The exact amplification factor of a point-mass lens

This chapter builds on why a lens delays a gravitational wave. You need complex exponentials, the Gaussian integral $\int_{-\infty}^{\infty}e^{i a u^2/2}\,\mathrm du=\sqrt{2\pi/|a|}\,e^{i\pi\,\mathrm{sgn}(a)/4}$, and the idea of a Fourier transform.

Adding up every path

Huygens' principle says each point of the lens plane re-emits the wave. A path through the point $x$ arrives late by $t_d(x)$, so at frequency $f$ it contributes a phase $e^{2\pi i f t_d(x)}$. Summing all points with equal weight and dividing by the same sum without a lens defines the amplification factor. In the dimensionless variables of the last chapter,

with $w=8\pi GM_{Lz}f/c^3$ and $T=\tfrac12|x-y|^2-\ln|x|-\phi_m(y)$. This is Eq. (11) of the 2003 paper written with $w$ and $T$. Takahashi and Nakamura 2003, Sec. 2

Two conventions sit inside that formula.

• Normalization. Without the lens, $T=\tfrac12|x-y|^2$, and the two Gaussian integrals give $2\pi i/w$. The prefactor cancels it, so $F=1$. Every departure from 1 is the lens.
• Sign of the phase. We take the Fourier transform of the strain as $\tilde h(f)=\int h(t)\,e^{+2\pi i f t}\,\mathrm dt$, the convention of Cutler and Flanagan. A signal delayed by $\Delta t$ then picks up the factor $e^{+2\pi i f\Delta t}$, which is why the exponent above is $+iwT$. The lensed spectrum is $\tilde h_{\rm L}(f)=F(f)\,\tilde h(f)$. Cutler and Flanagan 1994, Eq. 20

Recovering the two images

When $w$ is large, $e^{iwT}$ oscillates rapidly and almost every part of the integral cancels. Only the neighbourhoods of the points where $T$ is stationary survive. Near such a point $x_j$, expand $T$ to second order along its two principal directions:

Each direction is a Gaussian integral, $\sqrt{2\pi/(w|\lambda|)}\,e^{i\pi\,\mathrm{sgn}(\lambda)/4}$. Multiplying the two and the prefactor $w/(2\pi i)=(w/2\pi)\,e^{-i\pi/2}$ gives

The product $\lambda_1\lambda_2$ is the determinant of the second-derivative matrix of $T$, and its inverse is the magnification $\mu_j$. The last factor depends only on the kind of stationary point: 1 at a minimum, $e^{-i\pi/2}=-i$ at a saddle and $-1$ at a maximum. That is Eq. (15) of the 2003 paper,

and the same phase appears as Eq. 1 of the LIGO–Virgo–KAGRA lensing search, where $n_j$ is called the Morse index. LIGO–Virgo–KAGRA 2023, Sec. 3

A point lens has one minimum and one saddle, and no maximum. With the first image at $T=0$,

which is Eq. (18) of the 2003 paper.

Fringes in frequency

Square the magnitude. With $a=|\mu_+|^{1/2}$, $b=|\mu_-|^{1/2}$ and $\varphi=2\pi f\Delta t_d$,

The magnitude swings between $a-b$ and $a+b$, and one full swing takes $\Delta f=1/\Delta t_d$. The $-i$ of the saddle moves the peaks: they sit where $\sin\varphi=1$, a quarter cycle away from where two in-phase copies would add.

Worked example. For $y=0.5$, $\mu_+=1.591$ and $|\mu_-|=0.591$, so $a=1.261$ and $b=0.769$. The magnitude runs between 0.49 and 2.03. With $M_{Lz}=1000\,M_\odot$ the delay is 19.9 ms, so the peaks repeat every 50 Hz.

The exact formula

For an isolated point mass the integral can be done in closed form. The 2003 paper quotes the result, attributing it to Peters (1974):

with $\phi_m(y)=(x_m-y)^2/2-\ln x_m$ and $x_m=(y+\sqrt{y^2+4})/2$. Here $\Gamma$ is the gamma function and ${}_1F_1$ the confluent hypergeometric function. This is Eq. (17) of the 2003 paper. We did not take it on trust: the recompute guide evaluates the integral a second way, without the closed form, and the two agree to $4.2\times10^{-4}$ in real and imaginary parts on 20 test points.

A note on one sign. The paper's Eq. (12) prints $+\phi_m(y)$ in $T$, while its text says $\phi_m$ is chosen so that the earliest arrival is zero. We record this as an observation about notation. Eq. (17) as printed matches the numerical integral with the zero-minimum choice, $-\phi_m$, and that is the convention used throughout this course.

Long waves: why the factor tends to 1

For small $w$, expand each factor to first order. $\Gamma(1+\epsilon)\approx1-\gamma\epsilon$ with Euler's constant $\gamma=0.5772$, so $\Gamma(1-iw/2)\approx1+i\gamma w/2$. The hypergeometric factor is $1+(iw/2)(iwy^2/2)+\dots$, which differs from 1 only at order $w^2$. Collecting terms,

The leading change in magnitude is $|F|\approx1+\pi w/4$, the same for every $y$. A lens much smaller than the wavelength cannot form images. It only raises the amplitude slightly and adds a slow phase drift.

Worked example. A $15\,M_\odot$ lens at $f=20$ Hz has $w=1.24\times10^{-4}\times15\times20=0.037$, so $|F|\approx1+0.029=1.029$. The exact formula gives 1.029 at $y=0.5$, while the two-image formula claims 1.50. Ray optics overstates the effect by a factor of about 17 in $|F|-1$.

Short waves: approaching two images

For large $w$ the exact $F$ approaches $F_{\rm geo}$. How fast depends on $y$. In our reference runs the relative difference at $y=1$ falls from $2.6\times10^{-3}$ at $w=30$ to $2.6\times10^{-5}$ at $w=1000$. At $y=0.3$ it is $3.1\times10^{-2}$ at $w=30$, and on the way down it does not fall steadily: $6.5\times10^{-3}$ at $w=100$ but $8.7\times10^{-3}$ at $w=300$. The leftover difference itself oscillates with frequency, so a single comparison point can mislead. Small $y$ means a nearly aligned source and a saddle image close to the lens, where diffraction lasts longest.

Worked example. With $M_{Lz}=1000\,M_\odot$ and $y=0.5$, the exact and two-image magnitudes are 1.825 and 1.828 at 20 Hz, where $w=2.5$, and 1.411 and 1.438 at 100 Hz, where $w=12.4$. Close agreement at one frequency is luck of the fringe phase: at 31.5 Hz, near the first fringe minimum, they are 0.961 and 0.891.

Lab tasks

1. In the gravitational-wave lens Lab, choose the preset Waves too long to split. Read the solid and dashed curves of the amplification plot at the left edge, and check $1+\pi w/4$ with $w$ from the lens mass shown.
2. Choose the preset Two paths, clear fringes and count fringes between 50 and 200 Hz. Compare with $150\,\mathrm{Hz}\times\Delta t_d$.
3. Drag the lens toward $y=0.1$ and watch the phase plot. Where do the two curves part, and why does that happen at small $y$ first?

Exercises

1. Show that the prefactor $w/(2\pi i)$ makes $F=1$ when there is no lens.
2. Derive the phase factor $-i$ for a saddle point from the two one-dimensional Gaussian integrals.
3. For $y=1$, find the range of $|F_{\rm geo}|$ and the fringe spacing when $M_{Lz}=500\,M_\odot$.
4. Use the low-frequency formula at $y=1$, $w=0.1$ to estimate $F$. The exact value is $1.0722-0.0989i$. Where does the remaining difference come from?

Next: can a detector tell a lensed chirp from an ordinary one?