Is radial time a solution of Einstein's equations? 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?
 A: 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.
A: I believe that this is equivalent to a Universe which is closed spatially but not in the time dimension since the point at $r=0$ will transform to $t=-\infty$, the transform will be $r=e^t$. The singularity at $r=0$ merely represents the infinite past. We have merely 'squished' the time dimension up to make it appear it has a boundary.
