Does the curvature of spacetime occur in the fourth dimension in the form of a tesseract?

Question

I have heard contradictory ideas about spacetime and the fourth dimension. Some talk as though it is spatially tangent to all other dimensions and this is where the curvature of spacetime occurs. Is this true?

Answer 1

Imagine you found a spherical ball with a surface area equal to $4 \pi r^2$, for some $r$, but the distance between the centre of the ball and its surface was not equal to $r$. In that case the 3-dimensional space in which the ball is sitting would be 'warped'; the rules of Euclidean geometry are not applying. Or suppose that you draw three lines between three points, each line being the shortest possible between that pair of points, but the angles of the triangle constructed this way did not add up to 180 degrees. Again, the rules of Euclidean geometry are not working. This is another way in which a region of space can be warped. The technical mathematical name for such 'warping' is 'curvature'. If a space has properties like these then we say the space is curved.

The reason for the name is that one can get a lot of insight into this kind of curvature by looking at examples in fewer dimensions, and then there is a very nice analogy with the more ordinary use of the word. But when applied to three-dimensional space, and indeed to spacetime, the concept gets quite a lot more sophisticated.

So, to answer your question, the presence of curvature does not necessarily require the presence of extra dimensions.

Answer 2

There isn't a notion of time being "spatially tangent." Let us take for simplicity flat Minkowski spacetime $\mathbb{R}^{1,3}$. We can identify vectors tangent to a point $p \in \mathbb{R^{1,3}}$ with the vector in $\mathbb{R}^{1,3}$ itself, and we can choose a basis,

$$e_t = \begin{pmatrix} 1\\ 0\\ 0\\ 0 \end{pmatrix}, \quad e_x = \begin{pmatrix} 0\\ 1\\ 0\\ 0 \end{pmatrix}, \quad e_y=\begin{pmatrix} 0\\ 0\\ 1\\ 0 \end{pmatrix},\quad e_z=\begin{pmatrix} 0\\ 0\\ 0\\ 1 \end{pmatrix}$$

so all the coordinates, including time, are orthogonal to each other. As for curvature, there are two kinds - extrinsic and intrinsic. What appears in the Einstein field equation is intrinsic curvature, and it is a property of the spacetime manifold, which takes different values on the manifold. So, curvature is not something restricted to a temporal dimension.

If you had a space, say $M \cong \mathbb R^1 \times S^3$, when we say $M$ is curved, we mean it in the intrinsic sense, and it is a statement applied to the space as a whole. However, $\mathbb R^1$ itself is intrinsically flat and $S^3$ is intrinsically curved so you could say for example, it's because of the $S^3$ that $M$ is intrinsically curved but that's not a useful idea.

If we had a metric like $\mathrm d s^2 = e^{-|z|}\mathrm{dt}^2 - \mathrm{d}x^2 - \mathrm{d}y^2- \mathrm{d}z^2$, then you could say the submanifold corresponding to at $t=0$ has a flat metric, so you could loosely say it's curved in the time direction - this might be the sense intended, depending on where you read the statements in your question.

Answer 3

No.

The fourth dimension of spacetime is time. The time dimension is not “spatially tangent” to anything.

In general, curvature is not associated with any particular dimension. For example, the Riemann curvature tensor can have a component like $R_{xyzt}$ that involves all four dimensions of spacetime.

Tesseracts are irrelevant to curvature. If curvature occurs only in the time dimension, it cannot do so “in the form of a tesseract” since tesseracts are four-dimensional, not one-dimensional, shapes. And, in any case, curvature is not a shape.