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Physics LibreTexts

1: Euclidean Geometry

  • Page ID
    5099
  • One of the great privileges of teaching this class is the opportunity I have to blow your minds with a radically different understanding of the nature of space and time. The shift from a Euclidean/Newtonian understanding of space and time, to a Riemannian/Einsteinian one is centrally important to our understanding of cosmology. This chapter is entirely focused on the Euclidean geometry that is familiar to you, but reviewed in a language that may be unfamiliar. The new language will help us journey into the foreign territory of Riemannian geometry where space is curved. Our exploration of that territory will then help you to drop your pre-conceived notions about space and to begin to understand the broader possibilities -- possibilities that are not only mathematically beautiful, but that appear to be realized in the natural world. 

    According to Euclidean geometry, it is possible to label all space with coordinates x, y, and z such that the square of the distance between a point labeled by \(x_1\), \(y_1\), \(z_1\) and a point labeled by \(x_2\), \(y_2\), \(z_2\) is given by \(\left(x_1 -x_2\right)^2+\left(y_1 -y_2\right)^2+\left(z_1 -z_2\right)^2\). If points 1 and 2 are only infinitesimally separated, and we call the square of the distance between them \(d\ell^2\), then we could write this rule, that gives the square of the distance as

    \[d\ell^2 = dx^2 + dy^2 + dz^2 \label{eqn:EuclidCartesian} \]

    This rule has physical significance. The physical content is that if you place a ruler between these two points, and it is a good ruler, it will show a length of \(d\ell = +\sqrt{d\ell^2}\). Since it is difficult to find rulers good at measuring infinitesimal lengths, we can turn this into a macroscopic rule. Imagine a string following a path parameterized by \(\lambda\), from \(\lambda = 0\) to \(\lambda = 1\), then the length of the string is \(\int_0^1 d\lambda (d\ell/d\lambda)\). That is, every infinitesimal increment \(d\lambda\) corresponds to some length \(d\ell\). If we add them all up, that's the length of the string.

    Box \(\PageIndex{1}\)

    Exercise 1.1.1: Find the distance along a path from the origin to (x,y,z) = (1,1,1) where the path is given by

    \[ x(\lambda) = \lambda, y(\lambda) = \lambda, z(\lambda)= \lambda \]

    Answer

    Solution not available yet.

    There are many ways to label the same set of points in space. For example, we could rotate our coordinate system about the z axis by angle \(\theta\) (with positive \(\theta\) taken to be in the counterclockwise direction as viewed looking down toward the origin from positive z) to form a primed coordinate system with this transformation rule:

    IMG_6761.PNG\[z' = z\]

    \[y' = -x \sin\theta + y \cos\theta\]

    \[x' = x \cos\theta + y \sin\theta\]

    Under such a re-labeling, the distance between points 1 and 2 is unchanged. Physically, this has to be the case. All we've done is used a different labeling system. That can't affect what a ruler would tell us about the distance between any pair of points. Further, for this particular transformation, the equation that gives us the distance between infinitesimally separated points has the same form.

     

    Figure 1: A counterclockwise rotation of the coordinate system about the z axis by \( \theta \) creates a new coordinate system which we’ve labeled with primes. The \( z \) axis comes out of the screen and is identical to the \( z’ \) axis. As is true for any point in space, point 1 can be described in either coordinate system, by specifying \( (x_1,y_1,z_1) \) or \( (x’_1,y’_1,z’_1) \) with the relationship between the two given by the equations to the right.

    Box \(\PageIndex{2}\)

    Exercise 1.2.1: Show that the distance rule of Eq. \ref{eqn:EuclidCartesian} applied to the prime coordinates,

    \[\begin{equation*}
    \begin{aligned}
     (d\ell')^2 = (dx')^2 + (dy')^2 + (dz')^2
    \end{aligned}
    \end{equation*}\]

    gives the same distance; i.e, show that \(d\ell' = d\ell\). [Warning: \( \theta \) is not a coordinate here. It specifies the relationship between the coordinate systems. So, e.g., \( dx' = dx\cos\theta + dy \sin\theta \).] Because this distance is invariant under rotations of the coordinate system, we call it the invariant distance.

    Answer

    \[\begin{equation*}
      \begin{aligned}
            dx' &= \cos\theta dx - \sin\theta dy \\ dy' &= \sin\theta dx + \cos\theta dy \\ dz' &= dz
      \end{aligned}
    \end{equation*}\]

    \[\begin{equation*}
      \begin{aligned}
            (dx')^2 &= \cos^2\theta dx^2 - 2\cos\theta\sin\theta dxdy + \sin^2\theta dy^2 \\ (dy')^2 &= \sin^2\theta dx^2 + 2\cos\theta\sin\theta dxdy + \cos^2\theta dy^2 \\ (dz')^2 &= dz^2
      \end{aligned}
    \end{equation*}\]

    \[\begin{equation*}
      \begin{aligned}
            (d\ell')^2 &= dx^2 + dy^2 + dz^2 \\ d\ell' &= d\ell
      \end{aligned}
    \end{equation*}\]

    We want to emphasize that the labels themselves, x, y, z or x', y', z' have no physical meaning. All physical meaning associated with the coordinates comes from an equation that tells us how to calculate distances along paths. To drive this point home, note that we could also label space with a value of x, y, z at every point, but do it in such a way that we would have the distance between x, y, z and \(x+dx, y+dy, z+dz\) have a square given by

    \[d\ell^2 = dx^2 + x^2\left(dy^2 + \sin^2y dz^2\right) \]

    For many readers, this result would look more familiar if we renamed the coordinates \(r=x\), \(\theta = y\), and \(\phi = z\) so that we get another expression for the invariant distance,

    \[d\ell^2 = dr^2 + r^2\left(d\theta^2 + \sin^2\theta d\phi^2\right)\ \label{eqn:EuclidSphere} \]

    This is the usual spherical coordinate labeling of a 3-dimensional Euclidean space by distance from origin, r, a latitude-like angle, \(\theta\), and a longitudinal angle, \(\phi\). The transformation between the two coordinate systems is given by

    \[z=r\cos\theta\]

    \[y=r\sin\theta\sin\phi\]

    \[x=r\sin\theta\cos\phi\]

    Box \(\PageIndex{3}\)

    Exercise 1.3.1: Show that the invariant distance given by the equation \(d\ell^2 = dx^2 + dy^2\), the 2-D version of \(\ref{eqn:EuclidCartesian}\), and the invariant distance given by the equation \(d\ell^2 = dr^2 + r^2 d\phi^2\), the 2-D version of \(\ref{eqn:EuclidSphere}\), are consistent if the coordinates are related via:

    \[\begin{equation*}
    \begin{aligned}
    x &= r\cos\phi \\ \\ y &= r\sin\phi
    \end{aligned}
    \end{equation*}\]

    Hint: use the chain rule, so that, e.g., \( dx = dr \cos\phi - r \sin \phi d\phi \). (Note that the coordinate transformation equations here are obtained from the 3-dimensional case by setting \( \theta = \pi/2 \).)

    Answer

    \[\begin{equation*}
      \begin{aligned}
            dx &= \cos\theta dr - r\sin\theta d\theta \\ dy &= \sin\theta dr + r\cos\theta d\theta
      \end{aligned}
    \end{equation*}\]

    \[\begin{equation*}
      \begin{aligned}
            dx^2 &= \cos^2\theta dr^2 - 2r\cos\theta\sin\theta drd\theta + r^2sin^2\theta d\theta^2 \\ dy^2 &= \sin^2\theta dr^2 + 2r\cos\theta\sin\theta drd\theta + r^2\cos^2\theta d\theta^2
      \end{aligned}
    \end{equation*}\]

    \[\begin{equation*}
      \begin{aligned}
            d\ell^2 &= dx^2 + dy^2 \\ d\ell^2 &= dr^2 + r^2d\theta^2
      \end{aligned}
    \end{equation*}\]

    We will now use Equation \(\ref{eqn:EuclidSphere}\) to derive a familiar result. Our motivation is to introduce to you the techniques that you can use when the geometry is not Euclidean, but to do so first in the familiar setting of Euclidean geometry.

    Box \(\PageIndex{4}\)

    The familiar result we are going to derive is that the circumference of a circle is \(2\pi\) times the radius. There are essentially two steps: calculate the circumference (length of the set of points all at equal distance from the center of the circle) and then calculate the radius (length of shortest path from the center to any point on the circle). We get finite distances (lengths) by integrating up infinitesimal ones. For paths parameterized by an independent variable \(\lambda\) (we could call it anything but we have to make some specific choice so are just calling it \(\lambda\)) that means length = \(\int d\lambda (d\ell/d\lambda) \). 

    Exercise 1.4.1: One way to parameterize the path of a circle is as follows: \(z=0\), \(x=r_1 \cos\lambda\) and \(y = r_1\sin\lambda\). This path is clearly periodic in \(\lambda\). Take the period to be \( 2\pi \). [Note that we will not manage to prove that \( \pi = 3.14159... \).] What is the length of the path over one period? Is this the circumference?

    Answer

    Using \(d\ell^2 = dx^2 + dy^2 + dz^2\) we get

    \[\begin{equation*}
      \begin{aligned}
            d\ell^2 = r_1^2sin^2\lambda d\lambda^2 + r_1^2cos^2\lambda d\lambda^2
      \end{aligned}
    \end{equation*}\]

    which simplifies to

    \[\begin{equation*}
      \begin{aligned}
            d\ell^2 = r_1^2 d\lambda^2 \quad \rightarrow \quad d\ell = r_1 d\lambda
      \end{aligned}
    \end{equation*}\]

    Now integrate both sides, with \(\lambda\) from \(0\) to \(2\pi\) (one period)

    \[\begin{equation*}
      \begin{aligned}
            \ell = \int_{0}^{2\pi} r_1 d\lambda \quad \rightarrow \quad \ell = 2\pi r_1
      \end{aligned}
    \end{equation*}\]

    This is the circles circumference.

    You already have in your mind that this path is that of a circle. But we should prove it. To do so, we need to know that every point on the path is the same distance from the origin. This is more easily done in spherical coordinates. Using spherical coordinates we can parametrize the path as follows:

    \[\begin{equation*}
    \begin{aligned}
    \phi &= \lambda, \\ \\ \theta &= \pi/2, \; {\rm and} \\ \\ r &= r_1.
    \end{aligned}
    \end{equation*}\]

    Exercise 1.4.2: Verify, using the Spherical to Cartesian coordinate transformation (equations 1.8-1.10), that this is indeed the same path, just expressed in different coordinates.

    Answer

    To Cartesian coordinates to Spherical:

    \[\begin{equation*}
      \begin{aligned}
            x &= rcos\phi sin\theta \\ y &= rsin\phi sin\theta \\ z &= rcos\theta
      \end{aligned}
    \end{equation*}\]

    substituting in \(\phi = \lambda\), \(\theta = \pi/2\), and \(r = r_1\), we get

    \[\begin{equation*}
      \begin{aligned}
            x &= r_1cos\lambda \\ y &= r_1sin\lambda  \\ z &= 0
      \end{aligned}
    \end{equation*}\]

    which is indeed the same path given in Exercise 1.3.1.

    Exercise 1.4.3: Use Equation \(\ref{eqn:EuclidSphere}\) and the above parameterization of the circle in spherical coordinates to show that the length of the path from \(\lambda = 0\) to \(2\pi\) is \(2\pi r_1\) which should be a confirmation of the result you got in Cartesian coordinate.

    Answer

    Substituting in the above to Equation 1.7 we get

    \[\begin{equation*}
      \begin{aligned}
            d\ell^2 = dr^2 + r_1^2(d\theta^2 + sin^2(\frac{\pi}{2})d\lambda^2)
      \end{aligned}
    \end{equation*}\]

    and since \(r\) and \(\lambda\) are constants the whole thing simplifies to

    \[\begin{equation*}
      \begin{aligned}
            d\ell^2 = r_1^2 d\lambda^2 \quad \rightarrow \quad  d\ell = r_1 d\lambda
      \end{aligned}
    \end{equation*}\]

    Now integrate both sides, again with \(\lambda\) from \(0\) to \(2\pi\)

    \[\begin{equation*}
      \begin{aligned}
            \ell = \int_{0}^{2\pi} r_1 d\lambda \quad \rightarrow \quad \ell = 2\pi r_1
      \end{aligned}
    \end{equation*}\]

    This is the same result as we got in Exercise 1.3.1.

    We have the circumference now (step one of our two essential steps), derived using two different coordinate systems (even though one would be sufficient). Let's use the spherical coordinate system for the next step: finding the radius. We are going to do so in a way that also sets us up for demonstrating that the radius is the same no matter which straight line we take from the center out to the circle; i.e., we'll also manage to show that our set of points is indeed a circle. Consider the path (or paths, really -- there's one for any given value of \(\lambda\) ) parameterized by \(\mu\) that runs from the origin of coordinates \(r=0\) out to \(r=r_1\) at fixed \(\phi = \lambda\) and \(\theta = \pi/2\). The paths are simply given by

    \[\begin{equation*}
    \begin{aligned}
    \phi &= \lambda, \\ \\ \theta &= \pi/2, \; {\rm and} \\ \\ r &= \mu.
    \end{aligned}
    \end{equation*}\]

    for \(\mu\) going from 0 to \(\mu = r_1\).

    Exercise 1.4.4: Show that \(\int_0^{r_1} (d\ell/d\mu) d\mu = r_1\). Is this the radius? How are radius and circumference related? Note that because the length between the center and the circle is indeed independent of which point on the circle (which value of \(\phi\) or \(\lambda\)) we have proved that this is indeed a circle.

    Answer

    Substituting in the above to Equation 1.7 we get

    \[\begin{equation*}
      \begin{aligned}
            d\ell^2 = d\mu^2 + \mu^2(d\theta^2 + sin^2(\frac{\pi}{2})d\lambda^2)
      \end{aligned}
    \end{equation*}\]

    and since \(\theta\) and \(\lambda\) are fixed the whole thing simplifies to

    \[\begin{equation*}
      \begin{aligned}
            d\ell^2 = d\mu^2 \quad \rightarrow \quad  d\ell = d\mu
      \end{aligned}
    \end{equation*}\]

    It's now easy to see that  \(\int_0^{r_1} (d\ell/d\mu) d\mu\) does indeed equal \(r_1\). This is the radius.

    You have now demonstrated that the path parametrized by \(\lambda\) is a circle, that it has radius \(r_1\) and it has circumference \(2\pi r_1\). This may seem tedious. All we have managed to achieve is to derive a simple geometrical fact that you already know. The value will come soon, when you will use the same techniques to calculate the relationship between radius and circumference in geometries for which this well-known result no longer holds.

    Before going on, we should take a little more care. We have shown that a particular path that takes us from the origin to \(r=r_1\), \(\theta = \pi/2\) and \(\phi = \lambda\) has distance \(r_1\) independent of \(\lambda\). But maybe there's a shorter path? Here we will demonstrate that there is not a shorter path; the one prescribed is the shortest path possible. To do so, we use a result from the calculus of variations. That result is as follows:

    For \(J = \int_1^2 d\mu f(q_i,\dot q_i , \mu)\) where \(\dot q_i \equiv dq_i/d\mu\), the path from point 1 to 2 that extremizes \(J\) satisfies these equations

    \[\dfrac{d}{d\mu}\left(\dfrac{\partial f}{\partial \dot q_i}\right) = \dfrac{\partial f}{\partial q_i}\]

    This is a mathematical result with more than one application. In mechanics, the action is given as an integral over the Lagrangian so that

    \[S = \int dt L(q_i, \dot q_i, t)\]

    with \(\dot q_i \equiv dq_i/dt\), and because a system passes from point 1 to point 2 along the path that minimizes the action, the path taken will satisfy

    \[\dfrac{d}{dt}\left(\dfrac{\partial L}{\partial \dot q_i}\right) = \dfrac{\partial L}{\partial q_i}\]

    which you know as the Euler-Lagrange equations.

    In the case at hand we have length = \(\int d\mu \dfrac{d\ell}{d\mu}\) where

    \[f = \dfrac{d\ell}{d\mu} = \sqrt{\dot r^2 + r^2\left(\dot \theta^2 + \sin^2\theta \dot \phi^2\right)}\]

    (note the overdot is differentiation with respect to the independent variable which here is \(\mu\) again) so the shortest-length path between any two points should satisfy

    \[\dfrac{d}{d\mu}\left(\dfrac{\partial f}{\partial \dot r}\right) = \dfrac{\partial f}{\partial r}\]

    \[\dfrac{d}{d\mu}\left(\dfrac{\partial f}{\partial \dot \theta}\right) = \dfrac{\partial f}{\partial \theta}\]

    \[\dfrac{d}{d\mu}\left(\dfrac{\partial f}{\partial \dot \phi}\right) = \dfrac{\partial f}{\partial \phi}\]

    These equations are kind of hairy, if you work them out in generality. However, we are testing to see if a particular path satisfies them, the path from the origin to \(r=r_1\), \(\theta = \pi/2\) and \(\phi = \lambda\) that proceeds at fixed \(\theta\) and \(\phi\). For this reason we have \(\dot \theta = 0\) and \(\dot \phi = 0\) which really simplifies the evaluation of the above equations. We will just do one term out of the first equation as an example, and leave evaluation of the rest of the terms as an exercise. In particular, we evaluate \(\partial f/\partial r = (r/f)(\dot \theta^2 + \sin^2 \theta \dot \phi^2) = 0\), where the last equality follows since \(\dot \theta\) and \(\dot \phi\) are both zero. The other terms rapidly simplify for the same reason.

    Box \(\PageIndex{5}\)

    Exercise 1.5.1: Evaluate the five other terms, in the three equations above, and verify that the given path does indeed satisfy these equations, thereby demonstrating that it is the shortest possible path.

    Answer

    Solution not available yet.

    Summary

    1. Space can be labeled with coordinates. The same space can be labeled with a variety of coordinate systems; e.g., Cartesian or Spherical. 
    2. The coordinate labelings themselves have no physical meaning. Physical meaning resides in the distances between coordinates, which one can calculate from a rule that relates infinitesimal changes in coordinates to infinitesimal distances. 
    3. Paths through a space can be parameterized by a single variable, and we saw several examples of this.
    4. The Euler-Lagrange equations can be used to prove that a particular path is (or is not) one with an extreme value of distance between a pair of points on the path. Usually the extreme is a minimum rather than a maximum. 

    Homework Problems

    Problem \(\PageIndex{1}\)

    Starting from \(d\ell^2 = dx^2 + dy^2\) prove the Pythagorean theorem that the squares of the lengths of two sides of a right triangle are equal to the square of the hypotenuse. Start off by proving it for a triangle with the right-angle vertex located at the origin, so all three vertices are at \( (x,y) =(0,0), (x_1,0),\) and \( (0,y_1) \). Be careful to use the distance rule to determine the length of each triangle, rather than your Euclidean intuition. Let's call the length of the side along the \(x\)-axis \(\ell_x\) and similarly the other lengths \(\ell_y\) and \(\ell_h\). Parameterize each path and perform the appropriate integral over the independent variable you used for the parametrization (like we did with \(\lambda\) in this chapter). Doing so, you should find that \(\ell_h^2 = \ell_x^2 + \ell_y^2\). Having proved the Pythagorean theorem for this specially located and oriented triangle, note that since translations and rotations of the coordinate system leave our invariant distance rule unchanged, you have effectively proved it for all right triangles.

    Problem \(\PageIndex{2}\)

    Prove that the hypotenuse, the straight line from \( (x_1,0) \) to \( (0,y_1) \) you described in 1.1, is the shortest path between those two points.

    Problem \(\PageIndex{3}\)

    Show that for a primed system that is rotated relative to the unprimed system so that

    \[\begin{equation*}
    \begin{aligned}
    x' &= x \cos\theta + y \sin\theta \\ \\ y' &= -x \sin\theta + y \cos\theta
    \end{aligned}
    \end{equation*}\]

    the square of the invariant distance is unchanged; i.e., \( dx^2 + dy^2 = (dx')^2 + (dy')^2 \).