# Show a geometry can be embedded in 3D euclidean space

Consider the 2D space with line element $$ds^2=\frac{dr^2}{1-\frac{2\mu}{r}}+r^2d\phi^2$$. How can I prove that this geometry can be embedded in a 3D Euclidean space and also find the equations for the corresponding 2-dimensional surface?

On a first attempt, I tried using the induced metric functions equation: $$h_{ij}=g_{ab}\frac{\partial x^a}{\partial u^i}\frac{\partial x^b}{\partial u^j}$$, so that parametrizing the surface in terms of $$r$$ and $$\theta$$, I could try to satisfy the relation between the euclidean metric tensor $$g_{ab}=\mathbb{I}$$ and the induced metric functions, which I found to be: $$h_{11}=\frac{1}{1-\frac{2\mu}{r}}$$ and $$h_{22}=r^2$$.

Anyway, I couldn't find any way forward, so any help will be much appreciated!

• I would try rotationally symmetric surface first. Mar 5, 2021 at 19:49
• One thing I notice is that, if $r<2\mu$, then the coefficient of $dr^2$ is negative. That suggests to me that the surface will only make sense for $r\geq 2\mu$. Mar 5, 2021 at 19:51

Using @Arctic Char' suggestion: it make sence to use the rotational symmetry.

If we embed the required surface into 3-dymensional Euclidean space, we can require the rotational symmetry over axis Z (the coordinate system in 3D space - this is our choice). We can also use the angle $$\phi$$ for parametrization in 3D, so this angle will be the same for both surface and 3D space. To define the surface it would be enough to draw its cross-section (a curve), for instance, in X-Z plane.

$$ds^2=\frac{dr^2}{1-\frac{r_s}{r}}+r^2d\phi^2$$, where $$r_s$$ is the "Schwarzschild radius".

In 3D space we introduce $$z$$ and $$\rho$$ coordinates: $$ds^2=dz^2+d\rho^2+\rho^2d\phi^2$$. On the other hand, if we stay in the X-Z plane, $$d\phi=0$$, so

$$ds^2=\frac{dr^2}{1-\frac{r_s}{r}}\Rightarrow dz^2+d\rho^2=\frac{1}{1-\frac{r_s}{r}}dr^2$$

$$(\frac{dz}{dr})^2+(\frac{d\rho}{dr})^2=\frac{r}{r-r_s}$$ - the parameterized equation for $$z(r)$$ and $$\rho(r)$$

Considering $$z(r)$$ as a function $$z(\rho(r))$$ we get

$$\Bigr((\frac{dz}{d\rho})^2+1\Bigl)(\frac{d\rho}{dr})^2=\frac{r}{r-r_s}$$, or $$\Bigr((\frac{dz}{d\rho})^2+1\Bigl)=\frac{r}{r-r_s}(\frac{dr}{d\rho})^2$$.

But we are allowed to transform the variable , switching from $$r$$ to $$r(\rho)$$. We can choose simply $$r=\rho$$.

$$(\frac{dz}{dr})^2+1=\frac{r}{r-r_s}\Rightarrow \frac{dz}{dr}=\pm\sqrt{\frac{r_s}{r-r_s}}$$

$$z(r)=\pm2\sqrt{r_s(r-r_s)}+const$$. A constant here defines the origine for the variable $$z$$ and can be chosen zero.

We got a parabola lying "on the side" with the beginning at the point $$r=r_s$$. Rotating this parabola around axis Z we get the required surface.

So, in 3D space we got the surface defined by the coordinates

Polar angle $$\phi$$

$$z$$ coordinate depending on $$r$$: $$z(r)=\pm2\sqrt{r_s(r-r_s)}$$

At $$r>>r_s$$ $$z(r)\approx2\sqrt{r_sr}<, and $$ds=\sqrt{1+(\frac{dz}{dr})^2}$$ $$dr\approx(1+\frac{r_s}{2r})$$ $$dr\approx dr$$, so at $$r\to\infty$$ the surface has two-dimension Euclidean metrics.

At $$r\to{r}_s$$ the surface is curved and $$ds\to\infty$$, but $$S=\int_{r_1}^{r_s}ds$$ is finite, since the singularity at $$r=r_s$$ is integrable.

If we suppose that the time in this system is absolute, a two-dimensional creature living on the surface and moving along the radius $$r$$ with a constant velocity $$v$$ will reach $$r_s$$ ($$r_1>r_s$$) in finite time

$$T=-\frac{1}{v}\int_{r_1}^{r_s}\sqrt{\frac{r}{r-r_s}}dr=\frac{1}{v}\Bigl(\sqrt{r_1(r_1-r_s)}+r_s\log\frac{\sqrt{r_1}+\sqrt{r_1-r_s}}{\sqrt{r_s}}\Bigr)$$

At $$r_1>>r_s$$ $$T=\frac{r_1}{v}(1-\frac{r_s}{2r_1})+\frac{r_s}{v}\log(2\sqrt{\frac{r_1}{r_s}})\geqslant\frac{r_1}{v}$$

At $$(r_1-r_s)< $$T=2\frac{\sqrt{r_s}}{v}\sqrt{r_1-r_s}>>\frac{r_1-r_s}{v}$$