---
title: "Why a lens delays a gravitational wave"
description: "Build the arrival-time surface of a point-mass lens, find its two images and their delay, and see when a wavelength is too long to split."
canonical_url: "https://sandboxphysics.com/stories/gravitational-wave-lensing/diffraction-and-the-lens-time-delay/"
language: "en"
page_type: "story"
alternate_languages:
  zh-CN: "https://sandboxphysics.com/zh/stories/gravitational-wave-lensing/diffraction-and-the-lens-time-delay/"
last_modified: "2026-11-12"
---

[Gravitational-wave lensing and frequency fringes](https://sandboxphysics.com/stories/gravitational-wave-lensing/)

# Why a lens delays a gravitational wave

Sandbox Physics · 2026-11-12

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](https://sandboxphysics.com/stories/gravitational-wave-lensing/one-chirp-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 $2\pi f\,t$ in a time $t$.

## Set up the geometry

Put the observer at the origin. The lens, a point mass $M$, sits at distance $D_{\rm L}$; the source plane is at distance $D_{\rm S}$, and the lens-to-source distance is $D_{\rm LS}$. Angles are measured from the observer. The source sits at angle $\beta$ from the lens; a candidate path crosses the lens plane at angle $\theta$. All angles are tiny, a few millionths of a radian or less, so $\sin\theta\approx\theta$ and $\cos\theta\approx1-\theta^2/2$.

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

$$
\Delta\ell_{\rm geom}=\frac{D_{\rm L}D_{\rm S}}{2D_{\rm LS}}\,(\theta-\beta)^2 .
$$

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 $1+z_{\rm L}$ from the lens redshift. [Wave effects in gravitational lensing of gravitational waves from chirping binaries, Sec. 2](https://arxiv.org/abs/astro-ph/0305055v2)

## Add the time spent in the mass's gravity

Clocks run slow near a mass, so a wave crossing the lens plane at distance $\xi=D_{\rm L}\theta$ from the mass is delayed by an amount that grows as $\xi$ shrinks. For a point mass this gravitational, or Shapiro, delay is $-(4GM/c^3)\ln\xi$ 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,

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

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

$$
t_d(x)=\frac{4GM_{Lz}}{c^3}\,T(x,y),\qquad T(x,y)=\tfrac12|x-y|^2-\ln|x|-\phi_m(y).
$$

The constant $\phi_m(y)$ 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 $\phi_m$, while its closed form, Eq. (17), and our independent numerical integral both follow the zero-minimum rule. The unit $4GM_{Lz}/c^3$ 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 $T$ to first order. Along the line through lens and source, set $\partial T/\partial x=0$:

$$
x-y-\frac1x=0\quad\Rightarrow\quad x_\pm=\frac{y\pm\sqrt{y^2+4}}{2}.
$$

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 $\mu$, is the ratio of image area to source area. For a lens with circular symmetry it is $\mu^{-1}=(y/x)\,(\mathrm dy/\mathrm dx)$. With $y=x-1/x$,

$$
\mu^{-1}=\Bigl(1-\frac1{x^2}\Bigr)\Bigl(1+\frac1{x^2}\Bigr)=1-\frac1{x^4},\qquad \mu_\pm=\frac12\pm\frac{y^2+2}{2y\sqrt{y^2+4}} .
$$

The saddle has $\mu_-<0$: its image is flipped. Notice that $\mu_+-|\mu_-|=1$ for every $y$.

## The delay between the images

Evaluating $T$ at the two roots and subtracting,

$$
\Delta t_d=\frac{4GM_{Lz}}{c^3}\left[\frac{y\sqrt{y^2+4}}{2}+\ln\frac{\sqrt{y^2+4}+y}{\sqrt{y^2+4}-y}\right].
$$

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

**Worked example.** Take $M_{Lz}=1000\,M_\odot$ and $y=0.5$. The time unit is $1000\times19.7\,\mu\mathrm s=19.7\,\mathrm{ms}$. The bracket is $0.5\times2.062/2+\ln(2.562/1.562)=0.515+0.495=1.010$, so $\Delta t_d=19.9\,\mathrm{ms}$. The magnifications are $\mu_+=1.59$ and $\mu_-=-0.59$, and the amplitude ratio of the two copies is $\sqrt{0.59/1.59}=0.61$. Fringes in frequency are spaced by $1/\Delta t_d\approx50\,\mathrm{Hz}$.

## 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 $2\pi f\,\Delta t_d$. Define the dimensionless frequency

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

where $R_{\rm s}=2GM_{Lz}/c^2$ is the Schwarzschild radius of the lens and $\lambda=c/f$ the wavelength. The image phase difference is $w$ times the bracket above. When $w\gg1$, the arrival-time surface is many wavelengths deep and the two images are sharp. When $w\lesssim1$, 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 $f=100\,\mathrm{Hz}$, $w=1.24\times10^{-4}\,(M_{Lz}/M_\odot)(f/1\,\mathrm{Hz})$ gives $w=12.4$ for $1000\,M_\odot$ and $w=0.19$ for $15\,M_\odot$. 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 $y=2$ to $y=0.1$ in [the gravitational-wave lens Lab](https://sandboxphysics.com/labs/gravitational-wave-lens-fringes/). Then drag it and read the delay.
2. 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.
3. Compare the readout for the loudness of the second copy with $\sqrt{|\mu_-|/\mu_+}$ from the formula above at $y=1$.

## Exercises

1. Derive the geometric excess length from the three Pythagorean path lengths, using $D_{\rm S}=D_{\rm L}+D_{\rm LS}$.
2. Show that $\mu_+-|\mu_-|=1$ for every $y>0$.
3. For $M_{Lz}=1500\,M_\odot$ and $y=1.2$, compute $\Delta t_d$ and the fringe spacing.
4. A lens of $30\,M_\odot$ sits at $y=0.3$. Above which frequency is $w>1$?

Answers

1. The three legs give $D_{\rm L}\theta^2/2$, $(D_{\rm S}\beta-D_{\rm L}\theta)^2/(2D_{\rm LS})$ and $-D_{\rm S}\beta^2/2$. Put them over $2D_{\rm LS}$ and use $D_{\rm L}+D_{\rm LS}=D_{\rm S}$ to collect $D_{\rm L}D_{\rm S}(\theta^2-2\theta\beta+\beta^2)$.
2. $\mu_+-|\mu_-|=\mu_++\mu_-$ because $\mu_-<0$, and the two terms $\pm(y^2+2)/(2y\sqrt{y^2+4})$ cancel, leaving $\tfrac12+\tfrac12=1$.
3. The bracket at $y=1.2$ is $1.2\times2.332/2+\ln(3.532/1.132)=1.399+1.138=2.537$. The time unit is $1500\times19.7\,\mu\mathrm s=29.6\,\mathrm{ms}$, so $\Delta t_d=75\,\mathrm{ms}$ and fringes are $13\,\mathrm{Hz}$ apart.
4. $w=1$ needs $f=1/(1.24\times10^{-4}\times30)\approx270\,\mathrm{Hz}$. Below that frequency diffraction dominates. The threshold does not depend on $y$; 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](https://arxiv.org/abs/2304.08393v1)

Next: [the exact amplification factor of a point-mass lens](https://sandboxphysics.com/stories/gravitational-wave-lensing/point-lens-amplification-factor/).

## Try it in the Lab

[Gravitational-wave lens fringes](https://sandboxphysics.com/labs/gravitational-wave-lens-fringes/): Read the delay as you move the lens toward the line of sight.

## Sources

1. [Wave effects in gravitational lensing of gravitational waves from chirping binaries, Sec. 2](https://arxiv.org/abs/astro-ph/0305055v2)
2. [Search for gravitational-lensing signatures in the full third observing run, Eq. 1](https://arxiv.org/abs/2304.08393v1)

## Keep reading

Gravitational-wave lensing and frequency fringes · Reading path

### From two paths to a lens search

1. [Article **What a detector hears when one chirp takes two paths**](https://sandboxphysics.com/stories/gravitational-wave-lensing/one-chirp-two-paths/)
2. [You are here **Why a lens delays a gravitational wave**](https://sandboxphysics.com/stories/gravitational-wave-lensing/diffraction-and-the-lens-time-delay/)
3. [Article **The exact amplification factor of a point-mass lens**](https://sandboxphysics.com/stories/gravitational-wave-lensing/point-lens-amplification-factor/)
4. [Article **Can a detector tell a lensed chirp from an ordinary one?**](https://sandboxphysics.com/stories/gravitational-wave-lensing/matched-filtering-a-lensed-chirp/)
5. [Article **Recompute point-lens amplification and test a lens search**](https://sandboxphysics.com/stories/gravitational-wave-lensing/recompute-the-point-lens-amplification/)
6. [Article **How to read the O3 search for lensed gravitational waves**](https://sandboxphysics.com/stories/gravitational-wave-lensing/reading-the-o3-microlensing-search/)