A loop of length ℓ repeats its motion every ℓ/2, so it radiates gravitational waves at the harmonics fn=2n/ℓ. Cusps, points that briefly reach the speed of light, and kinks, sharp bends, keep the high harmonics strong: the power in harmonic n falls only as n−4/3 or n−5/3. The total power ΓGμ2 does not depend on the loop's size, so the loop shortens at the constant rate ΓGμ until all of its rest energy has become gravitational waves.
This chapter follows the previous one, What the tension of a cosmic string measures. You need vectors, derivatives and a geometric series. Units have c=1, and the loop sits in flat space, which is accurate while it is much smaller than the distance light has travelled since the Big Bang.
A loop is two curves on a sphere
In a suitable gauge, the general solution of the Nambu–Goto equations in flat space is a sum of a left-moving and a right-moving wave,
X(σ,t)=21[X−(σ−)+X+(σ+)],σ±=t±σ,∣X±′∣=1.
For a loop of invariant length ℓ both functions are periodic with period ℓ. Their derivatives are unit vectors, so as σ± runs over one period each derivative traces a closed curve on the unit sphere. Because the loop's centre of mass is at rest, each curve averages to zero. Auclair et al., 2020, Appendix A.1
Differentiating gives the velocity and the tangent of the string:
∂tX=21(X−′+X+′),∂σX=21(X+′−X−′).
At a fixed point σ, after a time ℓ/2 the arguments σ− and σ+ have each moved by ℓ/2, and together they return the configuration to itself. The loop's period is T=ℓ/2, and it radiates at the harmonics
fn=Tn=ℓ2n,n=1,2,3,…
Cusps: where the two curves cross
Two closed curves on a sphere that both average to zero generically cross. At a crossing, X−′=X+′. There the tangent ∂σX vanishes and the speed ∣∂tX∣ equals one: the string doubles back on itself and moves at the speed of light. That event is a cusp. A kink is a jump in one of the two derivatives, left behind when strings reconnect.
Worked example: the loop drawn in the LabShow the answer
The Lab draws a Kibble–Turok loop with X−′(u)=(cosu,sinu,0) and X+′(v)=(cosv,cosψsinv,sinψsinv), using ψ=1.05. Both curves are great circles. The first lies in the horizontal plane, the second in a plane tilted by ψ about the first axis, so they cross at ±(1,0,0). Each crossing happens once per period, which gives two cusps per oscillation moving in opposite directions along that axis. The loop has ℓ=2π and T=π in these units. A direct numerical evaluation of ∣∂tX∣ at the crossing gives 1.0000.
At a cusp the emission is beamed. At harmonic n the burst is concentrated in a cone of opening angle θb≈[2/(3n)]1/3 around the cusp's direction of motion. Auclair et al., 2020, Appendix A.2
How the power is shared among harmonics
Averaged over the loops in a network, the share of power in harmonic n is modelled as a power law,
Pn=ζ(q)Γn−q,n=1∑∞Pn=Γ,
with q=4/3 for cusps, 5/3 for kinks and 2 for collisions between kinks. The Riemann zeta function ζ(q) makes the shares add up to Γ. Simulations of realistic loops find total powers sharply peaked near Γ≈50. Auclair et al., 2020, Section 4.1
Worked example: how much power sits in the low harmonicsShow the answer
ζ(4/3)=3.601 and ζ(5/3)=2.124. The fundamental carries 1/ζ(q) of the power: 27.8% for cusps and 47.1% for kinks. Summing the first ten terms of the series gives 62.0% for cusps and 85.3% for kinks. A kinky loop puts more of its energy into its lowest notes.
A computer has to stop the sum somewhere. Summing 105 cusp harmonics keeps 98.2% of the power and misses 1.8%; for kinks the first 105 terms keep 99.967%. The NANOGrav analysis also stops at 105 harmonics. NANOGrav Collaboration, 2023, Section 5.4
Steady shrinking and a finite life
The power ΓGμ2 contains no length. Energy conservation, d(μℓ)/dt=−ΓGμ2, then gives
dtdℓ=−ΓGμ,ℓ(t)=ℓi−ΓGμ(t−ti),τ=ΓGμℓi.
Integrating the power over the lifetime returns ΓGμ2τ=μℓi: all of the rest energy leaves as gravitational waves. As the loop shrinks, its harmonics rise in frequency. A small loop therefore sings ever higher until it vanishes.
Worked example: a loop born one second after the Big BangShow the answer
Take α=0.1, so the loop is born with ℓ=0.1 light-seconds, or 3.0×107m. Its fundamental frequency at emission is 2c/ℓ=20Hz. In the flat model with a constant number of particle species used in this course, the universe was one second old at 1+z=4.88×109. The wave arrives today at 20Hz/4.88×109=4.1×10−9Hz, inside the pulsar-timing band. With Gμ=10−10 the loop lives 2×107s, about eight months, so it radiates almost at the redshift where it was born.
Lab task
In the cosmic-string Lab, keep the default and watch two cusps flash in opposite directions each period. Then tick kinks instead of cusps. The flat high-frequency readouts hardly change, because the plateau depends on the total power Γ and not on how it is shared. The value at one cycle per year changes, because there the low harmonics of large loops matter.
Exercises
Find the fundamental frequency at emission of a loop one light-year long.
What is the cusp beaming angle at harmonic n=105?
Show that a loop born at time ti with ℓi=αti lives α/(ΓGμ) times its birth age.
Why does doubling the share of power in the fundamental leave the total radiated energy of a loop unchanged?
AnswersShow the answer
f1=2c/ℓ=2/(1yr)=6.3×10−8Hz.
θb≈[2/(1.732×105)]1/3=0.0226rad, about 1.3∘. At n=103 the cone is about 6∘ wide, so higher harmonics are beamed more tightly.
τ/ti=ℓi/(ΓGμti)=α/(ΓGμ). For α=0.1 and Γ=50 this is 0.002/Gμ.
The total power is fixed at ΓGμ2 and the lifetime at μℓi/(ΓGμ2), so the radiated energy is always μℓi. Changing q only moves that energy between harmonics.