How to determine if gravity is roughly linear?

Question

The Einstein field equations are famously nonlinear, which is one of the properties that makes them difficult to solve. I know (or at least I believe) that a linear system's behavior is roughly speaking the sum of the behaviors of all the subsystems, like $f(a+b)=f(a)+f(b)$. Electromagnetism is linear, so that two electromagnetic fields overlayed on one another have the effect of one field plus the effect of the other field, but the EFEs are nonlinear so if you overlay two gravitational fields on each other the resulting system is completely different. (Or I'm misunderstanding the concept of linearity/nonlinearity.)

It's clear though that there is some way to say that at a certain threshold where the masses are low enough that you can approximate reality with a linear combination of the two systems, i.e. adding a planet's gravity to the moon's gravity gives an accurate description of the gravity between the two bodies, even though the EFEs which govern gravity are nonlinear and normally wouldn't allow that.

The question is, how do you know when to make that approximation? Does it depend on the simplicity of the metric tensors of the two systems, or does it depend on the mass-energies involve, or..? I'm wondering specifically because I have a metric describing a very low-energy system, on the order of nanojoules, and am wondering if I can describe the system involving that low-energy system plus a small mass nearby as a linear combination of the low-energy system and the small mass.

Answer 1

Your understanding of linearity is correct. The EFEs can be linearized if the metric deviation $h_{\mu \nu}$ defined as

\begin{equation} h_{\mu\nu} = g_{\mu\nu} - \eta_{\mu\nu} \end{equation}

is small. Here $g_{\mu\nu}$ is the full spacetime metric and $\eta_{\mu\nu}$ is the Minkowski metric. The exact expression will be dependent on the mass-energy distribution of your configuration, but in general if the components of $h_{\mu\nu}$ are small enough (by which I mean they are much smaller than unity), the equation is well-treated linearly.

Answer 2

controlgroup wrote: "adding a planet's gravity to the moon's gravity gives an accurate description of the gravity"

That depends on how many digits of accuracy you need. For example with the combined gravitational time dilation of two bodies you won't see much of a difference if instead of the proper multiplication you simply do an addition when the factors are small:

$$dt/d\tau=\color{green}{(1+0.0001)(1+0.0002)-1=0.00030002} \to \color{blue}{0.0001+0.0002=0.0003}$$

but below where the gravitational time dilation factor is higher, like for example

$$dt/d\tau=\color{green}{(1+1)(1+2)-1=5} \neq \color{red}{1+2=3}$$

the difference becomes noticable, so you have to check if your numbers are small compared to $1$, or large. If your numbers are large you know the approximation is too far off.