### Decay of waves IIIb: tails for homogeneous linear equation on curved background

Now we will actually show that the specific decay properties of the linear wave equation on Minkowski space–in particular the strong Huygens’ principle–is very strongly tied to the global geometry of that space-time. In particular, we’ll build, by hand, an example of a space-time where geometry itself induces back-scattering, and even linear, homogeneous waves will exhibit a tail.

For convenience, the space-time we construct will be spherically symmetric, and we will only consider spherically symmetric solutions of the wave equation on it. We will also focus on the 1+3 dimensional case.

Spherically symmetric space-times and their wave equations

In a 1+3 dimensional spherically symmetric space-time, associated to each point (unless the point is on the symmetry axis; that is, a fixed point under the symmetry action) is a two-dimensional sphere, the orbit of the point under the symmetry. Much like the way we define spherical coordinates for 3-dimensional Euclidean space, we can define an analogous system of coordinates for such manifolds. (For more detailed and more involved discussion of this, see Szenthe, On the global geometry of spherically symmetric space-times, Math. Proc. Cam. Phil. Soc. (2004) 137:741–754.)

The spherical symmetry in particular implies that, away from the symmetry axis, our space-time take the form of a warped product $M\setminus \Gamma = Q \rtimes_r \mathbb{S}^2$ where $M$ is the space-time, $\Gamma$ is the axis, $Q$ is the two-dimensional quotient manifold, $\mathbb{S}^2$ is given the standard metric on the unit sphere, and the warping function $r$ is the area-radius of the symmetry spheres. (That is, $r$ is defined to be the square root of $(4\pi)^{-1}$ times the surface area of the orbit of the spherical symmetry.) So we can build a coordinate system on $M\setminus\Gamma$ by taking an arbitrary coordinate system on the 1+1 dimensional Lorentzian manifold $Q$ and joining to it some coordinate system on the standard sphere.

Now, any 1+1 Lorentzian manifold admits a (in fact a large number of) double null coordinate system. That is, we can always find a pair of functions $u,v:Q\to \mathbb{R}$ so that they are linearly independent functions, with $\langle du, du\rangle = \langle dv,dv\rangle$ under the induced Lorentzian metric on $Q$ (this follows from the fact that at each point $q\in Q$, the tangent space $T_qQ$ admits two such preferred directions, and that Frobenius theorem is trivial in two dimensions). Under an assumption of time-orientability, we can find some function $\Omega:Q\to \mathbb{R}$ such that the metric on $M$ can be expressed as

Equation 27
$\displaystyle ds^2 = -\Omega^2(u,v) du\tilde{\otimes}dv + r^2(u,v) d\omega^2$

where $d\omega^2$ is the standard metric on the sphere.

Let us now consider a spherically symmetric solution to the wave equation on such a space-time.

Equation 28
$\Box \phi = 0$

Expanding the d’Alembertian operator as the Laplace-Beltrami operator for the Lorentzian manifold $M$, we get that

Equation 29
$\displaystyle \Box = -\frac{1}{2r^2\Omega^2}(\partial_u r^2\partial_v + \partial_v r^2 \partial_u) + \frac{1}{r^2} \triangle_{\mathbb{S}^2}$

which in particular implies that the function $\psi := r \phi$ will solve the following equation

Equation 30
$\displaystyle \frac{1}{\Omega^2} \partial^2_{u,v}\psi = \frac{\partial^2_{u,v}r}{r\Omega^2}\psi$

So in particular $\psi$ will solve a 1+1 dimensional wave equation with source term on manifold $Q$ (which we can partially complete by adding the axis $\Gamma$ and imposing the Dirichlet boundary condition there). This will be the starting point of our construction (recall the support propagation properties of the wave equation from Part IIb of this series).

Consider the following partial Penrose diagram

Figure 5

where we assume that a background double-null coordinate system $(u,v)$ is fixed, and so for a solution whose initial data is strictly supported inside the draw region, its future evolution will not depend on how we (smoothly) complete the space-time.

To prescribe the geometry we do so by describing the functions $\Omega,r$ as functions of the coordinates; we will build the space-time such that in the yellow regions of Figure 5 the space-time is Minkowskian, and that we have a perturbation in the red region. The red region we will assume to be bounded by $v\in (0,\infty)$ and $u\in (0,u_0)$. This we will take to be a poor-man’s version of a radiating space-time, where the red region represents some sort of out-going gravitational disturbance. Of course, here we prescribe the metric freely, rather than try to solve for it using some sort of the dynamical model, so the actual form of the metric is somewhat aphysical. Nevertheless this should give some indication of why it is not feasible to expect too good a decay of the solutions of linear wave equations on arbitrary space-times.

On the yellow region near future time-like infinity, we will insert in the Minkowski space solution in the usual parametrization. That is, we assume that there exists some $R_0$ such that $r = \frac{1}{2}(v - u) + R_0$, and that the function $\Omega = 1$ identically.

In the red region, we choose the function $r = \frac{1}{2}(v-u) + R$ so that it joins smoothly to the first yellow region where $u > u_0$. To be more precise, $R$ will be taken to be equal to satisfy $\partial^2_{u,v}R = \frac{\mu(u)}{v-u + 2R_0}$ with $\mu\in C^\infty_c((0,u_0))$ and $\mu\geq 0$. By prescribing initial conditions on the bottom left boundary of the figure, we can assume that in the region $u < 0$ we have $\partial_u R = 0$. On the other hand, we necessarily have that there $\partial_v R = - \int_0^{u_0}\frac{\mu(u)}{v-u + 2R_0} du$ is a function of $v$ alone. By assuming that $\mu, u_0$ are sufficiently small, we can guarantee that $\partial_v r > 0$ in the region, and so we can define a change of variables $v\to v'$ so that in the $(u,v')$ coordinate system, we have that $\partial_u r = - \partial_{v'}r = -1$. We can then set $\Omega$ in the region $u < 0$ so that $\Omega^2 du\tilde{\otimes}dv = du\tilde{\otimes}dv'$, thereby enforcing that the region is also Minkowskian. Lastly we take an arbitrary positive, smooth choice of $\Omega$ in the red region to join the values we have now prescribed for the two yellow regions.

Tail for homogeneous linear wave equation

On this space-time we just constructed, let us consider an initial data for the wave equation such that $\psi$ is an incoming wave-packet. That is, we can assume that for $0 > u > u_1$, we have $\partial_u \psi = 0$, and that $\psi \geq 0$, and $\psi|_{v \leq 0} = 0$. Such solutions are fairly generic in the space of compactly supported Cauchy data. Then using the fact that $\mu(u) \geq 0$, we can integrate Equation 30 to obtain, inductively, that $\psi \geq 0$ for $-R_0 > u_0 > u > u_1$. Furthermore, integrating the expression for $\partial_u\partial_v R$, we have that $|R| = \|\mu\|_\infty O(\ln v)$ in the red region, which implies, for $\mu$ sufficiently small, that $r \sim v$ in the red region.

Therefore the coefficient on the right hand side of Equation 30 decays like $v^{-2}$, so is in particular integrable in the red region. So for small enough $\mu$, we can iterate the fundamental solution representation for the 1+1 dimensional wave equation to get a convergent series representation of the solution $\psi$, with all terms non-negative. The first term of the iteration guarantees that $\psi$ is bounded below inside $u_0 \geq u \geq \epsilon > 0$ for any fixed $\epsilon$. Which then implies that $\partial_v \psi$ is bounded below by $v^{-2}$ when $v$ is sufficiently large along $u = u_0$.

Propagating forward into the yellow region, we have that in the yellow region you have $\partial_v \psi$ decaying like $v^{-2}$. We can transfer this to an estimate on $\phi$: in the yellow region $u > u_0$, we have $\psi(u,v) = \int_{u-2R_0}^v \partial_v \psi(u,s) ds$ using that $\psi$ vanishes on the center axis. But $\partial_v\psi$ is conserved along $u$, so we actually have that $\frac{v-u+2R_0}{(u-2R_0)v} \lesssim \psi(u,v) < \int_{u-2R_0}^\infty \partial_v\psi(u_0,s)ds \lesssim u^{-1}$. Dividing through by $r$ we get a lower bound that, within the eventual Minkowski region, the solution $\phi$ to the original linear, homogeneous wave equation exhibits a $t^{-2}$ tail.

A more physical decay rate

[Thanks to Mihalis for reminding me of this!] One sees from the argument above that the tail decay rate is intimately tied to the rate of decay of $\partial^2_{u,v} r$ in the wave region. And indeed, we can run through the exact same argument as above with the rates

$\displaystyle \partial^2_{u,v}R = \frac{\mu(u)}{(v-u + 2R_0)^p}$

for any power $p \geq 1$. This will lead to the appearance of a corresponding tail of size

$\displaystyle \phi \sim t^{-p-1}$.

The above construction shows that in arbitrary Lorentzian manifolds, one can not do better than a $t^{-2}$ decay. But in the context of general relativity, this heuristic leads to $t^{-3}$ being the more physical decay rate. This has to do with the fact that for physical space-times appearing in relativity theory, we expect the space-time to have a well-defined Bondi mass (which is the limit of the Hawking mass at null infinity. Formally we can read off the Bondi mass from the $r^{-1}$ term in the asymptotic expansion of $g(\nabla r,\nabla r) = \Omega^{-2} \partial_u r \partial_v r$. For the choice of $\partial^2_{u,v} r$ given above, with $2 > p$ the Bondi mass blows up at infinity (the asymptotic expansion is not well defined). For $p > 2$, the Bondi mass is zero. So heuristically we expect the case $p = 2$ to correspond to the physically interesting case where gravitation has non-trivial long range effects. And for this rate we have a cubic decay law expected (as consistent with the so-called “Price’s Law” in general relativity).