- #1

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Not going anywhere with this, I am just curious as to how such a metric could look like, mathematically.

Thanks in advance !

- Thread starter Markus Hanke
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- #1

- 259

- 45

Not going anywhere with this, I am just curious as to how such a metric could look like, mathematically.

Thanks in advance !

- #2

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The "curling-up" of dimensions is a global property of the space, and so (at least in the category of Riemannian manifolds) it is not reflected by the metric, which is local. There are geometric structures you can put on a manifold that, I think, do force a connection between the global and local structures. But I'm only starting to learn about that, so I'll let someone else jump in and say whether that's the right way to think about it.

- #3

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I am not an expert in this ( which is why I asked the question in the first place ), but somehow that statement does not appear to make sense. My thinking is that the extra dimension(s) would be a direct product between some "normal" manifold and the extra one, in which case they must appear in the metric tensor in some form.The "curling-up" of dimensions is a global property of the space, and so (at least in the category of Riemannian manifolds) it is not reflected by the metric, which is local.

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- #4

HallsofIvy

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The simplest example would be an infinitely long cylinder with a very small radius.

Take the axis of the cylinder to be the x-axis, the radius R, and [itex]\theta[/itex] the angle a line, pependicular to the axis, from the axis to a point on the cylinder makes with the z-axis. Then we can write [itex](x, y, z)= (x, R cos(\theta), R sin(\theta))[/itex].

Then [itex]dy= -Rsin(\theta)d\theta[/itex] and [itex]dz= Rcos(\theta)d\theta[/itex] and then [itex]ds^2= dx^2+ dy^2+ dz^2= dx^2+ R^2cos^2(\theta)d\theta+ R^2sin^2(\theta)d\theta= dx^2+ R^2d\theta^2[/itex].

The metric tensor is

[tex]\begin{bmatrix}1 & 0 \\ 0 & R^2\end{bmatrix}[/tex]

and the "curling up" will be for R very small.

More generally, given any metric tensor, you can "add" curled up dimensions by adding rows and columns with all 0s except that the values on the diagonal are very small numbers-**almost** 0.

Take the axis of the cylinder to be the x-axis, the radius R, and [itex]\theta[/itex] the angle a line, pependicular to the axis, from the axis to a point on the cylinder makes with the z-axis. Then we can write [itex](x, y, z)= (x, R cos(\theta), R sin(\theta))[/itex].

Then [itex]dy= -Rsin(\theta)d\theta[/itex] and [itex]dz= Rcos(\theta)d\theta[/itex] and then [itex]ds^2= dx^2+ dy^2+ dz^2= dx^2+ R^2cos^2(\theta)d\theta+ R^2sin^2(\theta)d\theta= dx^2+ R^2d\theta^2[/itex].

The metric tensor is

[tex]\begin{bmatrix}1 & 0 \\ 0 & R^2\end{bmatrix}[/tex]

and the "curling up" will be for R very small.

More generally, given any metric tensor, you can "add" curled up dimensions by adding rows and columns with all 0s except that the values on the diagonal are very small numbers-

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