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The Einstein angular radius is the angular form of the Einstein radius: $R_E=D_l{\theta }_E$ in the lens plane. Its displayed value applies to a point-mass gravitational lens with the angular diameter distances defined above. It depends on source distance as well as lens mass and position. At fixed distances, multiplying $M$ by four doubles ${\theta }_E$.

At exact alignment, the vector thin gravitational lens equation is $0=(1-{\theta }_E^2/r^2)\mathbf{\theta }$. For a nonzero image direction it requires $r={\theta }_E$ and leaves the direction unrestricted: all azimuths form an Einstein ring. The signed equation's roots $\theta =\pm {\theta }_E$ are only a diameter cut through that circle. A source of finite angular size makes an annulus or arcs with finite received light; the infinite point-source lensing magnification at alignment is an idealization. An Einstein ring expresses rotational symmetry and source alignment, not a circular arrangement of emitting stars.

A critical curve is a locus in the image plane where the Jacobian matrix of the thin gravitational lens equation becomes singular. Its formal point-source lensing magnification diverges, while finite source size regularizes the measured flux. Its mapped locus in the source plane is a gravitational-lensing caustic.

The trace of the angular derivative of the reduced deflection defines convergence in the standard thin gravitational lens equation convention. It equals projected surface mass density divided by the critical surface density, ${\Sigma }_{\mathrm{crit}}=c^2{D}_s/(4\pi G{D}_d{D}_{ds})$. If deflection is instead defined as source minus image angle, the divergence relation has a minus sign.

The signed magnification is the reciprocal determinant of the Jacobian matrix of the thin gravitational lens equation. Its absolute value gives the flux and solid-angle amplification; its sign gives image parity. Gravitational lensing preserves surface brightness in the geometric-optics approximation.

The general thin gravitational lens equation is a two-component angular relation,For a spherical lens, image, source and lens centre lie in one plane containing the line of sight. Choose its sky direction as a signed axis. Then $\alpha ,\theta ,\beta$ can be signed scalars andThey are not fundamentally scalar quantities for a general lens; the reduction uses this symmetry and axis choice. An exactly aligned source requires the full ring geometry, not a unique preferred sky axis.

In general, $D_S\ne D_L+D_{LS}$. For example, in a spatially flat expanding cosmology, comoving radial distances ${\chi }_L,{\chi }_S$ giveThus $D_S-D_{LS}={\chi }_L/(1+z_S)$ differs from $D_L$ when the redshifts differ. Ordinary Euclidean addition applies only when the same static physical-distance convention is appropriate.

Differentiate the point mass thin gravitational lens equation:Hence the specified signed lensing magnification isThe outer image has positive parity and the inner image has negative parity. Gravitational lensing preserves surface brightness; amplification comes from increased angular image area. The observed flux magnifications are $|{\mu }_{+}|$ and $|{\mu }_{-}|$; a negative determinant reverses orientation, not flux. Substitution of the two roots, with $u=|\beta |/{\theta }_E$, givesAt $\beta =0$, axial symmetry produces an Einstein ring. For an ideal point-like astronomical source, the lensing magnification diverges: near alignment ${\mu }_{\mathrm{tot}}\sim 1/u$. Finite source size gives a finite observed flux and a ring of finite thickness.

For $0<\beta <{\theta }_E$, there are two images on opposite sides. Their separation is $\sqrt{{\beta }^2+4{\theta }_E^2}$, approaching $2{\theta }_E$ at alignment. For $\beta \ll {\theta }_E$, their positions are ${\theta }_{\pm }\simeq \pm {\theta }_E+\beta /2$ and ${\mu }_{\pm }\simeq 1/2\pm {\theta }_E/(2\beta )$: both images are bright, with the outer one slightly brighter. Moving away from alignment reduces the total amplification and makes the inner image relatively fainter.

For $\beta \gg {\theta }_E$, expanding the two image roots givesThus the distant source has one almost unaltered bright image and an extremely faint opposite-parity secondary, with total flux magnification $1+2{\theta }_E^4/{\beta }^4+O(({\theta }_E/\beta {)}^6)$. The secondary persists for every finite nonzero source offset in the point mass model.

Use the small-angle, thin gravitational lens equation geometry. An undeflected ray seen at angular position $\theta$ would have transverse source position $D_s\theta$. Bending that ray through physical angle $\hat{\alpha }$ at the lens changes its source intercept by $D_{ds}\hat{\alpha }$. Thusand the required physical deflection isAlso $\beta =\eta /D_s$. Define the reduced deflection $\alpha$ as the angular difference $\theta -\beta$. Dividing the first equation by $D_s$ givesThese are angular diameter distances; in a cosmological spacetime one must not replace $D_s$ by $D_d+D_{ds}$ as if the distances were ordinary collinear Euclidean lengths.

Put $r=|\mathbf{\theta }|$ and write the vector thin gravitational lens equation as $\mathbf{\beta }=(1-\bar{\kappa }(r))\mathbf{\theta }$. A tangential displacement at fixed radius has eigenvalue ${\lambda }_t=1-\bar{\kappa }$. A radial displacement has eigenvalue ${\lambda }_r=1-{\alpha }^{\prime }(r)$. From the definition of the mean lensing convergence,Since $\alpha =r\bar{\kappa }$, this gives ${\alpha }^{\prime }=2\kappa -\bar{\kappa }$. Consequently,The two requested gravitational-lensing critical curve conditions are thereforeThey correspond respectively to a tangential critical curve of an axisymmetric lens and a radial critical curve of an axisymmetric lens.

Choose a signed angular coordinate along the source-lens line. The singular isothermal sphere lens equation isFor $\beta >0$, the positive image is ${\theta }_{+}=\beta +{\theta }_E$. A negative image would require ${\theta }_{-}=\beta -{\theta }_E<0$, which is possible only for $\beta <{\theta }_E$. Here $\beta =4$ arcmin and ${\theta }_E=1$ arcmin, so there is only one image, at $\theta =5$ arcmin. Formally using the negative-image formula would produce a positive value, inconsistent with the branch from which it was obtained.

The paper defines its vector deflection as source minus image angle, ${\mathbf{\alpha }}_{\mathrm{paper}}=(\beta -\theta )\hat{\mathbf{\theta }}$. This is the negative of the usual reduced deflection in $\mathbf{\beta }=\mathbf{\theta }-{\mathbf{\alpha }}_{\mathrm{standard}}$. Therefore its relation to lensing convergence isFor the SIS, ${\mathbf{\alpha }}_{\mathrm{paper}}=-{\theta }_E\hat{\mathbf{\theta }}$. The two-dimensional polar-coordinate divergence givesUsing a plus sign with the printed inward deflection would incorrectly give negative convergence.

For a small intrinsically circular source, differentiate the thin gravitational lens equation. The radial and tangential source-image Jacobian matrix eigenvalues areThus the image is stretched tangentially around the lens centre, with no radial stretch in this idealized model. At $\theta =5$ arcmin, ${\lambda }_t=4/5$; the image is an ellipse with tangential-to-radial axis ratio $5/4$. It is not a complete ring. This local description assumes a source small enough that the Jacobian is nearly constant across it.

The signed lensing magnification isSince the lensing shear magnitude equals $\kappa$, it reduces to $\mu =(1-2\kappa {)}^{-1}$. At the image, $\kappa ={\theta }_E/(2\theta )=1/10$, henceThe image has positive parity and is $25\%$ brighter for the same intrinsic source. Gravitational lensing preserves surface brightness, so this flux amplification follows from its larger apparent solid angle. In evaluating the convergence, use the image angle $5$ arcmin, not the source angle $4$ arcmin.

For the singular isothermal sphere, integrate its projected surface mass density inside a circle of proper radius $\xi$:With $\xi =D_d\theta$ this gives the reduced deflection magnitudeindependent of $\theta$. Hence the singular isothermal sphere lens hasThe spherical three-dimensional mass $2{\sigma }^{2}r/G$ would give the wrong coefficient here; the thin gravitational lens equation needs the projected cylindrical mass.

Using the same ${\theta }_E=3\times 10^{-4}$ and $D_{ds}/D_s=0.5$, the one-dimensional velocity dispersion isThus $\sigma \simeq 2100\mathrm{km}{\mathrm{s}}^{-1}$ within the idealized SIS model.

A complete Einstein ring requires an axisymmetric gravitational lens and exact alignment of the source with the lens centre and observer: $\beta =0$. A finite aligned circular source gives a ring of finite width. The background source must lie behind the lens, so the lensing distance factor is positive.

With $\beta =0$ the thin gravitational lens equation givesThe Einstein radius here is angular, and the mass is projected inside the corresponding aperture. For an extended mass distribution the equation is implicit because the enclosed mass depends on ${\theta }_E$.

Using the supplied approximations, $D_{ds}/(D_s{D}_d)=1/(3\times 10^{25}\mathrm{m})$ and ${\theta }_E=3\times 10^{-4}$, the inferred mass isThusThis measures the projected aperture mass without requiring the singular isothermal sphere model used subsequently.

A point-mass gravitational lens is the ideal gravitational lens obtained by concentrating a positive mass $M$ at one location. Use geometrical optics, the weak-field approximation, small angles in radians, and a stationary lens whose deflecting region is short compared with the observer-lens and lens-source distances. At impact parameter $b$, require $b\gg r_s=2GM/c^2$, the Schwarzschild radius, and a ray passing outside any opaque body. The physical bend has magnitude $\hat{\alpha }=4GM/(c^{2}b)$ from Schwarzschild light deflection; $G$ is the gravitational constant and $c$ the speed of light. Multiple images can occur within this weak-field approximation: strong gravitational lensing does not require a ray to approach a black hole closely.^[ref]

Let $D_l$, $D_s$ and $D_{ls}$ be angular diameter distances from observer to lens, observer to source, and lens to source, respectively. They refer to the background geometry used in the thin gravitational lens equation. In nearby Euclidean geometry, $D_{ls}=D_s-D_l$; in cosmology, do not make that subtraction. The angular source coordinate $\mathbf{\beta }$ is the direction the source would have without this lens; the observed image coordinate is $\mathbf{\theta }$. Both are measured relative to the observer-lens optical axis.^[ref]

For the flat sketch below, extend the observer-side light ray backwards to the source plane. Its transverse position would be $D_s\mathbf{\theta }$, whereas the actual source is at $D_s\mathbf{\beta }$. Their separation is $D_{ls}\hat{\mathbf{\alpha }}$. Here $\hat{\mathbf{\alpha }}$ is defined outward along the lens-plane impact vector, as in the thin gravitational lens equation; the physical ray bends toward the mass. Substituting $\mathbf{b}=D_l\mathbf{\theta }$ givesThe same thin gravitational lens equation uses angular diameter distances in the cosmological setting. The factor $D_{ls}/D_s$ converts the physical bend into the reduced angular deflection; these are different angles.

Projected surface mass density is mass per unit projected area, obtained by integrating mass density along the line of sight. It determines the enclosed cylindrical mass used by a thin gravitational lens equation. It need not equal a physical disk's density, and its aperture mass differs from a spherical three-dimensional enclosed mass.

The projected density $\Sigma (R)={\sigma }^2/(2GR)$ gives cylindrical mass $\pi {\sigma }^{2}R/G$, so the reduced deflection magnitude is the constant displayed angular Einstein radius. The signed thin gravitational lens equation is $\beta =\theta -{\theta }_E\mathrm{sgn}\theta$. A source with $0<\beta <{\theta }_E$ has two images; one with $\beta >{\theta }_E$ has one. The lensing convergence and lensing shear magnitude are both ${\theta }_E/(2|\theta |)$.