Borehole Oscillators(gregegan.net)
gregegan.net
Borehole Oscillators
https://www.gregegan.net/SCIENCE/Borehole/Borehole.html
12 comments
Funnily enough I went into it with a background in math and was surprised about one specific claim that I couldn't quite understand, and it turns out it was subtly incorrect in such a way that it actually adds an interesting twist to the story (Greg Egan acknowledged it). I can't quite find the web page with the discussion (ETA: found it, it's the addendum at the end of the FAQ about the book [0]) but it's about <spoilers>the Garden of Eden configuration of the automaton.</spoilers>
ETA: I realize this sounds nitpicky and stickler-y so I just want to point out that I loved the book (and Greg Egan's work in general) and figuring out the automaton stuff was genuinely some of the most fun I've had out of a book.
[0] https://www.gregegan.net/PERMUTATION/FAQ/FAQ.html
ETA: I realize this sounds nitpicky and stickler-y so I just want to point out that I loved the book (and Greg Egan's work in general) and figuring out the automaton stuff was genuinely some of the most fun I've had out of a book.
[0] https://www.gregegan.net/PERMUTATION/FAQ/FAQ.html
In the interior of a sphere of uniform density, the gravitational strength is proportional to the radius, exactly like Hooke's law for an ideal spring. That's why the object in the bore hole undergoes harmonic motion.
Why is the graviational strength proportional to the radius?
Firstly, you have to know that the field strength is zero inside a hollow sphere. This is part of that shell theorem.
So for a point at a given depth inside the sphere, we can divide the sphere into a hollow sphere consisting of everything less deep, and a solid sphere consisting of everything deeper. Only the deeper sphere matters; we can ignore the hollow sphere.
So as we progress toward the centre, the attraction is due to a smaller and smaller sphere, whose mass is proportional to r^3. The radius is shrinking though, which has the effect of increasing gravity: the gravitational field strength is proportional to 1/r^2. Wen we combine these factors, we get r.
Why is the graviational strength proportional to the radius?
Firstly, you have to know that the field strength is zero inside a hollow sphere. This is part of that shell theorem.
So for a point at a given depth inside the sphere, we can divide the sphere into a hollow sphere consisting of everything less deep, and a solid sphere consisting of everything deeper. Only the deeper sphere matters; we can ignore the hollow sphere.
So as we progress toward the centre, the attraction is due to a smaller and smaller sphere, whose mass is proportional to r^3. The radius is shrinking though, which has the effect of increasing gravity: the gravitational field strength is proportional to 1/r^2. Wen we combine these factors, we get r.
It’s really odd how often and where damped spring motion comes up
Spring motion is the motion of systems where the force is proportional to the distance.
Many interesting systems (like springs) are near equilibrium, which means that the potential energy is at a local minimum. A spring is an example, but also a pendulum.
When the potential is at a local minimum, its gradient is zero. So if you Taylor expand it you only get second-order contributions. For a spring, the potential energy looks like V(x) = V(0) + k * x * 2 where x is the displacement and k is a constant.
Differentiating, you get harmonic motion: F(x) = k * x
Broadly speaking, this applies to all systems near equilibrium, simply from Taylor expanding the energy. And it's not only in classical mechanics, but in all branches of physics. Sydney Coleman [0] is often quoted as saying something like "QFT is simple harmonic motion taken to increasing levels of abstraction." [1]
[0] https://en.wikipedia.org/wiki/Sidney_Coleman
[1] https://physics.stackexchange.com/questions/355487/qft-is-si...
Many interesting systems (like springs) are near equilibrium, which means that the potential energy is at a local minimum. A spring is an example, but also a pendulum.
When the potential is at a local minimum, its gradient is zero. So if you Taylor expand it you only get second-order contributions. For a spring, the potential energy looks like V(x) = V(0) + k * x * 2 where x is the displacement and k is a constant.
Differentiating, you get harmonic motion: F(x) = k * x
Broadly speaking, this applies to all systems near equilibrium, simply from Taylor expanding the energy. And it's not only in classical mechanics, but in all branches of physics. Sydney Coleman [0] is often quoted as saying something like "QFT is simple harmonic motion taken to increasing levels of abstraction." [1]
[0] https://en.wikipedia.org/wiki/Sidney_Coleman
[1] https://physics.stackexchange.com/questions/355487/qft-is-si...
Isn't that just a second order differential equation?
FYI Greg Egan is practically his own genre of ultra hard math heavy sci-fi that I highly recommend to anyone who knows what partial differential equations are.
"Daedalus"- who did the back pages of New Scientist, is said to have got the UK patent office to issue a patent for an entirely passive metro system based on this theory.
I believe it was a protest against beaurocracy, and to prove a point about it being illegal to patent perpetual motion machines. It wasn't (a perpetual motion machine) but it was based off "free energy" -it comes to a halt eventually.
I believe it was a protest against beaurocracy, and to prove a point about it being illegal to patent perpetual motion machines. It wasn't (a perpetual motion machine) but it was based off "free energy" -it comes to a halt eventually.
It's legitimate perpetual motion, but not a machine. In a vacuum the test mass will never stop moving, but you can't extract any work from the system if you want it to keep going forever.
Technically you can't even look at it, because that requires bouncing photons off its surface. The resulting radiation pressure will slow it down eventually.
Technically you can't even look at it, because that requires bouncing photons off its surface. The resulting radiation pressure will slow it down eventually.
Because of the shell theorem mentioned in the article, any straight tunnel between two points on the surface of a sphere would take the exact same amount of time to traverse under gravitational acceleration (assuming no air resistance and uniform density of the sphere). In the case of the Earth, this time would be approximately 42 minutes.
That explains why the Earth was created to compute the question of life, the universe, and everything.
It's a one-dimensional orbit, right?
If that sounds interesting, I recommend not reading too much about the book before starting it; there are spoilers in most synopses.
https://en.wikipedia.org/wiki/Permutation_City
You don't necessarily need a background in programming and theoretical computer science to enjoy it. But you'll probably like it better if you already have some familiarity with computational thinking.