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# 4. Curvature

Before continuing with spacetime, we temporarily retreat to the study of space alone. We do so to introduce the notion of "curvature.'' We can label space with coordinates, for example, we could label every point in a 2-dimensional space with an $$x$$ value and a $$y$$ value. These coordinates are just labels, with no physical meaning, until we also say something about the distance between infinitesimally separated pairs of points. For example, in a two-dimensional Euclidean space with which you are familiar, the square of the distance between $$x,y$$ and $$x+dx,y+dy$$ is given by:

$ds^2 = dx^2 + dy^2 \label{eqn:EuclidCartesian}$

The physical interpretation of $$ds^2$$ is as follows:
The length of a ruler with an end on each of the two points is $$\sqrt{ds^2}$$.

Expressions for the distance are not always as simple as the one above. Even for the same physical space, with different coordinate schemes the equation for $$ds^2$$ will look different. For example, we could choose polar coordinates instead of Cartesian ones and then distances would be given by:

$ds^2 = dr^2 + r^2 d\phi^2 \label{eqn:EuclidPolar}$

This is the same space, just described with different coordinates.

Note that one can find the distance along any path through the space by calculating $$\int ds$$ along the path.

Box $$\PageIndex{1}$$

Exercise 4.1.1: Use Equation $$\ref{eqn:EuclidPolar}$$ to show that the distance along the path with constant $$\phi$$ that goes from the origin to $$r = r_1$$ is simply given by $$r_1$$. (You've already done this, or something very similar, in Chapter 1. If you like, you can just go back now and review what you did. Or you can just do it again, it should be quick.)

Box $$\PageIndex{2}$$

Exercise 4.2.1: Consider the set of points all at $$r = r_1$$. These define a circle because they are the same distance from a common point (in this case the origin). Assuming that $$r, \phi$$ and $$r, \phi+2\pi$$ are the same point, use Equation $$\ref{eqn:EuclidPolar}$$ to show that the circumference of this circle is $$2\pi r_1$$.  (You've already done this, or something very similar, in Chapter 1. If you like, you can just go back now and review what you did. Or you can just do it again, it should be quick.)

The space we are describing in this chapter so far (and its higher- and lower-dimensional versions) we call "Euclidean" because these spaces are consistent with geometry as described by Euclid. It turns out though that space is not Euclidean (although a Euclidean description is often a very good approximation). This is a bit startling. If you are not startled by it, you don't understand it yet. But don't worry; you will. And hopefully your mind will be blown.

From Einstein we learned that space can be quite different from Euclidean. To begin to "free your mind" from its Euclidean constraints, we begin with a 2-dimensional space labeled with coordinates $$r$$ and $$\phi$$ and with distances given by:

$ds^2 = \frac{dr^2}{1-kr^2} + r^2 d\phi^2 \label{eqn:two-sphere}$

Box $$\PageIndex{3}$$

Exercise 4.3.1: How long would a path be that stretches from $$r=0$$ to $$r=r_1$$ at constant $$\phi$$? Call the length $$\ell$$ and express it as an integral that depends on $$r_1$$ and $$k$$. Assume that $$r_1$$ is much less than $$\sqrt{1/k}$$. Make a Taylor expansion that approximates the integrand so that it contains the first order corrections due to $$k \ne 0$$. After this approximation, do the integral.

Box $$\PageIndex{4}$$

Exercise 4.4.1: Consider the set of points all at $$r= r_1$$ with all values of $$\phi$$. Is this a circle? What is the circumference of this object? Again, assume that $$r,\phi$$ and $$r,\phi+2\pi$$ are the same point. First find the circumference as a function of $$r_1$$ and then use your result from the previous problem to express it as a function of $$k$$ and $$\ell$$. [Don't get too hung up on solving for $$\ell$$ as a function of $$r_1$$. The trick to doing this quickly is realizing that to our level of approximation $$k \ell^2$$ = $$kr_1^2$$, as is explicitly demonstrated in our solutions.]

Exercise 4.4.2: Discuss the result from Exercises 4.3.1 and 4.4.1 and what it means qualitatively for the circumference-radius relationship for circles in spaces with $$k < 0$$, $$k > 0$$ and $$k=0$$. [Note that you can do this even if you did not manage to get $$\ell$$ as a function of $$r$$ in the above exercise.]

Box $$\PageIndex{5}$$

Exercise 4.5.1: If you were a two-dimensional creature, and could travel around this space with a measuring tape, describe in a few sentences at least one way for measuring the value of $$k$$.

#### Embedding

Sometimes it is possible to visualize a non-Euclidean space, such as the one with invariant distance rule given by Equation \ref{eqn:two-sphere}, by embedding it in a higher-dimensional Euclidean space. Such an embedding, into a space with one extra dimension, is shown in the figure for $$k>0$$. When using such an embedding diagram it is important to keep in mind that there are extra dimensions in the diagram whose sole purpose is visualization -- they have no physical significance. In the figure here, the space we are describing is the two-dimensional sphere; the radial dimension is fictional, included here only for purposes of visualization.

Note that every point on the sphere can be labeled by the latitude-like coordinate $$\theta$$ and the longitudinal coordinate (not shown) $$\phi$$. Also, at every point one can convert $$\theta$$ and $$\phi$$ to $$r$$ and $$\phi$$. In a homework problem you will derive Eq. \ref{eqn:two-sphere} starting from the assumption that the three-dimensional space used for the embedding is Euclidean.

Box $$\PageIndex{6}$$

Exercise 4.6.1: Observe the circle at constant coordinate value $$r$$ in the embedding diagram. The distance from the origin (top of the sphere) to any point in the circle is $$\ell$$. Is the circumference of the circle greater than, equal to, or less than $$2\pi \ell$$? Compare to the relevant result in the boxes above.

Not all spaces can be embedded by placing them in just one extra dimension. For example, the $$k<0$$ space requires two extra dimensions for embedding. Part of the $$k < 0$$ space is sometimes shown embedded in just one extra spatial dimension, with the two-dimensional surface having a shape similar to a saddle. One can start to see the problem here because if the diagram were extended, the saddle would curve into itself. Such self intersection can only be avoided by introduction of yet another fictional extra dimension.

#### Three-dimensional Homogeneous and Isotropic Spaces

Let us now take things up one dimension into 3-D.

Previously we asserted that one could label a 3-dimensional Euclidean space with coordinates $$r$$, $$\theta$$, and $$\phi$$ such that points separated by $$dr$$, $$d\theta$$, and $$d\phi$$ would be separated by a distance (as one would measure with a ruler) with square given by

$ds^2 = dr^2 + r^2\left(d\theta^2 + \sin^2\theta d\phi^2\right)$

A space that can be labeled in this way is homogeneous (invariant under translations) and isotropic (invariant under rotations). The easiest way to see this is to remember that there's a coordinate transformation to Cartesian coordinates for which

$ds^2 = dx^2 + dy^2 + dz^2$

Now the homogeneity is more evident, since transforming $$x$$ to $$x' = x + L$$ would clearly leave the distance rule unchanged. We've also already seen that rotations leave the distance rule unchanged. So, the space is homogeneous and isotropic. If we choose to label it with spherical coordinates about a particular origin, our labeling obscures the homogeneity and isotropy, but the space itself is still homogeneous and isotropic.

It turns out that whether one can label space in this way or not is a matter to be settled by experiment. It's not necessarily true. Even if we restrict ourselves to completely homogeneous and isotropic geometries, we can mathematically describe spaces that cannot be labeled in this way.

What is generally true is that all three-dimensional homogeneous and isotropic spaces can be labeled with coordinates $$r$$, $$\theta$$, and $$\phi$$ such that

$ds^2 = \dfrac{dr^2}{1-kr^2} + r^2\left(d\theta^2 + \sin^2\theta d\phi^2\right)$

for $$k$$ a constant that can be positive, negative or zero. Euclidean space is a special case with $$k=0$$.

Exercise $$\PageIndex{7}$$

Exercise 4.7.1: You know that in a Euclidean space the relationship between radius and area of a sphere is $$A = 4\pi r^2$$ with $$r$$ specifying the radius. Note that the angular ( $$d\phi$$ and $$d\theta$$ ) parts of the invariant distance equation are unchanged by having $$k \ne 0$$. Therefore we still have $$A = 4\pi r^2$$. I also claim that the relationship between sphere area and radius does depend on $$k$$. How can these statements both be true?

HOMEWORK Problems

For the homework problems we will be considering a two-dimensional space labeled with coordinates $$r$$ and $$\phi$$ with invariant distance given by

\begin{equation*} \begin{aligned} d\ell^2 = \frac{dr^2}{1-kr^2} + r^2 d\phi^2 \end{aligned} \end{equation*}

In the following, assume $$k > 0$$ unless otherwise specified.

Problem $$\PageIndex{1}$$

How long would a path be that stretches from $$r=0$$ to $$r=r_1$$? Call the length $$\ell$$ and express it as a function of $$r_1$$ and $$k$$. Assume that $$r_1 < \sqrt{1/k}$$. Unlike in the exercises, do not use a Taylor series approximation to $$1/(1-kr^2)$$.

Problem $$\PageIndex{2}$$

Consider the set of points all at $$r= r_1$$ with all values of $$\phi$$. This set is a circle since it is the set of all points located a particular distance away from another point ( $$r=0$$ ). What is the circumference of the circle? First find the circumference as a function of $$r_1$$ and then use your result from the previous problem to express it as a function of $$k$$ and $$\ell$$.

Problem $$\PageIndex{3}$$

If you were a two-dimensional creature, and could travel around this space with measuring tape, describe in a few sentences at least
one way for measuring the value of $$k$$.

Problem $$\PageIndex{4}$$

Use what you know about Euclidean geometry to show that the square of the length of the path between $$r,\phi$$ and $$r+dr, \phi+d\phi$$, constrained to lie in the sphere, is indeed given by $$\frac{dr^2}{1-kr^2} + r^2 d\phi^2$$ for the appropriate choice of $$k$$ as a function of $$R$$.  Also specify that function.

What I want you to take away from this is that the embedding is purely for visualization purposes. The radial direction is entirely a fiction. If the $$r, \phi$$ space is some physical space, that's all there is. You don't need that fictitious radial coordinate in order to calculate things, such as ratio of circle circumference to radius, as demonstrated in these homework problems. It's a very useful visualization tool, so we will use it. But remember that the radial dimension has no physical reality.

Problem $$\PageIndex{5}$$

Fun with the Schwarzschild solution. The spacetime outside of a central spherically symmetric mass distribution can be labeled with coordinates $$t,r,\theta,\phi$$ such that

\begin{equation*} \begin{aligned} ds^2 = -c^2 (1-r_s/r)dt^2 + (1-r_s/r)^{-1} dr^2 + r^2\left(d\theta^2 + \sin^2\theta d\phi^2\right) \end{aligned} \end{equation*}

where $$r_s = 2GM/c^2$$ is the "Schwarzschild radius" and $$M$$ is the total mass of the mass distribution. For the Earth, $$r_s$$ is about 9 mm. We are just going to work with the spatial part of it in this problem.

Consider two concentric circles in the same plane with centers at the center of the mass distribution ( $$r=0$$ ). If the spatial geometry were Euclidean, the differences in their circumferences would be $$\Delta C = 2\pi \Delta \ell$$ where $$\Delta \ell$$ is the radial distance between the two circles. But the presence of the mass means the spatial geometry is not Euclidean, and instead given by the Schwarzschild solution. If the two circles are at coordinate values $$r_1$$ and $$r_2$$, show that, for $$r_s$$ much smaller than $$r_1$$ or $$r_2$$ one instead gets $$\Delta C = 2\pi \Delta \ell\ - \frac{1}{2}r_s \ln(r_2/r_1)$$. Hints: 1) Choose the circles so that it's just $$\theta$$ that's changing, not $$\phi$$ and 2) Taylor expand to first order so $$\sqrt{(1-r_s/r)^{-1}}$$  = $$1+\frac{1}{2}\frac{r_s}{r}$$.

Note that if we take $$r_2$$ to be 42,000 km (about the distance to geostationary orbit from the center of the Earth) and $$r_1$$ to be 6,000 km (the distance from center of Earth to the surface), the correction to the difference in the circumferences is $$\frac{1}{2}r_s \ln(r_2/r_1)$$= 1.8 cm. A very small correction! The spatial geometry around Earth is very close to Euclidean.