Metric of spacetime with zero speed of light
From a mathematical perspective this is a question about signatures of quadratic forms. On $\mathbb{R}^4$ with space coordinates $(x, y, z)$ and time coordinate $t$ we can consider
- A Euclidean metric, corresponding to the quadratic form $x^2 + y^2 + z^2 + (ct)^2$, with signature $(+, +, +, +)$
- A Lorentzian / Minkowski metric, corresponding to the quadratic form $x^2 + y^2 + z^2 - (ct)^2$, with signature $(+, +, +, -)$ (or the negative of this, depending on your conventions)
- A Galilean metric ($c \to \infty$), corresponding to the quadratic form $-t^2$, with signature $(0, 0, 0, -)$; we get this by dividing the previous expression by $c^2$ and just taking the limit in the most obvious sense
- A $c = 0$ metric, corresponding to the quadratic form $x^2 + y^2 + z^2$, with signature $(+, +, +, 0)$.
The physical significance of this is not entirely clear to me; I don't have much experience thinking about relativity, but here are some speculations off the top of my head.
If we think of these quadratic forms as describing spacelike vs. timelike vs. lightlike directions, then in Galilean spacetime every direction is timelike or lightlike, while in $c = 0$ spacetime every direction is spacelike or lightlike. I guess we can think of $c$ as describing the "slope of the light cone," in which case the Galilean limit $c \to \infty$ corresponds to everything being in your light cone, while the $c \to 0$ limit corresponds to nothing being in your light cone.
I guess your conception of the Galilean limit $c \to \infty$ is that it describes a universe where "time is infinitely more important than space," so first we have the quadratic form $-t^2$ dividing up time slices but then in each timeslice we have the usual 3d Euclidean metric, reflecting the fact that an infinite speed of light means we can in principle reach any point in space from any other point in space at a given time, but we still can't e.g. travel backwards in time. If so, then the corresponding $c \to 0$ limit describes a universe where "space is infinitely more important than time," so we can think of it as divided up into isolated points of 3d space, each of which has a timeline with a 1d time metric, reflecting the fact that a zero speed of light means nothing can move.
Perhaps we should call the $c = 0$ case Zenoan spacetime.
Edit, 9/4/20: The $c \to 0$ limit is briefly discussed in Freeman Dyson's Missed Opportunities; he calls the corresponding automorphism group $G$ the "Carroll group," after Lewis Carroll:
"A slow sort of country," said the Queen, "Now, here, you see, it takes all the running you can do, to keep in the same place."
I don't know what would happen with the spherical shell; I'm not proficient with General Relativity. But I guess that nothing would move (or that everything "moves" in the same timelike direction), and I generally agree with Qiaochu Yuan's answer.
The standard Lorentzian metric $\eta$ applies to a pair of vectors $u,v$ as
$$\eta(u,v)=u^xv^x+u^yv^y+u^zv^z-c^2u^tv^t.$$
Sometimes it's better to think of $\eta$ as a function from vectors to covectors, in which case $\eta(u,v)$ should be written as $\eta(u)(v)$, with
$$\eta(u)=\eta(u^xe_x+u^ye_y+u^ze_z+u^te_t)=u^x\varepsilon^x+u^y\varepsilon^y+u^z\varepsilon^z-c^2u^t\varepsilon^t.$$
$e_\mu$ and $\varepsilon^\nu$ are the basis vectors and covectors, related by $\varepsilon^\nu(e_\mu)=\delta_\mu^\nu$. This form of $\eta$ has an inverse, a function from covectors to vectors:
$$\eta^{-1}(\omega)=\eta^{-1}(\omega_x\varepsilon^x+\omega_y\varepsilon^y+\omega_z\varepsilon^z+\omega_t\varepsilon^t)=\omega_xe_x+\omega_ye_y+\omega_ze_z-\frac{1}{c^2}\omega_te_t.$$
The Galilean metrics come from taking $c\to\infty$ :
$$\eta_\infty^+(u)=\lim_{c\to\infty}\frac{1}{c^2}\eta(u)=-u^t\varepsilon^t$$
$$\eta_\infty^-(\omega)=\lim_{c\to\infty}\eta^{-1}(\omega)=\omega_xe_x+\omega_ye_y+\omega_ze_z.$$
These have signature $(0,0,0,-)$ on vectors, and $(+,+,+,0)$ on covectors. (And the limit has broken their inverse relationship.) Now we look at zero lightspeed:
$$\eta_0^+(u)=\lim_{c\to0}\eta(u)=u^x\varepsilon^x+u^y\varepsilon^y+u^z\varepsilon^z$$
$$\eta_0^-(\omega)=\lim_{c\to0}c^2\eta^{-1}(\omega)=-\omega_te_t.$$
These have signature $(+,+,+,0)$ on vectors, and $(0,0,0,-)$ on covectors. In this sense, $c\to0$ and $c\to\infty$ are dual to each other.
Another way to express $\eta$ uses the tensor product:
$$\eta=\eta_{\mu\nu}\varepsilon^\mu\otimes\varepsilon^\nu=\varepsilon^x\otimes\varepsilon^x+\varepsilon^y\otimes\varepsilon^y+\varepsilon^z\otimes\varepsilon^z-c^2\varepsilon^t\otimes\varepsilon^t.$$
You asked in the comments (but not in the OP) about covariant derivatives. Let's require the metric to be "constant", differentiating with respect to $e_\xi$ :
$$0=\partial_\xi\eta=(\partial_\xi\eta_{\mu\nu})\varepsilon^\mu\otimes\varepsilon^\nu+\eta_{\mu\nu}(\partial_\xi\varepsilon^\mu)\otimes\varepsilon^\nu+\eta_{\mu\nu}\varepsilon^\mu\otimes(\partial_\xi\varepsilon^\nu)$$
(each $\eta_{\mu\nu}$ is just a number, $1$ or $0$ or $-c^2$)
$$=(0)+\eta_{\mu\nu}(-\Gamma^\mu_{\pi\xi}\varepsilon^\pi)\otimes\varepsilon^\nu+\eta_{\mu\nu}\varepsilon^\mu\otimes(-\Gamma^\nu_{\pi\xi}\varepsilon^\pi)$$
(renaming summation variables in the last term)
$$=-\eta_{\mu\nu}\Gamma^\mu_{\pi\xi}\varepsilon^\pi\otimes\varepsilon^\nu-\eta_{\pi\mu}\Gamma^\mu_{\nu\xi}\varepsilon^\pi\otimes\varepsilon^\nu$$
$$=-\Big(\eta_{\mu\nu}\Gamma^\mu_{\pi\xi}+\eta_{\mu\pi}\Gamma^\mu_{\nu\xi}\Big)\varepsilon^\pi\otimes\varepsilon^\nu.$$
All of the tensors $\varepsilon^x\otimes\varepsilon^x,\;\varepsilon^x\otimes\varepsilon^y,\;\varepsilon^y\otimes\varepsilon^x,\cdots$ are independent, so this sum vanishing means that each coefficient vanishes:
$$0=\eta_{\mu\nu}\Gamma^\mu_{\pi\xi}+\eta_{\mu\pi}\Gamma^\mu_{\nu\xi}$$
(and the sum over $\mu$ reduces to a single term, because most of $\eta_{\mu\nu}$ are $0$)
$$=\eta_{\nu\nu}\Gamma^\nu_{\pi\xi}+\eta_{\pi\pi}\Gamma^\pi_{\nu\xi}.$$
This is to be true for all $\xi,\pi,\nu$. Cycling the names,
$$0=\eta_{\pi\pi}\Gamma^\pi_{\xi\nu}+\eta_{\xi\xi}\Gamma^\xi_{\pi\nu}$$
$$0=\eta_{\xi\xi}\Gamma^\xi_{\nu\pi}+\eta_{\nu\nu}\Gamma^\nu_{\xi\pi}$$
and using $\Gamma^\nu_{\mu\xi}=\Gamma^\nu_{\xi\mu}$ (no torsion), these 3 equations have the form $a=-b=c=-a$ which implies $a=0$, that is,
$$\eta_{\nu\nu}\Gamma^\nu_{\pi\xi}=0.$$
Finally, with the Lorentzian metric (finite $c$), each $\eta_{\nu\nu}\neq0$, and thus all Christoffel symbols must vanish: $\Gamma^\nu_{\pi\xi}=0$. This means that the "straight lines" are the obvious ones.
Applying the same process to $\eta^{-1}$ instead, we would get
$$0=\eta^{\nu\nu}\Gamma^\pi_{\nu\xi}+\eta^{\pi\pi}\Gamma^\nu_{\pi\xi}$$
but we'd need to multiply by $\eta^{\xi\xi}$ (losing information in the non-Lorentzian cases) to combine the 3 cycled equations. The result is
$$\eta^{\nu\nu}\eta^{\xi\xi}\Gamma^\pi_{\nu\xi}=0$$
which, with the Lorentzian metric, again gives $\Gamma^\pi_{\nu\xi}=0$.
For the Galilean metrics, making $\eta_\infty^+$ constant gives only $\Gamma^t_{\pi\xi}=0$.
Making $\eta_\infty^-$ constant gives $\Gamma^\pi_{\nu\xi}=0$ for the spatial coordinates $\{\xi,\pi,\nu\}\subseteq\{x,y,z\}$, and $\Gamma^x_{xt}=\Gamma^y_{yt}=\Gamma^z_{zt}=\big(\Gamma^x_{yt}+\Gamma^y_{xt}\big)=\big(\Gamma^x_{zt}+\Gamma^z_{xt}\big)=\big(\Gamma^y_{zt}+\Gamma^z_{yt}\big)=0$.
There's still some freedom; for example $\Gamma^x_{tt}\neq0$, which says that a timelike geodesic may accelerate in the $x$ direction. Even if all other Christoffel symbols vanish, one component of the Riemann curvature is $R^x_{txt}=\partial_x\Gamma^x_{tt}$ which is not necessarily zero. (It is zero if $\Gamma^x_{tt}$ is uniform across space, which would happen if the Galilean spacetime is deformed in the sense of Cavalieri's principle, sliding the spacelike layers over each other while maintaining each layer's shape; this is equivalent to using an accelerating reference frame.) See also Schuller's lecture "Newtonian spacetime is curved".
For zero-lightspeed, making $\eta_0^+$ constant gives $\Gamma^x_{\pi\xi}=\Gamma^y_{\pi\xi}=\Gamma^z_{\pi\xi}=0$.
Making $\eta_0^-$ constant gives $\Gamma^t_{t\xi}=\Gamma^x_{t\xi}=\Gamma^y_{t\xi}=\Gamma^z_{t\xi}=0$ (but those last three are special cases of the previous line, with $\pi=t$).
Again there's some freedom; for example $\Gamma^t_{xx}\neq0$. Similar comments apply.