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# S01. Euclidean Geometry - SOLUTIONS

[ "article:topic", "Incomplete", "Box Solutions", "authorname:knoxl" ]

Box $$\PageIndex{1}$$

Exercise 1.1.1: Show that this distance rule:

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

applied to the prime coordinates, gives the same distance; i.e, show that $$d\ell' = d\ell$$. Because this distance is invariant under rotations of the coordinate system, we call it the invariant distance.

Solution $$\PageIndex{1.1}$$

\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*}

Box $$\PageIndex{2}$$

Exercise 1.2.1: Show that the invariant distance given by the equation $$d\ell^2 = dx^2 + dy^2$$, the 2-D version of Equation 1.1, and the invariant distance given by the equation $$d\ell^2 = dr^2 + r^2 d\theta^2$$, the 2-D version of Equation 1.7, are consistent if the coordinates are related via:

\begin{equation*} \begin{aligned} x &= r\cos\theta \\ \\ y &= r\sin\theta \end{aligned} \end{equation*}

Solution $$\PageIndex{2.1}$$

\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*}

Box $$\PageIndex{3}$$

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 1.3.1: 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 $$\lambda$$. What is the length of the path over one period? Is this the circumference?

Solution $$\PageIndex{3.1}$$

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 1.3.2: Verify, using the Spherical to Cartesian coordinate transformation, that this is indeed the same path, just expressed in different coordinates.

Solution $$\PageIndex{3.2}$$

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 1.3.3: Use Equation 1.7 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 coordinates.

Solution $$\PageIndex{3.3}$$

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 1.3.4: 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.

Solution $$\PageIndex{3.4}$$

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.

Box $$\PageIndex{4}$$

Exercise 1.4.1: 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.

Solution $$\PageIndex{4.1}$$

Solution not available yet.