Cosmology · Clusters and gravitational lensing
From shear to projected mass
Compare two supplied projected-density profiles that preserve reduced shear over one source plane while changing aperture mass and magnification.
Preparation: ratios, powers of ten and ordinary arithmetic. Physical quantities and notation are introduced as they are needed.
Back to the chapterAfter reading, you can
- Distinguish convergence, shear and reduced shear in a stated lens geometry.
- Calculate projected aperture mass for a supplied circular profile.
- Explain a finite-field mass-sheet ambiguity without claiming a unique three-dimensional mass.
Lensing first constrains projected information
A galaxy image has an intrinsic shape as well as lensing distortion. One observed ellipticity is therefore not an exact shear measurement. Population methods and calibration are needed before shape information becomes a lensing constraint.
For a thin lens and one specified source plane, define convergence κ = Σ/Σ_crit, where Σ is projected mass per physical area and Σ_crit = c²D_s/(4πGD_lD_ls). The distances are angular-diameter distances in a supplied geometry.
Shear γ describes an anisotropic part of the local mapping. Reduced shear is g = γ/(1−κ). A shape-based constraint generally involves this ratio rather than an independently known κ and γ.
For a circular lens, tangential shear satisfies γ_t = κ̄(<R) − κ(R). The mean convergence inside a projected radius enters, so interpreting a local distortion needs more than the surface density at that point.
An explicit profile gives an explicit aperture mass
Choose an arbitrary positive physical lens-plane length R_s and q = R/R_s. Supply κ(q) = 0.4exp(−q²), with a fixed positive Σ_crit. This analytic profile is a teaching input rather than a fitted cluster.
Its projected aperture mass is M_proj(<R) = 2πΣ_crit∫₀ᴿκ(R′)R′dR′. Dividing by πΣ_critR_s² gives the dimensionless quantity m(q) = 0.4[1−exp(−q²)].
Thus κ̄(<R) = m(q)/q² and γ_t = m(q)/q² − κ(q). Evaluate g_t = γ_t/(1−κ) on the finite annulus 0.5 ≤ q ≤ 3.
This aperture mass is a projected plane or cylinder integral. It is not the mass inside a three-dimensional sphere of radius R, and it cannot be substituted directly for a spherical-overdensity mass without a deprojection model.
A density rescaling plus a sheet leaves reduced shear unchanged
Choose λ = 0.8 and define κ′ = λκ + 1−λ and γ′_t = λγ_t. Then 1−κ′ = λ(1−κ), so g′_t = γ′_t/(1−κ′) = g_t at every displayed q.
The transformed projected mass is m′(q) = λm(q) + (1−λ)q². At q = 1, the original value is about 0.252848 and the transformed value about 0.402279 in units of πΣ_critR_s².
The equal reduced-shear columns therefore coexist with unequal convergence and aperture mass. This is a mass-sheet transformation under the stated single-source-plane assumptions.
No independent absolute source position, size or flux calibration is supplied. The corresponding unknown source coordinates can rescale consistently with this transformation; shape constraints alone do not choose one mass column.
| q | κ | κ′ | g_t = g′_t | m | m′ |
|---|---|---|---|---|---|
| 0.5 | 0.311520 | 0.449216 | 0.0615827 | 0.0884797 | 0.120784 |
| 1 | 0.147152 | 0.317721 | 0.123933 | 0.252848 | 0.402279 |
| 1.5 | 0.0421597 | 0.233728 | 0.122025 | 0.357840 | 0.736272 |
| 2 | 0.00732626 | 0.205861 | 0.0915126 | 0.392674 | 1.114139 |
| 3 | 0.0000493639 | 0.200039 | 0.0443918 | 0.399951 | 2.119961 |
Equal shape response does not mean equal magnification
The determinant of the lens mapping scales by λ² under this transformation. Signed magnification therefore changes as μ′ = μ/λ². With λ = 0.8, the multiplicative factor is 1/0.64 = 1.5625.
For example, at q = 1 the original magnification is about 1.396300 and the transformed value about 2.181719. Equal g does not make every possible lensing observable equal.
The added sheet is uniform throughout q ≤ 3 in the construction. Its continuation outside the finite field is not inferred. It can taper outside that region without changing the inner circular deflection; no infinite sheet is being declared an isolated finite halo.
Independent magnification information, source calibration, additional redshift information or boundary assumptions may constrain the ambiguity. Each adds its own assumptions and uncertainty, and the single-plane invariance should not be silently extended to arbitrary source populations.
Mass reconstruction is a conditional inference
Real images require correction for the point-spread function, detector response, intrinsic shape populations and orientation-dependent selection. Source-redshift estimates enter Σ_crit and may correlate with other calibration choices.
Even a calibrated reduced-shear profile does not automatically fix absolute projected mass over a finite field. The supplied counterexample identifies one reason a reconstruction needs additional information or conditions.
Converting projected mass into a spherical three-dimensional mass requires a geometric model and a matched aperture convention. Matter along the line of sight and departures from spherical structure can also affect that interpretation.
A useful lensing mass result states the measured quantities, source geometry, calibration, boundary assumptions and deprojection. Those definitions are necessary before comparing it with a gas-based or population-based mass estimate.
Worked reasoning
Check the q = 1 transformation
Why can the q = 1 reduced shear remain unchanged while projected mass increases?
- At q = 1, κ = 0.4/e ≈ 0.147152 and m = 0.4(1−1/e) ≈ 0.252848.
- Both γ and 1−κ are multiplied by λ = 0.8, leaving their ratio g unchanged.
- The transformed mass is 0.8m + 0.2 ≈ 0.402279, because the added sheet contributes projected mass inside the aperture.
Reduced shear is a ratio invariant under this single-plane transformation, while aperture mass includes the added sheet.
Try the reasoning
Is M_proj(<R) automatically the mass inside a three-dimensional sphere of radius R?
Show the answer and reasoning
No.
It integrates projected surface density over an aperture; spherical mass requires additional deprojection assumptions.
Does g′ = g imply μ′ = μ in this transformation?
Show the answer and reasoning
No.
Magnification changes by 1/λ², which is 1.5625 for the supplied λ.
Does the finite-field example establish an infinite uniform sheet as an isolated cluster?
Show the answer and reasoning
No.
Only the stated inner field is constrained; the exterior continuation is left unspecified and may taper.
What to carry forward
- Reduced shear combines shear and convergence, with source geometry and calibration attached.
- A finite-field shape profile can be compatible with different projected aperture masses.
- Absolute mass and three-dimensional comparisons require additional information, boundaries and deprojection assumptions.
References
Sources supporting the explanations, physical distinctions and examples in this lesson.