> **username_0**: Second, If even $\vec\theta$ means <$\theta_1$, $\theta_2$>, Cartesian Coordinate, is it right adding the factor '1/'? I understood the $\vec\theta$ is not theta in polar coordinate.  
> **username_3**: How do you derive the 2D Delta function is a separate question (for Maths SE). Second question. I have no idea, you put no definitions of coordinates or symbols in your question, so I am guessing that the definition of divergence is incorrect, or your switch from a 2D to a 3D integral. My derivation is correct. @BAO  
> **username_0**: Thank you. I fully understood, so I edit in my question with what I realized. Could I ask one more question? Why did you say "$\vec\nabla\cdot\vec\alpha(\vec\theta)=\frac{1}{\pi}\int_{\mathbb{R}^2} \kappa(\vec\theta')\frac{1}{\mid\vec\theta-\vec\theta'\mid} \nabla \frac{\vec\theta-\vec\theta'}{\mid\vec\theta-\vec\theta'\mid^2}\ d\theta'^2$"? is the two dimensional divergence's form like $\frac{1}{\mid\vec\theta-\vec\theta'\mid}\nabla$, not just $ \nabla$ ?  
> **username_3**: @BAO I am not sure it is correct (other than it seems to lead to the correct solution) and have deleted that part of my answer. However, I don't see why your original version is correct at all. Divergence is not a gradient.  
> **username_0**: Okay, your answer is really helpful!  

---

