Good afternoon. My name is Andrew Love (andrew.love@jhuapl.edu). In my work at Johns Hopkins University's Applied Physics Laboratory I give presentations on a variety of dry topics. This time, though, I'm talking about a subject I really enjoy: the science, and particularly the physics that can be found and played with in various science fiction stories. I was inspired to create this talk after attending the "SF and Education" panels at the 1998 Worldcon; since then I've given versions of this talk at two college campuses and also put it on the web.
Before I start, I'd like to mention my background. I have both a bachelors and a masters degree in electrical engineering, and a masters in applied physics as well, but most of the concepts I'll be discussing require only high school physics to understand and only slightly more advanced mathematics to actually calculate the answers. I have found however that thinking about these SF cases often helps me get more intuition about the physics and sometimes by understanding the physics better I get a better feel for the story too.
Normally when I give this talk I can't assume that my audience knows much about science fiction, so I say a few words about the role of science in science fiction before starting the main body of the talk - i.e. the part of the talk with equations. With a more experienced audience I can skip through those slides pretty quickly.
There is no commonly accepted definition of Science Fiction - all of the above have been proposed. I'm going to stick with the third and fourth, but the other two aren't as silly as they seem. These are taken from a book called "Turning Points: Essays on the Art of Science Fiction" edited by Damon Knight (1977). Number 1 is attributed to Norman Spinrad, number 2 is an obvious generalization of number 1 (and I'm sure I've seen it used somewhere), number 3 is attributed to Theodore Sturgeon, and number 4 is Reginald Bretnor's definition, paraphrased by Robert Heinlein. "Extrapolation" and "future" are two words also strongly associated with SF.
The sciences usually associated with Science Fiction are physics, biology and sociology (broadly understood as including economics and group psychology), but one also runs across chemistry and linguistics and more unusual choices.
Today I'll be talking about stories with structures and events that can be analyzed with current, easy physics (by easy, I mean of course, physics that I can write and solve equations about).
The themes I'm going to touch on are, first, gravity, stability and tides, which can produce some nicely counterintuitive phenomena, second, space travel and orbits, and third, some weird stuff that doesn't quite fit in any of the categories but is too interesting to leave out. If I don't get to finish all of the topics I at least hope to interest you all enough that you read further - I've printed out the slides of this talk and the last few pages give plenty of references for further reading (and sources for the ideas I will have talked about). You'll notice that many of my detailed examples come from the works of Larry Niven. This is not because he's the only author who deals with concepts from physics, but merely because many structures in his stories are relatively simple to analyze and because he's an author whose work I'm very familiar with (I participate in a list-serve discussion group about his work and am often asked questions about the physics of Ringworld, etc.)
In a couple of places I've found accidental errors in a book's physics/mathematics. Science fiction writers seem uniquely accessible by fans - I sent Mr. Niven an e-mail asking about the errors I thought I'd found, and he replied, telling me to trust my equations. Of course in one very famous case, Mr. Niven actually wrote a sequel to one of his novels to remedy an oversight in the earlier book (I'll talk about that later).
Here's a big object: The Smoke Ring. It is a torus of gas circling an old neutron star, created from the atmosphere of the planet within the Ring. Its radius is about 26,000 kilometers and it gets light from the star that both it and the neutron star circle. It holds life. By the way, there is a gas torus around Saturn which one of Saturn's satellites (Titan) orbits within (see Niven's N-Space for details on how he was inspired to create the Smoke Ring).
Being so close to the high gravity of the neutron star, things in the Smoke Ring orbit the star pretty quickly. Niven gives a minimum period of 2 hours, while I get a typical period of a little less than 2 minutes (using this equation). I don't know what causes the discrepancy, unless it's a mistake on the author's part (this is the mistake I mentioned e-mailing Mr. Niven about). The mass of Voy is implied to be 0.5 solar mass which is about 1033 grams, but to get the right answer, the actual mass of Voy would have to be much less (about 1/10000th of a solar mass), which is a pretty implausible value for the mass of a neutron star mass.
The first time I gave this talk a student asked how a planet could have survived the supernova that created Voy - it's in the book that the planet came later and also that Voy is old enough that it is not rotating anymore and therefore not sweeping a radiation beam through the Smoke Ring periodically. Good questions though.
One of the life forms common in the Ring is the Integral tree. It's pretty big - people actually live on it. Now when people are simply in the Smoke Ring they're in free fall, but what if they live on a tree? It turns out that tides give an equivalent to gravity. The gravity of the neutron star is so great that the ends of the tree should be in significantly different orbits, but instead they're constrained to orbit together. Therefore anything attached to the upper end of the tree is moving too fast for its orbit and thus feels an acceleration outward.
Meanwhile, anything attached to the lower end is moving too slowly for its orbit and is accelerated inward towards Voy. The middle of the tree remains in free fall (otherwise the whole tree would accelerate into Voy or out of the Ring) - it's in the right orbit after all. Another way to think about it is that the whole tree is being swung around Voy, which should result in a centrifugal force outward on any object on the tree, but the gravitational force of Voy cancels out the centrifugal force exactly at the center of the tree, too much on the inner tuft and not enough at the outer tuft.
The equation for the acceleration experienced by something attached to the tree is remarkably simple - it's just 3 times square of the rotation rate of the tree around Voy times the distance of the object from the center of the tree. In other words, it's just the force that would be felt if Voy didn't exist, and the tree were being rotated around its center at the rate omega, plus the differential force due to change in Voy's gravity over the length of the tree.
Unfortunately while the book reports a maximum force of about 1/5 g at the ends of a tree, I get a force of almost 30 gees - probably because of the same miscalculation that caused the error in the period of objects in the Smoke Ring. However, that problem is easily resolved - just make the tree shorter. By the way, the same tidal force that creates pseudo-gravity for objects attached to the tree makes the orientation of integral trees very stable - each end of the tree is being pulled so that it lines up with the radial vector. If the tree ever got out of that orientation, it would oscillate a bit, but then settle down due to the damping effect of air resistance. We'll talk a little more about stability later, but I should note that man-made satellites orbiting the earth sometimes use tidal stabilization to remain oriented with one axis pointing towards the earth.
I'm sure all of you are familiar with Niven's Ringworld, which he described as halfway between a normal planet and a Dyson sphere (Freeman Dyson, the mathematician who established that Feynman's and Schwinger's approaches to quantum electrodynamics were equivalent, suggested in 1959 that a technological species would eventually need all its sun's energy, and that it could get that energy by surrounding the sun with an immense number of planetoids, so that all the light from the sun was captured. Nowadays, people usually use the phrase to mean a solid sphere surrounding the sun). Ringworld is a less audacious concept than a Dyson sphere but it's still so immense a construct that most minds boggle even after quite a lengthy exposure to the idea. Basically, Ringworld is a huge ribbon circling a star; the inside of the ribbon is habitable. An example of the scale of the Ringworld is this: on Ringworld, there is an ocean large enough that a full scale map of earth is a tiny archipelago separated from other such island groups by an ocean voyage of up to a half million miles, or twice the distance from the earth to the moon.
So what's it like on Ringworld? It's a big place - the surface area is about 3 million times the Earth's. Shadow squares (see diagram) provide a day-night cycle of 30 hours, but when the sun is not shadowed, it's always directly overhead - it never rises or sets. Unlike in a Dyson sphere you can see the stars, but the biggest feature of the sky is the Arch, the other side of the Ring from where you are. Somehow, the Ringworld became inhabited by primates half a million years ago or so and now there are dozens of species of human-like but non-human creatures filling the Ring, some intelligent, some not.
In several places in the book, the tangential velocity is given as 770 miles/sec, but 740 gives the right answer for the given period of rotation and the given surface gravity, than 770 does. This difference doesn't really affect much though, except trivia contests I suppose.
What are some implications of the how the Ring is laid out? There is surface "gravity" caused by the Ring's rotation and somewhat attenuated by the sun's gravity - in fact the surface gravity is almost identical to Earth's. However, the cost of gaining standard gravity with rotation is that it's not easy to get off the Ring. Orbital velocity near earth's surface is only about 5 miles per second and escape velocity is only about 7 miles a second - velocities that can be achieved with chemical rockets. On Ringworld, however, any object lofted away from the surface must achieve a velocity of 740 miles/second in the right direction or the velocity it already has by being on the Ringworld will slam the object into the Ringworld's surface with great force. On the other hand, if there were an way to get to the outside of the Ring, simply letting go would give you enough velocity to escape the star's gravity entirely. For that matter, of course, even landing on the Ringworld would be tricky, since matching the surface's velocity would be required and that velocity is far above the natural velocity of an object orbiting the Ringworld's sun at the Ringworld's radius. In the novel, the heroes crashland but this is not recommended for anyone without an impervious stasis field.
Exercise for the student of materials - how strong is the Ringworld's material? Very strong indeed to survive the tension induced by the tremendous spin, but can you calculate how strong it would have to be? How would you assemble the Ringworld? Can you calculate the trajectory of a rocket launching from the inside of Ringworld, relative to the Ringworld? Relative to inertial space?
Here's a demonstration of the difficulty in taking off from Ringworld. At 8 miles/second a rocket would leave earth easily, but on Ringworld at the same velocity a projectile crashes badly and is moved to the left of the takeoff point (assuming that the ring is rotating rightward). Aiming slightly left or right causes no real change except that the same angle of launch results in slightly different maximum altitudes and displacements on the ground (aiming against the spin improves the height by about 36 miles over aiming with the spin - not very useful).
I got a little carried away with calculating the differences between a planet's gravity and the pseudo-gravity on Ringworld. Here's a plot of the time a rocket would spend in flight if launched straight up with various initial speeds - not too much different with low speeds.
But quite different at only slightly higher speeds.
The same goes with maximum height achieved by the rocket. Not much difference at first -
A lot different later.
Two interesting not-obvious implications of the Ringworld's layout are the lack of natural resources and the lack of seasons. Resources are lacking because Ringworld just isn't that thick and was not formed by geological processes - there are no oil wells to tap or veins of metal to mine. If civilization falls, it will be tricky to revive. But what about seasons?
What causes seasons on earth? The change in length of day and angle of incidence of sunlight as the earth revolves around the sun changes the amount of heat the sun provides to a particular place on the earth. The distance from the sun to the earth does change but plays no significant effect on seasons - in fact the Earth is closest to the sun during January (northern hemisphere's winter).
On Ringworld, the length of day is constant. While this could be changed by adjusting the distance between the shadow squares (note that the shadow squares aren't in orbit around the sun either - they're connected by improbably strong wires and are spinning faster than necessary to stay in orbit), it's hard to imagine getting a useful amount of day length change over a long period and the mechanism to do so would be pretty complicated. Changing the distance to the sun is also possible (but see what happens two slides further), but since the Ringworld rotates with a period of 9 days or so, seasons would be very short.
But it turns out that there is one other thing to change about Ringworld.
Here's a concept called the Oscillating Ringworld. If we move the Ringworld so that the Sun is no longer in the plane of the Ringworld, gravity will act as a restoring force. With no friction, oscillation occurs and goes on indefinitely, changing the angle of incidence of light on the Ring - Seasons! The equations go as follows. There is a gravitational force on the Ring and an equal and opposite force on the sun. The Ring mass is about 1/1000th of the Sun, so the Sun's oscillation is a thousand times smaller than the oscillation of the Ring. The period turns out to be about 377 days, giving a reasonable length to the seasons. This analysis relies on a small angle approximation, but would be reasonably accurate for angles up to 10 or so degrees, which would give significant seasonal effects (insolation would vary by 1.5% due to the angle, and 3% due to the increase in distance from the sun, adding up to a variation of about 5%, which is significant enough - compare with the change in insolation over a year due to earth's orientation - another exercise for the student!). Note that there are two winters and two summers in each cycle - an "up winter," a "down winter," an "up-going summer" and a "down going summer". Also note that the period of 377 days is just the same as the period for a planet moving in the Ringworld's orbit. There are only so many ways to combine the gravitational constant, the mass of the sun and the mass of an object to get an answer in units of time, so it's not surprising that the same answer turns up in a few different places. By the way, the mass of the Ringworld is actually significant even compared to the huge mass of the sun, so for the calculation of the oscillation's period I need to include both masses, instead of neglecting the smaller mass, which can be done in most planetary orbit calculations.
Now for the bad news. Although moving the Ring so the sun is out of plane leads to a nice gravitational restoring force, moving the Ring so the sun is off center doesn't. This can be demonstrated without calculating the whole force on the Ring, just by calculating the forces due to gravity from small arcs of the Ring. On the near side, the arc contains a mass proportional to (R-x) and the effect of this mass is divided by a distance of (R-x) squared. On the far side, the opposite arc has mass proportional to R+x and the effect is divided by (R+x) squared. These two effects don't balance - any slight offcentering will get worse and worse. The fact that the Ring is rotating doesn't make anything worse or better - I think (someday I will do a full-up analysis to make sure though).
Note that a Dyson sphere is much better off. The mass contained in an arc at distance (R-x) is proportional to (R-x) squared, so when this is divided by (R-x) squared to get a force, that exactly balances the force due to the opposite arc. The sun can be placed anywhere inside the sphere and will feel no force moving it elsewhere. In Ringworld Engineers Larry Niven revealed that the Ringworld has attitude jets along its edge which convert solar wind into thrust, making the Ringworld stable by providing a restoring force that increases properly as the off-centering gets bigger. The only problem is that someone has stolen most of the jets.
Let's talk a little more about stability, since I mentioned it.
Stability is not the same as equilibrium - for a system to be in equilibrium all forces need to balance, but to be stable, the system has to be robust to small perturbations. A pencil balanced on its point is in equilibrium, but not stable. One on its side is in equilibrium and stable.
This chart shows the different kinds of stability. The lines are potential energy functions. The oscillating Ringworld is as stable in the vertical direction as a ball in a bowl is. Any perturbation simply changes the magnitude of the oscillation. In the horizontal direction though, the Ringworld is completely unstable - any perturbation causes a complete disaster as the Ring tends to get more and more off center. The Dyson sphere is neutrally stable. Any perturbation just moves the Sphere slightly but gives it no tendency to move further along or to return to the original position. Finally, with attitude jets, the Ringworld is probably meta-stable - it will recover from small perturbations, but be vulnerable to a large perturbation. Most systems in the real world are metastable; they will stay put, if they're not pushed too far - just like most people.
It's worth noting that if the impact of light on the shadow squares causes a greater outward force than gravity produces an inward one, then the shadow squares are stable horizontally but unstable vertically. In other words, because the dominant force is outward, not inward, all the signs reverse making stable situations for the Ringworld into unstable ones for the shadow squares and vice versa.
Here's the real Ringworld potential function, calculated by integrating the total force on the Ring - note the rapid drop off. That's why even with attitude jets, the Ring is probably only metastable - a big perturbation might overwhelm the jets' ability to restore the Ringworld to its proper place.
In Greg Benford's Tides of Light, a character named Killeen is dropped through a planet (long story). More recently, in John Cramer's Twistor, some valuable equipment takes the same kind of trip, when accidentally converted to a form of matter that is affected by no force but gravity. The question of what would happen in such a situation is interesting. It turns out that if we assume the earth has a uniform density and isn't rotating, that travel through a straight, frictionless tunnel is very simple to analyse. The force on the object in the tunnel is given by this equation, but it is only the force parallel to the tunnel's length that matters, so we get this equation. What simplifies this equation is that once the object is at (say) this point along its path, the gravitational force of this part of the earth completely cancels out. Only the mass here matters (this fact was discovered by Newton himself). When this fact is incorporated into the equation, it simplifies to the equation for simple harmonic motion. The length of the tunnel drops out entirely, so the total travel time is constant regardless of whether the endpoints are Baltimore and DC or Baltimore and L.A. The duration of a round trip depends only on the acceleration of gravity at the earth's surface and the radius of the earth and equals 84.4 minutes. This turns out to be a period that turns up frequently. It's also the period of a satellite orbiting just above the earth's surface (even at 500 km the period is only 94 minutes) and the period of a pendulum whose length is the radius of the earth. Not surprisingly the maximum speed of something dropped through the tunnel is about 5 miles per second, which is orbital velocity at the surface of the earth.
It's interesting to see how the answer changes if you take into account the change in the density of the earth as a function of the earth's radius. When I first wondered about the effects of such variation I concentrated on the two obvious special cases:
By the way, I used to think that Arthur C. Clarke had gravity transport in mind in City and the Stars/Against the Fall of Night, but when I checked it out recently, it's not entirely clear that this is true.
Deciding which mass distribution results in the quickest trip is a good test of intuition. For the black hole case, the object accelerates down a gravity well that gets very deep, but spends a great deal of time in the very flat part of the well. In the hollow shell case, all the acceleration happens during the trip through the shell and the rest of the trip is at a constant speed - the question is what that constant speed is. It turns out that the delta velocity due to the trip through the shell gets smaller as the shell gets thinner, so the total time to get across the shell can get arbitrarily long (the velocity gained during the trip through a thin shell is the square root of the acceleration of gravity at the surface of the shell times the thickness of the shell - but watch the units).
For the black hole case, the funny thing is that the longer the trip, the shorter the time it takes. For a very short trip, the roundtrip time is only slightly shorter than the 84 minutes we arrived at for the solid earth case, because even for the solid earth case, almost all the mass of the earth can be treated as if it were compressed at the earth's center. For longer trips (trips that go closer to the center to the earth), the object gains a lot of velocity by cutting close to the infinite gravity well - simulations indicate that by going within a couple of miles of the black hole, the length of the round trip declines to about an hour.
It should be noted that the hollow earth case is one that a lot of authors get wrong. There is no gravitational force felt inside a really hollow sphere - so stories in which people are standing on the inside surface are wrong. On the other hand, if the sphere has an internal atmosphere, then there will be a small gravitational force pulling inwards, as Hal Clement noted in Still River
By the way, the size of the black hole holding all the earth's mass is pretty small (2GM/c/c=0.44 cm), so missing it by a few miles keeps you well away from its event horizon and keeps your maximum speed well less than light speed (no relativistic effects), but tides might be a problem (see tide equations on previous pages).
Now some of you are probably wondering about the effect of the Earth's rotation on this method of travel. There is one, but it's not as bad as it might seem. Tunnel 1 runs parallel to the earth's rotational axis, so the any acceleration due to the rotation is always perpendicular to the motion of Killeen's body, and so has no effect on that motion (as long as we assume the tunnel is completely frictionless). Tunnel 2 goes through the earth's rotational axis, so the acceleration due to the earth's rotation is always outward along the path of the tunnel. This requires some further analysis. The rotational velocity at the surface of the earth at latitude x is given by the distance of the surface from the axis and the period of the earth's rotation. The rotational velocity of a point in the tunnel is linearly related to that and to the distance of the object from the earth's axis. Combining these two factors we get a rotational velocity in terms of distance along the tunnel and the period of the earth's rotation. The acceleration due to this rotational velocity is given by this familiar equation, giving us the outward acceleration due to the earth's rotation (frictionless walls are really needed here, because the object would touch the walls throughout its trip). Combining this with the gravitational acceleration derived earlier, our differential equation is still simple and the solution is still harmonic motion, just with a different period. The new period of Killeen's oscillation through the earth is only 8 seconds longer than it was before. So, a trip that goes to an opposite latitude while staying at a given longitude is acceptable, a trip that goes to an opposite longitude at a given latitude is also acceptable, and in fact, any trip with a constant latitude (even if it did not go through the axis) would also be acceptable and have the same period as derived above. What's not acceptable? Well, any trip that changes latitude, but doesn't go to an opposite latitude would cause a rotational acceleration that was always in the same direction - Killeen would rocket out of the tunnel or perhaps get stuck part way through. I calculated one specific example - imagine a tunnel going from the North pole to the equator. It turns out that the final velocity of Killeen as he left the tunnel at the equator would be about 3 tenths of a mile per second - pretty fast. If the earth rotated as fast as Hal Clement's Mesklin (Mission of Gravity) the rate of egress would be a lot faster, but I wouldn't be justified in assuming a spherical Earth any more.
These are some of the handiest equations for dealing with orbital mechanics in a science fictional environment. The total velocity of an object at any point in its orbit is equal to a constant times the difference between twice the reciprocal of the current distance from the primary and the reciprocal of the average distance from the primary. What this gives you (among other things) is the delta-v for transfer orbits. If you want to travel from earth to Mars, simply calculate the current orbital velocity of earth, and then the orbital velocity of an object currently at earth's radius but with a maximum distance that just reaches Mars' orbit. That gives you the dV to apply. Once you reach Mars the same equation gives the dV that will put the spacecraft in the same orbit as Mars. What makes this equation particularly useful is that you can pick your units for your convenience. When I was taking courses I used units of astronomical units for r and a, and units of miles per second for velocity whenever possible, so that for calculations in the solar system I could determine GM by taking the earth case - all I needed to remember was the length of the year and the size of earth's orbit and I could determine the earth's orbital velocity, and since the distance term in my units was (2-1), GM was easy to solve for and easy to use for any subsequent solar system problem without the pain of memorizing G or M.
The applications to SF for these equations are pretty obvious, but one thing I'd like to highlight is that at any point in an orbit, increasing velocity increases your orbit's semi-major axis, and decreasing velocity decreases it. The trick is that for any given orbit, a nearby orbit with a larger semi-major axis will have a smaller velocity (this time r is changed as well as a), so if you attempt to "catch up" with something in a higher orbit by speeding up, you'll end up falling behind. As John Brunner said in Shockwave Rider, "see you later, accelerator"
There's an old saying in the Smoke Ring.
Is it true? Why? Imagine yourself in the Smoke Ring with a jet pack at your command and think about the equations on the previous page:
If you apply a thrust eastward (in the direction of your orbit), you put yourself in a higher orbit (you move Out);
If you apply a thrust Westward (against the direction of your orbit), you put yourself in a lower orbit (you move In);
If you apply a thrust outward, you put yourself in a slower orbit, so you move West relative to everything else in your old, faster orbit;
If you apply a thrust inward, you put yourself in a faster orbit and you move East relative to everything else in your old, slower orbit.
If you thrust to the left or right, you move to an orbit that is at an angle to your original orbit, but which intersects your original orbit at two points, one of which is the point at which you applied the thrust - hence you come back to where to started after one orbit.
The "old saying" compressed all that knowledge into a few sentences.
Now for that promised weird stuff that didn't quite fit in to the other categories.
Let's take a simple premise - that it is possible for a technological species to create vehicles with relativistic velocities. Within a few decades, we might be able to create such vehicles. They have the following characteristics:
- Quick travel times - with a few decades of travel one could reach dozens or hundreds of other star systems.
- High destructive potential - if one hit a planet, it would be worse than a huge asteroid strike, just because of the speed of the impact.
- No conceivable defense - Defense against such a vehicle is very difficult because the huge momentum such an object has would be impossible to deflect, and mere detection would be difficult because the object is right behind the its own light.
Killing Star (by Pellegrino and Zebrowski) suggests that the possibility of such vehicles might lead all intelligent life forms to try to destroy all other intelligent life as soon as it's detected due to simple prudence. This is one of the key strategies of SF - taking a simple premise and finding the most outrageous (and horrifying) but plausible consequences.
Here's another simple premise: Teleportation - it's not very likely but what if it were possible?
What are the implications? Some are obvious (no airplanes or cars except for fun), some are less so.
A velocity damper is needed to absorb the potential energy difference that is produced when you teleport from a mountain to a valley and the velocity difference that occurs when you teleport from Canada to Florida (different tangential velocities due to different latitudes) or New York to LA (a vertical velocity due to the rotation of the earth).
A flash crowd is a crowd that appears whenever and wherever something interesting is happening. If we assume that teleportation is cheap and easy, a crowd can grow very quickly, since it draws on the whole world's population of possibly interested parties. We see some similar effects with popular web sites - "teleporting" to a web site is easy, so an overwhelming number of hits can happen to any site that is attractive.
No good reason, just a favorite quote, which captures the idea that knowing the causes of something does not dilute one's pleasure in seeing it.
