PHYS 555

From Tabitha

Tuesday, January 23 - Introduction

Introduction

Newton worked out the idea of bending of light by gravity but according to GR it is twice as large. Einstein thought that it wouldn't be observable but Zwicky thought that it would be -- Zwicky was right. Lensing was first detected in 1979. Weak lensing by LSS was detected in 2000 - probe of dark matter and dark energy.

What is it good for?

Mass and mass distribution where there is no light.
Number density - mass distirbution (microlensing)
Cosmological parameters
Gravitational telescope

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Lensing Geometry

$S$ is source. $L$ is lens. $ξ$ is the impact parameter. $β$ is unlensed position of source. $θ$ is lensed position of source . $\hat \alpha$ is bending angle.

$D L S$ is the distance between source and lens.. $D L$ is the distance to the lens and $D S$ is the distance between the source.

The lensing equation is $\beta = \theta - \hat \alpha \frac{D_{LS}}{D_S} = \theta - \alpha$ where $α$ is the scaled deflection angle.

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Sources

We asssume that

$\hat \alpha = \frac{4 G M}{c^2 \xi}$

where $M$ is the mass of the lens.

Let's imagine the deflection due to a bunch of lenses.

$dm = \rho (\vec r) dV$

The location of the light ray is $(ξ, z)$ and $d m (ξ', z')$ . Approximate the path of the light ray as a bunch straight lines -- all of the bending occurs at the point of closest approach (Born approximation). The impact parameter is the distance in the plane of the sky between the lens and the ray.

This allows us to sum up all of the deflections vectorially, so we have

$\hat \alpha = \frac{4G}{c^2} \int d^2 \vec \xi' \int dz' \rho ( \vec \xi', z') \frac{\vec\xi-\vec\xi'}{|\vec\xi-\vec\xi'|^2}.$

We can define the surface mass density as

$\Sigma(\vec \xi') = \int d z' \rho ( \vec \xi, z )$

yielding the simpler equation

$\hat \alpha = \frac{4G}{c^2} \int d^2 \xi' \Sigma ( \vec \xi' ) \frac{\vec \xi-\vec\xi'}{|\vec\xi-\vec \xi'|^2}.$

This is true to cluster lensing and lensing by galaxies -- for mass distributions that are localized in a plane (thickness is small compared to distance).

Let's switch from the plane of the lens to the observer plane to get

$\alpha(\vec \theta) = \frac{4G}{c^2} \frac{D_{LS}}{D_S} D_L \int d^2\theta' \Sigma(\vec \theta') \frac{\vec \theta-\vec \theta'}{|\vec \theta-\vec \theta'|^2}$

and

$\alpha(\vec \theta) = \frac{1}{\pi} \int d^2 \theta' \kappa ( \vec \theta' ) \frac{\vec \theta-\vec \theta'}{|\vec \theta-\vec \theta'|^2}$

where we have defined the convergence and the critical mass density

$\kappa = \frac{\Sigma}{\Sigma_{\rm crit}} = \Sigma \left ( \frac{c^2}{4\pi G}\frac{D_S}{D_{LS} D_L} \right )^{-1}.$

We can also define a lensing potential as

$\alpha(\vec \theta) = \nabla \Psi (\vec \theta)$

and

$\Psi(\vec \theta) = \frac{1}{\pi} \int d^2 \theta' \kappa (\vec \theta') \ln (\vec \theta -\vec \theta')$

and

$\kappa(\vec \theta) = \frac{1}{2} \nabla^2 \psi (\vec \theta) = \frac{1}{2} \left ( \Psi_{,11} - \Psi_{,22} \right )$

Lensing conserves the surface brightness of an image, so if the image is magnified we will receive more light. The magnification is defined as

$\mu = \frac{\rm Image Area}{\rm Source Area} = \frac{\delta \theta^2}{\delta \beta^2}.$

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Tuesday, Janaury 30 - the Lensing Jacobian

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Recap

Using the standard diagram above we can look at the magnification of a source for a particular image. We have

$\alpha(\vec \theta) = \frac{1}{\pi} \int d^2\theta' \kappa(\vec \theta') \frac{\vec \theta -\vec \theta'}{|\vec \theta - \vec \theta'|} = \nabla \Psi \left (\vec \theta \right )$

and

$\kappa(\vec \theta) = \frac{1}{2} \nabla^2 \Psi(\vec \theta) = \frac{\Sigma}{\Sigma_{\rm crit}}$

where

$\Sigma = \int \delta \rho( \xi , z ) d z.$

$δρ$ is the overdensity.

The size of the source unmagnified is $δβ 2$ and in the image it is $δθ 2$ so the magnification is

$\mu = \frac{\delta \theta^2}{\delta \beta^2}.$

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Lensing Jacobian

$A_{ij} = \frac{\partial \beta_i}{\partial \theta_j} = \left [ \begin{matrix} 1-\kappa-\gamma_1 & -\gamma_2 \\ -\gamma_2 & 1-\kappa+\gamma_1 \end{matrix} \right ]$

where $i$ and $j$ count over the coordinates in the plane of the sky and

$\gamma_1 = \frac{1}{2} \left ( \Psi_{,11} - \Psi_{,22} \right ), \gamma_2 = \Psi_{,12}.$

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Proof

We have

$A_{ij} = \frac{\partial \beta_i}{\partial \theta_j} = \frac{\partial}{\partial \theta_j} \left ( \theta_i - \alpha_i \right ) = \delta_{ij} - \frac{\partial \alpha_i}{\partial \theta_j} = \delta_{ij} - \Psi_{,ij}$

Remember that

$\kappa = \frac{1}{2} \left ( \Psi_{,11} + \Psi_{,22}\right )$ .

Component by component we have

$A 11 = 1 - Ψ,11 = 1 - κ - γ 1, A 12 = A 21 = - Ψ,12 = - γ 2, A 22 = 1 - Ψ,22 = 1 - κ + γ 1$

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Shear

The shear is a pseudovector. A forty-five degree rotation in real space yields a ninety degree rotation in shear space. We can write the shear as a magnitude and a phase

$γ 1 = γcos2φ,γ 2 = γsin2φ$

or as a complex field

$γ = γ 1 + i γ 2 .$

We can define the reduced shear

$g = \frac{\gamma}{1-\kappa}$

which simplifies the Jacobian to be

$A = (1-\kappa) \left [ \begin{matrix} 1 - g_1 & -g_2 \\ -g_2 & 1 - g_1 \end{matrix} \right ]$ .

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Circular Source

What happens to a circular source that gets lensed?

$A = (1-\kappa) \left [ \begin{matrix} 1 & 0 \\ 0& 1 \end{matrix} \right ] - \gamma \left [ \begin{matrix} \cos2\phi & \sin2\phi \\ \sin2\phi & -\cos2\phi \end{matrix} \right ].$

Let's take $γ = 0$ we have

$\left [ \begin{matrix} \beta_1 \\ \beta_2 \end{matrix} \right ] = (1-\kappa) \left [ \begin{matrix} \theta_1 \\ \theta_2 \end{matrix} \right ]$

Therefore if $\beta_1^2 + \beta_2^2 = r^2$ then

$\theta_1^2+\theta_2^2 = \frac{r^2}{\left (1 - \kappa\right)^2}$ .

For the general Jacobian we can look at the eigenvalues of matrix to get that

$A x = λ x$

where

$\lambda = 1 - \kappa \pm \gamma$

and the eigenvectors point toward

$\left [ \begin{matrix} \sin\phi \\ -\cos\phi \end{matrix} \right ]$ and $\left [ \begin{matrix} \cos\phi \\ \sin\phi \end{matrix} \right ]$ for the $+$ and $-$ respectively.

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Shear and Convergence

We have

$\gamma = \frac{1}{\pi} \int d^2 \theta' D \left ( \vec \theta - \vec \theta' \right ) \kappa(\vec \theta')$

where

$D(\vec \theta ) = - \frac{1}{\left ( \theta_1 - i \theta_2 \right )^2}$

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Proof

Let's start with the definition of the lensing potential

$\Psi(\vec \theta) = \frac{1}{\pi} \int d^2 \theta' \kappa (\vec \theta') \ln (\vec \theta -\vec \theta')$

and calculate the gradient

$\nabla\Psi = \frac{1}{\pi} \int d^2 \theta' \kappa (\vec \theta') \frac{\vec \theta -\vec \theta'}{(\vec \theta -\vec \theta')^2}$

and the second derivative

$\Psi_{,ij} = \frac{1}{\pi} \int d^2 \theta' \kappa (\vec \theta') \left [ \delta_{ij} \frac{1}{(\vec \theta -\vec \theta')^2} - 2 ( \theta - \theta')_i ( \theta - \theta')_j \frac{1}{(\vec \theta -\vec \theta')^4} \right ].$

Let's write out the shear

$\gamma_1 = \frac{1}{2} \left ( \Psi_{,11} - \Psi_{,22} \right ) = -\frac{1}{\pi} \int d^2 \theta' \kappa (\vec \theta') \frac{( \theta - \theta')^2_1 - ( \theta - \theta')^2_2}{(\vec \theta -\vec \theta')^4}$

and

$\gamma_2 = \Psi_{,12} = -\frac{2}{\pi} \int d^2 \theta' \kappa (\vec \theta') \frac{( \theta - \theta')_1 ( \theta - \theta')_2}{(\vec \theta -\vec \theta')^4}$

We can write

$\gamma = \gamma_1 + i \gamma_2 = -\frac{1}{\pi} \int d^2 \theta' \kappa (\vec \theta') \frac{\overline{ \left ( \theta_1 - i \theta_2 \right )^2 }}{\overline{ \left ( \theta_1 - i \theta_2 \right )^2}\left ( \theta_1 - i \theta_2 \right )^2 }$

where we have written $\theta_1 = \left ( \theta -\theta'\right)_1$ to be concise. Working this out we get

$\gamma = -\frac{1}{\pi} \int d^2 \theta' \kappa (\vec \theta') \frac{(\theta-\theta')^2_1 - (\theta-\theta')^2_2 + 2 i (\theta-\theta')_1 (\theta-\theta')_2}{(\theta-\theta')^4}$

proving the theorem.

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Mass-sheet Degeneracy

Let's take

$\beta_\lambda = \vec \theta - \lambda \alpha (\vec \theta )$

and

$\alpha_\lambda (\vec \theta ) = (1-\lambda) \vec \theta + \lambda \alpha (\vec \theta ).$

We can write

$\vec \beta = \frac{1}{\lambda} \vec \beta_\lambda = \vec \theta - \alpha (\vec \theta).$

Therefore, the convergence is degenerate with changes the scale of the source plane

$\kappa_\lambda ( \vec \theta ) = ( 1 - \lambda ) + \lambda \kappa (\vec \theta )$

and so is the shear $γ λ = λγ$ , but the reduced shear

$g_\lambda = \frac{\gamma_\lambda}{1-\kappa_\lambda} = \frac{\gamma}{1-\kappa}$

is preserved. You can add a sheet of mass behind or in front of the cluster without changing the observable signal.

You can break the degeneracy by looking at strong and weak lensing and sources at different distances behind the lens.

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Tuesday February 6 - Concepts from strong lensing

Critical curves lie in the lens plane where $det A (θ) = 0$ and caustics are the corresponding curves in the source plane; this occurs where $| γ | = 1 - κ$
Critical curves divide regions of magnification from regions of demagnification, and when a source crosses a caustics two new images (one magnified and one demagnified) are created or two images are annihilated.
Generally if $κ > 1$ , you have multiple images

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Circularly Symmetric Source

Let's look at a symmetric lens where $\Sigma(\vec \xi) = \Sigma (|\vec \xi|)$ or $Σ(x, y) = Σ(r)$ . In this case we have

$\vec \alpha = \nabla \Psi = \hat r \frac{\partial \Psi}{\partial r}$

and

$\vec \kappa = \frac{\nabla^2 \Psi}{2}$

$2\kappa = \frac{1}{r}\frac{\partial}{\partial r} \left ( r \frac{\partial \Psi}{\partial r} \right )$

$\alpha(\vec r) = \frac{2 \hat r}{r} \int_0^r r' \kappa(r') dr' = \frac{2\hat r}{r\Sigma_{\rm crit}} \int_0^r r' \Sigma(r') d r' = \frac{M(<r)}{\pi \Sigma_{\rm crit}} \frac{\hat r}{r}$

where N.B. $α$ , the scaled deflection angle, has the units of length.

There is no net deflection due to a circular shell of matter than lies outside image in the lens plan; consequently for a circularly symmetric lens the properties of the images only tell us about the mass enclosed within the radius of the images.

Let's define a mean value of $κ$ where we have

$\vec \alpha = \bar \kappa = \frac{\vec \theta}{\theta}$

and

$\bar \kappa = \frac{2}{\theta^2} \int_0^\theta d\theta' \theta' \kappa(\theta ') .$

Using the lensing equation

$\vec \beta = \vec \theta - \vec \alpha (\theta) = \left [ 1 - \bar \kappa(\theta) \right ] \vec \theta$

we can show that

$A(\theta) = \left ( 1 - \bar \kappa(\theta) \right ) \left [ \begin{matrix} 1& 0 \\ 0 & 1 \end{matrix} \right ] - \frac{\bar \kappa'(\theta) }{\theta} \left [ \begin{matrix} \theta_1^2 & \theta_1 \theta_2 \\ \theta_1 \theta_2 & \theta_2^2 \end{matrix} \right ]$

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Proof

This is easiest to prove using index notation. We have

$A_{ij} = \frac{\partial \beta_i}{\partial \theta_j} = \frac{\partial}{\partial \theta_j} \left \{ \left [ 1 - \bar \kappa(\theta) \right ] \theta_i \right \}$

$= \left [ 1 - \bar \kappa(\theta) \right ] \delta_{ij} - \frac{\bar \kappa' (\theta)}{\theta} \theta_i \theta_j$

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Where are the critical curves of this lens?

$\frac{d\bar \kappa}{d \theta} = \frac{d}{d\theta} \left ( \frac{2}{\theta^2} \int_0^\theta d\theta' \theta' \kappa(\theta') \right )$

$= -2 \frac{2}{\theta} \int_0^\theta d\theta' \left [ \theta' \kappa(\theta') \right ] + \frac{2}{\theta^2} \theta \kappa (\theta) = \frac{2}{\theta} \left[ \kappa(\vec \theta) - \bar \kappa(\theta) \right ].$

Let's write out

$A_{11} = 1- \kappa - \gamma_1 = 1 - \bar \kappa - 2 \frac{\kappa - \bar \kappa}{\theta^2} \theta^2 \cos^2 \phi = 1 - \kappa - \left ( \kappa - \bar \kappa \right) \cos 2\phi$

so in general we have

$\gamma = \left ( \kappa - \bar \kappa \right ) e^{2i\phi}.$

Now let's find where the determinant is zero,

$\det A = 0 = \left ( 1 -\kappa \right )^2 - | \gamma |^2 = \left ( 1 - \kappa \right )^2 - \left ( \kappa - \bar \kappa \right )^2 = ( 1 - \bar \kappa ) (1 - \bar \kappa - 2 \kappa ) .$

The two solutions are called

Tangential critical curve $1-\bar \kappa=0$ : strong shear in the tangential direction; helpful to measure enclosed mass
Radial critical curve $1-\bar \kappa - 2 \kappa=0$ : strong shear in the radial direction (really close in); helpful to measure the slope of the mass distribution

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Einstein Radius

Looking at the tangential critical curve we have

$\bar \kappa = 1 = \frac{M(<\theta)}{\pi \Sigma_{\rm crit}} \frac{1}{\theta^2 D_L^2}$

and we define this angular distance as the Einstein radius

$\theta_E^2 = \frac{4 GM(<\theta)}{c^2} \frac{D_{LS}}{D_S D_L}$

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Tuesday, February 13

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The Singular Isothermal Sphere

We have

$\rho(r) = \frac{\sigma_v^2}{2\pi G r^2}$

Note that as $r \rightarrow 0$ , $\rho\rightarrow \infty$ , hence the name singular. Let's calculate the surface mass density

$\Sigma(\xi) = \int_{-\infty}^\infty dz \frac{\sigma_v^2}{2\pi G (z^2+\xi^2)}=$

The convergence is

$\kappa = \frac{\Sigma}{\Sigma_{\rm crit}} = \frac{\sigma_v^2}{2G} \frac{1}{\theta D_L} \frac{4\pi G}{c^2} \frac{D_{LS} D_L}{D_S}$

$= \frac {2\sigma_v^2\pi}{\theta c^2} \frac{D_{LS}}{D_L}$

Remember that

$\bar \kappa(\theta) = \frac{2}{\theta^2} \int_0^\theta d\theta' \theta' \kappa(\theta') = \frac{4\sigma_v^2 \pi}{c^2 \theta} \frac{D_{LS}}{D_S}=2\kappa(theta)$

and that the shear is

$\gamma(\theta) = \left [ \kappa(\theta) - \bar \kappa(\theta) \right ]^{2i\phi}=-\frac{2\pi\sigma_v^2}{\theta c^2} \frac{D_{LS}}{D_S} e^{2i\phi}$

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The NFW Model

They found that the density of DM halo could be modelled by

$\rho(r) = \frac{\delta_c \rho_{\rm cr} (z)}{\frac{r}{r_s} \left ( 1 + \frac{r}{r_s} \right)^2}$

where the critical density of the universe is

$\rho_{\rm cr} = \frac{3 H^2(z)}{8\pi G}$

and

$\delta_c = \frac{200}{3} \frac{e^2}{\ln (1 + c ) - \frac{c}{1+c} } .$

The concentration parameter is

$c = \frac{r_{200}}{r_s}$

and at the radius $r 200$ we have $ρ(r 200) = 200ρ c r (z)$ .

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Weak-Lensing by Large-Scale Structure

The power spectrum of the convergence depends on the three-dimensional power spectrum of matter. Looking at density fluctuations we have

$\delta(\vec x) = \frac{\rho(\vec x) - \bar \rho}{\bar \rho}.$

The fluctuations will grow as

$\delta(a) \propto a^2$

during radiation domination and and

$\delta(a) \propto a$

during matter domination if we have $Ω m = 1,Λ = 0$ When we have $\Lambda\neq 0$ and $\Omega_m\neq 1$ ,

$\delta(a) = \delta_0 a \frac{g(a)}{g(a_0)}$

and

$g(a) = \frac{s}{2} \Omega_m(a) \left[ \Omega_m^{4/7}(a) - \Omega_\Lambda (a) + \left ( 1 +\frac{\Omega_m(a)}{2} \right ) \left ( 1 + \frac{\Omega_\Lambda}{70} \right ) \right ]^{-1}.$

Furthermore, the perturbations do not grow during radiation domination if they lie within the horizon.

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Power Spectrum and Correlation Function

We have a homogeneous, isotropic, Gaussian random field, the correlation function is

$C(r) = C(|\vec x - \vec x'|) = \left \langle g(\vec x) g^*(\vec x') \right \rangle.$

Let's take the Fourier transform of the field

$\hat g(\vec k) = \int d^n x g(\vec x) e^{i \vec x \cdot \vec k}$

and

$g(\vec x) = \frac{1}{(2\pi)^n} \int d^n k \hat g (\vec k) e^{-i \vec x \cdot \vec k }.$

We have

$\left \langle \hat g(\vec k) \hat g^*(\vec k') \right \rangle = \int d^n x e^{i \vec x \cdot \vec k }\int d^n x' e^{-i \vec x' \cdot \vec k' } \left \langle g(\vec x) g^*(\vec x') \right \rangle$

$\left \langle \hat g(\vec k) \hat g^*(\vec k') \right \rangle = \int d^n x e^{i \vec x \cdot \vec k }\int d^n r e^{-i \vec r' \cdot \vec k' } C(r) = (2\pi)^n \delta_D(\vec k-\vec k') P(|k|) .$