At every point of the 4-D space-time, it's metric, being a symmetric 2-tensor, has $\frac{D(D+1)}{2}=10$ independent components. From this we can subtract four degrees of freedom according to the four coordinate transformations; Hence, we have six independent degrees of freedom at every point. Thus, an arbitrary 4-D space-time, can be embedded locally as a four dimensional hyperplane into a (4+6) dimensional, Minkowski space-time. Similarly, we can think of the Schwarzschild metric as the induced metric on a four dimensional hypersurface embedded in a flat six-dimensional space-time with the following line element:

$$ \mathrm ds^{2}_{6}=-\mathrm dZ^{2}_{1}+\mathrm dZ^{2}_{2} +\mathrm dZ^{2}_{3} +\mathrm dZ^{2}_{4} +\mathrm dZ^{2}_{5}+\mathrm dZ^{2}_{6} $$


$$ \begin{aligned} Z_{1}&=4GM\sqrt{1-\frac{2GM}{r}}\sinh\left(\frac{t}{4GM}\right),\\ Z_{2}&= 4GM\sqrt{1-\frac{2GM}{r}}\cosh\left(\frac{t}{4GM}\right),\\ Z_{3}&=\pm\int\left[\frac{2GM}{r}+\left(\frac{2GM}{r}\right)^{2}+\left(\frac{2GM}{r}\right)^{3}\right]^{1/2}dr,\\ Z_{4}&=r\sin{\theta}\cos{\phi},\\ Z_{5}&=r\sin{\theta}\sin{\phi},\\ Z_{6}&=r\cos{\theta}. \end{aligned} $$

I have obtained the metric, which is:

$$\mathrm ds^{2}_{6}=-\left(1-\frac{2GM}{r}\right)\mathrm dt^{2}+\left[1+\frac{2GM}{r}+\left(\frac{2GM}{r}\right)^{2}+\left(\frac{2GM}{r}\right)^{3}\right]\mathrm dr^{2}+r^{2}\mathrm d\Omega^{2}$$

By just taking the time slice, with zero polar angle and $\theta=\pi/2$, I can see the relation of the plots between this metric and the normal Schwarzschild metric, moreover the $dr^{2}$ term is just the Taylor expansion of that of the Schwarzschild term at $r=\infty$. Although this six-dimensional picture is difficult, we must be able to visualize this using embedding. I am facing difficulties in obtaining the embedding diagram, could you please help me in doing so. What deeper insights into the Schwarzschild metric do we obtain from the embedding scheme? Also, how would various physical phenomenon appear when viewed in the six-dimensional space, e.g., particle motion?

My attempt at particle motion in 6-D space: (I might be horribly wrong!) I am aware that the Schwarzschild metric, being symmetric, can be embedded into six-dimensional flat space using the Kruskal-Szekeres coordinate. Moreover, while drawing the Penrose-Carter diagram, we take the 2-D part of the Schwarzschild metric in Kruskal-Szekeres coordinates and do a conformal mapping from $(U,V)$ non-compact space to $(\psi,\xi)$ compact space. Hence, this means that the Penrose-Carter diagram is nothing but the conformal mapped diagram of a 4-D hyperplane embedded in a six-dimensional space-time. thus, particle trajectories in the 6-D space-time must be similar to that represented in the Penrose diagram.

  • $\begingroup$ Hi. I don't think your procedure actually represents the Schwarzschild BH as a hypersurface embedded in a six-dimensional Minkowskian space(time). Because upon substituting the restricted values of the $Z$s (that you have described in the "Using:" section), I would expect $dS_6^2$ to directly produce the exact Schwarzschild BH metric - instead there are some Taylor-expansion-look-alike terms (which I won't really call Taylor expansion like because of the missing leading term $1$ and the missing factors) in the radial-radial component of the metric. Also, I don't understand what is $x$. $\endgroup$ – Dvij D.C. Jul 15 '17 at 12:23
  • $\begingroup$ I think you don't claim the induced metric to be exactly identical to the BH metric anyway but I commented just in order to ensure I understood your scheme correctly and to point out that the Taylor expansion argument doesn't make much sense to me - especially in the absence of my knowledge of what your $x$ is. $\endgroup$ – Dvij D.C. Jul 15 '17 at 12:32
  • $\begingroup$ x is actually r, I forgot to edit that part. Its the Laurent series, i.e., expansion of $\left(1-\frac{2GM}{r}\right)$ at $r=\infty$ $\endgroup$ – Spoilt Milk Jul 15 '17 at 12:32
  • $\begingroup$ Please check this: wolframalpha.com/input/… $\endgroup$ – Spoilt Milk Jul 15 '17 at 12:36
  • $\begingroup$ Oh, okay, of course. My bad. But still, the issue of the leading order $1$ persists. $\endgroup$ – Dvij D.C. Jul 15 '17 at 12:39

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