## Deep Future of Gravitational Waves – Part 1

In a Newtonian Universe orbital motion lasts forever, although dynamical chaos can make its evolution in the long term increasingly unpredictable as time passes. In Einstein’s Universe, where General Relativity applies, even the simplest orbital system of binary stars loses energy and angular momentum as gravitational waves, contracting orbital separations over trillions of trillions of years.

First, a review of the underlying physics.

For a pair of binary stars of masses $$M_1$$ & $$M_2$$ separated by distance $$r$$, the important features are the total energy of the system $$E$$ and the angular velocity $$\omega$$. These are calculated by:

$$E(r) = -\frac{G M_1 M_2}{2 r}$$

$$\omega^2(r) = \frac{r^3}{G(M_1+M_2)}$$

Another useful figure is the Schwarzschild Radius, which is the radius of a black hole’s event horizon, but defined for any mass as:

$$r_s = \frac{2GM}{c^2}$$

What follows is a simplified version of the relevant equations, with some important limitations on its applicability.

The power $$P$$ radiated away as gravitational waves by the binary is computed by:

$$P = \frac{dE}{dt} = -\frac{128}{5}\left(\frac{G}{c^5}\right)\left(\frac{E}{\omega}\right)^2$$

Using the Chain Rule expand the first time derivative of Energy as follows:

$$\frac{dE}{dr}.\frac{dr}{dt}$$

The first term we can evaluate as:

$$\frac{dE}{dr} = \frac{G M_1 M_2}{2 r^2}$$

Which gives the following:

$$\left(\frac{G M_1 M_2}{2 r^2}\right) \frac{dr}{dt} = \left(\frac{E}{r}\right) \frac{dr}{dt} = -\frac{64}{5}\left(\frac{G}{c^5}\right)\left(\frac{E}{\omega}\right)^2$$

From which we get:

$$\frac{dr}{dt} = -\frac{64}{5}\left(\frac{G}{c^5}\right)\left(\frac{Er}{\omega^2}\right)$$

This allows us to compute the time it takes for the binary system to shrink from an initial radius $$r_o$$ to a final radius $$r_f$$ as follows:

$$\int_{r_o}^{r_f} -\frac{5}{64}\left(\frac{c^5}{G}\right) \frac{\omega^2}{E r} dr = \int_{t_o}^{t_f} dt$$

Which gives:

$$(t_f – t_o) = \left[\frac{5}{128}\left(\frac{c^5}{G}\right) \frac{\omega^2}{E}\right]_{r_f}^{r_o}$$

Remembering that the total Energy and the angular velocity are functions of radius, we can substitute those functions back in and we get:

$$(t_f – t_o) = \left[\frac{5}{256}\left(\frac{c^5}{G^3}\right)\frac{r^4}{(M_1 M_2)(M_1+M_2)}\right]_{r_f}^{r_o}$$

If we assume that $$M_1 = M_2 = M$$,$$t_o = 0$$ and $$r_f = 0$$ we get:

$$t_f = \left[\frac{5}{512}\left(\frac{c^5}{G^3}\right)\frac{r^4}{M^3}\right]_0^{r_o}$$

Substituting in the expression for the Schwarzschild Radius,$$r_s$$, we get the following:

$$t_f = \left[\frac{5}{64}\left(\frac{r}{c}\right)\left(\frac{r}{r_s}\right)^3\right]_0^{r_o}$$

For a non-rotating Black Hole, the point of no return isn’t the Event Horizon, but the Innermost Stable Circular Orbit (ISCO) which is three times larger:

$$r_{ISCO} = 3 r_s$$

Which gives the following equation for the in-spiral time from ISCO as:

$$t_f = \frac{135}{64}\left(\frac{3 r_s}{c}\right)$$

In reality the equations of General Relativity become highly non-linear past this point and this simple calculation is somewhat inaccurate. Unless our binary system is two black holes, it’s also never achieved as the stars will have collided.

So what does it all mean? Part Two will explore that question.