# Metric of spacetime with zero speed of light

Despite references to physical spaces, this question is purely mathematical on differential geometry.

While deriving coordinate transformations based on common assumptions of homogeneity and uniformity of space and time (see A Primer on Special Relativity or Nothing but Relativity), typically three logical options are considered:

1. Finite speed of light - Minkowski spacetime.

2. Infinite speed of light - Galilean spacetime, ruled out by observation.

3. Imaginary speed of light (negative square) - Euclidean spacetime, ruled out by causality.

There exist however the forth logical option:

1. Zero speed of light.

At first it appears unreal, but in fact there exists a conceptual case asymptotically close to this scenario.

Imagine a thin spherical shell approaching its Schwarzschild radius. Spacetime inside this empty shell is locally flat Minkowski with the same time dilation as at the shell. As the shell approaches an infinite time dilation at the Schwarzschild radius, the speed of light at and inside the shell approaches zero.

What is the metric structure of this space in the limit of the speed of light being exactly zero?

Metrics for other three cases are well known. The Euclidean and Minkowski metrics don't require an introduction. The Galilean structure is described here: What is a mathematical definition of the Maxwellian spacetime?

Would the $$c=0$$ spacetime collapse simply to a 3D Euclidean space with no time or would it have two separate metrics for space and time like the Galilean spacetime?

• It's a good question, I don't know. However, Minkowski metric with $c=0$ is degenerate. This may cause difficulties e.g. with volume element or different isomorphisms widely used (e.g. this between contravariant and covariant vectors). – Paweł Czyż Aug 1 '19 at 22:08
• @PawełCzyż If by "degenerate" you mean a zero determinant, then the Minkowski metric $dt^2-\dfrac{dr^2}{c^2}$ is also degenerate with $c=\infty$ (please correct me if I am wrong). However, this is the case of the Galilean spacetime with separate spatial and temporal metrics. Physically though, yes, $c=0$ seems to imply no energy, no matter, no time, no motion, nothing happening ever. Still this space may have some interesting properties. For example, how can it be connected to a pseudo-Riemannian manifold (e.g at the event horizon)? – safesphere Aug 2 '19 at 6:25
• "in the limit of the speed of light being exactly zero". does this mean "the speed of light is zero"? – mathworker21 Sep 16 '19 at 16:34
• A physical space that does not transmit transverse waves (physically equivalent to $c=0$) could perhaps be formed as a result as of a phase change e.g. a "solid" like space changing to a "fluid" like space. A physical space which does not transmit light cannot have a metric defined with reference to light. It will be interesting what you conclude in regard to mathematical spaces. – James Arathoon Sep 16 '19 at 19:20
• @mathworker21 I am not sure, this is a part of a question. For example, is $x\cdot\sin{(x)}$ exactly zero at $x=0$? I am not a mathematician, so I am just trying to be careful with my statements. My primary interest is, yes, $c=0$. However, in case this option is trivial or gives a different result, please also consider the following option as well: $$\lim_{t\to\infty}c=0$$ where t is the Schwarzschild coordinate time, e.g. the time of a thin massive shell collapsing to the event horizon, but never exactly reaching it. In this case $c$ is the speed of light asymptotic to zero inside the shell. – safesphere Sep 16 '19 at 19:53

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.

• Thank you! A fantastic answer that I plan to accept shortly, although it would take me a bit longer to fully comprehend. Meanwhile, a quick question. I understand I cannot do a coordinate transformation (velocity boost) in this spacetime, because all velocities are zero. However, what if this spacetime is spatially limited, say, to a ball of a certain size with a gradient of the speed of light outside reaching zero at the border? If I look at this ball from outside, I can change to a coordinate system, in which the ball is moving (still with no relative motion inside of course). ... [TBC] – safesphere Sep 20 '19 at 21:49
• Intuitively I don't see any contradiction in this setup, but I wonder if there are any caveats or anything to say or clarify about the description of this spacetime when it is looked at from an outside moving frame (where by "looked" I just mean coordinates, but not light rays, as no light can come from there of course)? Thanks again! – safesphere Sep 20 '19 at 21:57
• With $c\to\infty$, a velocity boost becomes a shear transformation preserving spacelike directions. With $c\to0$, a velocity boost becomes a shear transformation preserving the timelike direction. As for the sphere, maybe you could get an answer at physics.stackexchange.com . – mr_e_man Sep 23 '19 at 0:20

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

1. A Euclidean metric, corresponding to the quadratic form $$x^2 + y^2 + z^2 + (ct)^2$$, with signature $$(+, +, +, +)$$
2. 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)
3. 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
4. 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."

• Your description that "space is infinitely more important than time" matches my intuition and "it as divided up into isolated points of 3d space, each of which has a timeline" confirms my expectation of "two separate metrics for space and time". This is a spacetime with absolute space as opposed to Galilean absolute time. So the two metrics are the same as Galilean, right? And the difference is in the coordinate derivative operator specifying geodesics, as defined in 3.1 here: philsci-archive.pitt.edu/11265/1/spacetimestructure.pdf - If so, what is the expression for this operator? – safesphere Sep 16 '19 at 22:22
• As a side (relativity) note, this spacetime would occupy a (spherical) region inside a thin and heavy collapsing shell (plus some other cases). Being very much finite in its spatial extent, it is important how this spacetime is described from outside. E.g. I can boost my frame to some velocity, so my coordinates now are (x',y',z',t'). So how does my relative space and time reconcile with the absolute space and "infinitely less significant" time inside the shell? Having equations would help with this understanding, so I wonder if you could please check the link in my comment above. Thanks much! – safesphere Sep 16 '19 at 22:38
• Also, a related (but less important) question on the connection of this spacetime to infinity outside both having the same spatial scale without an obvious reason: math.stackexchange.com/questions/3352423/… – safesphere Sep 16 '19 at 22:42
• @safesphere: I'm not familiar enough with physics or with differential geometry here to be sure I know what I'm talking about here. If I understand correctly, I think a geodesic in Zenoan spacetime is only allowed to move in time and can't move in space. Does that answer your question about the covariant derivative? – Qiaochu Yuan Sep 16 '19 at 23:14
• I think a better approach to Galilean spacetime is to put the $000+$ (or $000-$) metric on vectors, and $+\!+\!+0$ on covectors rather than spacelike vectors. (That's implied, but not explained, in the linked "Maxwellian" question, by using upper indices on $h^{ab}$.) For the $c\to0$ spacetime, we could put the $+\!+\!+0$ metric on vectors, and $000+$ on covectors; so $c\to0$ and $c\to\infty$ are dual to each other. I haven't thought about the spherical shell or covariant derivatives. – mr_e_man Sep 18 '19 at 23:01