# Predicting gravitational waves

There has been a lot of excitement recently with the news that physicists at LIGO have directly detected gravitational waves. Many wonderful people have done popular science blog posts or videos about what gravitational waves are, but I haven’t really seen anyone talk much about the mathematics of it. For example, why do Einstein’s equations mean that gravitational waves exist? How did Einstein predict gravitational waves?

It turns out that starting with Einstein’s equations and a few simplifying assumptions, it’s relatively easy to derive the necessity of gravitational waves. That’s what this post will try and do.

Starting with the Einstein Field Equations (EFEs):

$R_{\mu \nu} - \frac{1}{2}R g_{\mu \nu} = 8 \pi T_{\mu \nu}$

we would like to show that there are solutions for the metric, $g_{\mu \nu}$, which propagate in a wave-like manner. In fact, we’re going to simplify things a bit first. We’re going to assume that there isn’t any matter or energy floating around, that is, $T_{\mu \nu} = 0$. The EFEs then become simpler, and are known as the vacuum Einstein Field Equations. They describe the shape of spacetime when you don’t have any planets or stars or galaxies floating around bending things.

For those who have done an elementary course in electromagnetism, this should be familiar. Starting with Maxwell’s equations in a vacuum, it is quite easy to show that there exist solutions which are waves which move at the speed of light – light IS electromagnetic waves.

It’s a mathematical fact that the left hand side of the EFEs vanish if and only if the Ricci tensor, $R_{\mu \nu}$, vanishes. That is, vacuum solutions of the Einstein Field Equations are obtained by solving the equation $R_{\mu \nu} = 0$. The full, non-linear force of the Einstein Field Equations are typically overkill when we are talking about regions of space where there is little matter/energy content. In these cases, the curvature is quite small and we can approximate general relativity with a linearised form of the theory. We’re interested in ripples of spacetime moving through empty, intergalactic space , and so this approximation works pretty nicely for us. Furthermore, we’re going to make an ansatz about the form of the metric. We are going to look for solutions which are “almost flat”. That is, we’re going to consider a metric of the form $g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$, where $\eta_{\mu \nu}$ is the flat Minkowski metric familiar from special relativity, and $h_{\mu \nu}$ is a small perturbation. Since the point of this post isn’t to talk about the finer technicalities of metric perturbation theory, I’m being intentionally imprecise when I say things like “small perturbation”.

Ok, so here is our recipe. We want to solve the linearised form of $R_{\mu \nu} = 0$, with the “almost flat” metric $g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$. If we’re lucky, this will tell us what type of perturbations we can have. The calculation is pretty brute force, but I’ll run you through the salient features of it. Recall the description of the Christoffel symbols in terms of derivatives of the metric:

$\Gamma^{\sigma}_{\mu \nu} = \frac{1}{2} g^{\sigma \rho} (g_{\rho \mu , \nu} + g_{\rho \nu , \mu} - g_{\mu \nu , \rho} )$

We are ignoring terms of order $h^2_{\mu \nu}$, so with our ansatz this becomes

$\Gamma^{\sigma}_{\mu \nu} = \frac{1}{2} \eta^{\sigma \rho} (h_{\rho \mu , \nu} + h_{\rho \nu , \mu} - h_{\mu \nu , \rho} )$

Then, we need to substitute these expressions for the Christoffel symbols into the expression for the Ricci tensor:

$R_{\mu \nu} = \Gamma^{\rho}_{\nu \mu , \rho} - \Gamma^{\rho}_{\rho \mu , \nu} + \Gamma^{\rho}_{\rho \lambda} \Gamma^{\lambda}_{\nu \mu} - \Gamma^{\rho}_{\nu \lambda} \Gamma^{\lambda}_{\rho \mu} = \Gamma^{\rho}_{\nu \mu , \rho} - \Gamma^{\rho}_{\rho \mu , \nu}$

where in the second equality we have ignored terms of order $h_{\mu \nu}^2$. If we make this substitution then, after not too much tedious calculation, we arrive at the following expression for the Ricci tensor:

$R_{\mu \nu} = \frac{1}{2} \left( h^{\alpha}_{\mu , \nu \alpha} + h^{\alpha}_{\nu , \mu \alpha} - h_{\mu \nu , \alpha} \,^{, \alpha} - h^{\alpha}_{\alpha , \mu \nu} \right)$

where we raise and lower indices with the flat space metric $\eta_{\mu \nu}$. We’re almost there now. For notational simplicity, we define the “trace-reversed perturbation”:

$\bar{h}_{\mu \nu} = h_{\mu \nu} - \frac{1}{2}h \eta_{\mu \nu}$

In this notation, a tensor without indices refers to the trace of that tensor with respect to the flat space metric. Why “trace-reversed perturbation”, I hear you ask. Well, let’s compute the trace of $\bar{h}_{\mu \nu}$:

$\bar{h} = \eta^{\mu \nu} \bar{h}_{\mu \nu} = \eta^{\mu \nu} h_{\mu \nu} - \frac{1}{2} h \eta^{\mu \nu} \eta_{\mu \nu} = h - 2h = -h$

where we remember that we’re working in 4-dimensional spacetime, so that $\eta = 4$. Ok, so the name makes sense, that’s always nice. What is the Ricci tensor with respect to this trace-reversed perturbation? It is

$R_{\mu \nu} = \frac{1}{2} \left( \bar{h}^{\alpha}_{\mu , \nu \alpha} + \bar{h}^{\alpha}_{\nu , \mu \alpha} + \frac{1}{2}\eta_{\mu \nu} \bar{h}_{,\alpha} \,^{,\alpha} - \bar{h}_{\mu \nu , \alpha} \,^{,\alpha} \right)$.

This is where we get to the unsatisfying part of the calculation, at least as far as this blog post is concerned. It turns out, although it’s too much of a calculation to put into this post, that one can do a gauge transformation in such a way that the form of the Ricci tensor simplifies considerably (see Appendix 4 of here for more details) . By choosing what’s known as the Lorentz gauge (the similarity to electromagnetism is no accident), the Ricci tensor simplifies to

$R_{\mu \nu} = - \frac{1}{2} \bar{h}_{\mu \nu , \alpha} \,^{, \alpha}$

We’re now finally in a position to finish the calculation. Remember that we wanted to solve $R_{\mu \nu} = 0$, so we have

$\eta^{\alpha \beta} \bar{h}_{\mu \nu , \alpha \beta} = 0$.

In standard Cartesian coordinates $(ct, x, y, z)$, this expression is simply

$\left( - \frac{1}{c^2} \frac{\partial^2 }{\partial t^2} +\frac{\partial^2 }{\partial x^2} + \frac{\partial^2 }{\partial y^2} + \frac{\partial^2 }{\partial z^2} \right) \bar{h}_{\mu \nu} = 0$

This is simply the wave equation for the trace-reversed perturbation:

$\frac{\partial^2 \bar{h}_{\mu \nu}}{\partial t^2} = c^2 \nabla^2 \bar{h}_{\mu \nu}$.

The solutions of this equation describe wavey perturbations of spacetime which travel at the speed of light – gravitational waves!