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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.

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.

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 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.

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. 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 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.