When we derive the equations for propagating waves in GR, we have to make a gauge choice to get something unique. I understand that in electromagnetism, the gauge is not in general something measurable, only the force or at least some line integral $\int\!dx^\mu A_\mu$. But in GR, isn't the metric always measurable? After all, 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, I don't see how there can be any remaining 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?

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    $\begingroup$ you measure $g_{\mu\nu}\mathrm dx^\mu\mathrm dx^\nu$, not $g_{\mu\nu}$ itself. $\endgroup$ – AccidentalFourierTransform Feb 16 at 13:20
  • $\begingroup$ yes and if i divide by $dx^\mu$ (the size of my meter stick in my local frame) an any point, I get $g_{\mu\nu}$ $\endgroup$ – Eric David Kramer Feb 17 at 12:07
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    $\begingroup$ You cannot divide by $\mathrm dx^\mu$. It is a vector. Division by vectors is not defined. $\endgroup$ – AccidentalFourierTransform Feb 17 at 12:36
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    $\begingroup$ No, that is not called a derivative. If anything, a directional derivative; and look up its definition -- you never divide by a vector, but by a scalar instead. In any case, this discussion feels pretty pointless at this point. Cheers! $\endgroup$ – AccidentalFourierTransform Feb 17 at 12:51
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    $\begingroup$ $dx^\mu$ (meaning nothing mathematically more rigorous than a small displacement) is a scalar and not a vector for any fixed $\mu$. As such this is a perfectly fine method of reproducing the coefficients of a tensor, at least locally. That's not where this argument fails $\endgroup$ – Kasper Feb 21 at 11:49

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.

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  • $\begingroup$ Sounds intelligent but I have a little trouble following. If you could add some details I might do better... $\endgroup$ – Eric David Kramer Feb 19 at 13:00
  • $\begingroup$ @EricDavidKramer: Where do you need details? I need specific questions to say more. $\endgroup$ – Arnold Neumaier Feb 19 at 13:39
  • $\begingroup$ Thanks! I'm undecided between your answer and knzhou's. Does harmonic gauge ensure that the waves die off at infinity? $\endgroup$ – Eric David Kramer Feb 20 at 8:41
  • $\begingroup$ @EricDavidKramer: Spherical waves (i.e., what can be emitted locally) die off because their energy is diluted with time over more and more volume. - In my opinion, knzhou's answer is not quite correct since he argues that the metric is fixed by the meters. This is why he obtains a contradiction. In contrast, I just obtain the harmonic approximation since in fact only the coordinate system is fixed: One can extend the local coordinate system by the exponential map and does not need any meters far away; everything measured in LIGO is local and far away curvature effects are still measurable. $\endgroup$ – Arnold Neumaier Feb 21 at 11:34

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.

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  • $\begingroup$ Your conditions don't fix a flat metric. For that you also need to specify $g_{ij}=\delta_{ij}$. $\endgroup$ – Arnold Neumaier Feb 19 at 10:19
  • $\begingroup$ I agree that if I fix g=minkowski everywhere then I am forbidding gravitational waves. But I know that I can do this locally and I would like to somehow piece together these local conditions to get good coordinates where my I can read off my measurement. After all, the laser wave in LIGO is somehow making such measurements which is what I am observing in the interference signal. $\endgroup$ – Eric David Kramer Feb 19 at 13:12
  • $\begingroup$ @EricDavidKramer But the whole point is that you can't! For example, maybe you want to put Cartesian coordinates on the surface of the Earth, by laying down orthogonal metersticks. It won't work, because the Earth is curved. What it will concretely look like is that when you try to link up different sets of metersticks, they won't fit. You'll have to bend or squeeze them a bit to make it work, and the freedom in choosing how to do that is the gauge freedom we're talking about here. $\endgroup$ – knzhou Feb 19 at 20:36
  • $\begingroup$ @EricDavidKramer The exact same intuition applies to LIGO, except that we're talking about the curvature of spacetime rather than space. If you assume you can piece together local frames without any weird mismatches when they overlap, you have assumed you're in flat spacetime. $\endgroup$ – knzhou Feb 19 at 20:38
  • $\begingroup$ @knzhou Sorry, btw. I forgot to award the bounty and it auto-awarded at 50%. I selected Arnold Neumaier's answer because I thought it was also good and I wanted him to get some credit too. But I'm glad you got the +200. $\endgroup$ – Eric David Kramer Mar 2 at 12:28

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.

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  • $\begingroup$ I like your last paragraph. But could you elaborate? Is transverse traceless gauge compatible with my detector? What about harmonic gauge? Can you think of a gauge this is incompatible? $\endgroup$ – Eric David Kramer Feb 19 at 13:05
  • $\begingroup$ TT gauge connects you directly to strain, which is then very closely (if not exactly) what you're measuring on your personal LIGO. Other gauge choices are not necessarily "wrong" but may not be as directly applicable to questions related to your detector. Of course if your personal LIGO works significantly differently than the real LIGO, a different gauge may suit you better. en.wikipedia.org/wiki/Linearized_gravity#Transverse_gauge @EricDavidKramer $\endgroup$ – Brick Feb 19 at 17:08
  • $\begingroup$ A point about a "poor" gauge: You could conceivable choose a "poor" gauge where the perturbations that you consider satisfy a wave equation but those "waves" are purely coordinate effects. I guess you could probably do that even for a space that's Minkowski, although I don't have a construction off-hand. $\endgroup$ – Brick Feb 19 at 20:02

Let me present this argument from a more fundamental side. Gauge invariance should fundamentally be thought of a redundancy in our description of the physics. In GR, this gauge freedom is the redundancy in the coordinates $x^\mu$, which we use to label points on the spacetime. Choosing a particular coordinate frame removes the gauge freedom.

The physics should be independent of which coordinates we use, and a change of coordinates $x^\mu \to x'^\mu$ should not alter the physics. The physical quantities in GR are therefore precisely those that do not depend on the coordinates we use. In the language of differential geometry, these are called tensors. The metric tensor $g$ in particular is independent of the coordinates, but can be expanded in a given coordinate basis $x^\mu$ as $$g = g_{\mu \nu} (x) \, dx^\mu \odot dx^\nu,$$ where the functions $g_{\mu \nu}(x)$ are the components of the metric in that particular basis. If we change the coordinates $x^\mu \to x'^\mu$ then the components $g_{\mu \nu}(x)$ of the metric change, but the metric tensor $g$ itself is unchanged, as are all physical tensors.

In fact, general relativity can be described purely in terms of the (differential) geometry of spacetime, and all physical quantities can (in principle) be calculated without resorting to a particular coordinate system (choice of gauge). For example, the proper time along a geodesic is physical and does not depend on the coordinates you use to parametrise the geodesic. However, choosing a clever coordinate system (gauge) undoubtedly simplifies the calculations most of the time, although it obsures the intrinsic geometric nature of the physics.

Funnily enough, we can use the same language of differential geometry to describe the gauge freedom of electromagnetism. The gauge field $A$ here is also a tensor, which can be written in a particular set of coordinates $x^\mu$ as $A = A_\mu(x)\, dx^\mu.$ The physical quantity associated to this is the Maxwell field strength tensor $F = dA$, where $d$ is the exterior derivative. In components, this is $$F_{\mu \nu} = (\partial_\mu A_\nu(x) - \partial_\nu A_\mu(x)) dx^\mu \wedge dx^\nu.$$ This is where gauge invariance enters. The exterior derivative satisfies $d^2 = 0$, which means that $F$ is unchanged under the gauge transformation $$A \to A + d \alpha \qquad F = dA \to dA + d^2 \alpha = dA = F.$$ In components, this is the familiar gauge transformation $$A_\mu(x) \to A_\mu(x) + \partial_\mu \alpha(x).$$ (The Aharonov-Bohm effect shows us that $A$ is (at least quantum mechanically) a truly physical quantity, but I will leave that for you to ponder.)

One last thing about the relative importance of Lorentz invariance and gauge invariance. If we assume Lorentz invariance and QM, we can show that any massless helicity-1 field $A_\mu$ (e.g. the photon) must (in four dimensions) show the gauge redundancy above. Similarly, any massless helicity-2 field (i.e. the graviton $h_{\mu \nu}$) must show the gauge redundancy of (linearised) diffeomorphism invariance $$h_{\mu \nu} \to h_{\mu \nu} + \partial_\mu \alpha_\nu + \partial_\nu \alpha_\mu.$$ As such, gauge invariance is implied by Lorentz invariance. All the beauty of the geometry of general relativity (and also Yang-Mills theory, fibre bundles, etc.) all originates in Lorentz invariance!

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  • $\begingroup$ Thank you. I am not satisfied by your answer because although you present a very nice explanation of the gauge redundancy in GR, you did not explain the connection with choosing a local frame and with the experimentally measured signal in an experiment like LIGO. (I think you want a wedge in $F_{\mu\nu}$, I'm sure it was just a typo). $\endgroup$ – Eric David Kramer Feb 23 at 9:32

Thank you for all your great answers! I'm just writing an answer of my own here, to summarize what I've gathered from all your responses. Comments welcome.

  1. Physical lengths (defined as the length of some spacelike curve) are coordinate invariant. For example, in LIGO, the length of my detector arm (suppose it lies along the x-direction), is $$\ell = \int_{\ell_1}^{\ell_2} g_{xx}^{1/2} dx$$ where $\ell_1$ and $\ell_2$ are the two endpoints of the arm, at some equal time coordinate that I define. Even if I change coordinates to $x'$ by defining $x=f(x')$ for some legal $f(x')$, the metric will change in exactly the right way to compensate this. In some sense, this is the definition of the metric.

  2. I can choose proper time and proper distance as my coordinates locally. Then my metric will look like the Minkowski metric locally (e.g. at one of the endpoints of the detector arm). But I cannot do this globally because the space is curved. Thus eventually, once I am outside of some neighborhood of the endpoint of my arm, I cannot use $dx$ to integrate the length of the arm; I need to used $g^{1/2}_{xx}dx$.

  3. All of the physical quantities that I calculate e.g. in the case of gravitational waves are some kind of spacetime interval or other invariant quantity. As such, it really does not matter which coordinates I use, as long I stick to invariant quantities.

  4. A gauge choice does not completely define the coordinates, but it restricts me to a class of coordinates. It also makes the calculations doable. So it's a convenient way of doing the calculation without loosing too much generality (which I would certainly lose by choosing specific coordinates).

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  • $\begingroup$ Noting that Euclidean space is a Riemann manifold, in essence, you've warped the space-time metric into a space and time metric - or the direct product of $2$ Euclidean manifolds - where time is now arbitrary. Although you have a LIGO signal, you just don't know where it came from. $\endgroup$ – Cinaed Simson Mar 16 at 22:15

Not a very long answer so I was going to post in the comments but I guess it's better this way.

The metric isn't just a function of spacetime, it depends on your coordinate system. $g_{\mu \nu}=\hat{e}_{\mu}\cdot\hat{e}_\nu$. Choose different coordinates (spherical, rectangular) and the inner product of basis vectors will also change $\implies g_{\mu\nu}$ will change, though nothing physical is different. There is physical information encoded in the metric; you can write all the curvature tensors in terms of $g_{\mu\nu}$, but each of the components is not directly measurable. This is true in SR and GR. Check out the spherical coordinate components of the metric in flat space for example: in rectangular coordinates they are (-1,1,1,1), but in spherical coordinates they are not (see here) even though both situations have identical (flat) spacetimes. Note the link I used only gives $g_{rr}, g_{\theta\theta}, g_{\phi\phi}$. A 4d metric would also have $g_{tt}=-1$ and all other components $0$.

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  • $\begingroup$ Only one of those coordinate systems (up to Lorentz transformations) corresponds to what I measure with a clock and three orthogonal meter sticks. $\endgroup$ – Eric David Kramer Feb 19 at 12:56
  • $\begingroup$ The positions on those meter sticks can be expressed in any coordinates - spherical, rectangular, cylindrical or otherwise. The choice of those coordinates is left entirely to the theorist, who tries to describe the situation on paper. If the theorist analyzes the problem in rectangular coordinates instead of spherical, it only affects what he writes on the paper, not the physical meter sticks or what numbers they show. In spherical coordinates, for example, the location of the tick for "2cm" on a meter stick is at a coordinate $(\rho_2, \theta_2, \phi_2)$. $\endgroup$ – doublefelix Feb 19 at 13:13
  • $\begingroup$ Which coordinate system the theorist chooses to express the position of the ticks on the meter sticks has no effect on the physical position of the ticks - or any other physical observable. It is therefore immeasurable. $\endgroup$ – doublefelix Feb 19 at 13:14
  • $\begingroup$ Transverse-traceless gauge corresponds to having no longitudinal modes. What does that have to do with spherical coordinates? $\endgroup$ – Eric David Kramer Feb 19 at 13:24

To make any kind of physics in a lab (for example: measuring some gravitational waves), you need to select a reference frame. Any frame will do, but you still need to choose one. This is your "gauge freedom".

For any local frame, you need to built four spacetime "arms", i.e. axes represented by four orthogonal and normalised 4-vectors $\mathbf{e}_a$ such that \begin{equation}\tag{1} \mathbf{e}_a \cdot \mathbf{e}_b \equiv g_{\mu \nu}\, e_a^{\mu} \, e_b^{\nu} = \eta_{ab}, \end{equation} where $\eta_{ab} = \mathrm{diag}(1, -1, -1, -1)$ are the Minkowski metric components in an inertial basis. This tetrad $\{ \mathbf{e}_0, \mathbf{e}_1,\mathbf{e}_2, \mathbf{e}_3 \}$ defines a locally inertial frame, at some point $\mathcal{P}$ in spacetime. The 4-vector $\mathbf{e}_0$ represents your time axis, while $\mathbf{e}_i$ represents your three orthogonal spatial axes. In your local frame, the metric is Minkowskian: \begin{align} ds^2 = g_{\mu \nu} \, \mathbf{d}x^{\mu} \otimes \mathbf{d}x^{\nu} &\equiv \eta_{ab} \, \mathbf{e}^a \otimes \mathbf{e}^b \notag \\[12pt] &= \mathbf{e}^0 \otimes \mathbf{e}^0 - \mathbf{e}^1 \otimes \mathbf{e}^1 - \mathbf{e}^2 \otimes \mathbf{e}^2 - \mathbf{e}^3 \otimes \mathbf{e}^3. \tag{2} \end{align} This metric is invariant under any local Lorentz transformation: \begin{equation}\tag{3} \tilde{\mathbf{e}}_a = \Lambda_a^{\; b} \, \mathbf{e}_b, \end{equation} which is a gauge transformation, i.e a change of inertial frame at $\mathcal{P}$.

This has nothing to do with the coordinates $x^{\mu}$ used to define the metric components $g_{\mu \nu}$. A local reference frame is not the same as a coordinates system. A coordinates system could be interpreted as a collection of observers standing at different locations in spacetime and moving in various ways, while the reference frame is a local concept. Your gauge freedom is the ability to select any frame at the same location.

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  • $\begingroup$ The gauge freedom is more than that: it is general diffeomorphism invariance. The Lorentz invariance at any point is of course much more trivial and does not bother me. $\endgroup$ – Eric David Kramer Feb 19 at 12:55
  • $\begingroup$ @EricDavidKramer, it's the reverse! Democracy of coordinates is trivial, any theory could be made independant of coordinates. Coordinates don't have any physical meaning and are arbitrary. It is the Lorentz invariance which makes relativity a relativistic theory. $\endgroup$ – Cham Feb 19 at 14:37
  • $\begingroup$ I agree with that. But with regard to my question about the gauge invariance of the LIGO interference signal, the Lorentz invariance is the the trivial one, while the coordinate invariance is the thing I was confused about. But I think I'm getting less confused as I keep reading yours and others' answers. $\endgroup$ – Eric David Kramer Feb 20 at 8:18

In flat (Minkowski) spacetime, the metric in Cartesian coordinates is $$ ds^2 = -dt^2 + dx^2 + dy^2 + dz^2 $$ $$ (\textrm{that is}, \quad g_{\mu\nu}= diag(-1,1,1,1)),$$ but in spherical coordinates in the same space the metric is $$ ds^2 = -dt^2 + dr^2 + r^2 d\theta^2 + r^2 \sin^2 \theta \, d\phi^2 $$ $$ (\textrm{that is}, \quad g_{\mu\nu}= diag(-1,1,r^2,r^2 \sin^2 \theta)) .$$

These are two descriptions of the exact same physical space. Therefore clearly it can't be possible to directly measure the metric components. The metric components depend both on the physical metric, and on a choice of how the metric is expressed.

The "gauge freedom" is basically the freedom to express the same physical metric in many ways, i.e., the freedom to choose a coordinate system.

Regarding trying to measure the metric with clocks and meter sticks: every observer has a locally inertial frame which looks like the Minkowski metric on small enough scales. What matters is how the metric varies over space and time, which determines curvatures. So you'd need a whole family of observers to measure it. Where things get non-trivial is in patching together all those observers local measurements.

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  • $\begingroup$ I think the first metric is the one I measure using three orthogonal meter sticks, not the second. $\endgroup$ – Eric David Kramer Feb 19 at 12:53
  • $\begingroup$ I agree. But as I mentioned in the last paragraph, that is also the metric that everyone always measures with their meter sticks, even in curved spacetime, on short enough scales. It is the local Lorentz frame. But the question is how the local Lorentz frames are stitched together. $\endgroup$ – Joe Schindler Feb 19 at 15:25
  • $\begingroup$ If you define the coordinate system as "numbers measured by my clock and sticks", you are choosing the local Lorentz frame as coordinates. This is indeed a choice of gauge. In this gauge the metric is (-1,1,1,1) to first order. But even in this "best" local gauge, second derivatives of the metric won't vanish. So as you extend your meter sticks out further away from yourself, the metric will vary spatially. $\endgroup$ – Joe Schindler Feb 19 at 15:29
  • $\begingroup$ So suppose you do that experiment. For sufficiently long meter stick distances away from yourself, the metric in your coordinate system will inevitably (in general) pick up off-diagonal terms. How are you going to measure those terms with your sticks and clock? $\endgroup$ – Joe Schindler Feb 19 at 15:32
  • $\begingroup$ Yes when there are gravitational waves that's true. Thanks. By the way, in a static spacetime you don't pick up those terms if you choose your time coordinate correctly. But you do pick up spatial curvature. $\endgroup$ – Eric David Kramer Feb 23 at 9:34

You already got a lot of answers, and I think they are good quality, but no-one seems to have mentioned the following simple fact. The metric tensor is not like other tensors, in that, in the linearized theory, under a gauge change, the metric DOES change but other tensors do not. To be precise: in the linearized theory, when we make an appropriate first-order change in coordinates so as to preserve tensors, we find that under the change of gauge, the functional forms of all scalars, vectors and tensors are unaltered, with the exception of the metric tensor. I have not checked whether this also applies to a more general gauge change, but it already makes a useful point.

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