5.2 Gravitational Waves

A long time ago, in a galaxy far far away, two black holes collided. Here a “long time ago” means 1.3 billion years ago. And “far far away” means a distance of about 1.3 billion light years.

To say that this was a violent event is something of an understatement. One of the black holes was roughly 35 times heavier than the Sun, the other about 30 times heavier. When they collided they merged to form a black hole whose mass was about 62 times heavier than the Sun. Now $30+35\ne 62$ . This means that some mass, or equivalently energy, went missing during the collision. In a tiny fraction of a second, this pair of black holes emitted an energy equivalent to three times the mass of the Sun.

That, it turns out, is quite a lot of energy. For example, nuclear bombs convert the mass of a handful of atoms into energy. But here we’re talking about solar masses, not atomic masses. In fact, for that tiny fraction of a second, these colliding black holes released more energy than all the stars in all the galaxies in the visible universe put together.

But the most astonishing part of the story is how we know this collision happened. It’s because, on September 14th, 2015, at 9.30 in the morning UK time, we felt it. The collision of the black holes was so violent that it caused an enormous perturbation of spacetime. Like dropping a stone in a pond, these ripples propagated outwards as gravitational waves. The ripples started 1.3 billion years ago, roughly at the time that multi-cellular life was forming here on Earth. They then travelled through the cosmos at the speed of light. The ripples hit the outer edge of our galaxy about 50,000 years ago, at a time when humans were hanging out with neanderthals. The intervening 50,000 years gave us just enough time to join together into hunter-gatherer tribes, develop cohesive societies bound by false religions, invent sophisticated language and writing, discover mathematics, understand the theory that governs the spacetime continuum and, finally, build a machine that is capable of detecting the ripples, turning it on just in time for the gravitational wave to hit the south pole and pass, up through the Earth, triggering the detector.

The purpose of this section is to tell the story above in equations.

5.2.1 Solving the Wave Equation

Gravitational waves propagate in vacuum, in the absence of any sources. This means that we need to solve the linearised equation

$$\square {\bar{h}}_{\mu \nu }=0\text{\hspace{1em}(5.17)}$$

One solution is provided by the gravitational wave

$${\bar{h}}_{\mu \nu }=\mathrm{Re}\left(H_{\mu \nu }{e}^{i{k}_{\rho }{x}^{\rho }}\right)\text{\hspace{1em}(5.18)}$$

Here $H_{\mu \nu }$ is a complex, symmetric polarisation matrix and the wavevector $k^{\mu }$ is a real 4-vector. Usually when writing these solutions we are lazy and drop the Re on the right-hand side, leaving it implicit that one takes the real part. This plane wave ansatz solves the linearised Einstein equation (5.17) provided that the wavevector is null,

$$k_{\mu }{k}^{\mu }=0$$

This tells us that gravitational waves, like light waves, travel at the speed of light. If we write the wavevector as $k^{\mu }=(\omega ,\mathbf{k})$ , with ω the frequency, then this condition becomes $\omega =\pm |\mathbf{k}|$

Because the wave equation is linear, we may superpose as many different waves of the form (5.18) as we wish. In this way, we build up the most general solution to the wave equation.

Naively, the polarisation matrix $H_{\mu \nu }$ has 10 components. But we still have to worry about gauge issues. The ansatz (5.18) satisfies the de Donder gauge condition ${\partial }^{\mu }{\bar{h}}_{\mu \nu }=0$ only if

$$k^{\mu }{H}_{\mu \nu }=0\text{\hspace{1em}(5.19)}$$

This tells us that the polarisation is transverse to the direction of propagation. Furthermore, the choice of de Donder gauge does not exhaust our ability to make gauge transformations. If we make a further gauge transformation $h_{\mu \nu }\to h_{\mu \nu }+{\partial }_{\mu }{\xi }_{\nu }+{\partial }_{\nu }{\xi }_{\mu }$ 2 then

$${\bar{h}}_{\mu \nu }\to {\bar{h}}_{\mu \nu }+{\partial }_{\mu }{\xi }_{\nu }+{\partial }_{\nu }{\xi }_{\mu }-{\partial }^{\rho }{\xi }_{\rho }{\eta }_{\mu \nu }$$

This transformation leaves the solution in de Donder gauge ${\partial }^{\mu }{\bar{h}}_{\mu \nu }=0$ provided that

$$\square {\xi }_{\nu }=0$$

In particular, we can take

$${\xi }_{\mu }={\lambda }_{\mu }{e}^{i{k}_{\rho }{x}^{\rho }}$$

which obeys $\square {\xi }_{\mu }=0$ because $k_{\rho }{k}^{\rho }=0$ . A gauge transformation of this type shifts the polarisation matrix to

$$H_{\mu \nu }\to H_{\mu \nu }+i(k_{\mu }{\lambda }_{\nu }+k_{\nu }{\lambda }_{\mu }-k^{\rho }{\lambda }_{\rho }{\eta }_{\mu \nu })\text{\hspace{1em}(5.20)}$$

Polarisation matrices that differ in this way describe the same gravitational wave. We now choose the gauge transformation ${\lambda }_{\mu }$ in order to further set

$$H_{0\mu }=0\mathrm{and}{H}_{\mu }^{\mu }=0\text{\hspace{1em}(5.21)}$$

These conditions, in conjunction with (5.19), are known as transverse traceless gauge. Because H is traceless, this choice of gauge has the advantage that $h_{\mu \nu }={\bar{h}}_{\mu \nu }$

At this stage we can do some counting. The polarisation matrix $H_{\mu \nu }$ has 10 components. The de Donder condition (5.19) gives 4 constraints, and there are 4 residual gauge transformations (5.20). The upshot is that there are just $10-4-4=2$ independent polarisations in $H_{\mu \nu }$

(There is a similar counting in Maxwell theory. The polarisation of $A_{\mu }$ seemingly has 4 components. The Lorentz gauge ${\partial }^{\mu }{A}_{\mu }=0$ kills one of them, and a residual gauge transformation kills another, leaving the 2 familiar polarisation states of light.)

An Example

Consider a wave propagating in the z direction. The wavevector is

$$k^{\mu }=(\omega ,0,0,\omega )$$

The condition (5.19) sets $H_{0\nu }+H_{3\nu }=0$ . The additional constraint (5.21) restricts the polarisation matrix to be

$$H_{\mu \nu }=\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& H_{+}& H_X& 0\\ 0& H_X& -H_{+}& 0\\ 0& 0& 0& 0\end{array}\right)\text{\hspace{1em}(5.22)}$$

Both $H_{+}$ and $H_X$ can be complex; we take the real part when computing the metric in (5.18). Here we see explicitly the two polarisation states $H_{+}$ and $H_X$ . We’ll see below how to interpret these two polarisations.

5.2.2 Bobbing on the Waves

What do you feel if a gravitational wave passes you by? Well, if you’re happy to be modelled as a pointlike particle, moving along a geodesic, then the answer is simple: you feel nothing at all. This follows from the equivalence principle. Instead, it’s all about your standing relative to your neighbours.

This relative physics is captured by the geodesic deviation equation that we met in Section 3.3.4. Consider a family of geodesics $x^{\mu }(\tau ,s)$ , with s labelling the different geodesics, and $\tau$ the affine parameter along any geodesic. The vector field tangent to these geodesics is the velocity 4-vector

$$u^{\mu }={\frac{\partial x^{\mu }}{\partial \tau }|}_s$$

Meanwhile, the displacement vector $S^{\mu }$ takes us between neighbouring geodesics,

$$S^{\mu }={\frac{\partial x^{\mu }}{\partial s}|}_{\tau }$$

We previously derived the geodesic deviation equation (3.36).

$$\frac{D^2{S}^{\mu }}{D{\tau }^2}=R_{\rho \sigma \nu }^{\mu }{u}^{\rho }{u}^{\sigma }{S}^{\nu }$$

We’ll consider the situation where, in the absence of the gravitational wave, our family of geodesics are sitting happily in a rest frame, with $u^{\mu }=(1,0,0,0)$ . As the gravitational wave passes, the geodesics will change as

$$u^{\mu }=(1,0,0,0)+\mathcal{O}(h)$$

Fortunately, we won’t need to compute the details of this. We will compute the deviation to leading order in the metric perturbation $h,$ but the Riemann tensor is already $\mathcal{O}(h)$ , which means that we can neglect the corrections in the other terms. Similarly, we can replace the proper time $\tau$ for the coordinate time t. We then have

$$\frac{d^2{S}^{\mu }}{d{t}^2}=R_{00\nu }^{\mu }{S}^{\nu }$$

The Riemann tensor in the linearised regime was previously computed in (5.3)

$$R_{\rho \sigma \nu }^{\mu }=\frac{1}{2}{\eta }^{\mu \lambda }({\partial }_{\sigma }{\partial }_{\rho }{h}_{\nu \lambda }-{\partial }_{\sigma }{\partial }_{\lambda }{h}_{\nu \rho }-{\partial }_{\nu }{\partial }_{\rho }{h}_{\sigma \lambda }+{\partial }_{\nu }{\partial }_{\lambda }{h}_{\sigma \rho })$$

Using $h_{\mu 0}=0$ , the component we need is simply

$$R_{00\nu }^{\mu }=\frac{1}{2}{\partial }_0^2{h}_{\nu }^{\mu }$$

Our geodesic deviation equation is then

$$\frac{d^2{S}^{\mu }}{d{t}^2}=\frac{1}{2}\frac{d^2{h}_{\nu }^{\mu }}{d{t}^2}{S}^{\nu }\text{\hspace{1em}(5.23)}$$

We see that the gravitational wave propagating in, say, the z direction with polarisation vector (5.22) affects neither $S^0$ nor $S^3$ . The only effect on the geodesics is in the $(x,y)$ plane, transverse to the direction of propagation. For simplicity, we will solve this equation in the $z=0$ plane.

$H_{+}$ Polarisation: If we set $H_X=0$ in (5.22), then the geodesic deviation equation (5.23) becomes

$$\frac{d^2{S}^1}{d{t}^2}=-\frac{{\omega }^2}{2}{H}_{+}{e}^{i\omega t}{S}^1\mathrm{and}\frac{d^2{S}^2}{d{t}^2}=+\frac{{\omega }^2}{2}{H}_{+}{e}^{i\omega t}{S}^2$$

We can solve these perturbatively in $H_{+}$ . Keeping terms of order $\mathcal{O}(h)$ only, we have

$$S^1(t)=S^1(0)\left(1+\frac{1}{2}{H}_{+}{e}^{i\omega t}+\dots \right)\text{ and }{S}^2(t)=S^2(0)\left(1-\frac{1}{2}{H}_{+}{e}^{i\omega t}+\dots \right)\text{\hspace{1em}(5.24)}$$

where, as we mentioned previously, we should take the real part of the right-hand-side. (Recall that $H_{+}$ can also be complex.)

From these solutions, we can determine the way in which geodesics are affected by a passing wave. Think of the displacement vector $S^{\mu }$ as the distance from the origin to a neighbouring geodesic. We will consider a family of neighbouring geodesics corresponding to a collection of particles which, at time $t=0,$ , are arranged around a circle of radius R. This means that we have initial conditions $S^a(t=0)$ satisfying $S^1(0{)}^2+S^2(0{)}^2=R^2$

The solutions (5.24) tell us how these geodesics evolve. The relative minus sign between the two equations means that when geodesics move outwards in, say, the $x^1=x$ direction, they move inwards in the $x^2=y$ direction, and vice-versa. The net result is that, as time goes on, these particles will evolve from a circle to an ellipse and back again, displaced like this:

$H_X$ Polarisation: If we set $H_{+}=0$ in (5.22), then the geodesic deviation equation (5.23) becomes

$$\frac{d^2{S}^1}{d{t}^2}=-\frac{{\omega }^2}{2}{H}_X{e}^{i\omega t}{S}^2\mathrm{and}\frac{d^2{S}^2}{d{t}^2}=-\frac{{\omega }^2}{2}{H}_X{e}^{i\omega t}{S}^1$$

Again, we solve these perturbatively in $H_X$ . We have

$$S^1(t)=S^1(0)+\frac{1}{2}{S}^2(0)H_X{e}^{i\omega t}+\dots \mathrm{and}{S}^2(t)=S^2(0)+\frac{1}{2}{S}^1(0)H_X{e}^{i\omega t}+\dots$$

The displacement is the same as previously, but rotated by $45^{\circ }$ . (To see this, note that the displacements $S^1(t)\pm S^2(t)$ have the same functional form as (5.24).) This means that this time the displacement of geodesics looks like this:

We can also take linear combinations of the polarisation states. Adding the two polarisations above gives an elliptic displacement whose axis rotates. This is analogous to the circular polarisation of light.

The displacements due to gravitational waves are invariant under rotations by π. This contrasts with polarisation of light which is described by a vector, and so is only invariant under $2\pi$ rotations. This reflects the fact that graviton has spin 2, while the photon has spin 1.

Gravitational Wave Detectors

Gravitational wave detectors are interferometers. They bounce light back and forth between two arms, with the mirrors at either end playing the role of test masses.

If the gravitational wave travels perpendicular to the plane of the detector, it will shorten one arm and lengthen the other. With the arms aligned along the x and y axes, the maximum change in length can be read from (5.24),

$$L^{\prime }=L\left(1\pm \frac{H_{+}}{2}\right)\Rightarrow \frac{\delta L}{L}=\frac{H_{+}}{2}$$

To get a ballpark figure for this, we need to understand how large we expect $H_{+}$ to be from any plausible astrophysical source. We’ll do this in Section 5.3.2. It turns out it’s not really very large at all: typical sources have $H_{+}\sim 10^{-21}$ . The lengths of each arm in the LIGO detectors is around $L\sim 3$ km, meaning that we have to detect a change in length of $\delta L\sim 10^{-18}\mathrm{m}$ . This seems like a crazy small number: it’s smaller than the radius of a proton, and around $10^{12}$ times smaller than the wavelength of the light used in the interferometer. Nonetheless, the sensitivity of the detectors is up to the task and the LIGO observatories detected gravitational waves for the first time in 2015. For this, three members of the collaboration were awarded the 2017 Nobel prize. Subsequently, the LIGO and VIRGO detectors have observed a large number of mergers involving black holes and neutron stars.

5.2.3 Exact Solutions

We have found a wave-like solution to the linearised Einstein equations. The metric for a wave moving in, say, the positive z direction takes the form

$$d{s}^2=-d{t}^2+({\delta }_{ab}+h_{ab}(z-t))d{x}^{a}d{x}^b+d{z}^2\text{\hspace{1em}(5.25)}$$

where the $a,b=1,2$ indices run over the spatial directions transverse to the direction of the wave. Because the wave equation is linear, any function $h_{ab}(z-t)$ is a solution to the linearised Einstein equations; the form that we gave in (5.18) is simply the Fourier decomposition of the general solution.

Because gravitational waves are so weak, the linearised metric is entirely adequate for any properties of gravitational waves that we wish to calcuate. Nonetheless, it’s natural to ask if this solution has an extension to the full non-linear Einstein equations. Rather surprisingly, it turns out that it does.

For a wave propagating in the positive z direction, we first introduce lightcone coordinates

$$u=t-z,v=t+z$$

Then we consider the plane wave ansatz, sometimes called the Brinkmann metric

$$d{s}^2=-dudv+d{x}^{a}d{x}^a+H_{ab}(u)x^a{x}^{b}d{u}^2$$

Note that our linearised gravitational wave (5.25) is not of this form; there is some (slightly fiddly) change of coordinates that takes us between the two metrics. One can show that the Brinkmann metric is Ricci flat, and hence solves the vacuum Einstein equations, for any traceless metric $H_{ab}$

$$R_{\mu \nu }=0\iff H_a^a(u)=0$$

The general metric again has two independent polarisation states,

$$H_{ab}(u)=\left(\begin{array}{cc}{H}_{11}(u)& H_{12}(u)\\ H_{12}(u)& -H_{11}(u)\end{array}\right)$$

It is unusual to find solutions on non-linear PDEs which depend on arbitrary functions, like $H_{11}(u)$ and $H_{12}(u)$ . The Brinkmann metrics are a rather special exception.

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正文：英文 · OCR 机器稿 · 待校对

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