Traceless general relativity

I was just introduced to the concept of a "traceless version" of general relativity through a pre-print of John Barrow and Spiros Cotsakis, whose viability coincidentally was something that I was wondering about last week. So this paper came at a very convenient time for me to enlighten me on some of the existing work that has been done in this subject.

Why traceless?

The main issue is cosmological, and has to do with the problem of scale. We start by recalling Einstein's equation for general relativity, which originally reads \begin{equation} G_{ab} := \mathrm{Ric}_{ab} - \frac12 R g_{ab} = T_{ab} \end{equation} (where we suppressed physical constants of proportionality). A basic observation is the following: the Einstein tensor $G_{ab}$ that appears on the left of the equation and describes the geometry of space-time is scale invariant, in the sense that replacing the metric $g_{ab}$ by $\lambda g_{ab}$ leaves both the Ricci curvature (as a $(0,2)$ tensor) and the product $R g_{ab}$ of the scalar curvature and the metric unchanged. Frequently, this is compatible with a global rescaling of the matter fields that also leaves $T_{ab}$ unchanged. (For example, if $(M,g,\phi)$ is a solution to the Einstein-scalar-field system, where the stress energy tensor is given by \begin{equation} T_{ab} = \nabla_a \phi \nabla_b \phi - \frac12 g_{ab} g^{cd} \nabla_c \phi \nabla_d \phi, \end{equation} then one sees that $(M,\lambda g,\phi)$ is also a solution.)

This scale-free property is further represented in the interpretation that the Weyl curvature of the space-time captures gravitational radiation (while the Ricci curvature gives the compatibility with the matter fields). Accepting this interpretation of the Weyl tensor, and recalling that the Weyl curvature is in fact conformally invariant (as a $(1,3)$ tensor, the Weyl curvature of $\Omega^2 g_{ab}$ and $g_{ab}$ are identical, here we can even allow $\Omega^2$ to not only be a constant, but be any non-zero function on $M$), one gets the sense that gravitational radiation being agnostic to scale, but sensitive to shape.

The scaling property is broken, however, once one admits the necessity of a cosmological constant. With the cosmological constant, Einstein's equation now reads \begin{equation} G_{ab} + \Lambda g_{ab} = T_{ab} \end{equation} where the number $\Lambda$ is the cosmological constant. This now forces a particular scale to our space-time: returning to the Einstein-scalar-field example, when $\Lambda \neq 0$ and $0 < \lambda \neq 1$, in general when $(M,g, \phi)$ is a solution to the equation with cosmological constant, then $(M,\lambda g,\phi)$ is not.

The breaking of the scaling property is severe enough that Einstein¹ considered modifying GR to be a traceless theory.

How traceless?

The basic idea is to simply remove the trace from both sides of Einstein's equation. In a $d$-dimensional space-time, given a symmetric $(0,2)$ tensor $X$ we can define its tracefree part to be \[ \hat{X}_{ab} = X_{ab} - \frac{1}{d} g_{ab} g^{cd} X_{cd}.\] The idea then is to replace Einstein's equation with the tracefree part on both sides: \begin{equation}\label{eq:tfEE} \hat{G}_{ab} = \hat{T}_{ab}. \end{equation}

An obvious question to ask is then: how is the dynamics defined by this new system different from the original?

First, the obvious: a lot more space-times verify this new equation. In particular, by virtue of our construction all space-forms solve the vacuum version (meaning $T_{ab} = 0$) of this equation. In fact, all vacuum solutions with some cosmological constant solve $\eqref{eq:tfEE}$.

On the other hand, the "a lot more" is not really that much, at least in the vacuum case. Suppose $(M,g)$ is a solution to the vacuum version of $\eqref{eq:tfEE}$, this means that the traceless part of the Einstein tensor (and hence the tracefree part of the Ricci tensor) vanishes identically. In particular, this means that the Ricci tensor has to be proportional to the metric $g$. It is however a well-known homework exercise that when the space-time is dimension at least 3, such a proportionality implies that $(M,g)$ is Einstein, and therefore at least in the vacuum case $\hat{G}_{ab} = 0$ is equivalent to $G_{ab} + \Lambda g_{ab}$ for some, unspecified, constant $\Lambda$.

Einstein² was able to show that the same holds true for Einstein-Maxwell theory in 4 space-time dimensions: namely that due to the Maxwell field being conformal in 4 dimensions, any solution of the Einstein-Maxwell version of $\eqref{eq:tfEE}$ is a solution of the "standard" Einstein-Maxwell equation with cosmological constant.

Given these results, what have we gained? What we gained is that the constant $\Lambda$ that arises now can be considered as "only" a constant of integration (something that comes out of solving the equation) and not a fundamental constant of the laws of physics. This metaphysical passing of the buck then allows one to focus more on what the value of $\Lambda$ is (something that can be determined by experiments) instead of why that value of $\Lambda$ is what it is, and the equations of motion retain its scaling property.

Now let us dive a bit deeper into the above question about dynamics. Admitting more solutions is not a problem in general. For physics, however, what is important is whether the theory can be used to predict the future evolution of the universe. In other words, whether the corresponding initial value problem is well-posed. Our discussion above indicates that in the case of Einstein-vacuum and Einstein-Maxwell theories, the answer is in the affirmative.

In the general case, however, one can ask whether discarding the trace part from the system can lead to additional phenomena. The easiest way to analyze this situation is to do so in local wave/harmonic coordinates³. Recalling that with respect to harmonic coordinates, we have that the Ricci curvature can be expressed as \[ \mathrm{Ric}_{\mu\nu} \sim - \frac12 g^{\lambda\sigma} \partial^2_{\lambda\sigma} g_{\mu\nu} \] where $\sim$ denotes equality up to terms that can be expressed in terms of the metric $g$, its first coordinate derivatives, and the inverse metric $g^{-1}$. A standard computation then yields \[ R \sim - \frac{1}{\sqrt{|\det(g)|}} g^{\lambda\sigma} \partial^2_{\lambda\sigma} \sqrt{|\det(g)|} .\] This implies that, in 4 space-time dimensions, \begin{equation}\label{eq:reducedTFe} \hat{\mathrm{Ric}}^{\mu\nu} = \hat{G}^{\mu\nu} \sim \frac12 |\det(g)|^{-1/4} g^{\lambda\sigma} \partial^2_{\lambda\sigma} \left[ |\det(g)|^{1/4} g^{\mu\nu} \right]. \end{equation} Equation \eqref{eq:reducedTFe} implies that \eqref{eq:tfEE}, in wave coordinates is a quasi-diagonal system of quasilinear wave equations for the quantities \[ |\det(g)|^{1/4} g^{\mu\nu} \] and hence has a well-posed initial value problem formulation thereof (subject to constraint equations).

We however see one major difference: in the class of the standard Einstein equations, the unknown in the quasi-diagonal system is \[ \tilde{g}^{\mu\nu} = |\det(g)|^{\color{red}{1/2}} g^{\mu\nu}. \] One can unambiguously reconstruct $g$ from $\tilde{g}$ by noting that \[ \det(\tilde{g}) = \frac{1}{\det(g)} \] and hence we can solve \[ g^{\mu\nu} = |\det(\tilde{g})|^{-1/2} \tilde{g}^{\mu\nu} \] from the solution to the quasi-diagonal system.

In the case of the tracefree equations, however, we must have \begin{equation} \left| \det\bigl( |\det(g)|^{-1/4} g \bigr) \right| = 1 \end{equation} This implies that from the tracefree equations, while we can recover the shape of the metric $g$, we cannot recover the scale. (This is "as expected" as the motivation is to make the equations scale free.)

In the case of the Einstein-Maxwell and Einstein-vacuum, what we used is the fact that the stress energy tensor itself is trace-free. This means that since $T_{ab}$ is divergence free, so is $\hat{T}_{ab}$. In these cases, by taking the divergence of \eqref{eq:tfEE} and considering the Bianchi identities ($\nabla^a G_{ab} = 0$) one automatically gets \[ \nabla_b R \equiv 0.\] We emphasize that this conclusion only holds for conformal matter fields. This statement asserts that the space-time scalar curvature must be constant on the manifold, and for initial data sets that satisfy the constraints, this value can be found from the initial data. So returning to our equation in wave coordinates, we can use \[ R \equiv R_0 \] as an equation of motion, noting that the principal part of $R$ is given by our wave operator acting on $\sqrt{|\det(g)|}$, precisely the missing scale factor.

We can try to reproduce this for general matter fields. Denote by $T$ the trace of the stress energy tensor, we have that \[ \nabla^a \hat{T}_{ab} = - \frac14 \nabla_b T.\] This implies, again using the Bianchi identities, that \[ \nabla_b R = - \nabla_b T \] or that \[ R = -T + \lambda \] where $\lambda$ is an arbitrary constant. This constant must be fixed by an initial data that satisfies the constraint condition. Repeating the argument above we find that we can solve for the scale using the quasilinear equation for $\sqrt{|\det(g)|}$ given by the scalar curvature.

Conclusion

The "trace-free" version of general relativity has a well-posed initial value formulation, and in fact is identical to standard general relativity, except that the value of the cosmological constant is not fixed as a universal physical constant, but as a constant defined through compatible initial data constraints.

The reason that cosmologists seem to be enamoured with this theory seems to be the fact that this gives them an excuse to consider scalar-fields (which model the inflaton field) with arbitrary potentials. From the point of view of the stress energy tensor, adding a nonlinear potential $V$ so that the equation of motion changes from \[ \Box \phi = 0 \Longrightarrow \Box\phi = V'(\phi) \] would also involve changing the stress energy tensor by adding a term $-\frac12 g_{ab} V(\phi)$. However, as this term is pure trace, from the point of view of the trace-free theory, the "effective" stress energy $\hat{T}$ is formally the same as that of the free scalar field. As we have seen above, however, the underlying mathematics and physical predictions are really no different.