The Distance between two points in a hypothetical universe.

Question

I have a hypothetical universe where the distance between two points in spacetime is defined as:$$ds^2 =−(\phi^2 t^2)dt^2+dx^2+dy^2+dz^2$$Where $\phi$ has units of $km\space s^{-2}$. The space in this universe grows quadratically with time (and, as I understand it, probably isn’t Minkowski spacetime). A particle travelling at the speed of causality, c, will follow this contour from point O to point P.

Given the time between point O and point Q, is it possible to find the distance from point P to point Q (e.g. does a function exist such that $f(\Delta t) = d_L$)? If so, what is that formula?

EDIT: I understand that in Minkowski space $ds^2 = -c^2\space dt^2 + dx^2 + dy^2 + dz^2$, but in this metric, space grows quadratically. In Minkowski space, if you were $3 \times 10^6\space km$ away from an emitter and it sent out a photon, the photon would arrive in exactly one second (following a null geodesic). In this metric, as the photon travels to you, the space in between you and the emitter expands - like a moving sidewalk - so the photon will arrive in something more than a second. As this universe gets older and the space between the objects gets larger, the problem is exacerbated. In fact, if the universe gets old enough, travel between two distant objects becomes impossible, so I have a tough time seeing how the speed of light is related to a null geodesic in this universe.

Answer 1

I am not a physicist nor a mathematician that knows about Minkowski Spaces, but here is a stab anyway.

It seems that the speed of causality for your universe is $c(t)=\phi t$. So the speed of the particle is $c(t)= dx/dt$. We simply get

$$dx/dt= \phi t$$ $$x_P-x_O= \phi (t_P^2-t_O^2)/2$$

So the distance travelled is merely $\phi (t_P^2-t_O^2)/2$ where $t_O$ is the time at point O and $t_P$ is the time at point P.

Answer 2

Here is my maiden attempt. Consider $$ds_{OP}^2 =−(\phi^2 t^2)dt^2+dx^2+dy^2+dz^2$$ dividing by $dt^2$, we get $$\frac{ds_{OP}^2}{dt^2} = −(\phi^2 t^2) + \frac{dx^2}{dt^2} + \frac{dy^2}{dt^2} + \frac{dz^2}{dt^2}$$, Now notice that all the rate of change is just $c$, the speed of causality. So we have, $$c^2 = −(\phi^2 t^2) + c^2 + c^2 +c^2$$ $$=>\phi^2 t^2 = 2c^2 $$ $$=>t = +\frac{\sqrt{2}c}{\phi} or -\frac{\sqrt{2}c}{\phi}$$ So the time taken to traverse a distance for a particle travelling at $c$ from any two points is based only on $\phi$.

Now consider for points A and B, $$-\phi^2 \int_{t_A}^{t_B}\int_{t_A}^{t_B}{t^2.dt.dt}$$ $$=>-\frac{\phi^2}{3} (\int_{t_A}^{t_B}{t_B^3}.dt - \int_{t_A}^{t_B}{t_A^3}.dt)$$ $$=>-\frac{\phi^2}{3} ({t_B^3}(t_B - t_A) - {t_A^3}(t_B - t_A))$$ Simplifying, $$=>-\frac{\phi^2}{3} ((t_B - t_A)({t_B^3} - {t_A^3}))->(1)$$

Now integrating both sides of eqn. $$ds_{OQ}^2 =−(\phi^2 t^2)dt^2+dx^2+dy^2+dz^2$$ with $$\int_{t_O}^{t_Q}\int_{t_O}^{t_Q}$$, w.r.t $dt$ we get from (1) $$ds_{OQ}^2 =-\frac{\phi^2}{3} ((t_Q - t_O)({t_Q^3} - {t_O^3}))+dx^2+dy^2+dz^2 + C_{1}$$ Now, Choose $$t_O = 0$$ to be the start time of a particle starting from O towards Q, then $$ => \Delta{t} = t_Q-t_O => \Delta{t} = t_Q$$, now consider another particle reaching Q from P at exactly the same time $$t_Q = \Delta{t}$$ with possibly different velocity but starting at time $$t_P = 0$$, so $$ds_{PQ}^2 =-\frac{\phi^2}{3} ((t_Q - t_P)({t_Q^3} - {t_P^3}))+dx^2+dy^2+dz^2 + C_{2}$$ $$=>ds_{PQ}^2 =-\frac{\phi^2}{3} (({t_Q^4}))+dx^2+dy^2+dz^2 + C_{2}$$ $$=>ds_{PQ}^2 =-\frac{\phi^2}{3} ((\Delta{t}^4))+dx^2+dy^2+dz^2 + C_{2}$$ $$=>d_L^2 =-\frac{\phi^2}{3} ((\Delta{t}^4))+\int_{x_p}^{x_q}\int_{x_p}^{x_q}dx^2+\int_{y_p}^{y_q}\int_{y_p}^{y_q}dy^2+\int_{z_p}^{z_q}\int_{z_p}^{z_q}dz^2$$ $$=-\frac{\phi^2}{3}(\Delta{t}^4)+(x_p-x_q)^2+(y_p -y_q)^2 + (z_p - z_q)^2$$