differential geometry - How can I trace the path of a photon on space-time defined by this metric?

I have a hypothetical universe where the relation between space and time is defined by this metric:

$$ds^2=-\phi^2dt^4+dx^2+dy^2+dz^2$$ Where $\phi = 1.94 \times 10^{-14}\space km\space s^{-2}$ (note that all terms are rank-2 differential form). Ignoring one of the dimensions of space, the surface of this space-time looks like this:

I would like to describe how a photon of light travelling at c would travel on this surface, but I lack the basic skills to perform the operation using differential geometry. Note that the size of the universe is accelerating with the passage of time. I know it can be done using integration, geometry and a lot of algebra, but I'm looking for a more elegant way to solve it using the metric.

My question is, if I know the current time and the amount of time the photon has been travelling at c, can I calculate the amount of space it's traveled?

Answer

First, your metric must be corrected to be rank-2. Evidently you are trying to write a metric for the universe expanding with the acceleration $\phi$ $$\,r\sim \phi t^2$$

Your corrected metric equation would be

$$ds^2=-\phi^2t^2dt^2+dr^2$$

Where

$$dr^2\equiv dx^2+dy^2+dz^2$$

Null geodesics (trajectories of light) would be found by setting $\,ds=0$

$$r=\dfrac{\phi}{2}(t-t_o)^2$$

As you can easily see, for $\phi=0$ time in this universe stops. A better metric is

$$ds^2=-(\phi t+c)^2dt^2+dr^2$$

With the trajectories of light given by

$$r=c(t-t_o)+\dfrac{\phi}{2}(t-t_o)^2$$

Where $r(t_o)$ and $t_o$ are the coordinates of the light emission event and the negative solution is dropped as the time reversal. The coordinate distance $r$ traveled by light is

$$r=\sqrt{x^2+y^2+z^2}$$

EDIT based on OP's requirements in the comments

The features that need to be captured are

A) $x=\phi t^2$ (the diameter of the universe is proportional to the square of the time)

B) The space is closed (travel far enough in a straight line and you arrive where you started)

C) A metric that can be compared to FLRW, and hopefully one that provides a differential solution for the path of a photon

We start with a generic metric for an expanding space assuming its homogeneity and isotropy

$$ds^2 = - c^2 dt^2 + a(t)^2 d\mathbf{\Sigma}^2$$

Where $\mathbf{\Sigma}$ ranges over a 3-dimensional space of a uniform positive curvature and $a(t)$ is a scale factor. According to the requirement (A) where $\phi$ is acceleration

$$r=\dfrac{\phi}{2}t^2$$

While the topology of this space may vary, the requirement (B) along with OP's comments and chart imply that the space is a 3-sphere $S^3$ whose round metric in the hyperspherical coordinates is

$$d\mathbf{\Sigma}^2=r^2 d\mathbf{\Omega}^2$$

Where

$$d\mathbf{\Omega}^2=d\psi^2 + \sin^2\psi\left(d\theta^2 + \sin^2\theta\, d\varphi^2\right)$$

Combining the formulas we obtain the final metric

$$ds^2 = - c^2 dt^2 + (\dfrac{\phi}{2}t^2)^2 d\mathbf{\Omega}^2$$

The null geodesics equation is $ds=0$ or for $\theta=\varphi=0$

$$\dfrac{2c}{\phi}\dfrac{dt}{t^2} = \pm d\psi$$

Solving for the path of the photon

$$\psi=\psi_{\text{o}}\pm\dfrac{2c}{\phi}\left(\dfrac{1}{t_{\text{o}}}-\dfrac{1}{t}\right)$$

We can set $\psi_{\text{o}}=0$ by the choice of coordinates and drop the negative direction

$$\psi=\dfrac{2c}{\phi}\left(\dfrac{1}{t_{\text{o}}}-\dfrac{1}{t}\right)$$

Where $t$ is the current age of the universe and $t_{\text{o}}$ is the age of the universe at the moment of the emission of the photon. Here $\psi$ represents the arc angle of the photon traveling around the expanding hypersphere of the universe. A full round trip is $\psi=2\pi$

$$2\pi=\dfrac{2c}{\phi}\left(\dfrac{1}{t_{\text{o}}}-\dfrac{1}{t}\right)$$

By solving for $t$ we obtain the time $t_{\text{r}}=t_{2\pi}-t_{\text{o}}$ required for the round trip

$$t_{\text{r}}=\dfrac{t_{\text{o}}^2}{t_{\infty}-t_{\text{o}}}$$

Where $t_{\infty}$ is the age of the universe, after which a round trip at the speed of light is no longer possible, because the universe is expanding too fast

$$t_{\infty}=\dfrac{c}{\pi\phi}$$

Finally, for the requirement (C), this metric essentially is in the same form as FLRW, just with a different scale factor, so comparing them should be pretty straightforward.

EDIT - The Hubble Law

The instantaneous $(t=const.)$ local coordinate $x$ distance along the direction of $\psi$ from a chosen location at $x_o=0$ is

$$x=\psi r$$

Where $r$ is the current radius of the universe and $\psi=const.$ in the expansion. We can define the recession speed of the point $x$ relative to $x_o$ as

$$v=\dfrac{dx}{dt}=\psi\dfrac{dr}{dt}=\psi\,\phi\,t=\dfrac{2x}{t}$$

Where $t$ is the current age of the universe. Thus the Hubble law is

$$v=Hx=\dfrac{2x}{t}$$

With the Hubble parameter showing the age of the universe as double the Hubble time

$$H=\dfrac{2}{t}$$

The radius of the observable universe (cosmic horizon) at the time $t_{\text{o}}$ is the distance, from which the photons will never reach back. From the path of the photon above for $t=\infty$ we have

$$x=\psi r=\dfrac{2cr}{\phi t_{\text{o}}}=ct_{\text{o}}$$

Which matches the intuition. The radius of the universe when at $t_{\infty}$ it becomes larger than the observable universe

$$r_{\infty}=\dfrac{c^2}{2\pi^2\phi}=\dfrac{\phi}{2}t_{\infty}^2$$