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enter image description here

From the review paper Bartelmann and schneider, 1999) Weak Gravitational Lensing, 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).

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    $\begingroup$ Might this be better suited to Mathematics Stack Exchange? $\endgroup$ Jan 21, 2022 at 22:19
  • $\begingroup$ Yes, I will do. Thank you. $\endgroup$
    – BAO
    Jan 23, 2022 at 1:51
  • $\begingroup$ @GrapefruitIsAwesome astrophysicists know about and use math as well $\endgroup$
    – uhoh
    Jan 23, 2022 at 6:20
  • $\begingroup$ @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! $\endgroup$
    – uhoh
    Jan 23, 2022 at 6:23
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    $\begingroup$ @uhoh Thank you for the comment. I will do it. $\endgroup$
    – BAO
    Jan 23, 2022 at 9:49

1 Answer 1

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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.

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    $\begingroup$ 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 "math.stackexchange.com/questions/1932451/…", but I cannot understand) $\endgroup$
    – BAO
    Jan 23, 2022 at 15:03
  • $\begingroup$ Second, If even $\vec\theta$ means <$\theta_1$, $\theta_2$>, Cartesian Coordinate, is it right adding the factor '1/𝜃'? I understood the $\vec\theta$ is not theta in polar coordinate. $\endgroup$
    – BAO
    Jan 23, 2022 at 15:16
  • $\begingroup$ 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 $\endgroup$
    – ProfRob
    Jan 23, 2022 at 18:24
  • $\begingroup$ 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$ ? $\endgroup$
    – BAO
    Jan 24, 2022 at 0:38
  • $\begingroup$ @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. $\endgroup$
    – ProfRob
    Jan 24, 2022 at 8:12

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