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Imagine a (flat) 4D space where we measure time outwards in a radial direction from the origin.

So that 3D space at a given time would consist of a spherical shell. (As such this would be a closed Universe.)

In a far distant time the spherical shells at any given position would be essentially flat and the shells so big as to make the Universe appear infinite.

Light rays would have to only cross the spherical shells at 45 degree angles. Hence we could impose a partially ordered set on the events. And the lightcones at every point in this 4D space would be well defined.

In a sense this 4D space-time would have no boundary, but in another sense we have defined the origin as a special point at which time "begins". Light rays would kind of spiral out from the origin.

Is the space-time as I've described it a solution of GR? Is there anything special about it - to me it seems like the second most obvious way to imposee a partial order to the set of events into a space that is not simple Minkowski space.

Is this a solution of GR? In which case what would the metric be?

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    $\begingroup$ This is a choice of coordinates for $\mathbb{R}^4$, but with a coordinate singularity at the origin. Since $\mathbb{R}^4$ satisfies the Einstein equations with canonical coordinates, you can be sure that the Ricci tensor will also vanish with this choice of coordinates. If you are instead working in Lorenztian signature, this would correspond to a choice of "light-cone" coordinates. $\endgroup$
    – hulsey
    Commented Mar 22, 2021 at 21:45
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    $\begingroup$ @hulsey Your argument cannot be quite right. Minkowski space, which is topologically $\mathbb{R}^4$, is indeed a solution, but you also have to keep track of the direction of time-like vector fields, which is an entirely coordinate independent thing. Minkowski space does not, in fact, have time-like vector fields pointing radially away from the origin of $\mathbb{R}^4$. I imagine you could probably Wick rotate from Euclidean $\mathbb{R}^4$, which is a solution to the Euclidean equations, but I think this is not trivial. $\endgroup$
    – Richard Myers
    Commented Mar 23, 2021 at 2:04
  • $\begingroup$ I imagine if you punched a hole in the origin and transformed the space into a tube. This would look like Minkowski space-time on a tube. Then you would need to adjust the metric to get back to what you started with. $\endgroup$
    – user84158
    Commented Mar 23, 2021 at 2:57
  • $\begingroup$ To clarify what you mean by "radial time": If you started with Minkowski spacetime in Minkowski coordinates $(t,x,y,z)$, are talking about what happens if you do a coordinate transformation such that $t \mapsto t' = \sqrt{t^2+x^2+y^2+z^2}$ and the spatial coordinates are unchanged? $\endgroup$
    – Brick
    Commented Jun 28, 2021 at 16:07

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In your question you specify in the beginning a flat 4D space. So that already defines all of the physics. In other words, this is just flat Minkowski spacetime with some odd coordinates.

This is a (trivial) solution to GR, and nothing in the coordinates changes any of the topology of the universe nor any of the physics. It is just standard flat spacetime.

I am assuming that the coordinate transform that you are thinking of is something like this. For simplicity I will use just 1+1D, the extension to full 4D hyperspherical coordinates is left as an exercise for the reader. Starting with standard Minkowski spacetime in natural units with $ds^2 = -dt^2 + dx^2$ we can define the following coordinate transformations to your coordinates:

$$T=\sqrt{t^2+x^2}$$ $$\theta=\tan^{-1}(t,x)$$

And the inverse transform from your coordinates:

$$t=T \cos(\theta)$$ $$x=T \sin(\theta)$$

Now, the issue is that your $T$ is labeled as "time" but it is no longer timelike everywhere. It is a perfectly valid coordinate, but it physically does not match what is measured by clocks. In fact, we can write the metric in your coordinates by simple substitution:

$$ds^2 = -dt^2 + dx^2 = -d(T\cos(\theta))^2 + d(T\sin(\theta))^2$$ $$ ds^2 = -\cos(2\theta) \ dT^2 + T^2 \cos(2 \theta) \ d\theta^2 + 2 \sin(2\theta) \ dT d\theta$$ Notice, in particular, if we choose a constant $\theta$ then $d\theta = 0$ and so $ds^2 = -\cos(2\theta)dT^2$. Whenever $\cos(2\theta)<0$ this expression becomes positive, meaning that $T$ stops being something that is measured with clocks. We call it time, but that is just a label, physically it is not something measured with a clock in some parts of spacetime.

In particular, when $\pi/4<\theta<3\pi/4$ then $\cos(2\theta)$ is negative and $T$ is spacelike. This corresponds exactly to the region outside of the light cone in our standard spacetime diagrams, so this is the expected and desired result.

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  • $\begingroup$ No this is (cleaerly) not what I mean. I mean time flowing outwards from a point at the origin. Not sure why this answer got upvoted. It is simply a change of coordinates from normal Minkwoski spacetime where time flows along an cartesian axis. $\endgroup$
    – user84158
    Commented Jun 30, 2021 at 2:42
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    $\begingroup$ @zooby Well, clearly your (clearly) is not as clear as you thought. Since you specified that it is flat that means that it is Minkowski spacetime with a different coordinate chart. There simply isn’t another way to read your question given that description. Both hulsey and Brick also understood it the same way I did. This answer was upvoted because it was a correct and careful answer to the question you actually asked. I am sorry to hear that you thought you had asked a different question. But at least the question you did ask was interesting to work through $\endgroup$
    – Dale
    Commented Jun 30, 2021 at 3:17
  • $\begingroup$ Read the first sentence of my question again. It's very very clear indeed. Thanks anyway. But I appreciate that maybe 'flat' in the context of SR might mean something different than flat in terms of just a 4D Euclideans space, in that it may presuppose that all lightcones align in a particular direction. $\endgroup$
    – user84158
    Commented Jun 30, 2021 at 20:29
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    $\begingroup$ @zooby in relativity “flat” means Minkowski spacetime. I have read your sentence many times and I honestly see no other way to interpret what you mean. You think your meaning is clear, but if it is not what I said then your meaning is in fact exceptionally unclear. Others agree $\endgroup$
    – Dale
    Commented Jun 30, 2021 at 20:34
  • $\begingroup$ @zooby if you did not say “flat” then you could have meant a metric like $ds^2=-dt^2 + t^2 dx^2 + t^2 dy^2 + t^2 dz^2$ which would be analogous to polar coordinates where $ds^2=dr^2 + r^2 d\theta^2$ except the radial coordinate is timelike and it is 4D. The problem is that this metric is not flat. Because you specified “flat” we are left with new coordinates on the Minkowski metric, as I described. It is a correct answer to the question you actually asked. $\endgroup$
    – Dale
    Commented Jun 30, 2021 at 23:17

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