
The Ring of Steel - ColinWright
https://www.solipsys.co.uk/new/TheRingOfSteel.html?sb13h
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ganonm
This reminds me of a similar problem I came across whilst studying quantum
mechanics in my undergrad. The problem is stated as follows 'imagine a
perfectly upright pencil balancing on its perfectly sharpened tip. How long
does it take to fall over?'. At first this seems surprising. Classically
speaking, the pencil would never fall over, but knowing that we live in a
quantum world, we can conclude that it would in fact fall over. The reason for
this is the famous Heisenberg Uncertainty Principle which says that the
product of the uncertainty in a particle's momentum and its position is always
greater than or equal to hbar/2\. You can derive a similar expression for
uncertainty in angular momentum and angular position.

Now, these expressions don't imply it's impossible for the angle (theta) and
angular momentum (L) both to be simultaneously zero, just that it's infinitely
improbable. If you plug in reasonable values for both parameters, say by
assuming that Δtheta = ΔL = sqrt(hbar) and that the actual values took a
figure corresponding to an expected value assuming these uncertainties, you
can work out that the pencil will actually fall over in a few seconds.

~~~
bsder
I don't find this quantum argument convincing.

We know that the better you isolate an atom, the longer it takes for an
excited electron to come back to ground state. The issue is that the electron
must dump a full quantum of energy, and it tends to do so as part of an
exchange with the outside world.

If you isolate an atom really well, that electron can take a _really_ long
time to decay.

See: "Collective Electrodynamics" [https://mitpress.mit.edu/books/collective-
electrodynamics](https://mitpress.mit.edu/books/collective-electrodynamics)

Brownian motion and standard thermodynamics almost certainly makes a stronger
case.

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jeffwass
I was literally repeating to myself “The Ringworld Is Unstable” as I read the
article, glad to see the last section where he mentions Niven’s book.

~~~
lmilcin
The Ringworld was kept stable artificially if I remember.

Also it rotated extremely quickly which prevented it from getting squashed
like the steel ring in the article. The rotation was so high that it created
artificial gravity and so was also high enough to counteract star's gravity.
In effect, the ringworld was in constant tension.

Since it was made from some pretty exotic stuff it is difficult to say
anything about the tensile strength of the material.

~~~
sliken
I believe this is pretty impractical, the required velocity was relativistic
and created too much tension for any plausible material.

~~~
ColinWright
For reference, using the Earth as a measure, we are about 150 x 10^9 metres
from the Sun, acceleration due to gravity is about 10 m/s^2, and a=v^2/R.

So the speed required is about sqrt( 10 x 150 x 10^9 ) which is about 1.22 x
10^6 m/s. Light speed is about 3 x 10^8, so this is about 0.4% of c.

So yes, we're coming into the realm of relativity.

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Symmetry
> What happens next?

It falls to the ground, of course. At the scale of something that long steel
isn't any more rigid than rope is.

~~~
ColinWright
The idea is that we're in "Puzzle World". This is one of the things that most
people take for granted, that the way this is set up has the ring being rigid,
even though that's not genuinely feasible. We assume that in the same way as
we are assuming that the Earth is a perfect, homogeneous sphere.

I might add a note to that effect - thanks for highlighting this.

 _Added in edit: I 've now added a note to the page to address this
question/issue. And when I say "address", I really mean "try to avoid"._

~~~
Symmetry
There's nothing wrong with an infeasible thought experiment. Frictionless
surfaces and spherical cows are a staple of learning physics. The problem is
when the author describes them as "greased surfaces" or "watermellon shaped
cows" and asks you to treat them as _perfectly_ frictionless or spherical.

Describing the rope as being "steel" instead of "rigid" is just needlessly
confusing.

~~~
ColinWright
If you're inside the world of physics, "rigid" carries exactly the meaning you
say it has. But it's a code word, it's got a technical meaning, and for people
_not_ in the world of physics, using the word "rigid" would, I think, be
confusing.

So I've updated the side-box to try to make it clearer. You might like to
check it out to see what you think.

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pintxo
Never heard of the Shell Theorem, great way to explain it With a very
simplified example: a picture frame

