I have a surface defined by the quadratic relation:$$0=\phi^2t^4-x^2-y^2-z^2$$Where $\phi$ is a constant with units of $km$ $s^{-2}$, $t$ is units of $s$ (time) and x, y and z are units of $km$ (space). The surface (ignoring one of the dimensions of space) looks like this:
Since the formula depends on the absolute value of $t$, is it possible to create a metric? If so, what would the metric be (it seems to me that the metric would have to be a function of time such that $ds^2(t)=
Answer
You're looking at an embedded submanifold,(*) and it sounds like you're asking for the metric that is induced by restricting the metric of the larger space — which apparently has coordinates $t, x, y, z$ — onto the submanifold. [This is what @marmot was saying in the comments on the question.] Of course, the answer depends on the metric of the larger space, but I'm guessing that you assume the larger space has a Minkowski metric: \begin{equation} ds^2_{l} = c^2 dt^2 - dx^2 - dy^2 - dz^2. \end{equation} If you dislike this signature, feel free to just flip all the signs here and in the result below. Also remember that there's another branch of the solution with negative values for $t$, which is not shown in your diagram.
To simplify the algebra, let's back up and define a different set of coordinates $t, r, \vartheta, \varphi$ on the larger manifold. In that case, the surface is defined implicitly by the equation \begin{equation} 0 = \phi^2 t^4 - r^2, \end{equation} and the original metric in these coordinates is just \begin{equation} ds^2_{l} = c^2 dt^2 - dr^2 - r^2 d\vartheta^2 - r^2 \sin^2 \vartheta d\varphi^2. \end{equation} (This is basically the familiar metric in spherical coordinates, plus the time term.)
Now, your equation implicitly defines the surface, but it doesn't define coordinates to use in parametrizing that surface. Since there's just one equation and the larger manifold is four-dimensional, we know that your submanifold is three-dimensional (except at $t=0$), so we hope to be able to use three coordinates to parametrize it everywhere. Indeed, away from $t=0$, we can basically use $t, \theta, \phi$. It might be a little clearer to define the new coordinates $T, \Theta, \Phi$ on the surface, even though the most obvious definition for these coordinates will give them very simple relationships to the original coordinates. Now we can give the surface in the larger space in terms of these three coordinates by this expression: \begin{align} t &= T, \\ r &= \phi T^2, \\ \vartheta &= \Theta, \\ \varphi &= \Phi. \end{align} The $t, \vartheta, \varphi$ expressions are basically definitions, whereas the $r$ expression is a consequence of those three definitions and your definition of the surface.
Next, we can use the expression from the Wikipedia entry to find the induced metric: \begin{align} ds^2_{s} &= \left[c^2 - \left( \frac{\partial r} {\partial T} \right)^2 \right] dT^2 - \phi^2 T^4 d\Theta^2 - \phi^2 T^4 \sin^2 \Theta d\Phi^2 \\ &= \left[c^2 - 4 \phi^2 T^2 \right] dT^2 - \phi^2\, T^4\, d\Theta^2 - \phi^2\, T^4\, \sin^2 \Theta\, d\Phi^2. \end{align} This result aligns with the intuition that the metric should "depend on time", even though it is explicitly given only in terms of the intrinsic coordinates on the submanifold.
(*) Actually, there's a little problem at $t=0$, where your "submanifold" pinches off. This disqualifies it from even being a differentiable manifold. But let's just ignore that point, because everything works properly everywhere else. You have a true metric singularity at that point anyway.
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