Typically the gravitational wave (GW) strain is calculated (e.g. by the Quadrupole formula) as some tensor object $h^{ij} (t)$.

However, I often read of the strain as some scalar time series, i.e. $h(t)$. From this time series signal-to-noise ratios and the like can be calculated.

My question is just what is the relation between $h^{ij} (t)$ and $h(t)$? Is it simply a contraction with the Minkowski metric?


2 Answers 2


In brief, $h$ is essentially just some component of $h^{ij}$ — as Gary said. But there's a bit more to it than that. I'll show precisely what that means, and why it is important.

First, you need to understand the much more general notion of "strain", which is just the change in some length divided by the original length: \begin{equation} h = \frac{\Delta L}{L} = \frac{L - L'}{L}. \tag{1} \end{equation} The emphasis here is that this is a very general definition of strain — which Wikipedia tells us is also called the "Cauchy strain" or "engineering strain". It just describes the change caused by any deformation, induced by any type of force, applied to any length in some chosen direction (which is chosen implicitly in the definition). Gravitational waves are just one particular type of deformation that happens to be adequately described as a strain. So our task is to understand how this type of strain is related to the quantity $h^{ij}$.

Recall that we usually start with some definition like(*) \begin{equation} h^{\alpha \beta} = \eta^{\alpha \beta} - g^{\alpha \beta}, \tag{2} \end{equation} where $g^{\alpha \beta}$ is the full dynamical metric that contains the gravitational waves as small perturbations, and $\eta^{\alpha \beta}$ is the Minkowski metric — which we assume describes the unperturbed "background" spacetime.(**) Basically, the metric measures the length of a single vector and angles between two vectors. Angles don't really come into gravitational waves so much, especially if we're just interested in the strain, which compares lengths. So we just need to look at how the metric describes the length of a single vector. Let's imagine that we have some vector $\vec{v}$ representing the separation between two objects (or maybe the distance between two ends of an object — its length). If there were no gravitational wave, we would have $g^{\alpha \beta} = \eta^{\alpha \beta}$, and the length of this vector would be described by the metric according to \begin{equation} L^2 = v_\alpha v_\beta \eta^{\alpha \beta}; \tag{3} \end{equation} when there is a gravitational wave its length is given by \begin{equation} L'^2 = v_\alpha v_\beta g^{\alpha \beta}. \tag{4} \end{equation} At least as far as detections here on earth go, we can safely assume that the gravitational wave is causing a very small strain, so that $\Delta L \ll L$, which allows us to derive \begin{equation} \frac{L^2 - L'^2}{L^2} = \frac{L^2 - (L-\Delta L)^2}{L^2} = \frac{L^2 - L^2 + 2L \Delta L - (\Delta L)^2}{L^2} \approx 2\frac{\Delta L}{L} = 2h. \end{equation} Note that we dropped only the factor of $(\Delta L)^2$ because we are assuming that $(\Delta L)^2 \ll L \Delta L$. So we can rearrange this a little, then plug in Eqs. (3) and (4), and then Eq. (2) to get \begin{equation} h \approx \frac{1}{2} \frac{L^2 - L'^2}{L^2} = \frac{1}{2} \frac{v_\alpha v_\beta \eta^{\alpha \beta} -v_\alpha v_\beta g^{\alpha \beta}}{v_\alpha v_\beta \eta^{\alpha \beta}} = \frac{1}{2} \frac{v_\alpha v_\beta h^{\alpha \beta}}{v_\alpha v_\beta \eta^{\alpha \beta}}. \end{equation} Now, we usually talk about gravitational waves in "transverse-traceless" gauge, which is purely spatial, so $v^\alpha$ is implicitly a purely spatial vector in whatever coordinates we're using, which means we can just sum over spatial indices (replace $\alpha \beta$ with $ij$). Also, since the overall length of the vector drops out, we could just as well assume that it is normalized, so that we use $\hat{v}^i$ where $\hat{v}_i \hat{v}_j \eta^{i j} = 1$. With these assumptions, we finally get the simplest relation between the metric perturbation and the strain: \begin{equation} \boxed{h = \frac{1}{2} \hat{v}_i \hat{v}_j h^{ij}.} \tag{5} \end{equation} This is exactly how the "engineering strain" $h$ is related to the metric perturbation $h^{ij}$ — implicitly by way of some direction $\hat{v}$ that you choose. That's the complete answer to the question.

As Gary pointed out, certain components of $h^{ij}$ are usually given names like $h_+$ and $h_\times$, though some times with different factors of 2 — and as you can see from Eq. (5), you always need to be careful of those factors anyway.(***) This is because in transverse-traceless gauge, there are precisely two independent components of $h^{ij}$. But in general, when the vector that we care about isn't precisely lined up with that gauge, we can just contract as shown in Eq. (5). For example, when LIGO/Virgo are searching for gravitational waves, they (essentially) represent a leg of their detectors using a vector like this, and then contract with the metric perturbation to see how much the detector's leg should deform as the gravitational wave passes. Of course, they actually use the difference in the strain between the two legs, since that's a much clearer signature of a gravitational wave (as opposed to just noise). So they will use two perpendicular vectors to derive the actual signal they expect to measure — something like $\hat{x}_i \hat{x}_j h^{ij} - \hat{y}_i \hat{y}_j h^{ij}$, or maybe more general directions.

(*) Note that different references make lots of different choices here, and for the individual components, including signs and factors of 2. You may also see factors of $\sqrt{|\det g|}$, and various other modifications. At least to linear order in the size of the perturbation, they're all more or less equivalent (though not always equal).

(**) I want to caution you against referring to $h^{ij}$ as a "tensor" because it isn't really one; it's usually just called the "metric perturbation". Though there are multiple meanings of "tensor", the main one we use in physics is well described in the first paragraph of the Wikipedia entry:

a tensor is a geometric object that maps in a multi-linear manner geometric vectors, scalars, and other tensors... Geometric in this context is chiefly meant to emphasize independence of any selection of a coordinate system.

[Emphasis theirs.] While $h^{ij}$ does map things in a multi-linear manner, it is not "geometric". Looking at Eq. (2), we see the full metric $g^{\alpha \beta}$, which is certainly a geometric object. But we also see $\eta^{\alpha \beta}$, which is essentially just some indexed thing with certain components with respect to whatever basis you happen to be using. In GR, $\eta^{\alpha \beta}$ is not considered to be a geometric object, and therefore Eq. (2) shows us that neither is $h^{\alpha \beta}$. At best, they transform as tensors under spatial rotations, but the selection of "spatial" is not even geometrically meaningful, and they will not transform as tensors under any other transformation that is usually allowed in GR.

(***) If you happen to see $h$ in a numerical-relativity paper, it usually actually stands for something more like the complex quantity $h = h_+ - i h_\times$, rather than the strain along some particular direction.


I think the $h(t)$ just stands for any one component of the strain tensor $h^{ij}$ (like $h_+$ or $h_\times$). In the local Lorentz frame and for the gravitational wave going in the $z$ direction (i.e., the two strain polarizations are in the $x$-$y$ plane) $$ \eta_{ij} = \begin{bmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \qquad h^{ij}=\begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & h_+ & h_\times & 0 \\ 0 & h_\times &-h_+ & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}.$$

If $h=\eta_{ij}h^{ji}$ then $h=0$, so this scalar is not what $h(t)$ in your question stands for.


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