# Unbounded perturbed geometry due to analyticity

I have an interesting mystery for you, loosely inspired on work related with this question.

I have a certain Ansatz for a gravitational wave perturbation of the metric $h_{\mu \nu}$ that is nonzero near an axis of background flat Minkowski spacetime

The Ansatz has the following form:

$g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu} = \begin{bmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 + V(x,y,t) & U(x,y,t) & 0 \\ 0 & U(x,y,t) & 1 -V(x,y,t) & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$

The Ansatz has the following property:

$$h^{\mu}_{\mu}=0$$

I want the Ansatz to be also in the Transverse-Traceless gauge, which implies

$$\partial_{\mu} h^{\mu \nu} = 0$$

When I apply this condition on the Ansatz, I'm left with two nontrivial conditions:

$$\frac{\partial U}{\partial x}= \frac{\partial V}{\partial y}$$

$$\frac{\partial U}{\partial y}=- \frac{\partial V}{\partial x}$$

Oh but what coincidence, these are the Cauchy-Riemann equations!

Now, is well known that analytic complex functions are either constant or unbounded.

I am trying to interpret this correctly:

The Ansatz geometry does not seem to be able to become asymptotically Minkowski, if one asks that the metric is in the Transverse-Traceless gauge. For any far away region from the $x=0, y=0$ axis, $h_{\mu \nu}$ will become larger in magnitude than $\eta_{\mu \nu}$, which seems that is not our linear regime anymore, and would produce some large deformations

Is there an intuitive reason why the Transverse-Traceless gauge is not consistent with a perturbed metric that has this form? what if I would've tried a compact set, bounded on $z$ as well?

After adding dependence of $z$ to $U$ and $V$, I am left with the following terms for Ricci curvature:

$$R_{xy} = R_{yx} = -\frac{1}{2}( \frac{d^2 U}{dz^2} - \frac{d^2 U}{dt^2} )$$

so if $U$ satisfies the wave equation, the off-diagonals are flat. The diagonals are

$$R_{xx} = \frac{d^2 V}{dx^2} - \frac{d^2 V}{dy^2} + 2 \frac{d^2 U}{dy dx} + \frac{d^2 V}{dt^2} - \frac{d^2 V}{dz^2}$$

Which, if we apply the Cauchy-Riemann equations, and the fact that the z and t dependence satisfy the wave equation (a dependency like $e^{I(kz-\omega t)}$ would suffice) then it follows that the geometry is Ricci flat

After some consideration, Ricci flatness does not seem to be a big deal, as according to this, a manifold can be Ricci flat and still have gravitational waves. But the problem of unbounded perturbation remains: a holomorphic function (and $U$ and $V$ must constitute one) will be either constant along each ($t,z$) hyperslice, or will be unbounded for large $x$, $y$. This seems to be a big problem

• The eigenvalues of $g_{\mu\nu}$ are $-1,1,1\pm\sqrt{U^2+V^2}$. Three of them have to be positive (and one negative), so we must have $U^2+V^2<1$. As these functions are bounded, they are constant. By a linear change of variables, you can prove that $h_{\mu\nu}=0$. – AccidentalFourierTransform Dec 11 '15 at 0:14
• so that proves that gravitational waves with the shown symmetry do not exist? – lurscher Dec 11 '15 at 0:16
• what do you mean by shown symmetry? your gauge condition? or your ansatz? if you mean the latter, then yes: your ansatz is just flat spacetime. – AccidentalFourierTransform Dec 11 '15 at 0:20
• it doesn't seem to be flat: I computed the first order correction to some Ricci components. $R_{22}$ entry is proportional to $\ddot{V}(x,y,t)$ – lurscher Dec 11 '15 at 0:45
• @lurscher Doesn't $\Box h^{\mu\nu} = 0$ give $\Box V = 0$ and therefore $\ddot V = \Delta V = 0$ due to $V$ being spatially constant (at least in an actual solution of the linearized equations)? – Sebastian Riese Dec 11 '15 at 1:02

The Cauchy-Riemann equations that follow from the gauge condition indeed imply that $V$ and $U$ are constant along surfaces of constant $z$ and $t$.

This is proven by the fact, that two functions fulfilling the Cauchy-Riemann equations on all of $\mathbb R^2$ are the real and imaginary part of an entire function. If that function is bounded (which $U$ and $V$ must be if they do not violate the smallness of the perturbation $h^{\mu\nu}$, the boundedness also follows from the fact that $g^{\mu\nu}$ must have three positive and one negative eigenvalue, the eigenvalues of $g^{\mu\nu}$ are $-1$, $1$ and $1 \pm \sqrt{U^2 + V^2}$).

However, there is no restriction that $U$ or $V$ be constant along the $z$-direction. So the interpretation of this ansatz metric is a gravitational plane wave travelling in $z$-direction. In fact, $U$ and $V$ fulfil the one-dimensional wave equation (as they are constant in $x$ and $y$ direction the Laplace operator simplifies): $$\partial_t^2 f - \partial_z^2 f = 0.$$

Furthermore, it is not surprising that a wave that can only travel in $z$ direction must have infinite wave fronts: If the wave front were finite there would be diffraction effects, that would generate components of motion in the $x$- and $y$-direction, the polarization restrictions of gravitational waves (namely, that they are transversal in the chosen gauge), however, forbids such terms, so the solution must be an infinite wave front.

A gravitational wave solution in vacuum, which means away from any source (be it matter or electromagntic radiation), must be Ricci flat, because the field equations in vacuum reduce to: $$R^{\mu\nu} = 0.$$

If we choose $U$ and $V$ as $e^{i k^\mu x_\mu}$ (the wave equation then reduces to $k^\mu k_\mu = 0$) and add the rotated solutions (that is plane waves travelling in $x$ and $y$ direction), we get a complete basis set of wave states, from which we can construct physical solutions as wave packets. This is in complete analogy to the way physical wave packet solutions of the Maxwell equations can be Fourier developed in terms of unphysical plane wave solutions. A wave packet with finite extension in the $x$ or $y$ direction must include some non-zero $k_x$ and $k_y$ components by the uncertainty relation for the Fourier transform.