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 f advances its phase by 2πft in a time t.
Set up the geometry
Put the observer at the origin. The lens, a point mass M, sits at distance DL; the source plane is at distance DS, and the lens-to-source distance is DLS. Angles are measured from the observer. The source sits at angle β from the lens; a candidate path crosses the lens plane at angle θ. All angles are tiny, a few millionths of a radian or less, so sinθ≈θ and cosθ≈1−θ2/2.
In flat space with DS=DL+DLS, 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). The second leg and the straight path follow in the same way. Adding them and simplifying, the extra length is
Clocks run slow near a mass, so a wave crossing the lens plane at distance ξ=DLθ from the mass is delayed by an amount that grows as ξ shrinks. For a point mass this gravitational, or Shapiro, delay is −(4GM/c3)lnξ 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=c24GMDLDSDLS,
is the radius of the ring a perfectly aligned source would form. Write x=θ/θE and y=β/θE; y is the impact parameter, the source offset in Einstein radii. Substituting, the geometric term becomes (4GM/c3)⋅21∣x−y∣2, so both terms carry the same time unit. With the redshifted mass MLz=M(1+zL) the arrival time is
The constant ϕm(y) is chosen so that the earliest arrival is at T=0. The 2003 paper states this choice in words; its printed Eq. (12) shows a plus sign before ϕm, while its closed form, Eq. (17), and our independent numerical integral both follow the zero-minimum rule. The unit 4GMLz/c3 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 T to first order. Along the line through lens and source, set ∂T/∂x=0:
x−y−x1=0⇒x±=2y±y2+4.
The root x+>1 lies on the source's side, outside the Einstein ring. It is the minimum of T, the first image. The root x−<0 lies on the far side, inside the ring, where T rises in one direction and falls in the other: a saddle, the second image.
The brightening of an image, its magnification μ, is the ratio of image area to source area. For a lens with circular symmetry it is μ−1=(y/x)(dy/dx). With y=x−1/x,
μ−1=(1−x21)(1+x21)=1−x41,μ±=21±2yy2+4y2+2.
The saddle has μ−<0: its image is flipped. Notice that μ+−∣μ−∣=1 for every y.
The delay between the images
Evaluating T at the two roots and subtracting,
Δtd=c34GMLz[2yy2+4+lny2+4−yy2+4+y].
This is the delay quoted with Eq. (18) of the 2003 paper. It grows with y: a source far from alignment gives a long delay and a faint second image.
Worked example. Take MLz=1000M⊙ and y=0.5. The time unit is 1000×19.7μs=19.7ms. The bracket is 0.5×2.062/2+ln(2.562/1.562)=0.515+0.495=1.010, so Δtd=19.9ms. The magnifications are μ+=1.59 and μ−=−0.59, and the amplitude ratio of the two copies is 0.59/1.59=0.61. Fringes in frequency are spaced by 1/Δtd≈50Hz.
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 2πfΔtd. Define the dimensionless frequency
w=c38πGMLzf=4πλRs,
where Rs=2GMLz/c2 is the Schwarzschild radius of the lens and λ=c/f the wavelength. The image phase difference is w times the bracket above. When w≫1, the arrival-time surface is many wavelengths deep and the two images are sharp. When w≲1, 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=100Hz, w=1.24×10−4(MLz/M⊙)(f/1Hz) gives w=12.4 for 1000M⊙ and w=0.19 for 15M⊙. With the same geometry, the heavier lens splits the wave into two sharp images and the lighter one lets it diffract around.
Lab tasks
Predict how the delay changes when you drag the lens from y=2 to y=0.1 in the gravitational-wave lens Lab. Then drag it and read the delay.
Keep y=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.
Compare the readout for the loudness of the second copy with ∣μ−∣/μ+ from the formula above at y=1.
Exercises
Derive the geometric excess length from the three Pythagorean path lengths, using DS=DL+DLS.
Show that μ+−∣μ−∣=1 for every y>0.
For MLz=1500M⊙ and y=1.2, compute Δtd and the fringe spacing.
A lens of 30M⊙ sits at y=0.3. Above which frequency is w>1?
AnswersShow the answer
The three legs give DLθ2/2, (DSβ−DLθ)2/(2DLS) and −DSβ2/2. Put them over 2DLS and use DL+DLS=DS to collect DLDS(θ2−2θβ+β2).
μ+−∣μ−∣=μ++μ− because μ−<0, and the two terms ±(y2+2)/(2yy2+4) cancel, leaving 21+21=1.
The bracket at y=1.2 is 1.2×2.332/2+ln(3.532/1.132)=1.399+1.138=2.537. The time unit is 1500×19.7μs=29.6ms, so Δtd=75ms and fringes are 13Hz apart.
w=1 needs f=1/(1.24×10−4×30)≈270Hz. Below that frequency diffraction dominates. The threshold does not depend on y; how much the images are blurred near it does.