Physics – iRashida

Inside Out Universe – The (Legitimate) Hollow Earth Theory

Using projective geometry to invert the world as we know it.

The universe seems like a dauntingly large place. Our minds can barely comprehend distances between bodies within our own solar system – such as the 93 million miles separating the Earth and the sun – let alone the vast spaces between galaxies. Even when we use the fastest possible motion in the entire universe (the speed of light) to quantify these distances, they remain staggeringly large: it’s 2.5 million light-years to our next-door neighbor, the Andromeda Galaxy. And our Milky Way galaxy and the ones nearby are just the beginning…

Astronomers using the Hubble Space Telescope have captured the most comprehensive picture ever assembled of the evolving Universe — and one of the most colourful. The study is called the Ultraviolet Coverage of the Hubble Ultra Deep Field (UVUDF) project.

Images like this one from the Hubble Space Telescope are extremely popular because they illustrate the sheer vastness of the universe. Thousands of galaxies pepper just this one image, each one consisting of millions of solar systems, stars, and planets. Just look at this photoseries which zooms out from Earth until the observable universe is in view. Carl Sagan’s famous “pale blue dot” quote doesn’t just describe Earth, at the right scale it describes our solar system, our galaxy, and even our local supercluster. It’s hard not to feel insignificantly small on such an unimaginably large backdrop.

So imagine how surprised I was when I learned of a theory that the universe could be contained in a sphere the size of the Earth. It posits that the Earth is hollow and we live on the inside (along the perimeter) and all celestial bodies are contained within Earth’s roughly 8,000 mile diameter.

This seems utterly deranged at first; not even on par with flat-Earthers or conspiracy theories claiming the Earth is hollow and filled with secret “inner-Earth” societies. There are dozens of experiments that prove that the Earth is a rotating sphere hurtling through space, including Foucault’s pendulum and Eratosthenes’ shadow experiments. For that matter, when we look up its not like we see people from across the world looking back down at us!

However, this theory can be backed up mathematically and in fact living in this “nutshell Earth” (a term coined by Martin Gardner when examining and ultimately rejecting the theory) is indistinguishable from our current hypothesis of the structure of the universe (the Copernican model).

How can this be? Well, there exists a method for mapping all points on a plane into points in a circle of fixed radius. This is called inverting the circle. It works as follows:

Select a point P outside the circle
Draw a straight line between P and the center of the circle O
Find M, the midpoint of the line between P and O
Draw another circle with center M and going through P
Draw a straight line between N and N’, the points where the two circles intersect
P‘ is where OP and NN‘ intersect

In this manner, any point outside the plane maps to one and only one point within the circle. Reversing the process means that every point in the circle (except the exact center of the circle) maps to one and only one point outside the circle. The further away the original point is from the circle’s edge, the closer its inverse will be to the center of the circle. Only a point infinitely far away from the circle will map to the circle’s exact center.

If you want to mess with this a bit more and get a better idea of what circle inversion actually does, this applet allows you to see what your drawings look like after being inverted into a circle.

Essentially the same process of inversion can be generalized into three dimensions to map any point in a volume to a point within a fixed radius sphere. So if we imagine the Earth to be hollow and the universe to surround it, we can apply this perspective transformation to place every object in the observable universe within the hollow sphere. The heavenly bodies become miniscule, but each and every one can coexist in our nutshell Earth.

So there is a mathematical method for transforming all the universe into a sphere the size of the Earth without losing any information (as the transformation can be applied in reverse to regain the much larger Copernican model). The Earth’s surface maps to itself (with us living on the inside surface of the hollow sphere) and all of outer space becomes embedded within, with the farthest galaxies closest to the origin point of the sphere. The Egyptian mathematician Mostafa A. Abdelkader, the most sophisticated defender of the nutshell Earth theory, described such an inversion in his article “A Geocosmos: Mapping Outer Space Into a Hollow Earth” published in 1983.

After inversion, the moon, our closest celestial neighbor, maps to a sphere 955 meters across that circulates around the Earth’s axis from 6265 kilometers above Earth’s surface (all these observations are are from a perspective outside the nutshell and therefore outside the universe). The sun shrinks to about 2.5 meters across and recedes to a location a mere 253 meters from the origin point (which is the center of the universe). Pluto shrinks to the size of a single bacterium orbiting seven meters from the origin, while Alpha Centauri, the star closest to our own Sun, becomes an infinitesimally small speck situated a mere millimeter from the origin. Every other star and object in the cosmos, therefore, is contained in a sphere less than two millimeters across that hovers 6371 kilometers above our heads.

But even if you can accept this strange idea, what about the phenomena we observe that demonstrate the Earth’s movement through space? To give a fairly prominent example, how does this theory explain the daily sunrise and sunset?

Well, this is where it gets interesting. For a nutshell Earth to mesh with observed phenomena, the laws of physics must also be inverted. The behavior of gravitational, electromagnetic, and light waves is distorted to create a new, consistent “inverted physics” that explains the observations that conventionally lead us to a Copernican universe in a nutshell Earth.

The changed behavior of light rays are perhaps the most striking feature of the new model. In the Copernican cosmos, rays of light travel in straight lines. The position of the sun in relation to the observer determines what they will see. In the figure, an observer stationed under ray C will be experiencing solar noon while an observer stationed under ray A will be watching a sunset. As the Earth rotates on a 24 hour cycle, people on Earth experience the change from day to night.

So the sun moving across the sky is a point in favor of the Copernican model, right? Actually, no. The inverse mapping preserves angular relationships, so that observers positioned in the nutshell Earth would experience exactly the same phenomena as those in a Copernican universe.

Angular relationships are preserved during circle inversion: the angles in the triangle are the same as those in the inverted triangle

In the nutshell Earth, light rays can follow curved paths according to the inverted laws of physics. A maps into as ray a, and an observer positioned at ray h’s intersection point would observe the sun on the horizon. Because the Sun rotates around the origin, O, the observer would see it as setting, exactly as does the observer in the Copernican cosmos. Just as in the previous figure (representing the Copernican model), an observer intercepting ray c would be experiencing solar noon and a person observing b would see the sun as being somewhere between the horizon and the solar zenith.

Rays D and E do not intersect Earth in the Copernican universe and, assuming they do not intersect anything else, will continue traveling to infinity. In the nutshell Earth, however, d and e travel in arcs that lead back to the origin. The rays never actually reach the origin, however, because the inversion operation affects not only the direction of light rays, but their velocities as well. The speed of light is constant in the Copernican universe, but variable in the nutshell Earth, ranging from $3 * 10^8$ meters per second at the perimeter to zero meters per second at O.

The result of these conditions is that all observations and estimates of the size, direction and distance of any celestial object would lead to exactly the same result for an observer on the outside of Earth in a Copernican universe and his image observer inside the nutshell Earth, no matter where they are with respect to Earth’s surface.

Even the iconic “Earthrise” picture taken from the Apollo 8 manned mission to the Moon, which appears to so convincingly show a spherical Earth floating in a vast empty space, can be explained in the nutshell Earth theory. The curved light rays emanating from the sun illuminate half the Earth (creating the night/day cycle as the Sun revolves around the origin). Many of the curved light rays from the Moon are lost in the night of the center of the hollow Earth and only a portion of the Earth is visible.

Similarly, other phenomena such as the movement of Foucault pendulums are accounted for by other inverted laws of physics. explained conventionally as effects arising from Earth’s rotation about its axis. The direct isomorphism between the Copernican universe and the nutshell Earth means that its is impossible to refute it as a valid model based on empirical tests. Every possible observation made in a Copernican universe has its exact analogue in the nutshell Earth. This also means that it is impossible to prove the nutshell Earth theory. Evidence such as the Tamarack Mines plumb lines which is commonly used by conspiracy theorists to “prove” the surface of the Earth is concave don’t apply to the nutshell Earth theory because they don’t mesh with its inverted laws of physics.

So if we can’t prove or disprove the nutshell Earth theory, what exactly is it? Well, it’s really more of a thought experiment. We rarely look at the assumptions we make when describing the structure of the universe. Our observations match the Copernican idea of the universe, given we accept the untestable (though understandable) assumption that light always propagates in straight lines. But if we make a different set of untestable assumptions? We end up with the nutshell Earth.

I mentioned before that prominent physicist Martin Gardner rejected the nutshell Earth theory. He did this on the basis of Occam’s Razor – which holds that when given two choices with the same explanatory and predictive power, we should adopt the simpler one. Increased complexity is acceptable only when it yields a better theory in terms of explaining or predicting observations. For example, Einstein’s relativity added complexity to Newtonian physics but it also allowed us to explain certain natural phenomena, like small deviations in the orbit of Mars. The nutshell Earth requires significantly more mathematical complexity (see below) and exchange we get only the security blanket of believing that our little planet is important in the vast scheme of things.

A significant increase in mathematical complexity results if we move to this inverted model of the universe.

And even this security blanket is rather thin and possibly nonexistent. Gardner points out that the nutshell Earth model of the universe doesn’t require a nutshell Earth, it’s just as likely to be a nutshell Moon or Mars or Sun or Planet X. There are an estimated $10^{10}$ galaxies in the known universe. Assuming that each of these contains $10^{11}$ , as does our own galaxy, and that each of these stars is orbited by a mere ten other objects (planets, their moons, comets, asteroids, and small bits of rock or ice—any semi-spherical body will do), there are approximately $10^{22}$ objects in the universe to choose from. The probability that any one of them, including Earth, is the preferred body is only $\frac{1}{10^{22}}$ , which is vanishingly close to zero. Moreover, there is no reason why the inversion must be done in relation to a physical body at all. It is equally plausible to simply perform the inversion around an arbitrarily chosen spherical region of space, in which case the choice of regions and spheres is limitless. Regardless of which sphere we choose, if it is anything other than Earth, our planet becomes even smaller and less significant than ever after performing the necessary perspective transformation. We would be a minuscule speck, probably less than a fraction of a millimeter wide, floating near the center of some hollow celestial body.

So if the justification for selecting a nutshell Earth is only an artifact of our human tendency towards geocentrism, why not take it even further towards egocentricism and map the universe to your own mind? Inverting the universe with respect to your brain means that your skull contains every star, galaxy, asteroid, and speck of dust in the cosmos. That has got to relieve some of those feelings of cosmic insignificance. Feel content in the knowledge that it is empirically impossible to disprove the theory that you completely contain the universe. Lest others think you’re a bit crazy, I would suggest storing this knowledge safely along with the rest of the universe: in your own hollow nutshell of a mind.

Rainbows and Optics

Finding the reason for a rainbow by analyzing a single spherical drop of water.

In the optics unit of freshman year physics, we learned that any natural rainbow will form a circle $42^\circ$ wide, centered at the anti-solar point, like this:

This seems like a very cool phenomenon, because it means that the rainbow’s location is independent of both the size and location of the water droplets that form it. It is also pretty cool because this number, $42^\circ$ , can be derived from looking at the behavior of light inside a single droplet of water. Rainbows are typically formed out of thousands of drops of water, but we can understand something about them by examining just one.

So lets look at the path of sunlight through a raindrop, represented by the solid yellow line in this diagram.

The diagram assumes total internal reflection (meaning all the light that enters the raindrop is reflected back out) which isn’t the case because some light does pass through the raindrop completely. However, only the internally reflected light forms the rainbow so we can safely ignore everything else. Also, notice how we don’t need to take into account the size of the raindrop.

Looking at the quadrilateral highlighted in yellow, we get the equation $\delta + (360 - 2\alpha) + (\theta -\alpha) + (\theta - \alpha) = 360$ because all the interior angles of a quadrilateral must add up to $360^\circ$ . This equation can be simplified down to $\delta = 4\alpha - 2\theta$ . The relationship between $\theta$ and $\alpha$ is dependent on the refraction of light when it travels from one medium to another (in this case from air into water). Therefore, we can use Snell’s Law to calculate $\alpha$ in terms of $\theta$ .

Snell’s Law states that $c_a \sin\theta = c_w \sin\alpha$ , where $c_a$ is the speed of light in air and $c_w$ is the speed of light in water. Using the index of refraction of water ( $n = \frac{c_a}{c_w}$ ) simplifies the equation down to:

$\sin\theta = n\sin\alpha$
$\alpha = \arcsin(\frac{\sin\theta}{n})$

Plugging back in for $\delta$ :

$\delta = 4(\arcsin(\frac{\sin\theta}{n})) - 2\theta$

Now we can graph $\theta$ vs. $\delta$ and see if we can understand why the rainbow forms where it does.

We can see that the graph peaks at $\delta \approx 42$ but how does that relate to the formation of the rainbow? Well, since the sun is so far away, light rays enter each droplet of water parallel to each other. This means that every value of $\theta$ from $0^\circ$ up to $90^\circ$ has the same amount of incoming sunlight. However, the outgoing angle of the light ( $\delta$ ) varies greatly with $\theta$ for most of the graph which means that the light is scattered and therefore too diffuse to see. Near the peak of the graph, however, $\delta$ barely changes with $\theta$ . This means that the same amount of light is concentrated into a much narrower band. This is where you can see the rainbow.

Here’s a visualization of what’s happening when you see a rainbow (the pictured cross section of the water droplet creates one point on the rainbow, the other cross sections and other droplets create the other points):

But what explains the different colors of the rainbow? You probably already know that white light is a combination of all colors of light, but how does a water droplet split these colors? Well, the different colors of light are a result of the different wavelengths of light. Red light has a longer wavelength than violet light, which allows it to travel faster in water. When I said the index of refraction for water was 1.333, that was actually just an approximation. In reality, red light has a slightly lower index of refraction of 1.331. Violet light, on the other hand, has a higher index of refraction of 1.340 (these values come from observations). If you use the exact values of $n$ for the different colors of light when graphing, the graph of incoming vs. outgoing angles looks like this:

The difference in $\delta$ between red and violet light at the peak of the graph is $2^\circ$ , which means that any rainbow you see will have a thickness of $2^\circ$ all the way around.

Just for fun (and in case graphs aren’t precise enough for you), lets try and determine $\delta$ at the peak of the graph using algebraic methods. We know that at the maximum or peak of the graph $\frac{d\delta}{d\theta} = 0$ .

Using the equations from above:

$\delta = 4\alpha - 2\theta$
$\frac{d\delta}{d\alpha}=4\frac{d\alpha}{d\theta}-2=0$
$\frac{d\alpha}{d\theta}=\frac{1}{2}$

Add in Snell’s Law and take the derivative with respect to $\theta$ :

$c_a sin\theta = c_w sin\alpha$
$\sin\theta = n\sin\alpha$
$\cos\theta = n\cos\alpha\frac{d\alpha}{d\theta}$
$\frac{\cos\theta}{n\cos\alpha}=\frac{d\alpha}{d\theta}$
$\frac{\cos\theta}{n\cos\alpha} = \frac{1}{2}$
$\frac{\cos^2\theta}{n^2\cos^2\alpha} = \frac{1}{4}$

To simplify further we need to use a trig identity to rewrite $\cos^2\theta$ :

$\cos^2\theta + \sin^2\theta = 1$ (Pythagorean Identity)
$\sin\theta = n\sin\alpha$
$\sin^2\theta = n^2\sin^2\alpha$
$1- \sin^2\theta = 1-n^2\sin^2\alpha$
$\cos^2\theta=1-n^2\sin^2\alpha$

Plug back in:

$\frac{1-n^2\sin^2\alpha}{n^2\cos^2\alpha}=\frac{1}{4}$
$\frac{1-n^2\sin^2\alpha}{1-\sin^2\alpha} = \frac{n^2}{4}$

To make things a bit easier to write, use $x^2 = \sin^2\alpha$ :

$\frac{1-n^2x^2}{1-x^2} = \frac{n^2}{4}$

Cross-multiply:

$4-4n^2x^2 = n^2-n^2x^2$
$4-n^2=3n^2x^2$
$x^2 =\frac{4-n^2}{3n^2}$
$x = \sqrt{\frac{4-n^2}{3n^2}}$
$\sin\alpha = \sqrt{\frac{4-n^2}{3n^2}}$

Using $n = 1.333$ , we get $\alpha = 40.225$ .

We can use a very similar procedure to solve for $\theta$ (this is left to the reader) or just plug this value of $\alpha$ into Snell’s Law. Either way, we get $\alpha = 40.225$ . Using these values in our original $\delta = 4\alpha - 2\theta$ , we get $\delta = 41.729$ .

We can do this same procedure with each of the individual colors of light to determine the exact location of a rainbow in the sky.

Another interesting implication of these calculations is that no two people ever see the same rainbow. Where you see the rainbow is dependent on where your eyes are in relation to the sun, so every person will see the rainbow in a slightly different location. So next time you see a rainbow, know that you are experiencing a unique aspect of nature’s beauty! 🙂

Error and Multiple Measurements

Mathematically quantifying error a.k.a. being right about how wrong you are.

Recently in Physics Club, we have been preparing for the first test in the Physics Olympiad series, the F = ma contest. The majority of the problems can be solved using basic equations of motion, but occasionally a problem comes up that requires some intuition. And these “intuitive” problems can be much trickier to solve mathematically than they appear to be.

A perfect example is Problem 25 on the 2016 F = ma contest which can be simplified down to this:
Christina measures a 1.5 meter piece of string with a meter stick, making two measurements which each have an error of $\pm$ 0.1
Alice measures the same 1.5 meter length of string with a 2 meter stick, making one measurement which has an error of $\pm$ 0.2

Who makes the more accurate measurement?

Initially, the group came to a consensus that the two measurements are equally good because they both have the same maximum error of $\pm$ 0.2 and an expected error of 0 (because the positive and negative error cancels out). But then somebody brought up the point that Christina’s measurements could “cancel out” if one was positive and the other was negative. Christina’s two measurements can combine to have zero error in multiple ways (for example, an error of +0.05 and then -0.05) while Alice’s measurement can only be error-free if he randomly gets that exact value. So Christina’s measurement must be better than Alice’s. This is actually the correct answer but this line of reasoning isn’t particularly helpful for figuring out specifics. How much better is Christina’s measurement? What if each of her measurements had an error of 0.175? How would having a normally distributed range of errors (rather than randomly distributed) affect the results? To answer these questions we need a mathematical solution to the problem.

First, we have to determine a mathematical definition of accuracy. Since we are dealing with probability, accuracy should be some sort of expected value. However, the expected value for the error doesn’t help us: it is zero for both cases. Instead, we have to take the absolute value of the error and calculate the expected value of that. Now we can simplify this problem down even further.

Randomly pick a number $n$ randomly such that $-1 \leq n \leq 1$ (I’m using 1 instead of 0.1 because I prefer to work with whole numbers and it doesn’t really make a difference)
Randomly pick another number $m$ with the same constraints.
Calculate the expected value of $\mid n + m \mid$ and compare to the expected value of $\mid p \mid$ where p is randomly chosen such that $-2 \leq p \leq 2$

It is easy to see that the expected value of $\mid p \mid$ where p is randomly chosen such that $-2 \leq p \leq 2$ is 1. Since $p$ has an even probability distribution, all we need to show is that $\mid p \mid$ has an average value of 1. This is pretty clear from a graph of p vs. $\mid p \mid$ or $\int_{-2}^{2} \mid p \mid dp = 1$ .

Finding the expected value of $\mid n + m \mid$ is quite a bit trickier, especially when you realize that $\mid n + m \mid \neq \mid n \mid + \mid m \mid$ . We can apply the same process we used above but we have to graph in 3 dimensions because we now have an additional variable. Put n on the x axis, m on the z-axis, and the error on the y-axis. The function $\mid n + m \mid$ now defines a plane that looks something like this:

The area under this plane can be calculated by finding the volume of 2 right tetrahedrons using the volume of a right tetrahedron formula ( $\frac{abc}{6}$ where a, b, and c are the lengths of the sides that from the right angle).

So:

$v = \frac{(2)(2)(2)}{6}$
$2v = 8/3$
$expected\:value = \frac{8}{3} /4$ (4 is the area of the base of the cube)
$expected\:value = \frac{2}{3}$

So the average error in Christina’s measurement is 0.0667 compared to Alice’s average error of 0.1! Each of Christina’s measurements could be up to 0.15 off, and her measurement would be equal to or better than Alice’s in terms of accuracy. You can do the same thing to determine expected error if each of Christina’s measurements had a different error (for example, if her first measurement was accurate within $\pm 0.125$ while her second was only accurate to within $\pm 0.175$ ) by drawing distorted tetrahedrons and calculating their volumes. But what if Christina took 3 measurements? Or 10? There is no way we can imagine and calculate volumes in 11 dimensional space. To do this problem, we have to recognize that what we just did above was calculate a double integral.
Namely:

$\frac{1}{4} \int_{-1}^{1}\int_{-1}^{1} \mid n + m \mid dn \: dm = \frac{2}{3}$

If we want to do the same thing for 3 measurements (each accurate to within $\pm 1$ , we would simply take the triple integral. The number in front would change to $\frac{1}{8}$ as well. Why? Because that number represents the probability distribution function for the individual errors. When there was just one randomly selected variable such that $-2 \leq p \leq 2$ , it was $\frac{1}{4}$ since the distribution function of p looked like a horizontal line through (0, $\frac{1}{4}$ because every $p$ had the same chance of being chosen. When there were two variables, the coefficient was still $\frac{1}{4}$ because the distribution functions of both n and m were a horizontal line through (0, $\frac{1}{2}$ ) and $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ . This also answers the question of what happens when the probability distribution of the error is a normal distribution, you just plug the Gaussian into your integral before calculating. The calculus gets quite messy at that point, however, so I’m not going to do an example.

Basically, for some arbitrary number of measurements $n$ where $\varphi(x)$ is the probability distribution function for x, we get:

$\int_{-1}^{1} \dots \int_{-1}^{1} \varphi(x_1)\varphi(x_2)\dots \varphi(x_n)\:\times\mid x_1 + x_2 + \dots + x_n \mid dx_1 dx_2 \dots dx_n$

But to do the original problem we didn’t need calculus at all, we just needed (interestingly enough) tetrahedrons and an understanding of probability.

Buoyancy and Perpetual Motion

Debunking a convincing “free-energy” device

Recently, I was introduced to a perpetual motion machine that appeared, at first glance, to be workable. In fact, at the end of nearly an hour of discussion with a physics teacher and some other students, I was convinced it would work. It relied on buoyancy to supply seemingly limitless, free kinetic energy. Many of the arguments against it could be easily be eliminated by making the machine larger or improving the quality of the parts. Alas, upon further examination I realized that in this machine (just like every perpetual motion machine) there is a fundamental flaw that cannot be overcome.

I talk about the specific machine later on, but much of my confusion about why the machine didn’t work stemmed from a misunderstanding of buoyancy. Buoyancy is not free energy at all, there is nothing magical about it. It is caused by the simple fact that water pressure increases with depth. This is best explained in a picture:

The buoyant force on the right “water ball” is exactly enough to suspend that mass of water, so it makes sense that the buoyant force is equal to the weight of the water displaced. The equation for buoyancy is:

$F_{buoyancy} = \rho g V$

where $\rho$ is the mass density of the liquid, $g$ is gravity, and $V$ is the volume of the object. Remember, since water pressure is so crucial to buoyancy, it is important not to discount it when looking at buoyancy based “perpetual motion”.

Simple “Perpetual Motion” Machine that works based on buoyancy.

Before we get to the machine I was talking about earlier, let’s look at this simpler machine. (I don’t know whose original idea this was but I have seen several iterations of it online).

The argument for why this machine rotates is that the ping-pong balls on the right will have a buoyant force acting on them, causing them to rise. The balls on the left only have a gravitational force acting on them and therefore fall. This causes the machine to spin counterclockwise. Looking at only the two sides of the conveyor belt, this argument is pretty convincing, it makes sense for the balls on the left to fall and the balls on the right to float. However, when you look at the top and bottom of the machine the argument starts to fall apart. Any kid who has pushed a rubber duck or bath toy underwater know that it takes work to force a low density object into the water. And that is exactly what is happening at the bottom left of this machine. In fact, we can even prove that the amount of work that the water does on a ping-pong ball in order to get it to the surface is exactly the same as the amount of work the ping-pong ball has to do to get from the air into the water. The net work is zero. This means that once the ping pong ball has entered the water, there is no work left over to actually rotate the machine.

$Work = Force \times Distance$
$Work_{done \:by \:water \:on \:ball} = F_{buoyancy} \times height$
$F_{buoyancy} = \rho g V$
$Work_{done \:by \:water \:on \:ball} = \rho g V h$
$Work_{done \:by \:ball \:on \:water} = pressure \times V$
$pressure (at \:bottom) = \rho g h$
$Work_{done \:by \:ball \:on \:water} = \rho g V h$

More Complicated “Perpetual Motion” Machine:
Finally we arrive at the machine that inspired this write-up. It appeared in a footnote of Eric Rogers’ book, Physics for Inquiring Minds, and showing how it doesn’t work was left as a challenge for the reader.

This underwater machine is comprised of a conveyor belt with attached cups containing gas and a frictionless piston on top of the gas. Opposite cups have tubes connecting them so gas can pass between them. When a cup is on the left side, the piston is pulled away from the cup bottom by gravity, causing more gas to be pulled into the cup. When the cup is on the right side, the piston is pulled towards the cup bottom by gravity causing gas to be pushed out of the cup. Since the cups on the left side displace more water, they have a greater buoyant force acting on them than the cups on the right side. This buoyant force puts a torque on the conveyor belt and causes it to rotate clockwise.

Again, we will try and prove that the work done by the water while the cups are on the sides of the machine is equal to the work done on the water at the top and bottom of the machine, where the cups “cross-over”.

Since every cup has an opposite cup connected by a gas tube, it makes sense to think of the work done on a single pair of cups. First, let’s look at the work done by water to raise one left cup and lower the corresponding right cup.

$V_{1} - minimum \:volume \:of \:gas \:in \:cup$
$V_{2} - maximum \:volume \:of \:gas \:in \:cup$
$V = V_{2} - V_{1}$
$Work_{done \:by \:water \:on \:pair \:of \:cups} = F_{net \:buoyancy} \times height$
$F_{net \:buoyancy} = \rho g V$
$Work_{done \:by \:water \:on \:pair \:of \:cups} = \rho g V h$

This is exactly the same result as we got before (it just took a little more calculation because there were buoyant forces on both sides in this scenario).

Now we have to consider what happens as the cups make the transition from the left to right and vice versa. This diagram explains how the pistons move as the cup crosses over.

We can see that the bottom piston is moving outwards, against the force of water pressure, while the top piston is moving inwards, with the force of water pressure. It may seem logical that these two forces cancel out and the “crossing-over” has no effect on the rotation of the system. If this were true, we would actually have a perpetual motion machine. The cups would speed up every time they passed the straight sides of the machine and continue moving through the top and bottom. But before you worry about the ramifications of breaking the laws of thermodynamics, remember that you cannot forget pressure differences in buoyancy problems. Also, remember that pressure increases with depth. This means that the bottom piston is expanding in a higher pressure environment than the top piston is contracting. The work, therefore, does not cancel out. In order to transition from one side to another, the pistons must do some work on the water. Now all we have to do is show that the work the pistons do is equal to the work the water did on the cups.

$P_{1} - water \:pressure\:at \:top \:of \:machine$
$P_{2} - water \:pressure\:at \:bottom \:of \:machine$
$A - cross-sectional \:area \:of \:piston$

$F = pressure \times A$
$Work = Force \times distance$
$= pressure \times A \times l$
$= pressure \times V$
$Work_{done\:by\:top\:piston} = P_{1} V$
$Work_{done\:by\:bottom\:piston} = P_{2} V$
$Work_{done\:by\:pair\:of\:pistons} = (P_{2} - P_{1}) V$
$P_{2} - P_{1} = \rho g h$
$Work_{done\:by\:pair\:of\:pistons} = \rho g V h$

So we have shown that the work done by the water in moving a set of cups to the top and bottom respectively is exactly cancelled by the work the pistons need to do to expand and contract while crossing from one side to the other. No matter how many cups you add or how streamlined you make the cups, the machine will not rotate forever.