$$$$

28  Life in Curved Space

Its one thing to do geometry, the way we do in a geometry course - deriving relationships between triangles circle and geodesics. But its another thing to feel it: to try and imagine yourself in the world you are studying, and think hard about what that experience would be like. This is a challenging but rewarding exercise, often not requiring much new geometry but requiring a lot of deep thought, and a lot of actual calculation. In this final chapter, we will attempt to give a taste of what hyperbolic space is like, relying on the material we have developed in the course.

Just like spheres come in may different sizes, so do hyperbolic spaces: so, the first thing we must do in asking ourselves what its like is to decide which hyperbolic space we are talking about. This question is actually interesting at all different levels of curvature, as different effects become important at different curvatures. But here in this short chapter we will fix a curvature, and work out the consequences. A convenient way to fix the curvature is to fix the units that we measure space in, we can specify a distance $R$ called the radius of curvature, defined so that if we measure everything in units of $R$, we will determine the curvature to be $-1$. (This is equivalent to instead fixing ahead of time some units, and then considering the hyperbolic space of curvature $-1/\sqrt{R}$ in those units.)

28.1 The Size of San Francisco

Here we fix the radius of curvature to be approximately the radius of San Francisco, $R=5km$. This will allow us to compare the behavior of small things (like humans) to medium things (like cities) and large things (like planets), and see how in curved space, these different regimes behave quite differently!

In our exploration, our goal will be to ask simple questions about how the world would be, try to deduce what sort of geometry will be relevant to solving them, do the proper computations, and then try to interpret the result.

28.1.1 How Big is the Earth?

Theorem 28.1 (Volume and Surface Area) The surface area of a sphere of radius $r$ is $$\mathrm{SA}(r)=4\pi {\sinh }^2(r)$$ (compare this to $4\pi r^2$ in flat space). The volume of a sphere is the integral of surface area: $$V(r)={\int }_0^r\mathrm{SA}(r)dr=\pi (\sinh (2r)-2r)$$

To figure out how big the earth would be, we need to think a bit about what we mean by this question. The earth formed from a collection of rocks in the early solar system, so its volume is fixed by the volume of the rocks that was used to make it up. The earths radius and surface area are geometric consequences of this: once we know the volume of rock we simply find the sphere of that size, and that’s the earth!

Example 28.1 (Radius of the Earth) First, we look to find the true earths volume, from its radius of 6300km, or 1250SF. The volume of the earth is

$$\frac{4}{3}\pi r^3=\frac{4}{3}\pi (1250{)}^3=8,181,230,868.7S{F}^3$$

That is, the earth is a little over eight billion cubic San Franciscos! To find the radius of the sphere which has this same volume in hyperbolic space, we need to solve the following equation for $r$:

$$\pi (\sinh (2r)-2r)=8,181,230,868$$

This needs to be solved using numerical methods, but doing so yields a shockingly small answer: $$r=11.11.1987,SF$$

So, the earth is only 11 San Franciscos in radius! Remembering we took $SF=5km$ this comes out to 55.98 kilometers, or 34.78 miles. In hyperbolic space, its closer to get from $SF$ to the Earth’s core than it is to get to the south bay!

Why is the earth so small? It all has to do with exponentials: since the volume of a sphere grows exponentially with radius, whereas in flat space it grows quadratically. This means there is just so much more volume as the radius grows in hyperbolic space, that it doesn’t have to be that big to fit all the rocks that make up the earth! This exponential property is also shared by the formula for hyperbolic surface area, which has an amusing consequence:

Example 28.2 (Surface Area of the Earth) Given a radius of $11.19$ units, we can find the surface area of the earth by

$$SA=4\pi {\sinh }^2(11.19)=16,468,700,000S{F}^2$$

To understnad what it means that the hyperbolic earth has surface area of 16 billion San Franciscos, we should compare this to the actual surface area of the earth we live on. This is (using the Euclidean radius $6300km=1249SF$) $$4\pi (1249{)}^2=19,602,972S{F}^2$$

The real earth is only 19 million San Franciscos! So, in hyperbolic space the Earth has 838 times the surface area of our current planet!

This is a lot of extra real-estate! For me, one good way to conceptualize this number is to think about what the depth of the ocean would be. Here, the average depth of the ocean is 2300 meters; but spreading the same amount of water over 838 times more surface area yields a depth of only 4.389 meters, or just under 15 feet!

What about the moon? In Euclidean space the moon’s radius is about 27 percent that of the earth (a little over a quarter as big), which means its volume is 2 percent that of the earth (since volume grows with the cube of radius).

Example 28.3 The volume of the moon is $0.02$ that of the earth, which means in units of San Franciscos,

$$0.02\times 8,181,230,868.7=3624617S{F}^3$$

Solving for the radius of the hyperbolic sphere with this volume, we find $$r_{\mathrm{moon}}=9.8SF$$

So, the moon isn’t that much smaller than the earth at all! This gives us a very good sense of just how quick the exponential growth of volume is. THe difference in radii between the earth and moon is

$$11.19-9.8=1.39SF=4.3\mathrm{miles}$$ Since the moon is only 2 percent the earth’s volume, this means that 98 percent of the earths volume is contained in the outer shell of radius 4.3 miles, just the outer 38 percent of the radius! But things only get weirder from here, if we look at larger spheres, since everything is driven by exponential growth. What can we say about the sun?

Example 28.4 (Radius of the Sun) The volume of the sun is 1.3 million times that of the earth, which means in units of San Franciscos, the Sun is

$$1,300,000\times 8,181,230,868.7=1,063,560,012,934,045S{F}^3$$

Solving for the radius of a hyperbolic sphere that has this volume, we find that $r=18.225$ - that is, the sun is only eighteen San Franciscos, or 56 miles in radius! Remember, the Earth is 11.19 San Franciscos in this world, meaning that the sun is only $\frac{18.225}{11.19}=1.62$ times as big in radius.

28.1.2 How Much of the Earth is Visible?

We’ve already learned some rather interesting things about the Earth in negative curvature: its simultaneously much smaller (in radius) and much larger (in surface area) than we are accustomed to. But what does it look like?

Theorem 28.2 (Distance to the Horizon) Standing at height $h$ above a sphere of radius $R$, the horizon in Euclidean space lies at a distance $d$ of $$d=R\arccos \left(\frac{R}{R+h}\right)$$ and in hyperbolic space, the analogous formula is

$$d=\sinh (R)\arccos \left(\frac{\tanh R}{\tanh (R+h)}\right)$$

As a warm-up, we can use this formula to find the distance to the horizon here in flat space. At a height of 2 meters =0.002km above the ground the horizon is

$$6,300\cdot \arccos \left(\frac{6,300}{6,300.002}\right)=5.019km$$

Thus, standing on the beach we can see a little over 5 kilometers, or around 3 miles out to sea. As we know well from experience, moving up a little bit in height lets us see much more: from our classroom on the fourth floor of Harney we can easily see many miles out to sea. Quantitatively this is easy to confirm: if we were at the top of the Salesforce tower (326 meters tall), we could see

$$6,300\cdot \arccos \left(\frac{6,300}{6,300.326}\right)=64.09km$$

But, what about in hyperbolic space?

Example 28.5 (The Horizon at Different Heights) Measuring in units of San Franciscos, a 2 meter tall human is $2/5000=0.0004$ San Francisco’s tall. Using the hyperbolic radius $r=11.19SF$ of earth, we find the horizon lies at a distance of

$$d=\sinh (11.19)\arccos \left(\frac{\tanh 11.19}{\tanh 11.1904}\right)=0.0282786SF$$

In more useful units, two percent of a San Francisco is $141.39$ meters. You can’t see very far at all, only a couple hundred feet until the earth has curved enough out of the way to be below the horizon! Moving upwards helps a bit: from the top of the sales force tower (whose height is 0.0652 San Franciscos) the horizon lies at a distance of

$$d=\sinh (11.19)\arccos \left(\frac{\tanh 11.19}{\tanh 11.2552}\right)=0.349651SF$$

This is around 1.7 kilometers, or just a bit over a mile. From the top of the Salesforce tower you wouldn’t be able to see all the way to USF, or even very far out into the bay! But it gets weirder, as we continue to ascend. From the height of a commercial airliner (30,000ft, or 1.828 San Francsicos) passengers can see

$$d=\sinh (11.19)\arccos \left(\frac{\tanh 11.19}{\tanh 13.018}\right)=0.9869SF$$

Even from miles into the sky, we can only just almost see all of San Francisco. And in fact, this is a fundamental limit: no matter how far above the sphere you are, you can only see up to 5km in any direction before the horizon. Even from space, when you look down at the earth, you would see the city stretch all the way across the earths’ disk (though seeing the earth from such a height is another challenge entirely, that we will confront shortly).

Exercise 28.1 Prove this: that as the height limits to infinity you can only see 1 unit of distance along the sphere.

What area of the sphere is this? This question is actually a bit more complicated than it seems at first. We can’t just use the formula for the hyperbolic area of a disk, because we’re not looking at a disk - we’re looking at a region on a sphere!

Theorem 28.3 (Area of a Spherical Cap) Given a sphere of hyperbolic radius $r$, the area of a disk of radius $d$ drawn on its surface is given by

$$\mathrm{area}=2\pi \sinh (r{)}^2(1-\cos \frac{d}{\sinh r})$$

Proof. A sphere in hyperbolic space is still a sphere - and we understand the intrinsic geometry of spheres quite well! So, we’ll be able to put this to work here. Indeed, we know that on the unit sphere the area of a disk is $2\pi (1-cosd)$ and if the sphere’s Radius is $\rho$, then the area of a disk of radius $d$ drawn on its surface is $$2\pi {\rho }^2(1-\cos \frac{d}{\rho })$$

So, all we need to do is figure out the radius of our sphere. It’s tempting to say that this is just $r$: that’s the distance in hyperbolic space to its center after all - but this is not the notion we are looking for here. The radius showing up in the formula above is the radius the sphere would have, if it were embedded in Euclidean space, which is where we derived this formula. Since the sphere’s area is $4\pi {\sinh }^2(r)$, we see that in Euclidean space the radius would be $\rho =\sinh (r)$ to have the same area giving

$$\mathrm{area}=2\pi \sinh (r{)}^2(1-\cos \frac{d}{\sinh r})$$

Exercise 28.2 (Spheres from a large distance) Explain why if you look at a sphere of large radius from far away, you only see approximately $\pi$ square units of its surface area.

Hint: Use a taylor series for $\cos$ and explain why its justified to only take the first terms (why is the angle you are taking $\cos$ of small?)

The fact that a sphere’s horizon is so nearby has far reaching consequences: one of them being the affordability of cell phones.

Example 28.6 (Price of Cell Phones) Cell phones work by using cell towers to collect and re-broadcast signals from phones, but such a signal can’t propagate over the horizon!

The tallest modern cell towers are around 100 meters tall (with most cell towers much shorter). From the top of such a tower in flat space, the horizon is 35.4km away, meaning the tower is accessible to approximately $3,936k{m}^2$ of land area. But in hyperbolic space? From $100m$ high the horizon is only

$$d=\sinh (11.19)\arccos \left(\frac{\tanh 11.19}{\tanh (11.19+\frac{100}{5000})}\right)=0.198SF$$

or $0.99km$ away, and the area of such a disk is

$$\mathrm{area}=\pi {\sinh }^2(11.19)(1-\cos \frac{0.198}{\sinh 11.19})=0.1231S{F}^2$$

$0.123$ square units: equivalently $3.15k{m}^2$ or $1.21m{i}^2$. This is $\frac{3936}{3.15}=1249$ times less coverage. To get similar coverage, you need over a thousand times more towers, making the cell network over a thousand times more expensive.

But its even worse than this: remember the earth’s surface area has grown by a factor of 838! Thus cell companies are hit with a double whammy: they need $1249$ times more towers per fixed area, and they also have 838 times more area to cover! Overall them, the cell network needs to be $1249\times 838=1,047,100$ times larger to give the same coverage: expect your cell plan to go up in cost by a factor of one million to pay for this increased overhead!

28.1.3 What does the Earth Look Like?

Theorem 28.4 (Visual Size of a Sphere) From height $h$ above a sphere of radius $R$, the angle $\alpha$ that the sphere takes up in your vision in Euclidean space is $$\alpha =2\arcsin \left(\frac{R}{R+h}\right)$$ and in Hyperbolic space is

$$\alpha =2\arcsin \left(\frac{\sinh R}{\sinh (R+h)}\right)$$

Proof.

Again, its useful to do some calculations on the Earth in flat space to get our bearings. From a standing height of 2 meters, the earth takes up

$$2\arcsin \left(\frac{6300}{6300.002}\right)=3.1365\mathrm{rad}\approx {179.666}^{\circ }$$

If the earth were a flat plane it would take up half of our field of view, or 180 degrees. So, the earth is rather indistinguishable from an infinite plane at human-height (as searching the uninformed corners of the internet show unfortunately all too well). From the top of the sales force tower the earth takes up

$$2\arcsin \left(\frac{6300}{6300.326}\right)={178.833}^{\circ }$$

which is just slightly smaller. Even from the height of an airplane (30,000ft = 9.144 kilometers), earth takes up almost half our field of view.

$$2\arcsin \left(\frac{6300}{6309.144}\right)={173.83}^{\circ }$$

And from the international space station at 254mi = 408km high, the Earth still looms large, taking up a wider field of view than our eyes provide us (we can see approximately 114 degrees with binocular vision) $$2\arcsin \left(\frac{6300}{6708}\right)={139.8}^{\circ }$$

Now, what happens in hyperbolic space?

Example 28.7 (Size of Earth from Different Heights) At 2 meters above the ground, the earth looks slightly smaller than its Euclidean counterpart, but perhaps not noticeably so. $$\alpha =2\arcsin \left(\frac{\sinh 11.19}{\sinh 11.1904}\right)={176.8}^{\circ }$$

From the height of the the Salesforce tower, $$\alpha =2\arcsin \left(\frac{\sinh 11.19}{\sinh 11.2552}\right)={139.1}^{\circ }$$ That is - the earth looks slightly smaller from this skyscraper than it does in flat space from the actual space station. And it only gets wilder, due to the exponential nature of the hyperbolic sine. From the height of an airliner,

$$\alpha =2\arcsin \left(\frac{\sinh 11.19}{\sinh 13.018}\right)={18.5}^{\circ }$$ The earth is only eighteen degrees across in your vision! This is about the size of your hand held at arms length away. In hyperbolic space, your airplane flight is rather dark, and you have to look almost straight down to even catch a glimpse of the tiny earth below.

The height the space station orbits above the earth is quite unrealistic in hyperbolic space (for some reasons we’ll encounter shortly), but its a good benchmark to evaluate nonetheless, to really appreciate the unrelenting growth of the exponential. At a height of 252 miles = 81 San Franciscos, the earth would appear to be

$$\alpha =2\arcsin \left(\frac{\sinh 11.19}{\sinh 92.19}\right)=$$ $${0.0000000000000000000000000000000007609}^{\circ }$$

This is so fantastically small that not only would the earth be completely invisible to the stations inhabitants, but it would not be detectable by any form of future super-telescope. A ride to the space station would be quite terrifying as the earth rapidly shrinks below you, fading forever from view into the black.

So, when we move away from a sphere of fixed radius, it shrinks rapidly in our vision. But what happens if we stand at a fixed distance from a sphere of different radii?

In Euclidean space, if you were distance $h$ from the surface of a sphere of radius $x$, the larger the sphere the more and more it would appear to take up half your vision. Precisely, we can see this using our formula: $$\alpha =2\arcsin \left(\frac{x}{x+h}\right)$$

As $x$ grows the argument of the arcsine approaches $1$, and so the arcsin approaches $\pi /2$ and $\alpha$ approaches $\pi$, half your field of view. But what happens in hyperbolic space?

Example 28.8 (Visual Size of Growing Sphere) At a distance $h$ from its surface, the visual size of a sphere of radius $x$ in hyperbolic space is

$$\alpha =2\arcsin \left(\frac{\sinh x}{\sinh (x+h)}\right)$$

What is this behavior like when $x$ is sufficiently large? Well, $\sinh (x)$ rapidly approaches $\frac{1}{2}{e}^x$ for large inputs, so we can simplify the argument of arcsine in this approximation as

$$\frac{\sinh x}{\sinh (x+h)}\approx \frac{\frac{1}{2}{e}^x}{\frac{1}{2}{e}^{x+h}}=\frac{e^x}{e^x{e}^h}=\frac{1}{e^h}$$

Thus, for large spheres, how big they actually are is essentially irrelevant to how big they appear in your field of view. Even if you know the distance to a sphere well, its impossible to gauge its size visually: even spheres that differ in size by millions of times still take up the same area in your field of view: $$\alpha \approx 2\arcsin (e^{-h})$$

28.1.4 What’s the Gravity Like?

Remark 28.1. There is a precise way to do all of this, using the formulation of gravity in terms of a gravitational potential: if $\rho$ is the mass density in space, the gravitational potential $U$ solves $\mathrm{\Delta }U=\rho$, where $\mathrm{\Delta }$ is the laplacian differential operator (in Euclidean space, this is ${\partial }_x^2+{\partial }_y^2+{\partial }_z^2$). In hyperbolic geometry, gravity follows the same equation, where we simply replace $\mathrm{\Delta }$ with the hyperbolic laplacian. Solving this for a point mass gives the gravitational potential, whose (negative) gradient is the gravitational force. And finally finding the magnitude of this recovers the law stated below $GM/{\sinh }^{2}r$.

Theorem 28.5 (Inverse Area Law of Gravity) Netwon’s law of gravity is usually referred to as the inverse square law, but this is only true in flat space. In fact, a more careful reading of the law might be read to say that gravity, like light, spreads out evenly in all directions. Thus, a mass $M$ causes a gravitational acceleration on an object at distance $r$ away proportional to $M$ and inversely proportional to the surface area of the sphere. $$\frac{M}{\mathrm{SA}(r)}$$

So for Euclidean space we would have $a=\frac{kM}{4\pi r^2}$ where $k$ is the proportionality constant giving the strength of gravity. Traditionally, the $4\pi$ in the surface area of the sphere is absorbed into the constant which is then renamed $G=k/4\pi$, or Newton’s constant, giving

$$a=\frac{GM}{r^2}$$

In hyperbolic space, the area of a sphere $4\pi {\sinh }^2(r)$ also contains a $4\pi$ so we can continue to use Netwon’s constant, getting the adjusted formula

$$a=\frac{GM}{{\sinh }^{2}r}$$

Conceptually it will be more helpful often to speak of the relative difference of this from Euclidean gravity, than to speak of the absolute numbers. The ratio between these two quantities is

$$\frac{a_{\mathbb{H}}}{a_{\mathbb{E}}}=\frac{\frac{GMm}{{\sinh }^2(r)}}{\frac{GMm}{r^2}}=\frac{r^2}{\sinh (r{)}^2}$$

To use this to understand the experience of gravity on the Earth’s surface, we will use Newton’s other insight, that the gravity of a spherically symmetric body acts just like a point mass located at its center.

Example 28.9 (Gravity on Earth’s Surface) On the surface of the earth in Hyperbolic space ($r_{\mathbb{H}}=11.19$ SF), the gravitational acceleration felt by people is $$a=\frac{G{M}_{\mathrm{earth}}}{\sinh (r_{\mathbb{H}}{)}^2}$$

But to actually get the numerical value here we need to think about $units*: we should express Newtons constant not in meters, but in San - Franciscos! Instead of doing this, its much easier to just compute the ratio with Euclidean gravity, which will automatically measure our result in $g$-forces:

$$\frac{r_{{\mathbb{E}}^2}}{\sinh (r_{\mathbb{H}}{)}^2}$$

Because the hyperbolic component requires us to work in units of San Franciscos, we need to measure Euclidean radius in those units: $r_{\mathbb{E}}=6300km=1249SF$. Thus

$$\frac{r_{\mathbb{E}}^2}{\sinh (r_{\mathbb{H}}{)}^2}=\frac{(1249{)}^2}{\sinh (11.19)}=0.00119g$$

Thus, gravity on the hyperbolic earth is only $0.1\mathrm{\%}$ as strong as on earth!

This means that we are only held very weakly to the surface of the earth: you can multiply your weight here by $0.00119$ to find how much you’d weigh there! This has some pretty scary consequences: since you are just as strong as you are here in Euclidean space (nothing about you changed, just the gravitational pull of you to the earth), you can imagine that it’s pretty easy to jump very high. Too high - I’d say, it turns out even the feeblest jump will launch you into space, never to return again. To see this, we need to calculate the escape velocity.

Theorem 28.6 (Escape Velocity) On the surface of a gravitating object, when you jump you can either fall back to the ground, go into orbit, or escape forever into the void. The dividing line between these bound states (falling back, or getting stuck in orbit) and the unbound states (getting ejected to infinity) is the escape velocity, a speed where if you jump any slower you’ll end up bound to the planet, but any faster will send you away forever. Our goal here is to compute the escape velocity for a hyperbolic earth.

The first step is to figure out how much energy is required to escape. The gravitational field tries to pull us back down all the way along our trajectory, and the total work it does on us is the integral of its force along our path. Thus, if we have mass $m$ and were to escape all the way to infinity, this would be

$$\begin{array}{rl}W& ={\int }_{r_{\mathbb{H}}}^{\mathrm{\infty }}\frac{GMm}{{\sinh }^{2}r}dr\\ & =GMm{\int }_{r_{\mathbb{H}}}^{\mathrm{\infty }}{\mathrm{csch}}^2(r)dr\\ & =GMm(-\coth r{|}_{r_{\mathbb{H}}}^{\mathrm{\infty }})\\ & =GMm(\coth (r_{\mathbb{H}})-\underset{r\to \mathrm{\infty }}{lim}\coth (r))\\ & =GMm(\coth (r_{\mathbb{H}})-1)\end{array}$$

This number (which we still have to compute in the appropriate units of San Franciscos) tells us how much energy is needed to escape. But how fast to we need to go? The kinetic energy of an object with mass $m$ is $\frac{1}{2}m{v}^2$. To escape to infinity, we need to give ourselves enough kinetic energy to be able to cancel out the pull of the gravitational field: thus, we need $$W=\frac{1}{2}m{v}^2⟹v=\sqrt{\frac{2W}{m}}$$

Putting these together, we find the escape velocity

$$v_{\mathrm{escape}}=\sqrt{2GM(\coth (r_{\mathbb{H}})-1)}$$

To actually compute this quantity we would need to be careful about units: convert everything to San Franciscos, do the calculation, and then convert to something more reasonable to interpret it in the end. However, like before, its easier to instead compute the ratio of the hyperbolic to the Euclidean escape velocities, as all these annoying constants cancel out.

Exercise 28.3 (Euclidean Escape Velocity) In Euclidean space with $F=\frac{GMm}{r^2}$, show that the escape velocity is $\sqrt{\frac{GM}{r}}$

Example 28.10 (Ratio of Escape Velocities) Let $v_{\mathbb{H}}$ be the escape velocity from hyperbolic earth, and $v_{\mathbb{E}}$ be the escape velocity from Euclidean earth. Then

$$\frac{v_{\mathbb{H}}}{v_{\mathbb{E}}}=\frac{\sqrt{2GM(\coth r_{\mathbb{H}}-1)}}{\sqrt{\frac{2GM}{r_{\mathbb{E}}}}}=\sqrt{r_{\mathbb{E}}(\coth r_{\mathbb{H}}-1)}$$

Now we can plug in some actual numbers and get a value.

Example 28.11 (Escaping from Earth) Measuring all radii in San Franciscos, $r_{\mathbb{E}}=1249$ and $r_{\mathbb{H}}=11.19$, we get the ratio

$$\frac{v_{\mathbb{H}}}{v_{\mathbb{E}}}=\sqrt{1249(\coth (11.19)-1)}=0.000000476$$

Thus, it takes less than one ten-millionth the speed to escape the hyperbolic earth as it does its euclidean counterpart! Since the escape velocity here is 11.2km/s (or 6.95 miles per second), in hyperbolic space this becomes

$$0.000000476\ast 11.2=0.0000053km/s=0.533cm/s$$

Thus if you make any movement faster than half a centimeter per second, you’ll be immediately ejected from the earth, never to return!

Life on such a world would be very perilous indeed: it’s impossible to walk as the mere act of taking a step will launch you beyond orbit! Perhaps we would all live below ground so there was a solid roof above our heads at all times, or put plungers on our feet to hold ourselves fast to the ground.

28.1.5 Can we see the Sun or Moon?

As we live our strange lives on the hyperbolic earth, what do we see in the sky around us? Is it blue? Is there a familiar sun and moon marking the times of day and month? Or are we forever shrouded in blackness?

To answer this, we first need to think about how far the earth should be from the sun. Just as the size of the earth and sun have shrank in negative curvature, the size of orbits and solar systems would also shrink, as the suns gravity drops off much quicker. A reasonable model to investigate then is what would life be like if we set the earth-sun separation so the suns gravitational pull is equal to its true value in Euclidean space? Since Euclidean gravity is inverse square and hyperbolic gravity is inverse sinh-square, this amounts to finding the radius $r$ such that $\sinh (r)$ equals the Euclidean distance of 93 million miles, or

$$\sinh (r)=29933798.4SF$$

Taking arcsinh, we find $r=17.907$ - that is, the earth is less than 18 San Franciscos from the sun! This is more wild when we realize that the Sun is also 18 San Franciscos across: we are only one sun-diameter away from the sun!

So, at this distance, how big does the sun look in the sky? Calculating much as we did for the earth-size previously, we find

$$\alpha =2\arcsin \left(\frac{\sinh 18.225}{\sinh (18.225+17.904)}\right)={0.0000192}^{\circ }$$

This is absurdly small - the sun would appear star-like in the sky. This is too small to be interesting, so let’s ask another question: how far away would the earth have to be from the sun for it to be as big as we see it here in flat space?

Example 28.12 (Distance to the Sun) The angular diameter of the sun in the sky is about half a degree, so we are looking to solve for at which distance $h$ a sphere of radius $18.225$ appears to be half a degree, or 0.00872665radians. This requires solving

$$2\arcsin \left(\frac{\sinh 18.225}{\sinh (18.225+h)}\right)=0.0087$$

We can solve this with some algebra:

$$h=\mathrm{arcsinh}\left(\frac{\sinh 18.225}{\sin \frac{0.0087}{2}}\right)-18.225=5.43758SF$$

This is pretty wild - the earth is 11 SFs in radius, and the sun is 18SFs, but for us to be able to see the sun at normal size in our sky, we need to be just over 5 SFs away from it! But the story gets weirder: because the size of the earth is actually larger than the distance between the earth and the sun, the size of the sun in the sky varies throughout the day! At high noon, we are at the location on the earth closest to the sun, with only 18 San Franciscos separating us from its firey surface.

But, at sunrise or sunset we have rotated away, and are actually much farther from the sun: of course, the same is technically true on the earth, where we are approximately 6,000km closer to the sun at noon than at sunset. But this is absolutely negligible in comparison to the 93 million miles that separate us. In hyperbolic space these numbers are 11 SFs and 18SFs however, which are of the same order of magnitude!

Example 28.13 (The Size of the Sun) At noon we are at a distance of 4.74 SFs from the sun, which itself has radius 18.225, and appears in our sky to be half a degree across. How far are we away from the sun at sunrise or sunset? Drawing ourselves a picture, we see that this distance is the hypotenuse of a right triangle, and so we can use the hyperbolic pythagorean theorem:

$$d=\mathrm{arccosh}(\cosh (11.19)\cosh (5.43))=15.9345SF$$

This is much farther from the sun (though, still less than a single sun-radius away from its surface!). How big does the sun appear from here?

$$\alpha =2\arcsin \left(\frac{\sinh 18.225}{\sinh (18.225+15.9345)}\right)={0.0000069}^{\circ }$$

Over the course of the day, the sun has changed in size by a factor of $\frac{0.5}{0.0000069}=72,306$ times! It rises in the sky as an almost invisibly small star, and then midday quickly grows by tens of thousands of times to briefly bathe the world in light, before fading into the abyss once more.

Where’s the moon in this story? We calculated the size of the moon above to be $r=9.8SF$, and we can play the same game and ask what distance it must be from the earth so that it appears the same visual size in the sky (at least at some point in time). By sheer coincidence the the moon and sun are both half a degree in our skies, so we are looking to solve the same equation we did for the sun, just with a different input radius.

Example 28.14 (Distance to the Moon) The angular diameter of the moon in the sky is about half a degree, so we are looking to solve for at which distance $h$ a sphere of radius $9.8$ appears to be half a degree, or 0.00872665radians. This requires solving

$$2\arcsin \left(\frac{\sinh 9.8}{\sinh (9.8+h)}\right)=0.0087$$

We can solve this with some algebra:

$$h=\mathrm{arcsinh}\left(\frac{\sinh 9.8}{\sin \frac{0.0087}{2}}\right)-9.8=5.43758SF$$

This is the same distance, to five figures after the decimal point!

This seems very, very strange at first: the sun and moon are very different sizes, why should they orbit the earth at essentially the exact same distance to look the same in the sky? But this goes back to something we already calculated: the size a sphere appears in the sky is pretty much independent of its actual size (so long as it is large enough that $\sinh r\approx \frac{1}{2}{e}^r$), it only depends on the distance to it. So, for the moon and sun to look the same size they must be at the same distance!

This of course has disastrous consequences, as if we orbit the sun, and the moon orbits us, the moons orbit will pass directly through the center of the sun. Goodbye moon! But, before the moon is burned - its reasonable to ask what we would see if we looked at its surface from earth at “lunar noon” - when it is highest in the sky. The amount of the moon thats visible would only be a disk of radius

$$d=\sinh (9.8)\arccos \left(\frac{\tanh 98}{\tanh (9.8+5.43)}\right)=0.999990SF$$

As we expect, we can see only $d\approx 1SF=5km$ in radius across the moon - meaning when we look up in the sky we will see the lunar disk with just a single crater or two across its surface at a time!