I Can Curves Live Out of Embedding Diagrams?

[LCSphysicist]

The embedding diagram is well known for its qualitative representation of how the stress energy tensor curve the spacetime. We can construct a map from a general spherical metric to a cylindrical metric if we want to construct such diagrams.

Now, my confusion is if there exist curves out of the surface. It seems to me that curves can indeed exist out of the diagrams. Since they are only a 3D representation of the curvature of the spacetime, a curve living out of the surface does not means that it lives out the spacetime. Now, even so this line of thought seems right to me, i am not 100% of this!

For example, to construct a embbeding diagram for the Schwarzild metric, we do as follow:

Let $\theta = \pi/2, dt = 0$

$ds^2_{\text{Schwazrchild}} = ds^2_{\text{cylinder}} = dr^2 + r^2 d\phi^2 + dz^2$

So we get $r_{\text{Schwazrchild}} = r_{\text{cylinder}}$

And when we assume $z=z(r)$, we can get a whole expression for $z = z(r)$ and then construct our embedding diagram.

But we assumed $\theta = \pi/2$! So technically, curves can live out of it right?

 

[Ibix]

Well, Flamm's paraboloid is not an embedding of Schwarzschild spacetime. It's an embedding of a particular 2d spacelike slice of Schwarzschild spacetime. So timelike curves in the original spacetime would only appear as dots if they intersect it at all and, indeed, only spacelike curves that lie in the plane you happen to pick would appear as lines.

So you can certainly have curves in spacetime that lie outside the slice you chose to represent. I don't think they'd appear elsewhere in your diagram, though. They're just "off the edge of the map".

 

[PeterDonis]

they are only a 3D representation of the curvature of the spacetime

No, they are a 3D representation of the curvature of a 2-dimensional "slice" out of a spacelike surface of constant time. In the case of the Flamm paraboloid, the embedding diagram usually seen for Schwarzschild spacetime, the 2-dimensional "slice" is the equatorial plane $\theta = \pi / 2$ of a spacelike surface of constant Schwarzschild coordinate time.

So technically, curves can live out of it right?

It depends on what you mean. If you mean, are there curves in the spacetime that are not represented in the embedding diagram, of course there are, lots of them. If you mean, are there curves in the 3-d spacelike surface of constant time hat are not represented in the embedding diagram, of course that's true to.

But if you mean, are there curves in the 3-d "space" of the embedding diagram, that do not lie in the 2-surface represented in that diagram, but which represent curves in the spacetime (or in the 3-d spacelike surface of constant time), then no, there are none. The 3-d "space" of the diagram is unphysical; only the 2-surface represented in that diagram represents anything physical.

 

[Ibix]

I suppose it's possible to "stack" Flamm's paraboloids to produce a (2+1)d spacetime diagram. I don't think that's an embedding any more, and probably no more helpful than just mapping Schwarzschild $r,\phi,t$ coordinates to Euclidean $r,\phi,z$ ones, but it could be done. All lines that lie in the equatorial plane (spacelike, timelike and null) would appear on the map.

 

[PeterDonis]

I suppose it's possible to "stack" Flamm's paraboloids to produce a (2+1)d spacetime diagram.

No, because you would need at least a 4-dimensional embedding space and we can't either draw or visualize such a thing.

I don't think that's an embedding any more

It would have to be, because you need an embedding space in which to draw the Flamm paraboloids. Otherwise you don't have a diagram.