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Gravitational-wave lensing and frequency fringes

Why a lens delays a gravitational wave

Sandbox Physics ·

A wave that passes a mass off-centre travels a longer, bent path and also spends time climbing out of the mass's gravity. Adding the two effects for every point the wave could cross gives an arrival-time surface. Its lowest point and its saddle point are the two images, and the difference in their heights is the delay the detector hears.

Before this chapter, read what a detector hears when one chirp takes two paths. You need small-angle approximations, the Pythagorean theorem, natural logarithms and the idea that a wave of frequency ff advances its phase by 2πf t2\pi f\,t in a time tt.

Set up the geometry

Put the observer at the origin. The lens, a point mass MM, sits at distance DLD_{\rm L}; the source plane is at distance DSD_{\rm S}, and the lens-to-source distance is DLSD_{\rm LS}. Angles are measured from the observer. The source sits at angle β\beta from the lens; a candidate path crosses the lens plane at angle θ\theta. All angles are tiny, a few millionths of a radian or less, so sin⁡θ≈θ\sin\theta\approx\theta and cos⁡θ≈1−θ2/2\cos\theta\approx1-\theta^2/2.

In flat space with DS=DL+DLSD_{\rm S}=D_{\rm L}+D_{\rm LS}, compare the bent path with the straight one. The leg from the observer to the lens plane has length DL2+DL2θ2≈DL(1+θ2/2)\sqrt{D_{\rm L}^2+D_{\rm L}^2\theta^2}\approx D_{\rm L}(1+\theta^2/2). The second leg and the straight path follow in the same way. Adding them and simplifying, the extra length is

Δℓgeom=DLDS2DLS (θ−β)2.\Delta\ell_{\rm geom}=\frac{D_{\rm L}D_{\rm S}}{2D_{\rm LS}}\,(\theta-\beta)^2 .

Exercise 1 asks you to carry out the algebra. In an expanding universe the same form holds with angular-diameter distances, and every time picks up a factor 1+zL1+z_{\rm L} from the lens redshift. Wave effects in gravitational lensing of gravitational waves from chirping binaries, Sec. 2

Add the time spent in the mass's gravity

Clocks run slow near a mass, so a wave crossing the lens plane at distance ξ=DLθ\xi=D_{\rm L}\theta from the mass is delayed by an amount that grows as ξ\xi shrinks. For a point mass this gravitational, or Shapiro, delay is −(4GM/c3)ln⁡ξ-(4GM/c^3)\ln\xi plus a constant that is the same for every path. The minus sign means a path closer to the mass loses more time.

Measure lengths in Einstein radii

Two scales make the formulas clean. The Einstein angle,

θE2=4GMc2 DLSDLDS,\theta_{\rm E}^2=\frac{4GM}{c^2}\,\frac{D_{\rm LS}}{D_{\rm L}D_{\rm S}},

is the radius of the ring a perfectly aligned source would form. Write x=θ/θEx=\theta/\theta_{\rm E} and y=β/θEy=\beta/\theta_{\rm E}; yy is the impact parameter, the source offset in Einstein radii. Substituting, the geometric term becomes (4GM/c3)⋅12∣x−y∣2(4GM/c^3)\cdot\tfrac12|x-y|^2, so both terms carry the same time unit. With the redshifted mass MLz=M(1+zL)M_{Lz}=M(1+z_{\rm L}) the arrival time is

td(x)=4GMLzc3 T(x,y),T(x,y)=12∣x−y∣2−ln⁡∣x∣−ϕm(y).t_d(x)=\frac{4GM_{Lz}}{c^3}\,T(x,y),\qquad T(x,y)=\tfrac12|x-y|^2-\ln|x|-\phi_m(y).

The constant ϕm(y)\phi_m(y) is chosen so that the earliest arrival is at T=0T=0. The 2003 paper states this choice in words; its printed Eq. (12) shows a plus sign before ϕm\phi_m, while its closed form, Eq. (17), and our independent numerical integral both follow the zero-minimum rule. The unit 4GMLz/c34GM_{Lz}/c^3 is 19.7 microseconds for each solar mass.

Find the images

By Fermat's principle, waves add up where the arrival time is stationary: small changes of path do not change TT to first order. Along the line through lens and source, set ∂T/∂x=0\partial T/\partial x=0:

x−y−1x=0⇒x±=y±y2+42.x-y-\frac1x=0\quad\Rightarrow\quad x_\pm=\frac{y\pm\sqrt{y^2+4}}{2}.

The root x+>1x_+>1 lies on the source's side, outside the Einstein ring. It is the minimum of TT, the first image. The root x−<0x_-<0 lies on the far side, inside the ring, where TT rises in one direction and falls in the other: a saddle, the second image.

The brightening of an image, its magnification μ\mu, is the ratio of image area to source area. For a lens with circular symmetry it is μ−1=(y/x) (dy/dx)\mu^{-1}=(y/x)\,(\mathrm dy/\mathrm dx). With y=x−1/xy=x-1/x,

μ−1=(1−1x2)(1+1x2)=1−1x4,μ±=12±y2+22yy2+4.\mu^{-1}=\Bigl(1-\frac1{x^2}\Bigr)\Bigl(1+\frac1{x^2}\Bigr)=1-\frac1{x^4},\qquad \mu_\pm=\frac12\pm\frac{y^2+2}{2y\sqrt{y^2+4}} .

The saddle has μ−<0\mu_-<0: its image is flipped. Notice that μ+−∣μ−∣=1\mu_+-|\mu_-|=1 for every yy.

The delay between the images

Evaluating TT at the two roots and subtracting,

Δtd=4GMLzc3[yy2+42+ln⁡y2+4+yy2+4−y].\Delta t_d=\frac{4GM_{Lz}}{c^3}\left[\frac{y\sqrt{y^2+4}}{2}+\ln\frac{\sqrt{y^2+4}+y}{\sqrt{y^2+4}-y}\right].

This is the delay quoted with Eq. (18) of the 2003 paper. It grows with yy: a source far from alignment gives a long delay and a faint second image.

Worked example. Take MLz=1000 M⊙M_{Lz}=1000\,M_\odot and y=0.5y=0.5. The time unit is 1000×19.7 μs=19.7 ms1000\times19.7\,\mu\mathrm s=19.7\,\mathrm{ms}. The bracket is 0.5×2.062/2+ln⁡(2.562/1.562)=0.515+0.495=1.0100.5\times2.062/2+\ln(2.562/1.562)=0.515+0.495=1.010, so Δtd=19.9 ms\Delta t_d=19.9\,\mathrm{ms}. The magnifications are μ+=1.59\mu_+=1.59 and μ−=−0.59\mu_-=-0.59, and the amplitude ratio of the two copies is 0.59/1.59=0.61\sqrt{0.59/1.59}=0.61. Fringes in frequency are spaced by 1/Δtd≈50 Hz1/\Delta t_d\approx50\,\mathrm{Hz}.

When rays stop being a good picture

Images exist only if the wave can tell the two paths apart. The phase difference between them is 2πf Δtd2\pi f\,\Delta t_d. Define the dimensionless frequency

w=8πGMLzfc3=4π Rsλ,w=\frac{8\pi GM_{Lz}f}{c^3}=4\pi\,\frac{R_{\rm s}}{\lambda},

where Rs=2GMLz/c2R_{\rm s}=2GM_{Lz}/c^2 is the Schwarzschild radius of the lens and λ=c/f\lambda=c/f the wavelength. The image phase difference is ww times the bracket above. When w≫1w\gg1, the arrival-time surface is many wavelengths deep and the two images are sharp. When w≲1w\lesssim1, arrival times across the whole region within an Einstein radius differ by less than about one radian of phase, and the wave diffracts around the mass instead of splitting. The next chapter turns this into an exact formula.

Worked example. At f=100 Hzf=100\,\mathrm{Hz}, w=1.24×10−4 (MLz/M⊙)(f/1 Hz)w=1.24\times10^{-4}\,(M_{Lz}/M_\odot)(f/1\,\mathrm{Hz}) gives w=12.4w=12.4 for 1000 M⊙1000\,M_\odot and w=0.19w=0.19 for 15 M⊙15\,M_\odot. With the same geometry, the heavier lens splits the wave into two sharp images and the lighter one lets it diffract around.

Lab tasks

  1. Predict how the delay changes when you drag the lens from y=2y=2 to y=0.1y=0.1 in the gravitational-wave lens Lab. Then drag it and read the delay.
  2. Keep y=0.5y=0.5 and change the lens mass from 1000 to 100 solar masses. Predict the new fringe spacing before you look at the amplification plot.
  3. Compare the readout for the loudness of the second copy with ∣μ−∣/μ+\sqrt{|\mu_-|/\mu_+} from the formula above at y=1y=1.

Exercises

  1. Derive the geometric excess length from the three Pythagorean path lengths, using DS=DL+DLSD_{\rm S}=D_{\rm L}+D_{\rm LS}.
  2. Show that μ+−∣μ−∣=1\mu_+-|\mu_-|=1 for every y>0y>0.
  3. For MLz=1500 M⊙M_{Lz}=1500\,M_\odot and y=1.2y=1.2, compute Δtd\Delta t_d and the fringe spacing.
  4. A lens of 30 M⊙30\,M_\odot sits at y=0.3y=0.3. Above which frequency is w>1w>1?
Answers
  1. The three legs give DLθ2/2D_{\rm L}\theta^2/2, (DSβ−DLθ)2/(2DLS)(D_{\rm S}\beta-D_{\rm L}\theta)^2/(2D_{\rm LS}) and −DSβ2/2-D_{\rm S}\beta^2/2. Put them over 2DLS2D_{\rm LS} and use DL+DLS=DSD_{\rm L}+D_{\rm LS}=D_{\rm S} to collect DLDS(θ2−2θβ+β2)D_{\rm L}D_{\rm S}(\theta^2-2\theta\beta+\beta^2).
  2. μ+−∣μ−∣=μ++μ−\mu_+-|\mu_-|=\mu_++\mu_- because μ−<0\mu_-<0, and the two terms ±(y2+2)/(2yy2+4)\pm(y^2+2)/(2y\sqrt{y^2+4}) cancel, leaving 12+12=1\tfrac12+\tfrac12=1.
  3. The bracket at y=1.2y=1.2 is 1.2×2.332/2+ln⁡(3.532/1.132)=1.399+1.138=2.5371.2\times2.332/2+\ln(3.532/1.132)=1.399+1.138=2.537. The time unit is 1500×19.7 μs=29.6 ms1500\times19.7\,\mu\mathrm s=29.6\,\mathrm{ms}, so Δtd=75 ms\Delta t_d=75\,\mathrm{ms} and fringes are 13 Hz13\,\mathrm{Hz} apart.
  4. w=1w=1 needs f=1/(1.24×10−4×30)≈270 Hzf=1/(1.24\times10^{-4}\times30)\approx270\,\mathrm{Hz}. Below that frequency diffraction dominates. The threshold does not depend on yy; how much the images are blurred near it does.

The derivation follows the thin-lens treatment in Sec. 2 of the 2003 paper. The same image types, minimum and saddle, appear in the strong-lensing model of the LIGO–Virgo–KAGRA search, where the saddle image carries a quarter-cycle phase shift. Search for gravitational-lensing signatures in the full third observing run, Eq. 1

Next: the exact amplification factor of a point-mass lens.

Sources

  1. Wave effects in gravitational lensing of gravitational waves from chirping binaries, Sec. 2
  2. Search for gravitational-lensing signatures in the full third observing run, Eq. 1

Keep reading

What this page links to directly

Gravitational-wave lensing and frequency fringes · Reading path

  1. ArticleWhat a detector hears when one chirp takes two paths
  2. You are hereWhy a lens delays a gravitational wave
  3. ArticleThe exact amplification factor of a point-mass lens
  4. ArticleCan a detector tell a lensed chirp from an ordinary one?
  5. ArticleRecompute point-lens amplification and test a lens search
  6. ArticleHow to read the O3 search for lensed gravitational waves
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