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The minimal dimension $N$ of the flat space in which the Schwarzschild metric can be embedded is equal to six, and the Fujitani-Ikeda-Matsumoto embedding is a method for embedding the Schwarzschild metric in a six-dimensional spacetime (https://arxiv.org/pdf/1202.1204.pdf). There are two possibilities of the signature choices: $(++----)$ and $(--+---)$.The embedding function $y^{a}$ is as follows:

$$y_{0}=t\sqrt{1-\frac{R}{r}}$$ $$y_{1}=\frac{1}{\sqrt{2}\gamma}\left(\frac{\gamma^{2}t^{2}}{2}-1\right)\sqrt{1-\frac{R}{r}}+\frac{h(r)}{\sqrt{2}}$$ $$y_2=\frac{1}{\sqrt{2}\gamma}\left(\frac{\gamma^{2}t^{2}}{2}+1\right)\sqrt{1-\frac{R}{r}}+\frac{h(r)}{\sqrt{2}}$$ $$y_{3}=rsin\theta sin\phi,y_{4}=rsin\theta cos\phi,y_5=rcos\theta$$ where $$h(r)=\frac{\gamma r(2r+3R)}{4}\sqrt{1-\frac{R}{r}}+\frac{3\gamma R^{2}}{8}ln\left(\frac{2r}{R}\left(1+\sqrt{1-\frac{R}{r}}\right)-1\right)$$

In all the above formulae $R$ is the Schwarzschild radius and $\gamma=\sqrt{2}$. Following a metric signature of $(++----)$, i.e., $ds^{2}=dy_{0}^{2}+dy_{1}^{2}-dy_{2}^{2}-dy_{3}^{2}-dy_{4}^{2}-dy_{5}^{2}$, and substituting the metric for the above transformations I got the following result:

$$ds^{2}=\left(1-\frac{R}{r}\right)dt^{2}-\left(1-\frac{R}{r}\right)^{-1}\left(\frac{4r^{4}-2r^{3}R+r^{2}R^{2}t^{2}-R^{2}}{4r^{4}}\right)dr^{2}-r^{2}d\Omega^{2}$$ The embedded metric resembles the Schwarzschild metric except for the an additional term which is a function of $t$ and $r$, i.e., $f(r,t)=\left(\frac{4r^{4}-2r^{3}R+r^{2}R^{2}t^{2}-R^{2}}{4r^{4}}\right)$. How do I visualize this metric and represent this diagrammatically? And how is this a successful embedding since the original Schwarzschild metric is not embedded but rather a modified one is (also note that this embedding is not asymptotically flat).

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    $\begingroup$ Hi Naveen, please stop making trivial edits to bump this question into the front page. You asked this 7 hours ago, you have to be more patient. Thanks. $\endgroup$ Jun 25, 2018 at 21:10
  • $\begingroup$ The previous edit was important as there was a key defining word missing which nicked up the grammar a bit. $\endgroup$ Jun 25, 2018 at 21:13
  • $\begingroup$ It is not the schwarzschild metric unless there is a coordinate transformation that turns this into schwarzschild coordinates. $\endgroup$ Jun 25, 2018 at 21:23
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    $\begingroup$ Minor comment to the post (v6): In the future please link to abstract pages rather than pdf files $\endgroup$
    – Qmechanic
    Jan 24, 2019 at 10:18

1 Answer 1

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It seems like I have found an answer to the question after all, or at least partially.

Consider the time-slice of the metric $t=const.,\theta=\pi/2$ (with $R=2M$): $$ds^2=(1-2M/r)^{-1}\left(\frac{4r^{4}-4r^{3}M-4M^{2}}{4r^{4}}\right)dr^{2}+r^{2}d\phi^{2}=\frac{dr^{2}}{\rho(r)}+r^{2}d\phi^{2}\tag{1}$$

This can be thought of as a flat plane, $z=0$, in cylindrical coordinates. Of course, the metric describes the geometry of a gravitating object of non-zero mass, i.e., $M(r)≠0$, hence we can assume the metric to be describing a surface of revolution given be $z=z(r)$, with the $\phi$ component of the cylindrical coordinates having the range $0<\phi≤2\pi$. The metric of such a surface embedded in three dimensions is described by the following metric:

$$ds^{2}=dz^{2}+dr^{2}+r^{2}d\Omega^{2}=\left[\left(\frac{dz}{dr}\right)+1\right]^{2}dr^{2}+r^{2}d\Omega^{2}\tag{2}$$

Comparing $(1)$ and $(2)$, we have:

$$\left[\left(\frac{dz}{dr}\right)+1\right]=\rho(r)^{-1}=(1-2M/r)^{-1}\left(\frac{4r^{4}-4r^{3}M-4M^{2}}{4r^{4}}\right)\tag{3}$$ We can now integrate $(3)$, (after an approximation) as follows:

$$z(r)=\int_{0}^{r}\frac{-M(\frac{M}{2}-r^{3})}{M^{2}+\frac{r^{3}(M-r)}{2}}dr=-M \sum_{\omega:-\omega^{4}+M\omega^{3}+2M^{2}=0}\frac{M log(-\omega +r)-2\omega^{3}log(-\omega+r)}{-4\omega^{3}+3M\omega^{2}}\tag{4}$$

Taking a series expansion (generalized Puiseux expansion) the expression obtained is:

$$-2Mlog(r)+\frac{2M^{2}}{r}+\frac{3M^{3}+M^{2}(2M^{2}-1)}{3r^{3}}+\frac{M^{3}(2M^{2}+3)}{4r^{4}}+\frac{M^{4}(2M^{2}+7)}{5r^{5}}+O\left(\frac{1}{r}\right)^{6}\tag{5}$$ Setting $M=1$, for simplicity, and performing 3D-parametric plot, we obtain plots for the lower and upper halves as follows:

enter image description here enter image description here

Calculations here and here. Performing operations as described above for the Schwarzschild metric, we would obtain equations analogous to $(1),(2)$,&$(3)$, but the final integrated value obtained is: $z(r)=\sqrt{8M(r-2M)}$. Thus, the time-slice of the Schwarzschild metric is nothing but the quartic surface defined by the equation:

$$x^{2}+y^{2}=\left(\frac{z^{2}}{8M}+2M\right)^{2}\tag{6}$$

which is embedded in a three-dimensional Euclidean space with cartesian coordinates $(x,y,z)$. To see this, consider the following coordinate transformations: $x=rcosϕ,y=rsinϕ$, and $z=√(8M(r-2M))$. Applying these transformations to the three-dimensional Euclidean metric: $ds^2=dx^2+dy^2+dz^2$, we get back the Schwarzschild metric. Performing a 3D-parametric plots, we get:

enter image description hereenter image description here

Calculations here and here. The plots seem very similar, and would probably be the exact same if the series expansion $(5)$ wasn't approximated.

Thus, the Fujitani-Ikeda-Matsumoto embedding does reproduce the Schwarzschild geometry when the latter is embedded in a six-dimensional spacetime using the embedding functions as described in the question. I'm open to any valuable edits and corrections.

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    $\begingroup$ Hi Naveen, I see you have an interest and expertise in embedding GR. I have asked a relevant question a number of times in different forms just to find much confusion. Finally I think I was able to formulate the simplest version consistently, but still no response from the community. I wonder if you could possibly please take a look and provide an answer, comment, insight, or direction. I would greatly appreciate it. Thanks much in advance! math.stackexchange.com/questions/3131651/… $\endgroup$
    – safesphere
    Mar 15, 2019 at 7:05

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