# "Gauge Freedom" in GR

This is an issue that confused me when I was learning general relativity. In special relativity, we put so much emphasis on the physical meanings of coordinates, while in general relativity we treat them as being almost completely arbitrary. As you know, the point is that once you have spacetime curvature, you *can't* define the usual network of clocks and rulers everywhere, so the special relativity prescription fails. And once we lose that, we might as well allow for completely general coordinate systems, because that's what the math lets us do anyway.

I suppose your question is, once you toss out "unreasonable" coordinate systems, and focus on the ones that are like the ones in special relativity, and focus on weak fields, why is there any redundancy left? The point is that "measuring the metric" is a lot more physically ambiguous than it sounds.

For example, let's focus on one arm of LIGO. The tube containing the arm is a rigid object, so it's reasonable to fix $g_{ii} = 1$ throughout the arm. After all, that's what we mean by LIGO being a ruler, right? And the other arm is clearly perpendicular to it, so by treating it similarly, we moreover have $g_{ij} = \delta_{ij}$. And the laser pulses that feed into LIGO effectively function as extremely accurate clocks. This ought to synchronize the time between the two arms, so we should just have $g_{tt} = -1$ everywhere, right? And obviously because we can imagine the rulers synchronizing time *throughout* each arm, there shouldn't be $dt \, dx$ cross terms, so $g_{i0} = 0$.

Well, if you assume all of these things, then you have effectively assumed that $g_{\mu\nu} = \eta_{\mu\nu}$. In other words, you have assumed that you're in flat spacetime, and thereby forbidden any gravitational waves at all! The point is that you can't satisfy all these perfectly reasonable physical constraints simultaneously -- that is the core of what it *means* for there to be spacetime curvature. (Yes, you can approximately satisfy them if your setup's effective length is much smaller than the wavelength of the gravitational wave. That's exactly why LIGO is as big as it is: this argument shows that if it were much smaller, it wouldn't be able to see gravitational waves effectively.)

The fact that there is gauge redundancy just tells you, essentially, that you can choose which of these assumptions to drop, and get the same answer in each case. The standard and arguably least confusing choice is the transverse traceless gauge, where we keep $g_{00} = -1$ and $g_{0i} = 0$, but you could make another.

Only diffeomorphism invariant quantities are measurable. The metric does not have this property.

*''every observer has a local proper time that he/she can measure on their clock, and proper distances that they can measure with a meter stick much shorter than the radius of curvature. If I insist on using these coordinates to describe my space''*

These are not coordinates but functions of two spacetime positions. These positions have different numerical values in different coordinate systems. To get something diffeomorphism invariant you need to describe these two points in an invariant way, e.g., as extremal points of two scalar fields (say, of the norm of a density gradient, or of temperature).

More generally, to the extent that you can define your local frame in invariant terms (with enough matter locally near you you can do this approximately, this is what a GPS does) you break the diffeomorphism invariance to such a degree that you can make arbitrary fields locally observable.

*''I think the first metric is the one I measure using three orthogonal meter sticks, not the second.''* (from a comment on another post here)

This is an example for what I meant. Specifically, here you implicitly posit a rigid piece of matter made of three meter sticks orthogonally joint at an origin, and posit in addition that this defines a Cartesian coordinate system. This fixes the spatial coordinate system locally, and hence by extension via the exponential map on a chart valid as far as the curvature is not too large to make the exponential map nonunique. By working in the rest frame of this piece of matter, and by placing on it a clock and laser equipment for locating other matter you can extend this to a prescription for defining a particular frame Minkowski coordinate system for spacetime locally, surely extending throughout the solar system (where curvature is tiny). Thus you fixed a coordinate system and with it all gravitational gauge freedom.

*''More concretely, when "I" build LIGO and measure the signal, I get something unique. There is no gauge choice consciously being made. Why does GR give me something ambiguous?''*

As you can see, with enough input on the matter level, nothing ambiguous remains.

From here on you are just working in the harmonic gauge and a particular decomposition into space and time. Near the solar system, the field equations can be linearized, and gravitation reduces to coupled linear wave equations for the components of the metric field. These equations allow gravitational wave solutions which were detected by LIGO.

Note that the meters employed by LIGO only fix the local coordinate system, not the metric, which remains a dynamical object. By extending the local coordinate system by the exponential map, one does not need any meters far away; everything measured in LIGO is local (i.e., on Earth) but far away curvature effects are still indirectly measurable through the gravitational waves they produce.

I think your confusion is most concisely caught in this part of your question:

If I insist on using these coordinates to describe my space, I don't see how there can be any remaining gauge freedom.

Without getting into the argument that's emerged in comments to your question about whether your particular choice of coordinates makes sense, it is generally true that once you choose a set of coordinates, you do *not* have any gauge freedom left. The gauge freedom in general relativity is *exactly* the freedom to choose your coordinates. No more and no less.

I think that's most naturally viewed through an ADM decomposition to 3+1 dimensions. (The part of your question about gravitational waves also gets computed in this formalism, or a variant of it.) Assuming for simplicity at this point a vacuum solution, if $g_{\mu\nu}$ is the metric of your spacetime, Arnowitt, Deser, and Misner showed in the 1950s that if you take a foliation of spacelike hypersurfaces, you can decompose this into an induced 3-metric on the surfaces $\gamma_{ij}$ plus and "lapse" function $\alpha$ and a shift vector $\beta^i$. These are related by $$ \left( \begin{array}{cc} g_{00} & g_{0j} \\ g_{i0} & g_{ij} \end{array} \right) = \left( \begin{array}{cc} \beta_k \beta^k - \alpha^2 & \beta_j \\ \beta_i & \gamma_{ij} \end{array} \right) $$ where the index on the shift is lowered by the 3-metric, $\beta_k = \gamma_{ik} \beta^i$. There's also a conjugate field $\pi^{ij}$ in this formalism that allow (coordinate) time derivatives $\partial_t \gamma_{ij}$ to be expressed as first-order-in-time PDEs.

I'm not going to write out all of the equations here because they are readily available, but making this split divides the Einstein equations into two types of equations: There are two "evolution" equations for $\gamma_{ij}$ and for $\pi^{ij}$, and there are two constraint equations. The constraints are a scalar constraint called the Hamiltonian constraint and a 3-vector constraint, known as the momentum constraint. (This is analogous, respectively, in EM to the two Maxwell equations that have time derivatives and the two that do not.)

In both the 3+1 formulation of GR and the Maxwell equations, one can prove that the "constraints propagate". That means that if the constraints are satisfied at any given time, they are satisfied at all times. See, for example, the appendix in Wald, which treats this in great detail.

Now note that **the lapse and the shift are your "gauge" fields in the this formalism**. They have no physical meaning. In particular, they don't appear in the Hamiltonian constraint or in the momentum constraint, which is analogous to the fact that the EM gauge fields don't appear in the constraint equations $\nabla \cdot E = 0$ or $\nabla \cdot B = 0$ (again assuming vacuum). Not appearing in the constraint equations means that if you have any ($\gamma_{ij}$, $\pi^{ij}$) pair that satisfy the constraint equations at any given time, i.e. they are "physical" at any time, you have can construct a solution to the full Einstein equations *for any choice of lapse and shift* by solving the evolution equations as an initial value problem, due to the fact that the constraints propagate, as mentioned above.

Secondly, **note that the choice of lapse and shift are essentially equivalent to a choice of coordinates**. You can find figures of this in many references, including MTW and Wald, but basically you can show that between two "nearby" surfaces $\Sigma_0$ and $\Sigma_1$, a point $x$ on $\Sigma_0$ will have the same spatial coordinate as a point $y$ on $\Sigma_1$ if $y \approx x + \alpha n + \beta$, where $n$ is normal to $\Sigma_0$ at $x$. You also have that the lapse is proportional to the proper time (rather than coordinate time) between slices.

So, in some sense you are right in your complaint. Once you fix the coordinates, you don't have gauge freedom left. When you construct your "personal LIGO" ala the end of your question, you will have effectively made your coordinate choice since the "natural" coordinates for your configuration will be determined by how you layout your detector's arms, giving you a rectangular coordinate system on your "spatial slices". In the weak-field region of Earth, that will be completely compatible with unit lapse and zero shift, i.e. the Minkowski space that you would have linearized around to compute your wave solution at the detector.

Even with this understanding, you can still find the effects of gauge. When you eventually detect a wave on your personal LIGO, it will have a polarization, which you'll probably decompose into "plus" and "cross" polarizations. If you rotate your detector, which is effectively a different gauge choice, you'll mix those differently for a given signal. This probably the simplest of multiple factors that you need to "match" between your theoretical calculation of the wave - where you have to fix the gauge - and your actual configuration of the detector on the ground. This is the other end of your mistake / confusion. When you do the theory, if you want to match to your detector, you not only need to choose a gauge, you need to choose a gauge that's compatible with how you're going to read the results from your detector. You cannot not just choose any gauge that you like and try to use the results from the calculation directly with your detector.