Plane embedded in Non-Euclidean spacetime

Question

The AdS-Schwarzschild black hole metric is given by,

$$ds^2 = \frac{1}{z^2} \left( -f(z) dt^2 + \frac{dz^2}{f(z)} + dx^2 \right)$$

where $t$ is time, $z$ is the radial direction, $x$ is a transverse spatial direction, $f(z) = 1-m z^4$, and $m$ is a constant mass.

Now for my question, we can always transform the constant mass $m$ AdS-Schwarzschild spacetime into a time-dependent mass $m(u)$ spacetime which is given by the AdS-Vaidya black hole metric,

$$ds^2 = \frac{1}{z^2} \left( -f(u,z) du^2 - 2 du dz + dx^2 \right)$$

The transformation that allows this is given by,

$$u = t - \int \frac{dz}{f(u,z)}$$

where $u$ acts like the new time coordinate, $f(u,z) = 1 - m(u) z^4$ and $m(u) = 1000 e^{-3 u}$. The ranges are $t \in [\epsilon, 1]$, $u \in [\epsilon, 1]$, $z \in [\epsilon, 1]$, $x \in [\epsilon, 1]$, and $\epsilon = 10^{-2}$.

Clearly this spacetime is non-Euclidean so obviously this is a spacetime with warping effects. Now, is it possible to visualize a constant plane $u=x$ in this time-dependent spacetime in the ($t,z,x$) coordinates using some Graphics like ParametricPlot3D or Plot3D? In other words, given the values of $(u,z,x)$ I have the AdS-Vaidya spacetime, and we can visualize the constraint $u=x$ in this coordinate system. However, what I want is to visualize the AdS-Vaidya spacetime as well as the constraint $u=x$ in the $(t,z,x)$ coordinates.

The constraint $u=x$ is a flat plane with zero intrinsic curvature, but embedded in the $(t,z,x)$ coordinates it will definitely have some extrinsic curvature where the plane bends due to the warping effect of the given spacetime.

Answer 1

Here's the simple approach in case it helps (not a full answer), noting that we can phrase the constraint $u=x$ in terms of $t$ by substituting $x$ in for $u$ in $t = u + \int \frac{dz}{f(u,z)}$:

\[Epsilon] = 10^-2;

f[u_, z_] := 1 - (1000 E^(-3 u)) z^4

intIndefinite[u_, z_] := Evaluate@Integrate[1/f[u, z], z]

int[u_, zz_] := 
 Evaluate[intIndefinite[u, zz] - intIndefinite[u, \[Epsilon]]]

tPlane[x_, z_] := Evaluate[x + int[x, z]]

ContourPlot3D[t == tPlane[x, z],
  {t, \[Epsilon], 1}, {x, \[Epsilon], 1}, {z, \[Epsilon], 1},
  AxesLabel -> Automatic]

However, the artifacts are pretty terrible in the result. This could be reduced by increasing PlotPoints above its default, but that takes very long. Also something looks off—I wonder if we're the victims of a branch cut somehow, as logs and arctans appear in ?intIndefinite. Unfortunately (and strangely), Integrate[1/f[u, zz], {zz, \[Epsilon], z}] takes ages to finish, and I'm not sure it will. So, I wonder if it's worth working through the integral from $\epsilon$ to $z$ by hand, then using Compile on the result so that it's fast. But hopefully this partial answer helps get things started!