GW luminosity depends on the 3rd time derivative but quadrupole formula depends on the 2nd time derivative?

Question

The quadrupole formula for GW emission (see here) states that the metric perturbation is given by:

\begin{align} \bar{h}_{ij}(t,r) = \frac{2G}{c^4 r} \ddot{I}_{ij}(t - r/c) \end{align}

This depends on the second time derivative of the quadrupole moment. However, the power, or 'luminosity', of a GW is given by:

\begin{align} P = \sum_{ij} \frac{G}{5c^5} \left(\frac{d^3 I_{ij}}{d t^3}\right)^2 \end{align}

which depends on the third derivative. This means you can have a GW providing a non-zero contribution to the metric tensor, and hence the dynamics of a system, without actually carrying any energy (and hence without existing?). What am I missing here?

Answer 1

The quadrupole formula is an approximate result which is valid at distances $r \gg c \tau \gg d$, where $\tau$ is the timescale of variation and $d$ is the size of the system. So, it works when you're very far away from masses moving nonrelativistically.

If you had $\ddot{I}_{ij} \neq 0$ and $\dddot{I}_{ij} = 0$, then you would necessarily have a system which is getting unboundedly large over time (e.g. it could contain uniform accelerating masses going out to infinity), which would cause the formula to break down. Usually, the formula is applied to approximately periodic motion, where $\dddot{I}_{ij}$ is comparable in magnitude to $\ddot{I}_{ij} / \tau$, guaranteeing a nonzero power.

It's still interesting to consider the case $\ddot{I}_{ij} \neq 0$ and $\dddot{I}_{ij} = 0$, even though you can't apply the quadrupole formula here. This question is analogous to asking whether, in electromagnetism, a uniformly accelerated charge emits radiation. That is itself quite subtle and kind of depends on what you mean by "radiation". You can find a discussion of some viewpoints here.

Answer 2

This is in fact a pretty subtle issue. Instead of trivializing this question, let's rephrase it with a concrete example:

Because one can easily mimic sin cos functions with piece-wise quadratic functions, thus a binary can be mimicked by a motion that's piece-wise quadratic in time, e.g.

Here the mimickers are piece-wise quadratic functions of time, thus $\ddot I_{ij} \neq 0, \quad \dddot I_{ij} = 0 $, and nothing gets unboundedly large here.

Here there're only two pieces and already look pretty good. One can further split into more pieces, doing quadratic interpolation in between, and the mimickers will approach true binary. However, if we use the simple estimate of power we get 0 for the mimickers and nonzero for true binary.

It seems like we have a discrepancy. But in fact, the quadruple formula is still valid here. The subtlety is, between the pieces, the metric perturbation $\ddot I_{ij}$ is actually not continuous, thus the power $\dddot I_{ij}$, although is 0 almost everywhere, is a collection of delta-function pulses.

For the accelerating particle, we have similar discontinuities in metric as we enter/exit the light cone of the particle. It is true even for a massless particle. This is the gravitational shock waves studied by 't Hooft in the 80s: gravitational shock wave

If one wants to make it more fancy, we can even build traversable wormholes out of that: Maldacena, Stanford, Yang

Answer 3

Unless the strain has a time derivative then the metric ($\eta_{\mu \nu} + h_{\mu \nu}$) is time-independent and therefore would not represent a gravitational wave. But for the strain to have a time derivative requires $\ddot{I_{ij}}$ to have a time derivative and hence $\dddot{I_{ij}} \neq 0$.

Answer 4

So if I have 2 masses ($m=1$) at $\pm R=\pm 1$ rotating at $\omega=1$ in the $x-y$ plane, then:

$$ \vec r_1(t) = (\cos t)\hat x +(\sin t)\hat y $$ $$ \vec r_2(t) = -\vec r_1(t) $$

so

$$ \vec r_1\vec r_1 = \left[\begin{array}{ccc} \cos^2 t & \cos t \sin t & 0\\ \cos t \sin t & \sin^2 t & 0 \\ 0 & 0 & 0 \end{array}\right] = \vec r_2\vec r_2 $$

while:

$$ \delta \times r^2 = \left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$$

for a quadrupole moment:

$$ I \equiv 3\vec r_1\vec r_1 + 3\vec r_2\vec r_2 - 2\delta r^2 $$

being:

$$ I =2\left[\begin{array}{ccc} 3\cos^2t -1 & 3\cos t\sin t & 0\\ 3\cos t \sin t & 3\sin^2 t - 1 & 0 \\ 0 & 0 & -1 \end{array}\right]$$

and now I can take derivatives:

$$ \dot I =6\left[\begin{array}{ccc} \sin 2t & \cos 2t & 0\\ \cos 2t & -\sin 2t & 0 \\ 0 & 0 & 0 \end{array}\right]$$

$$ \ddot I =12\left[\begin{array}{ccc} \sin 2t & -\cos 2t & 0\\ -\cos 2t & -\sin 2t & 0 \\ 0 & 0 & 0 \end{array}\right]$$

$$ \dddot I =24\left[\begin{array}{ccc} -\cos 2t & -\sin 2t & 0\\ -\sin 2t & \cos 2t & 0 \\ 0 & 0 & 0 \end{array}\right]$$

Not sure that helped, since there's already a correct answer.