Minimal cosmography

Abstract

The minimal requirement for cosmography—a non-dynamical description of the universe—is a prescription for calculating null geodesics, and time-like geodesics as a function of their proper time. In this paper, we consider the most general linear connection compatible with homogeneity and isotropy, but not necessarily with a metric. A light-cone structure is assigned by choosing a set of geodesics representing light rays. This defines a “scale factor” and a local notion of distance, as that travelled by light in a given proper time interval. We find that the velocities and relativistic energies of free-falling bodies decrease in time as a consequence of cosmic expansion, but at a rate that can be different than that dictated by the usual metric framework. By extrapolating this behavior to photons’ redshift, we find that the latter is in principle independent of the “scale factor”. Interestingly, redshift–distance relations and other standard geometric observables are modified in this extended framework, in a way that could be experimentally tested. An extremely tight constraint on the model, however, is represented by the blackbody-ness of the cosmic microwave background. Finally, as a check, we also consider the effects of a non-metric connection in a different set-up, namely, that of a static, spherically symmetric spacetime.

Appendix

So far we have generalised the Robertson-Walker metric to the setting where the gravitational field is encoded in a torsionless connection. In absence of a metric, but in presence of matter, there is no natural generalisation of the Einstein equation. Therefore we had to restrict ourself to cosmography. Thanks to the six symmetries and some physical requirements we still were able to identify new degrees of freedom, that might improve our understanding of cosmological observations.

In this appendix we would like to explore the new degrees of freedom that we obtain when we try to describe the gravitational field of a static, spherically symmetric star by a torsionless connection. Here we only have four symmetries, but we are in vacuum where Einstein’s equation (with vanishing cosmological constant) still makes sense. Again we will add some physical requirements:

• parity conservation, which is automatic for the Levi-Civita connection,

• asymptotic flatness of the vacuum solution, which is automatic for the Levi-Civita connection. In this asymptotic space we assume special relativity to hold. I.e there we do have a Lorentzian metric of signature $+---$. The coordinate t singled out by the staticity requirement is supposed to be time-like with respect to this Lorentzian metric.

• We also want our geodesics to describe an attractive force admitting freely falling particles on circles.

Again our starting point is Eq. (2) describing the transformation of the connection under infinitesimal diffeomorphisms, i.e. vector fields. Here we look for all simultaneous solutions with $\xi =\partial _t$ and $\xi =J_i$. In polar coordinates the most general such solution, which is at the same time parity even, has the following non-vanishing components (up to symmetry in the last two indices):

$$\begin{aligned} {\Gamma ^t}_{tr}&=D, \quad {\Gamma ^r}_{tt}=E, \quad \quad {\Gamma ^r}_{rr}=F,\quad {\Gamma ^r}_{\theta \theta }= {\Gamma ^r}_{\varphi \varphi }/\sin ^2\theta =Y, \end{aligned}$$

(30)

$$\begin{aligned} {\Gamma ^\theta }_{r\theta }&= {\Gamma ^\varphi }_{r\varphi }=X,\quad {\Gamma ^\theta }_{\varphi \varphi }=-\sin \theta \cos \theta ,\quad {\Gamma ^\varphi }_{\theta \varphi }= \cos \theta /\sin \theta \end{aligned}$$

(31)

with five functions of the radius r only: $D,\,E,\,F,\,Y,$ and X. For the Levi-Civita connection of the metric $\mathrm{{d}}\tau ^2=B(r)\,\mathrm{{d}}t^2-A(r)\,\mathrm{{d}}r^2-r^2\mathrm{{d}}\theta ^2-r^2\sin ^2\theta \mathrm{{d}}\varphi ^2$, we have: $D={\textstyle \frac{1}{2}} B'/B,\ E={\textstyle \frac{1}{2}} B'/A,\ F={\textstyle \frac{1}{2}} A'/A,\ Y=-r/A,\ X=1/r$. The prime denotes the derivative with respect to r.

If we want an attractive force we must have E positive and for circles with constant angular velocity to exist Y must be negative.

Up to anti-symmetry in the last two indices, the Riemann tensor has the following non-vanishing components:

$$\begin{aligned} {R^t}_{rtr}&=-D'-\,(D-F)\,D, \quad {R^t}_{\theta t\theta }={R^t}_{\varphi t\varphi }/\sin ^2\theta =DY , \end{aligned}$$

(32)

$$\begin{aligned} {R^r}_{trt}&= E'-(D-F)\,E, \quad {R^\theta }_{ t\theta t}={R^\varphi }_{ t\varphi t}=EX , \end{aligned}$$

(33)

$$\begin{aligned} {R^r}_{\theta r\theta }&= {R^r}_{\varphi r\varphi }/\sin ^2\theta =Y'+(F-X)Y,\quad {R^\theta }_{ \varphi \theta \varphi }=\sin ^2\theta \,[1+YX], \end{aligned}$$

(34)

$$\begin{aligned} {R^\theta }_{ r\theta r}&= {R^\varphi }_{ r\varphi r} =-X'+(F-X)X,\quad {R^\varphi }_{ \theta \varphi \theta }= [1+YX]. \end{aligned}$$

(35)

The Ricci tensor has components:

$$\begin{aligned} R_{tt}&=E'-(D-F)\,E+2\, EX,\quad R_{\theta \theta }=Y'+(D+F)\,Y+1, \end{aligned}$$

(36)

$$\begin{aligned} R_{rr}&=-D'-(D-F)\,D-2X'+2\,( F-X)\,X,\quad R_{\varphi \varphi }=\sin ^2\theta \,R_{\theta \theta }. \end{aligned}$$

(37)

Before solving the vacuum Einstein equation, let us simplify the connection by two appropriate coordinate transformations. To this end we need the transformation law of the connection under a finite coordinate transformation $\bar{x}^{\bar{\mu }} (x)$ with Jacobi matrix

$$\begin{aligned} {\Lambda ^{\bar{\mu } }}_\mu :=\,\frac{\partial \bar{x}^{\bar{\mu } }}{\quad \partial x^\mu }. \end{aligned}$$

(38)

It reads:

$$\begin{aligned} {\bar{\Gamma } ^{\bar{\lambda } }}_{\bar{\mu } \bar{\nu }}= {\Lambda ^{\bar{\lambda } }}_\lambda { \left( \Lambda ^{-1\,T}\right) _{\bar{\mu }}}^\mu { \left( \Lambda ^{-1\,T}\right) _{\bar{\nu }}}^\nu \, {\Gamma ^\lambda }_{\mu \nu } +{\Lambda ^{\bar{\lambda } }}_\lambda { \left( \Lambda ^{-1\,T}\right) _{\bar{\mu }}}^\mu \partial _\mu { \left( \Lambda ^{-1\,T}\right) _{\bar{\nu }}}^\lambda . \end{aligned}$$

(39)

Consider the coordinate transformation with two functions f(r) and g(r),

$$\begin{aligned}&\bar{t}:=t+f(r), \quad \bar{r}:= g(r), \quad \bar{\theta } :=\theta , \quad \bar{\varphi } := \varphi , \end{aligned}$$

(40)

and with Jacobian

$$\begin{aligned} \begin{pmatrix}1&{}\quad f'&{}\quad 0&{}\quad 0\\ 0&{}\quad g'&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 1&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 1 \end{pmatrix}. \end{aligned}$$

(41)

In the barred coordinates, our connection has components given by the five functions:

$$\begin{aligned}&\bar{D}=\,\frac{1}{g'}\, D-\frac{(f')^2}{g'}\, E,\quad \bar{E}=g'E \end{aligned}$$

(42)

$$\begin{aligned}&\bar{F}=\,\frac{(f')^2}{g'}\, E+\frac{1}{g'}\,F-\frac{g''}{(g')^2}, \quad \bar{Y}=g'\,Y, \quad \bar{X}=\,\frac{1}{g'}X. \end{aligned}$$

(43)

Note that from ${R^\varphi }_{ \theta \varphi \theta }=[1+YX]$ and the negativity of Y we conclude that X is positive. Otherwise we could not impose flatness asymptotically, as $r\rightarrow \infty$. Then we use a first coordinate transformation with $f=0$ and an appropriate function g to achieve $\bar{X}(\bar{r})=1/\bar{r}$. Let us drop the bars and define the positive function $A:=-r/Y$

We use a second coordinate transformation with $g(r)=r$ and an appropriate function f. Then we obtain

$$\begin{aligned}&\bar{A}=A,\quad \bar{E}=E,\quad \bar{D}=D-(f')^2E,\quad \bar{F}=F+(f')^2E, \end{aligned}$$

(44)

which motivates the definition $C:=D+F$ with $\bar{C}=C$. We choose f such that

$$\begin{aligned} \bar{D}- \bar{F}=\,\frac{r_S}{\bar{r}\,(\bar{r}-r_S)} , \end{aligned}$$

(45)

with a positive constant $r_S$, the “Schwarzschild radius”. Again we drop the bars.

Thanks to the two coordinate transformations we remain with only three functions, $A,\,E,$ and C, the first two being positive. For D and F we have:

$$\begin{aligned}&D={\textstyle \frac{1}{2}} \left( C+\frac{r_S}{ r\,( r-r_S)}\,\right) ,\quad F={\textstyle \frac{1}{2}} \left( C-\frac{r_S}{ r\,( r-r_S)}\,\right) . \end{aligned}$$

(46)

Now we integrate the three Einstein equations in vacuum:

$$\begin{aligned} R_{tt}=E'-\left( \frac{r_S}{ r\,( r-r_S)}-\frac{2}{r}\, \right) E=0 \end{aligned}$$

(47)

yields

$$\begin{aligned} E={\textstyle \frac{1}{2}} \,k\,r_S\,\frac{r-r_S}{r^3}, \end{aligned}$$

(48)

with a dimensionless integration constant k. Likewise

$$\begin{aligned} R_{rr}=-{\textstyle \frac{1}{2}} \,C'-{\textstyle \frac{1}{2}} \,\frac{-2r+3r_S}{r\,(r-r_S)}\,C=0 \end{aligned}$$

(49)

yields

$$\begin{aligned} C=\,\frac{\lambda }{r_S^3}\,\frac{r^3}{r-r_S} , \end{aligned}$$

(50)

with another dimensionless integration constant $\lambda$. Substituting this solution we find

$$\begin{aligned} {R^t}_{rtr}=\,\frac{r_S}{r\,(r-r_S)}\, -{\textstyle \frac{1}{2}} \,\frac{\lambda }{r_S ^3}\,\frac{r^2}{r-r_S}\, , \end{aligned}$$

(51)

which tends to zero for large r only if $\lambda =0$. Therefore C vanishes identically. Finally

$$\begin{aligned} R_{\theta \theta }=Y'+1=0, \end{aligned}$$

(52)

yields $Y=-r+(1-\ell )r_S$ with another dimensionless integration constant $\ell$ and

$$\begin{aligned} A=\,\frac{1}{1-(1-\ell )\,r_S/r} . \end{aligned}$$

(53)

We still have to verify that, for this solution, all components of the Riemann tensor vanish in the limit of large r, which is true.

We recover the Levi-Civita connection of the (exterior) Schwarzschild solution by setting $k=1$ and $\ell =0$.

1.1 Classical tests

Of course we want to know how these two parameters are constrained by the classical tests of general relativity. To this end we have to integrate the geodesic equations. Without loss of generality, we work in the plane $\theta =\pi /2$ and using the abbreviation $B(r):= 1-r_S/r$ and the over-dot for the derivative with respect to the affine parameter p, we remain with:

$$\begin{aligned}&\ddot{t} + \,\frac{B'}{B}\, \dot{t}\,\dot{r}\,=0, \end{aligned}$$

(54)

$$\begin{aligned}&\ddot{r}+ \,\frac{k}{2}\, B'\,B\,\dot{t} ^2-{\textstyle \frac{1}{2}} \,\frac{B'}{B}\, \dot{r}^2- r\left( 1-(1-\ell )\,\frac{r_S}{r}\right) =0, \end{aligned}$$

(55)

$$\begin{aligned}&\ddot{\varphi } +\,\frac{2}{r}\, \dot{r}\,\dot{\varphi }\,=0. \end{aligned}$$

(56)

As in the metric case these three equations integrate once immediately:

$$\begin{aligned}&\dot{t}=1/B,\quad \dot{\varphi }=J/r^2, \end{aligned}$$

(57)

$$\begin{aligned}&\,\frac{1}{B}\, \dot{r}^2-\frac{1}{B}+J^2 \gamma = -\epsilon -(k-1)+\frac{k-1}{B} , \end{aligned}$$

(58)

with two integration constants J and $\epsilon$ and the shorthand,

$$\begin{aligned} \gamma (r):= \,\frac{1-\ell }{r^2}\, -\,2\,\frac{\ell }{r_S^2} \left[ \ln \left( 1-\,\frac{r_S}{r}\right) +\,\frac{r_S}{r}\right] . \end{aligned}$$

(59)

The third integration constant has been set to one by a proper choice of the units of time. As declared at the beginning of the appendix, we rescale the time variable in such a way that, asymptotically, $\frac{dr}{dt}\rightarrow 1$ for a radial light ray. From the above equation, this means that photons have $\epsilon =0$, and massive particles have positive $\epsilon$. We will have to compute the shape, $\varphi (r)$, of the trajectory. With $\mathrm{{d}}r/\mathrm{{d}}\varphi =\dot{r}/\dot{\varphi }$ we obtain

$$\begin{aligned} \left( \frac{\mathrm{{d}}r}{\mathrm{{d}}\varphi }\right) ^2\,\frac{J^2}{r^4B}\,= -\epsilon -(k-1) +\,\frac{k}{B}\, -J^2\gamma . \end{aligned}$$

(60)

1.2 Bending of light

Consider now the trajectory of photons, Eq. (60) with vanishing $\epsilon$. Let us trade the integration constant J for the perihelion $r_0$ characterised by $\mathrm{{d}}r/\mathrm{{d}}\varphi =0$. Using Eq. (60) and $\epsilon =0$ we have

$$\begin{aligned} J^2=\left( 1-k+\,\frac{k}{B_0}\right) /\gamma _0. \end{aligned}$$

(61)

We have set $B_0:=B(r_0)$ and $\gamma _0:=\gamma (r_0)$. We replace J in Eq. (60), solve for $\mathrm{{d}}\varphi /\mathrm{{d}}r$ and integrate:

$$\begin{aligned} \varphi =\pm \int \,\frac{\mathrm{{d}}r}{r^2\sqrt{B}}\, \left[ \frac{k/B+1-k}{k/B_0+1-k}\,\gamma _0-\gamma \,\right] ^{-1/2}. \end{aligned}$$

(62)

To do the integral, we linearize the integrand in $r_S/r_0$ and set $x:=r/r_0$. Using

$$\begin{aligned} \gamma \sim \,\frac{1}{r^2}\, \left[ 1+\,\frac{2\ell }{3}\,\frac{r_S}{r}\,\right] , \end{aligned}$$

(63)

we find:

$$\begin{aligned} \varphi&\sim \pm \int \,\frac{\mathrm{{d}}x}{x}\,\frac{1}{\sqrt{x^2-1}}\, \left[ 1+{\textstyle \frac{1}{2}} \,\frac{r_S}{r_0}\,\frac{1}{x}\,+ {\textstyle \frac{1}{2}} k\,\frac{r_S}{r_0}\,\frac{x}{x+1}\, - {\textstyle \frac{1}{3}} \ell \,\frac{r_S}{r_0}\,\frac{x^2+x+1}{x\,(x+1)}\, \right] \nonumber \\&=:\int I(x)\,\mathrm{{d}}x. \end{aligned}$$

(64)

Let us choose the incoming direction of the photons at $\varphi =\pi$, denote the scattering angle by $\Delta \varphi$ and integrate Eq. (64) on the left-hand side from $\varphi = \varphi _0$ to $\pi$ and on the right-hand side from $x=1$ to $\infty$. In this domain $\mathrm{{d}}\varphi /\mathrm{{d}}r$ is positive and we have:

$$\begin{aligned} \Delta \varphi =\pi -2\varphi _0=\pi -2\left( \pi -\int _1^\infty I(x)\,\mathrm{{d}}x\right) =\,\frac{r_S}{r_0}\,\left[ 1+k-\,\frac{4}{3}\, \ell \right] . \end{aligned}$$

(65)

We have used the following integrals:

$$\begin{aligned} \int \,\frac{\mathrm{{d}}x}{x\sqrt{x^2-1}}\,&=-\arcsin \,\frac{1}{x},\quad \int \,\frac{\mathrm{{d}}x}{x^2\sqrt{x^2-1}}=\,\frac{\sqrt{x^2-1}}{x}, \end{aligned}$$

(66)

$$\begin{aligned}&\displaystyle \int \,\frac{\mathrm{{d}}x}{(x+1)\sqrt{x^2-1}}=\sqrt{\frac{x-1}{x+1}},\nonumber \\&\displaystyle \int \,\frac{\mathrm{{d}}x}{\sqrt{x^2-1}}\,\frac{x^2+x+1}{x^2(x+1)} =\sqrt{x^2-1}\left( \,\frac{1}{x}+\frac{1}{x+1}\,\right) . \end{aligned}$$

(67)

2010 data from Very Long Baseline Interferometry [12] constrain the parameters at the $10^{-4}$ level:

$$\begin{aligned} (k-1)-\,\frac{4}{3}\, \ell =(-0.8\pm 1.2)\times 10^{-4}. \end{aligned}$$

(68)

On galactic scales, let us mention data from lensing by fifteen elliptical galaxies in the Sloan Digital Sky Survey [13]. They constrain our parameters at the 10 % level.

1.3 Time delay of light

Our starting point is again Eq. (58) with $\epsilon =1-k$ and J expressed in terms of the perihelion $r_0$, Eq. (61). Here we need t(r) noting that t is the proper time of an observer at rest at infinity:

$$\begin{aligned} \,\frac{\mathrm{{d}}t}{\mathrm{{d}}r}\, =\,\frac{\dot{t}}{\dot{r}}\, = \,\frac{1}{B}\, \left[ (k+(1-k)\,B)\,- (k+(1-k)\,B_0)\,\frac{B\gamma }{B_0\gamma _0}\,\right] ^{-1/2}. \end{aligned}$$

(69)

We linearize in $r_S/r_0$,

$$\begin{aligned} \,\frac{\mathrm{{d}}t}{\mathrm{{d}}r}\,\sim \,\frac{x}{\sqrt{x^2-1}}\, \left[ 1+\,\frac{1}{x\,(x+1)}\, \left\{ x+\,\frac{3}{2}\, -\,\frac{k-1}{2}\,x-\,\frac{\ell }{3}\,\right\} \,\frac{r_S}{r_0}\,\right] . \end{aligned}$$

(70)

Integrating we obtain the time of flight of the photon from a location r to the perihelion,

$$\begin{aligned} t(r,r_0)\sim&\sqrt{r^2-r_0^2}\nonumber \\&+r_S\left[ \left( \frac{1}{2}\, -\, \frac{k-1}{2}-\frac{\ell }{3}\right) \frac{\sqrt{x-1}}{\sqrt{x+1}}\, +\left( 1 -\,\frac{k-1}{2}\right) \ln (x+\sqrt{x^2-1}\,)\right] , \end{aligned}$$

(71)

where we have used the first integral in (67) and

$$\begin{aligned} \int \,\frac{x\,\mathrm{{d}}x}{(x+1)\sqrt{x^2-1}}\, =-\,\frac{\sqrt{x-1}}{\sqrt{x+1}}\, +\ln (x+\sqrt{x^2-1}\,). \end{aligned}$$

(72)

The time delay after a round trip of the photon from the earth at $r_e$ to the reflector at $r_r$ and back therefore is:

$$\begin{aligned} \Delta t&\sim 2\,r_S\left[ \left( \frac{1}{2}\, -\,\frac{k-1}{2}-\frac{\ell }{3}\right) \left\{ \frac{\sqrt{r_e^2-r_0^2}}{\sqrt{r_e^2+r_0^2}}+\frac{\sqrt{r_r^2-r_0^2}}{\sqrt{r_r^2+r_0^2}} \right. \right\} \nonumber \\&\quad \left. +\left( 1 -\,\frac{k-1}{2}\right) \left\{ \ln \,\frac{r_e+\sqrt{r_e^2-r_0^2}}{r_0}\, +\ln \,\frac{r_r+\sqrt{r_r^2-r_0^2}}{r_0}\right\} \right] . \end{aligned}$$

(73)

The most precise measurement of the time delay today comes from the Doppler tracking of the Cassini spacecraft in 2002 [12, 14], with $r_0=1.6\,R_\odot$ and $r_r=6.43$ AU. The precision of this measurement is at the $10^{-5}$ level and yields for our parameters:

$$\begin{aligned} 11.0\,(k-1)-1.3\, \ell =(-2.1\pm 2.3)\times 10^{-5}. \end{aligned}$$

(74)

1.4 Perihelion advance of Mercury

Let us denote by $r_+$ and $r_-=r_0$ the maximum and minimum distance between Mercury and sun. The orbit of Mercury is given by Eq. (60) with positive $\epsilon$. Let us trade the two integration constants J and $\epsilon$ for $r_+$ and $r_-$ characterised by $\mathrm{{d}}r/\mathrm{{d}}\varphi =0$. Equation (60) yields:

$$\begin{aligned} J^2=k\,\frac{1/B_+-1/B_-}{\gamma _+-\gamma _-}\,\quad \mathrm{and}\quad \epsilon =k\,\frac{\gamma _+/B_+-\gamma _+/B_-}{\gamma _+-\gamma _-}\,-(k-1), \end{aligned}$$

(75)

Replacing J and $\epsilon$ in Eq. (60) we see that the ks cancel. Solving for $\mathrm{{d}}\varphi /\mathrm{{d}}r$ and integrating we have:

$$\begin{aligned} \varphi (r_+)=:\varphi _+=\varphi _-+\int _{r_-}^{r_+}\,\frac{\mathrm{{d}}r}{r^2\sqrt{B}}\, \left[ \,\frac{(1-B/B_-)\gamma _+-(1-B/B_+)\gamma _-}{B/B_+-B/B_-}\, -\gamma \,\right] . \end{aligned}$$

(76)

As usual we linerize the integrand in $r_S/r_-$,

$$\begin{aligned} \varphi _+-\varphi _-=&\left[ \sqrt{r_+r_-}+\frac{3-2\ell }{6}\,\frac{r_++r_-}{\sqrt{r_+r_-}}\, r_S\,\right] \int _{r_-}^{r_+}\frac{\mathrm{{d}}r}{r\sqrt{(r_+-r)(r-r_-)}}\nonumber \\&+\,\frac{3-2\ell }{6}\,{\sqrt{r_+r_-}}\, r_S\, \int _{r_-}^{r_+}\frac{\mathrm{{d}}r}{r^2\sqrt{(r_+-r)(r-r_-)}}. \end{aligned}$$

(77)

Using the integrals,

$$\begin{aligned} \int \frac{\mathrm{{d}}r}{r\sqrt{(r_+-r)(r-r_-)}}=&\,\frac{2}{\sqrt{r_+r_-}}\, \arctan \sqrt{\,\frac{r_+}{r_-}\,\frac{r-r_-}{r_+-r}}, \end{aligned}$$

(78)

$$\begin{aligned} \int \frac{\mathrm{{d}}r}{r^2\sqrt{(r_+-r)(r-r_-)}}=&\,\frac{1}{r_+r_-}\, \frac{\sqrt{(r_+-r)(r-r_-)}}{r}\, \nonumber \\&+ \,\frac{r_++r_-}{\sqrt{r_+r_-}^3}\, \arctan \sqrt{\,\frac{r_+}{r_-}\,\frac{r-r_-}{r_+-r}}, \end{aligned}$$

(79)

we obtain

$$\begin{aligned} \varphi _+-\varphi _-=\pi \left[ \,1+\,\frac{3-2\ell }{4}\,\frac{r_++r_-}{r_+r_-}\,r_S\,\right] . \end{aligned}$$

(80)

Finally the perihelion shift $\Delta \varphi$ after one circumnavigation is given by [15]:

$$\begin{aligned} \Delta \varphi =2(\varphi _+-\varphi _-)-2\pi =\,\frac{3-2\ell }{2}\,\frac{r_++r_-}{r_+r_-}\,r_S. \end{aligned}$$

(81)

Today, the most precise measurement of the perihelion advance include 2013 data on the orbit of Mercury collected by the spacecraft Messenger that orbited Mercury [12, 16]. Comparison with Eq. (81) yields,

$$\begin{aligned} \ell =(-0.15\pm 1.3)\times 10^{-5}. \end{aligned}$$

(82)

1.5 Gravitational redshift

For this test we would need a proper time not only far away from the sun, but everywhere in space. We do not see a natural substitute for the metric, that would define this proper time using the connection only. However within $10^{-5}$, the other three classical tests constrain our connection to derive from a metric. Using this metric to define proper time is of course what Einstein did a century ago.