# Does it make sense to view all solutions to Einstein's field equations as $g_{\mu \nu}=\eta_{\mu \nu}+h_{\mu \nu}$, for a given $h_{\mu \nu}$?

I have a question about General Relativity that so far I have never been able to solve.

Let's take Einstein's field equations:

$$R_{\mu \nu}-\frac{1}{2}g_{\mu \nu}R=\frac{8 \pi G_{N}}{c^3}T_{\mu \nu}. \tag{1}$$

By imposing proper initial conditions and an energy-momentum tensor, one is able to find solutions for the metric tensor that include the Schwarzschild's metric, the Freedman-Robertson-Walker one and so on. Now, let's define the following $$g_{\mu \nu} \equiv \eta_{\mu \nu}+h_{\mu \nu}\tag{2}$$

and let's insert this expression into the previous Einstein's field equations.

My question is the following

"Given a proper $$T_{\mu \nu}$$, are all solutions in $$g_{\mu \nu}$$ and $$h_{\mu \nu}$$ equivalent?"

Is it possible that there exist certain solutions that cannot be expressed as a Minkowski background $$\eta_{\mu \nu}$$ plus fluctuations $$h_{\mu \nu}$$?

• You seem to have answered your own question: You have given a bijective correspondence (2) between $g_{\mu \nu}$ and $h_{\mu \nu}$. – Qmechanic Nov 28 '18 at 18:39
• Writing $g$ as Minkowski plus a bit is a great way to get understanding of the weak field limit. In that limit it yields all sorts of good physical intuition. More generally, though, it does not help, as answers here say. (It's not mathematically wrong, just not useful for insight). – Andrew Steane Nov 28 '18 at 19:12
• I would say that there is not a strong agreement around my question, so I add an important element. Remember that in order to make sense of perturbation theory, one has to impose that $|h_{\mu \nu}|<1$ in order to have a well defined inverse for the metric tensor. Given this, how is it possible to start from $\eta_{\mu \nu}+h_{\mu \nu}$ and to find strong-fields solutions as the Schwarzschild one? – Andrea.Phys Dec 6 '18 at 10:35

Is it possible that there exist certain solutions that cannot be expressed as a Minkowski background $$\eta_{\mu \nu}$$ plus fluctuations $$h_{\mu \nu}$$?

No. Take any arbitrary solution of the Einstein field equations, pick a coordinate chart, and define $$h_{\mu \nu} := g_{\mu \nu}-\eta_{\mu \nu}.$$ Then the metric on that chart is given by $$g_{\mu \nu} = \eta_{\mu \nu}+h_{\mu \nu}.$$

However, there is no guarantee that this is at all useful, that $$h_{\mu\nu}$$ is at all simple or small (over anything more than a local approximation on a chart chosen to be so small that the deviations don't have time to come in), that this will be possible globally instead of over only a limited chart, or that you're doing anything other than pure symbol-pushing.

Is it possible that there exist certain solutions that cannot be expressed in a useful way as a Minkowski background $$\eta_{\mu \nu}$$ plus fluctuations $$h_{\mu \nu}$$?

Yes, absolutely. This is the generic case - most metrics are not reducible to a perturbed Minkowski background. If you want a concrete example, start with the Schwarzschild metric.

• The OP isn't asking whether this is possible on some small chart, they're asking whether it's globally possible. It's trivially true that you can do this on a sufficiently small region of space, because spacetime is locally Minkowski. – Ben Crowell Nov 28 '18 at 17:05
• @Ben I don't see any discussion of charts on the question - it's obviously less than a global map, and perhaps it's worth emphasizing a bit, but the way the question is phrased it's important to distinguish the symbol-pushing from the actual physics. – Emilio Pisanty Nov 28 '18 at 17:41
• I agree with you @EmilioPisanty , but your last sentence might sound a bit ambiguous for some people. The Schwarzschild metric is a concrete example of a metric reducible to...or not reducible to....? – magma Nov 29 '18 at 0:40

Is it possible that there exist certain solutions that cannot be expressed as a Minkowski background $$\eta_{\mu \nu}$$ plus fluctuations $$h_{\mu \nu}$$?

Yes. There are links between curvature and topology, and some topologies are not consistent with a Minkowski metric. For example, closed FLRW spacetimes have the spatial topology of a 3-sphere, which is not compatible with a Minkowski metric.

Even for spacetimes that topologically could admit a Minkowski metric, there would be the question of what coordinate system to impose on the spacetime in order to identify some coordinates with the Minkowski coordinates. In general this coordinatization is not unique.

• I understand your point and I think you are right. However I have still the impression that I am not really understanding the whole logic. Indeed I have another question. What is a "background"? Do you think it is possible to give a sense to this notion? – Andrea.Phys Nov 28 '18 at 17:18
• For any spacetime that can admit a Minkowski metric, for any coordinatization the metric can be expressed as $g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$. This is not at all useful (particularly since the $h_{\mu\nu}$ will depend on what coordinates you choose) and it is indeed (arguably) an over-literalistic reading of the question text, but that's the point - the question text does not offer enough handles to read the question in any way other than that. – Emilio Pisanty Nov 28 '18 at 17:44
• Global topology and the form of the metric don't necessarily have anything to do with each other, though. For example, a spacetime with topology $R x T^{3}$ can naturally have a global euclidean metric, but would be topologically nontrivial – Jerry Schirmer Dec 3 '18 at 15:22
• @Andrea.Phys: In this context, a "background" is any exact solution to Einstein's equation. For example, Minkowski spacetime is an exact solution to Einstein's equation if $T_{\mu \nu} = 0$. You can then look at "small" perturbations to this background to try to figure out how solutions that are "close" to your background behave. For a more rigorous treatment of this, I'd recommend Section 7.5 of Wald's General Relativity (the flat background case is discussed earlier, in Section 4.4.) – Michael Seifert Dec 3 '18 at 16:26

I think you didn't understand perturbation theory in the context of GR. You split the metric tensor into an exact solution ($$\eta$$) of Einstein's equations, and a perturbation ($$h$$) as:

$$g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$$

If you substitute this into your equation $$(1)$$, you get an infinite series in $$h$$, on both sides of the equality sign. One then typically chooses to ignore perturbations of order higher than one, given they are negligible compared to the background ($$\eta$$ here). This is known as linearized GR. One can choose to stop at any order of perturbation they like, but then equations get messy quickly.

If $$g_{\mu \nu}$$ was an exact solution, you wouldn't split it into (background + perturbation) in the first place.

Also, you cannot fix $$h_{\mu \nu}$$ from the start. $$h_{\mu \nu}$$ is the dynamical variable in the perturbation theory you construct, and it is exactly whose equations of motion you need to solve for.

Sometimes, perturbing around flat space is not ideal, like above. For example, when dealing with perturbations around a Schwarzschild solution, you want to perturb around the exact Schwarzschild solution in GR. For instance, see this classic paper by Regge and Wheeler - Stability of a Schwarzschild Singularity. This gave birth to black hole perturbation theory, the fruits of which we are witnessing today from gravitational wave observations.

You may also be interested in the existence of the Kerr-Schild exact solutions to GR. These allow to have space-times with potentially large curvature, and treat them in a way that's similar to what you have in mind when you say "useful". You start decomposing the metric as $$g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$$ with $$h_{\mu\nu} = \phi k_\mu k_\nu$$, such that $$\phi$$ is a scalar and $$k_\mu$$ is a null vector for both $$g$$ and $$\eta$$. The metric is easy to inverse, you just pick up a sign in front of the "perturbation", and the Einstein equations linearize in terms of $$\phi$$. Many black holes spacetimes can be put in this form. For Schwarzschild, you'd have $$h_{\mu\nu} = \frac{2GM}{r} k_\mu k_\nu$$ with $$k^\mu =(1,x^i/r)$$, $$r=x^i x_i$$, $$i=1\dots3$$.