Null infinity, Bondi news and classical memory — Examples

The scalar $Y$ is smooth because it is the restriction of a polynomial on ${\mathbb{R}}^3$. By contraction, $q^{AB}{T}_{AB}=D^{2}Y-\frac{1}{2}(2)D^{2}Y=0$. In spherical coordinates at the equator $\theta =\pi \mathrm{/}2$, $D_{\theta }{D}_{\theta }Y=3$, $D_{\varphi }{D}_{\varphi }Y=0$, and $D^{2}Y=3$, so $T_{\theta \theta }=3\mathrm{/}2\mathrm{\ne }0$. Thus $T$ is a nonzero smooth trace-free tensor.

For dimensionless $z$, define $b(z)=e^{-1\mathrm{/}(1-z^2)}$ for $\mathrm{\mid }z\mathrm{\mid }<1$ and $b(z)=0$ for $\mathrm{\mid }z\mathrm{\mid }\ge 1$. Every derivative of the interior formula is a polynomial in $(1-z^2{)}^{-1}$ and $z$ times $e^{-1\mathrm{/}(1-z^2)}$, which tends to zero at both endpoints; hence the zero extension is smooth. With $Z={\int }_{-1}^{1}b(z)dz>0$, the change of variable $z=u\mathrm{/}\tau$ shows that $g_{\tau }(u)=b(u\mathrm{/}\tau )\mathrm{/}(\tau Z)$ is smooth, nonnegative, supported on $[-\tau ,\tau ]$, has units ${\mathrm{l}\mathrm{e}\mathrm{n}\mathrm{g}\mathrm{t}\mathrm{h}}^{-1}$, and has integral one. Its primitive ${\mathrm{\Phi }}_{\tau }(u)={\int }_{-\mathrm{\infty }}^u{g}_{\tau }(v)dv$ is smooth, equals $0$ for $u\le -\tau$, equals $1$ for $u\ge \tau$, and has derivative $g_{\tau }$.

The fields $C_{AB}=a{\mathrm{\Phi }}_{\tau }{T}_{AB}$ and $N_{AB}=a{g}_{\tau }{T}_{AB}$ are smooth and trace-free, and $N={\mathrm{\partial }}_{u}C$. Therefore $\mathrm{\Delta }{C}_{AB}=a{T}_{AB}$ by the normalization $\int g_{\tau }=1$. Define $m(u,x):=\frac{1}{4}{D}_A{D}_B(C^{AB}(u,x)-C^{AB}(-\tau ,x))-\frac{1}{8}{\int }_{-\tau }^u{N}_{AB}(v,x)N^{AB}(v,x)dv\mathrm{.}$ This is smooth, has $m(-\tau ,x)=0$, and direct differentiation verifies the imposed local coefficient equation [F2]. The metric construction [L1] gives a smooth Bondi-Sachs end with these coefficients, but does not impose the vacuum Ricci equations.

Let $X^A:=D_B{T}^{AB}$. In spherical coordinates, $D_A{X}^{A}d\mathrm{\Omega }=\{{\mathrm{\partial }}_{\theta }(\sin \theta X^{\theta })+{\mathrm{\partial }}_{\varphi }(\sin \theta X^{\varphi })\}d\theta d\varphi$. The $\varphi$ boundary terms cancel by periodicity, and smoothness of $X$ makes its meridional component $X^{\theta }$ bounded near either pole, so $\sin \theta X^{\theta }\to 0$ there. Hence ${\int }_{S^2}{D}_A{D}_B{T}^{AB}d\mathrm{\Omega }=0$. Integrating the explicit $m$ from step 2.1 over the sphere therefore yields $m_B(\tau )-m_B(-\tau )=-\frac{a^2}{32\pi }({\int }_{-\tau }^{\tau }{g}_{\tau }(u{)}^{2}du)\left({\int }_{S^2}{T}_{AB}{T}^{AB}d\mathrm{\Omega }\right)\mathrm{.}$ The time integral is positive because $g_{\tau }$ is continuous and has unit integral. The sphere integral is positive because $q$ is positive definite and $T$ is continuous and nonzero by step 1.1. Since $a\mathrm{\ne }0$, the mass change is strictly negative. This derivation uses only the imposed coefficient equation, not the smooth-vacuum mass-loss proposition.

The pulse has $N(-\tau )=N(\tau )=0$, so [F3] applies. Its integrated shear change is $\mathrm{\Delta }C=aT$, hence the formal first-order detector response is $\delta s^A(\tau )=\frac{a}{2r}{T}^A{}_B{s}_0^B+O(r^{-2}\mathrm{\mid }{s}_0\mathrm{\mid })$. The finite-radius qualification is the stated one; the formal calculation uses no nonlinear amplitude remainder.

If the normalized local family in the Example is supplied, apply the quantitative clause of [F3] with $u_1=-\tau$, $u_2=\tau$ and $\mathrm{\Delta }C=aT$. The news vanishes at both endpoints because the smooth bump and all its derivatives vanish there. The result is the displayed actual Jacobi displacement and velocity estimate, with exactly that theorem's fixed-family amplitude constants. This uses the imposed formal boundary coefficients and detector-family hypotheses, not an inference of a vacuum metric or a measurement.

The restriction $Y=\frac{1}{2}(3x_3^2-1)$ is smooth on the sphere because it is the restriction of a polynomial. Put $T_{AB}:=D_A{D}_{B}Y-\frac{1}{2}{q}_{AB}{D}^{2}Y$. Its trace is zero. At $\theta =\pi \mathrm{/}2$, $D_{\theta }{D}_{\theta }Y=3$, $D_{\varphi }{D}_{\varphi }Y=0$ and $D^{2}Y=3$, so $T_{\theta \theta }=3\mathrm{/}2$; hence $T$ is nonzero.

With $t=u+r$, the metric $g_0=-d{u}^2-2dudr+r^{2}q$ becomes $-d{t}^2+d{r}^2+r^{2}q$, the Minkowski metric in spherical spatial coordinates. Thus its Bondi data are exactly $C_{AB}=N_{AB}=m=0$. Since $f=aY$ is independent of $u$, [F2] gives $C_{AB}^{(f)}=-2a{D}_A{D}_{B}Y+a{q}_{AB}{D}^{2}Y=-2a{T}_{AB}$ and $N_{AB}^{(f)}=0$. We use only these as formal boundary data; we do not assert that they are the coefficients of a finite transformed vacuum metric.

Set $m(u,x)=0$ for the formal data. With $N_{AB}=0$, [F3] is satisfied and $m_B(u)=\frac{1}{4\pi }{\int }_{S^2}md\mathrm{\Omega }=0$, so $d{m}_B\mathrm{/}du=0$. The shear $C_{AB}^{(f)}=-2a{T}_{AB}$ is independent of $u$, hence $\mathrm{\Delta }{C}_{AB}=0$ on every interval. These conclusions follow directly from the stipulated data and constraint; neither vacuum mass-loss nor the memory proposition is invoked.