Assignment 9
Problems
Acceleration
We saw in class that the acceleration of a curve on the unit sphere is the projection of its \(\EE^3\) acceleration vector onto the tangent plane. In terms of the curve \(\gamma(t)\), that worked out to
\[\acc_\gamma(t)=\gamma^{\prime\prime}(t)-\left(\gamma^{\prime\prime}(t)\cdot\gamma(t)\right)\gamma(t)\]
Exercise 1 (Geodesic Curvature) The magnitude of the acceleration of a unit speed curve is called its geodesic curvature: its a way to measure how much that curve differs from a geodesic. This exercise, you will calculate the geodesic curvature of the circle of radius \(r\) on \(\SS^2\):
- Let \(C\) be a circle on \(\SS^2\) of radius \(r\) (to make calculations easy, let \(C\) be centered at the north pole if you like).
- Write down a parameterization of \(C\) (hint: you know the plane that \(C\) lies in, and its Euclidean radius in that plane!)
- Find a parameterization of \(C\) that has speed \(1\) (hint if you wrote down a parameterization above, what speed does it travel at? Can you adjust it so the new curve has speed \(1\)?)
- What is the acceleration felt on \(\SS^2\) if you go around a circle of radius \(r\) at unit speed? What is it’s magnitude?
The idea of measuring acceleration along a surface in \(\EE^3\) as the projection of the second derivative onto the tangent space is foundational to the study of surfaces beyond just the sphere (it is one of the fundamental concepts in differential geometry). When the acceleration of a curve \(\gamma\) is equal to zero on a surface, then we say that curve is a geodesic on the surface. So, the equations we get by setting the acceleration equal to zero give us a set of differential equations that tell us what the geodesics are! These are called the geodesic equations
In this next problem, we will get a small taste of what happens in differential geometry, when our space is not nice and symmetric like the plane or the sphere, and we have to resort to finding the geodesic equations and solving them (you won’t have to solve them! Just find them….)
The surface we will look at is the paraboloid of revolution, or \(z=x^2+y^2\). Here’s a quick graph of this (using Desmos’ new 3D features!)
'Exercise 2 (Geodesics on Surfaces) Let \(S\) be the surface \(z=x^2+y^2\), which is the output of the function \(F\colon\EE^2\to\EE^3\) given by \[F(x,y)=(x,y,x^2+y^2)\in \EE^3\].
Let \(p=F(x,y)\) be a point on \(S\).
- Use calculus to find two tangent vectors to the graph at a point \(p\): Hint: \(DF_{(x,y)}\langle 1,0\rangle\) and…?
- Find a normal vector \(n\) to the graph at \(p\) using these and the cross product.
- Write down the projection of a vector \(v=\langle a,b,c\rangle\) onto the normal vector \(n\).
- Write down the projection of \(v=\langle a,b,c\rangle\) onto the tangent plane \(T_{p} S\).
This is all the data we need to be able to compute acceleration along the surface \(S\)! Let \(\gamma(t)\) be a curve that lies on the surface, so \[\gamma(t)=F(x(t),y(t))=\left(x(t),y(t),x(t)^2+y(t)^2\right)\] for some function \(x(t)\) and \(y(t)\).
- Find \(\gamma^{\prime\prime}\)
- Find the acceleration of \(\gamma\) on the surface \(S\), in terms of \(x(t)\) and \(y(t)\).
- What are the geodesic equations for \(S\)?
Here’s a program that computes the geodesics on this surface, so you can see them!
'Curvature
We saw in class that the circumference of a circle of radius \(r\) on \(\SS^2\) is given by \[C(r)= 2\pi \sin(r)\]
Furthermore, we saw that the area is given by
\[A(r) =\int_0^r C(r)dr = 2\pi(1-\cos(r))\]
The idea of curvature is a
Exercise 3 Check this, that as \(r\to 0^+\) the following limits both exist, and are both equal to zero: \[\lim_{r\to 0}\frac{C_{\EE^2}(r)-C_{\SS^2}(r)}{r}=0\] \[\lim_{r\to 0}\frac{C_{\EE^2}(r)-C_{\SS^2}(r)}{r^2}=0\]
But
\[\lim_{r\to 0}\frac{C_{\EE^2}(r)-C_{\SS^2}(r)}{r^3}=\frac{\pi}{3}\] Hint: L’Hospital’s rule.
Because the first two limits here are zero, they tell us that the difference between the circumference of a Euclidean circle and a Spherical is very small indeed - they agree to first and second order, and their difference is only revealed at the next (cubic) level.
We use this to define the curvature of any surface, where we normalize things so that the curvature of the unit sphere comes out to be 1:
Definition 1 Let \(S\) be any surface, and \(C\) the circumference function for circles drawn on that surace at a point \(p\). Then the curvature of \(S\) at \(p\) is \[\kappa = \frac{3}{\pi}\lim_{r\to 0}\frac{C_{\EE^2}(r)-C(r)}{r^3}\]
We dont need to work with circumference however; its possible to measure the curvature of space using area as well!
Exercise 4 Which power \(n\) is the smallest such that \[\lim_{r\to 0}\frac{A_{\EE^2}(r)-A_{\SS^2}(r)}{r^n}\] has a nonzero value? For this \(n\), what is the value of the limit? Use this to write down a definition of curvature in terms of the area of circles, normalized so that the curavture of the unit sphere is 1.
Spheres of Different Sizes
Unlike the Euclidean plane, spheres have no notrivial similarities: in fact, if you apply a similarity of \(\EE^3\) to the sphere, it sends it to a larger or smaller sphere - not to itself! Because of this there is not just one spherical geometry like there was for the plane, but many. For each positive real number \(R\) we can define spherical geometry of radius \(R\), denoted \(\SS_R^2\), as follows.
Definition 2 (Spherical geometry of Radius \(R\).) Let \(\SS^2_R\) denote the set of points which are distance \(R\) from the origin in \(\EE^3\). For each point \(p\in\SS^2_R\), the tangent space \(T_p\SS^2_R\) consists of all points in \(\EE^3\) which are orthogonal to \(p\) (definition unchanged from the unit sphere), and the dot product for measuring infintiesimal lengths and angles is the standard dot product on \(\EE^3\) (also unchanged from the unit sphere).
The development of each of these spherical geometries is qualitatively very similar to that for \(\SS^2\): we can see without any change that \((x,y,z)\mapsto (x,y,-z)\) is an isomery so the equator is a geodesic, and orthogonal transformations are still isometries so all great circles are geodesics.
What changes is the quantitative details: the formulas for length area and curvature. In the next two problems, your job is to redo the calculations that I did for \(\SS^2\), for the geometry \(\SS^2_R\):
Exercise 5 (Circumference and area.) What is the formula for the circumference and radius of a circle of radius \(r\) on \(\SS^2_R\)?
Hint: base your circles at \(N=(0,0,R)\) and look back at our arguments from class to see what must change, and what stays the same.
Exercise 6 Using the definition of curvature as a limiting ratio of circumfereces (Definition def-curvature-surf), compute the curvature of \(\SS^2_R\).
Solutions
Acceleration
Geodesic Curvature
On the unit sphere, we derived in class that the circle of radius \(r\) lies in the plane \(z=\cos r\), and it followed from this that its Euclidean radius in that plane must be \(\sin r\).
Since the standard parameterization of the unit circle in \(\EE^2\) is \((\cos t,\sin t)\) we can rescale this by the radius to get \((\cos t \sin r, \sin t\sin r)\), and then place it at the right height in \(\EE^3\), giving
\[\gamma(t)=\left(\cos t \sin r, \sin t\sin r,\cos r\right)\]
This is one parameterization of this circle, but there are many others: we could go around the circle twice as fast, one third as fast, or even backwards by changing how we parameterize it. The problem asks us to modify this to make a unit speed parameterization, so a first step is to find out what the speed of the parameterization we have is! This is found by differentiating, and then taking the norm of the tangent vector:
\[c^\prime(t)=(-\sin t\sin r, \cos t \sin r, 0)\]
\[\|c^\prime(t)\|=\sqrt{\sin^2(r)(\sin^2 t+\cos^2 t)}=|\sin r|\]
Thus, our parameterization isn’t unit speed - which makes sense: \((\cos t,\sin t)\) was unit speed on the unit circle, but then we changed the circles radius! How do we make this unit speed? If our curve was going 2 times too fast, we could divide \(t\) by 2, and this would half its speed (to confirm, think about the chain rule). Since our curve is actually going \(\sin r\) times too fast, we can divide by this:
\[\gamma(t)=\left(\cos\left(\frac{t}{\sin r}\right)\sin r,\sin\left(\frac{t}{\sin r}\right)\sin r,\cos r\right)\]
Now, if we were to take the derivative, a factor of \(\frac{1}{\sin r}\) will pop out of each term, and cancel the \(\sin r\) from before - its unit speed!
Finally, the question asks us to find the acceleration of this unit speed curve, by plugging into the acceleration formula. This means we need the curves second derivative:
\[\begin{align*}\gamma^{\prime\prime}(t) &= \left(\frac{-1}{\sin^2 r}\cos\left(\frac{t}{\sin r}\right)\sin r,\frac{-1}{\sin^2 r}\sin\left(\frac{t}{\sin r}\right)\sin r,0\right)\\ &=\left(\frac{-1}{\sin r}\cos\left(\frac{t}{\sin r}\right),\frac{-1}{\sin r}\sin\left(\frac{t}{\sin r}\right),0\right) \end{align*}\]
The rest is just more direct computation, using the identity on \(\cos^2+\sin^2\) gives (after some algebra) \(\gamma^{\prime\prime}\cdot\gamma=-1\), so that
\[\mathrm{acc}=\gamma^{\prime\prime}-(\gamma^{\prime\prime}\cdot\gamma)\gamma=\gamma^{\prime\prime}+\gamma\]
\[=\left(\left(\sin r-\frac{1}{\sin r}\right)\cos\left(\frac{t}{\sin r}\right),\left(\sin r-\frac{1}{\sin r}\right)\sin\left(\frac{t}{\sin r}\right),\cos r\right)\]
This is the acceleration you feel when going around a circle of radius \(r\) on the sphere! The last thing we want to find is its magnitude, which is the geodesic curvature. Again using that \(\cos^2+\sin^2=1\) we can simplify this to
\[\begin{align*} \kappa &=\sqrt{\left(\sin r-\frac{1}{\sin r}\right)^2+\cos^2 r}\\ &= \sqrt{\sin^2r-2+\frac{1}{\sin^2 r}+\cos^2 r}\\ &=\sqrt{\frac{1}{\sin^2 r}-1}\\ \end{align*}\]
Phew! Now we just need a little good old fashioned trigonometry to simplify it all the way down:
\[\frac{1}{\sin^2 r}-1 = \frac{1-\sin^2 r}{\sin^2r}=\frac{\cos^2 r}{\sin^2 r}=\cot^2 r\]
So, taking the square root, we find
\[\kappa = \cot r = \frac{1}{\tan r}\]
Geodesics on Surfaces
Following the instructions laid out in the problem, we first compute the derivative, which is a map from \(\EE^2\) into \(\EE^3\):
\[DF=\pmat{1&0\\0&1\\ -2x&-2y}\]
If we apply this to vectors in the plane at a point \(p=(x,y)\), the result will be vectors tangent to the surface of the graph in \(\EE^3\) above that point. In particular we can get two tangent vectors by multiplying by the unit basis vectors \(\langle 1,0\rangle\) and \(\langle 0,1\rangle\), which simply return the first and second columns of the matrix:
\[DF_p\langle 1,0\rangle = \langle 1,0-2x\rangle\] \[DF_p\langle 0,1\rangle = \langle 0,1,-2y\rangle\]
What we are after is a normal vector to the surface. So we can take the cross product of these two tangent vectors:
\[n=\left|\pmat{i&j&k\\ 1&0&-2x\\0&1&-2y}\right|=\langle 2x,2y,1\rangle\]
Projecting a vector \(v=\langle a,b,c\rangle\) onto \(n\) is accomplished via the formula we derived in class:
\[\begin{align*}\proj_n(v)&=\frac{v\cdot n}{n\cdot n}n\\ &=\frac{2xa+2yb+c}{4x^2+4y^2+1}\langle 2x,2y,1\rangle \end{align*}\]
Finally, to project a vector \(v\) onto the tangent space \(T_pS\), we need to subtract its projection onto the normal direction. This leaves us with
\[\langle a,b,c\rangle - \frac{2xa+2yb+c}{4x^2+4y^2+1}\langle 2x,2y,1\rangle\]
The second half of this question asks us to use this projection that we just derived on the curve \(\gamma(t)=(x(t),y(t),x(t)^2+y(t)^2)\), to find a system of equations that \(x(t)\) and \(y(t)\) must satisfy for \(\gamma\) to be a geodesic on the surface. The first rule of being a geodesic is that the acceleration must be zero: that means, our equation above will be equal to zero, or after moving one term to the other side, we will have
\[\pmat{a\\ b\\ c} = \frac{2xa+2yb+c}{4x^2+4y^2+1}\pmat{2x\\2y\\1}\]
The second rule of being a geodesic is that the vector \(v\) we are projecting onto the tangent space isn’t arbitrary: its the *acceleration of our curve, \(v=\gamma^{\prime\prime}\). So all we need to do is compute this, and find \(a,b,c\). To lighten the notation, I will write \(x\) for \(x(t)\) and \(y\) for \(y(t)\) throughout - just remember that both \(x\) and \(y\) depend on \(t\)! Using the chain and product rules,
\[\gamma^{\prime}=\langle x^\prime,y^\prime, 2xx^\prime+ 2yy^\prime\rangle\]
\[\gamma^{\prime\prime}=\langle x^{\prime\prime},y^{\prime\prime}, 2(x^\prime x^\prime + xx^{\prime\prime}+y^\prime y^\prime + yy^{\prime\prime})\rangle\]
So, replacing \(v\) with \(\gamma^{\prime\prime}\) just means setting \(a=x^{\prime\prime}\), \(b=y^{\prime\prime}\) and \(c=2(x^\prime x^\prime + xx^{\prime\prime}+y^\prime y^\prime + yy^{\prime\prime})\).
Because \(xa=xx^{\prime\prime}\) and \(yb=yy^{\prime\prime}\) both show up inside of our expression for \(c\), we can simplify the numerator of the big fractioN \(2xa+2yb+c\) a bit to get \(4xx^{\prime\prime}+4yy^{\prime\prime}+2x^\prime x^\prime + 2y^\prime\prime\), and then writing out each of the three equations in our system of equations gives
\[x^{\prime\prime} = 2x\frac{4xx^{\prime\prime}+4yy^{\prime\prime}+2x^\prime x^\prime + 2y^\prime y^\prime}{4x^2+4y^2+1}\]
\[y^{\prime\prime} = 2y\frac{4xx^{\prime\prime}+4yy^{\prime\prime}+2x^\prime x^\prime + 2y^\prime y^\prime}{4x^2+4y^2+1}\]
\[2(x^\prime x^\prime + xx^{\prime\prime}+y^\prime y^\prime + yy^{\prime\prime})= \frac{4xx^{\prime\prime}+4yy^{\prime\prime}+2x^\prime x^\prime + 2y^\prime y^\prime}{4x^2+4y^2+1}\]
Solving these differential equations is a hard process! But if you can find the solutions, you’ll have found the geodesics to the paraboloid.
Curvature
Circumference Limits
The numerator of all limits we are considering here is the difference from Euclidean to spherical circumference, or \[C_{\EE^2}(r)- C_{\SS^2}(r)=2\pi r -2\pi \sin r=2\pi(r-\sin r)\]
In the first case we are looking at the limit of
\[\lim_{r\to 0^+}2\pi\frac{r-\sin r}{r}\]
Evaluating the numerator and denominator separately gives the indeterminate form \(0/0\), so we apply L’Hospital’s rule to get
\[=\lim_{r\to 0^+}2\pi \frac{(r-\sin r)^\prime}{r^\prime} =\lim_{r\to 0^+}2\pi\frac{1-\cos r}{1}\] Here we see the denominator is 1 and the numerator goes to zero, so this limit exists, and equals zero, as we were asked to confirm.
Following the same procedure, we may look at the difference quotient \[\lim_{r\to 0^+}2\pi\frac{r-\sin r}{r^2}\] Again this is an indeterminate form, and we have to apply L’Hospitals twice before it gives us anything:
\[\frac{(r-\sin r)^\prime}{(r^2)^\prime}=\frac{1-\cos r}{2r}\to \frac{0}{0}\] \[\frac{(1-\cos r)^\prime}{(2r)^\prime}=\frac{\sin r}{2}\to \frac{0}{2}=0\]
This limit also equals zero, as was claimed. Finally we check the third limit,
\[\lim_{r\to 0^+}2\pi\frac{r-\sin r}{r^3}\]
which we expect to come out to a nonzero value, but only after three applications of L’Hosiptals rule:
\[\frac{(r-\sin r)^\prime}{(r^3)^\prime}=\frac{1-\cos r}{3r^2}\to \frac{0}{0}\] \[\frac{(1-\cos r)^\prime}{(3r^2)^\prime}=\frac{\sin r}{6 r}\to \frac{0}{0}\] \[\frac{(\sin r)^\prime}{(6r)^\prime}=\frac{\cos r}{6}\to \frac{1}{6}\]
Thus, we know the value of our limit!
\[\lim_{r\to 0^+}2\pi\frac{r-\sin r}{r^3}=2\pi \frac{1}{6}=\frac{\pi}{3}\]
Area Limits
This next question asks us to do the same thing, but for areas, looking at the limit
\[\lim_{r\to 0^+}\frac{A_{\EE^2}(r)-A_{\SS^2}(r)}{r^n}\]
for the first \(n\) where this is nonzero. One way to approach this problem is to mimic exactly the calculations we did above, but with the area functions \(A_{\EE^2}(r)=\pi r^2\) and \(A_{\SS^2}(r)=2\pi (1-\cos r)\) that we derived in class. This will require lots of L’Hospitals rules, as we try bigger and bigger values of \(n\), doing L’Hospitals more and more times, but is not technically challenging.
However - there’s also a shortcut! We know that area is the integral of circumference! Thus, our numerator is really of the form
\[A_{\EE^2}(r)-A_{\SS^2}(r)=\int_0^r C_{\EE^2}(r)dr-\int_0^r C_{\SS^2}(r)dr\]
If we differentiate this once, by the fundamenatl theroem of calculus we are back to the numerator from the previous problem! And there we know what limit works; the denominator needs to be \(r^3\). So…in this case, we need the denominator to be \(r^4\) so that after one differentiation, we have reached the same situation as the last problem. But what’s the value? Well, doing that first application of L’Hospitals we see
\[\frac{(A_{\EE^2}(r)-A_{\SS^2}(r))^\prime}{(r^4)^\prime}=\frac{C_{\EE^2}(r)-C_{\SS^2}(r)}{4r^3}\]
We have an extra factor of four in the denominator! Thus, since we know the limit without this four is \(\pi/3\), the new answer will be
\[\frac{1}{4}\frac{\pi}{3}=\frac{\pi}{12}\]
This gives us the normalizing constant we need if we want to define curvature using area:
\[\kappa = \frac{12}{\pi}\lim_{r\to 0^+}\frac{A_{\EE^2}(r)-A_{\SS^2}(r)}{r^4}\]
Spheres of Different Sizes
Circumference
On a sphere of radius \(R\), distance is not equal to angle (that two points on the sphere make, thought of as vectors from the center) but rather \(R\) times angle, since we have scaled up the unit sphere with a similarity, and these scale distances uniformly by their scaling factor.
\[\mathrm{dist}=R\theta\]
So, if we can find \(\theta\), all we need to do is multiply by \(R\). Our goal here is to describe the circles of radius \(r\) about the north pole \(N=(0,0,R)\), so these are the points \(p=(x,y,z)\) on the sphere where we have \(r=R\theta\), or equivalently, \(\theta = r/R\).
Given a point \(p=(x,y,z)\), we can find the angle it makes with \(N=(0,0,R)\) in \(\EE^3\) using the dot product:
\[\cos\theta = \frac{p\cdot N}{\|p\|\|N\|}\]
Since both \(p\) and \(N\) lie on the sphere of radius \(R\), the denominator here is \(RR=R^2\), and the numerator is \((x,y,z)\cdot (0,0,R)=Rz\). Thus
\[\cos\theta=\cos\left(\frac{r}{R}\right) = \frac{Rz}{R^2}=\frac{z}{R}\]
Multiplying by \(R\) we can solve this for \(z\), and see that on the sphere of radius \(R\), a circle of radius \(r\) lies in the horizontal plane \(z= R\cos(r/R)\).
This is useful information to us becasue now we can find it’s Euclidean circumference in this horizontal plane, which we know to be \(2\pi\) times its Euclidean radius.
This Euclidean radius is just the distance that a point \(p\) on the circle lies from the \(z\) axis: we know that \(p\) lies distance \(R\) from the origin, and we also know its \(z\)-coordinate is \(R\cos(r/R)\). By trigonometry, this means it lies at a distance of \(R\sin(r/R)\) from the \(z-\)axis, and so its circumference must be
\[C(r)=2\pi R\sin\left(\frac{r}{R}\right)\]
Curvature
Now that we have found the circumference as a function of radius for circles on the sphere of radius \(R\), we can calculate its curvature! We just need to plug this function into our limit calculation, and plug away at L’Hospital! We know we need three differentiations to get rid of the \(r^3\) in the denominator, so differentiating our numerator three times:
\[\begin{align*} (C_{\EE^2}(r)-C(r))^{\prime\prime\prime}&=\left(2\pi r-2\pi R\sin\left(\frac{r}{R}\right)\right)^{\prime\prime\prime}\\ &=\left(2\pi - 2\pi R\frac{1}{R}\cos\left(\frac{r}{R}\right)\right)^{\prime\prime}\\ &=\left(0+2\pi R\frac{1}{R^2}\sin\left(\frac{r}{R}\right)\right)^{\prime}\\ &=2\pi R\frac{1}{R^3}\cos\left(\frac{r}{R}\right) \end{align*}\]
This gets divided by \((r^3)^{\prime\prime\prime}=6\) and multiplied by the normalizing factor of \(3/\pi\) to give
\[\begin{align*}\kappa &= \frac{3}{\pi}\lim_{r\to 0^+}\frac{2\pi\frac{1}{R^2}\cos\left(\frac{r}{R}\right)}{6}\\ &= \frac{3}{\pi}\frac{2\pi\frac{1}{R^2}}{6}\\ &=\frac{1}{R^2} \end{align*}\]
Thus, the curvature of a sphere of radius \(R\) is equal to \(1/R^2\).