# Under what conditions is Niven's Ringworld unstable? **URL:** \ **Category:** Factual Questions\ **Created:** [December 19, 2013, 7:26pm UTC](https://boards.straightdope.com/t/under-what-conditions-is-nivens-ringworld-unstable/676755 "2013-12-19T19:26:09Z")\ **Posts on this page:** 18\ **Page:** 3
**Author:** ![Finagle](https://avatars.discourse-cdn.com/v4/letter/f/b19c9b/32.png) [@Finagle](https://boards.straightdope.com/u/Finagle)\ **Post date:** [December 20, 2013, 9:58pm UTC](https://boards.straightdope.com/t/under-what-conditions-is-nivens-ringworld-unstable/676755/41 "2013-12-20T21:58:02Z")
> [@teela\_brown](#): > > Just land \*me \*on the thing, and it’ll be safe. As I recall – \* you’d\* be safe. But all you’d have to do would be to make a bet that the Ringworld is unstable and everyone else is screeewwwed. I’m thinking that the consequences of scrith being stretchy might be very unpleasant. I assume that scrith would heat up like a rubber band when stretched, except on an astronomical level. And given the amount of energy it would take to stretch a substance so strong, it might heat up a lot. Hello vulcanism and pyroclastic flows. And then the Ringworld would contract again, and winter would come. ---
**Author:** ![md2000](https://avatars.discourse-cdn.com/v4/letter/m/73ab20/32.png) [@md2000](https://boards.straightdope.com/u/md2000)\ **Post date:** [December 20, 2013, 10:25pm UTC](https://boards.straightdope.com/t/under-what-conditions-is-nivens-ringworld-unstable/676755/42 "2013-12-20T22:25:11Z")
> [@billfish678](#): > > Maybe it is, maybe it isn’t. > > Gravity on “outside” of ring. One unit of gravity towards the sun due to the stuff close by that is under you. Plus one unit of of gravity also towards the sun due to all the stuff far away. Total equals two units towards the sun. > > Inside the ring. Stuff under your feet (and away from the sun). One unit of gravity away from the sun. Rest of the stuff one unit towards the sun. Units cancel. > > You can’t say it does or doesn’t work till you do the math. And my gut/poor memory says its zero inside the ring and non-zero outside the ring. I know in physics we did the math for a hollow globe, and gravity evrywhere was zero. Gravity falls off as a square of distance, so the really close material underfoot is cancelled by the extremely large mass attraction much further away. I’m guessing by symmetry the same would apply for a stubby cylinder like ringworld, or actually a cylinder of any length. ---
**Author:** ![BMalion](https://avatars.discourse-cdn.com/v4/letter/b/e274bd/32.png) [@BMalion](https://boards.straightdope.com/u/BMalion)\ **Post date:** [December 20, 2013, 10:42pm UTC](https://boards.straightdope.com/t/under-what-conditions-is-nivens-ringworld-unstable/676755/43 "2013-12-20T22:42:48Z")
> [@teela\_brown](#): > > Just land \*me \*on the thing, and it’ll be safe. Wanna make a baby? ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [December 20, 2013, 11:35pm UTC](https://boards.straightdope.com/t/under-what-conditions-is-nivens-ringworld-unstable/676755/44 "2013-12-20T23:35:07Z")
> [@CalMeacham](#): > > I can’t visualize the Gaussian envelope you’re drawing. It’s a cylinder co-axial with the Ringworld. For simplicity, ignore the Ringworld sides and assume its surface is flat. Make the Gaussian cylinder as wide as the Ringworld width, and make the radius so it’s just above its surface. Gravity has to pull in on the flat sides of the cylinder, because all the mass is on one side of the surface. So gravity has to pull out on the surface just above the Ringworld. Therefore, you’d be pulled towards the inside surface of the Ringworld. ---
**Author:** ![OldGuy](https://avatars.discourse-cdn.com/v4/letter/o/3bc359/32.png) [@OldGuy](https://boards.straightdope.com/u/OldGuy)\ **Post date:** [December 21, 2013, 12:10am UTC](https://boards.straightdope.com/t/under-what-conditions-is-nivens-ringworld-unstable/676755/45 "2013-12-21T00:10:02Z")
> [@ZenBeam](#): > > It’s a cylinder co-axial with the Ringworld. For simplicity, ignore the Ringworld sides and assume its surface is flat. Make the Gaussian cylinder as wide as the Ringworld width, and make the radius so it’s just above its surface. Gravity has to pull in on the flat sides of the cylinder, because all the mass is on one side of the surface. So gravity has to pull out on the surface just above the Ringworld. Therefore, you’d be pulled towards the inside surface of the Ringworld. But all the mass is on one side of every point of the curved surface as well so gravity has to pull in there as well by your reasoning. In any case this ignores the sun at the center. At every point of the Ringworld, the sun is clearly pulling towards the sun and away from the inner Ringworld surface. ---
**Author:** ![Andy\_L](https://avatars.discourse-cdn.com/v4/letter/a/c67d28/32.png) [@Andy\_L](https://boards.straightdope.com/u/Andy_L)\ **Post date:** [December 21, 2013, 12:27am UTC](https://boards.straightdope.com/t/under-what-conditions-is-nivens-ringworld-unstable/676755/46 "2013-12-21T00:27:03Z")
> [@md2000](#): > > I know in physics we did the math for a hollow globe, and gravity evrywhere was zero. Gravity falls off as a square of distance, so the really close material underfoot is cancelled by the extremely large mass attraction much further away. > > I’m guessing by symmetry the same would apply for a stubby cylinder like ringworld, or actually a cylinder of any length. That would be true for an infinitely long cylinder, but not for a finite cylinder. Let me try to explain without drawing pictures. Inside the infinite cylinder, in every direction you look, you see the cylinder, and the force pulling towards the far side of the cylinder in every direction is equal to the force pulling towards the near side of the cylinder - because for an infinitesimal solid angle dtheta in a particular direction, the mass on the near side is dtheta_density_x_x (where x is the distance to the near side) and the force pulling towards the near side is dtheta_density_x_x divided by x_x - which leaves dtheta_density. Similarly, the mass pulling towards the far side is dtheta_density_y\*y (where y is the distance to the far side) and again the distance cancels out, so the forces balance. But in a finite cylinder, like the Ringworld, there are directions you look where there is no Ring - so the balance doesn’t work; there’s always much more Ring mass on the “near” side than can be balanced by the mass on the far side. ---
**Author:** ![Trinopus](https://avatars.discourse-cdn.com/v4/letter/t/2bfe46/32.png) [@Trinopus](https://boards.straightdope.com/u/Trinopus)\ **Post date:** [December 21, 2013, 12:33am UTC](https://boards.straightdope.com/t/under-what-conditions-is-nivens-ringworld-unstable/676755/47 "2013-12-21T00:33:50Z")
> [@Andy\_L](#): > > . . . But in a finite cylinder, like the Ringworld, there are directions you look where there is no Ring - so the balance doesn’t work; there’s always much more Ring mass on the “near” side than can be balanced by the mass on the far side. I’m almost certain this is wrong. You can simply ignore those directions, and worry only about the directions where there is mass. You can take it all the way down to a mathematical limit or infinitesimal. Inside finite cylinders, just like inside spheres, there is no “gravity.” It all cancels out. Inverse square law. Isaac Newton figured this all out. IANA Mathematician, but I was a maths major in college. ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [December 21, 2013, 12:35am UTC](https://boards.straightdope.com/t/under-what-conditions-is-nivens-ringworld-unstable/676755/48 "2013-12-21T00:35:27Z")
> [@OldGuy](#): > > But all the mass is on one side of every point of the curved surface as well so gravity has to pull in there as well by your reasoning. No it isn’t. What are you talking about? When you are standing on the Ringworld’s inner surface, there is mass below you, and mass above you. > [@](#): > > In any case this ignores the sun at the center. At every point of the Ringworld, the sun is clearly pulling towards the sun and away from the inner Ringworld surface. We are discussing the gravitational field of the just the Ringworld. ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [December 21, 2013, 12:47am UTC](https://boards.straightdope.com/t/under-what-conditions-is-nivens-ringworld-unstable/676755/49 "2013-12-21T00:47:26Z")
> [@Trinopus](#): > > I’m almost certain this is wrong. You can simply ignore those directions, and worry only about the directions where there is mass. You can take it all the way down to a mathematical limit or infinitesimal. > > Inside finite cylinders, just like inside spheres, there is no “gravity.” It all cancels out. Inverse square law. Isaac Newton figured this all out. > > IANA Mathematician, but I was a maths major in college. Well, you’re wrong here. What makes you think you can just ignore some directions? When you’re on the Ringworld, and look down, all you see is the Ringworld surface. When you look up, _most_ of what you see has no Ringworld. You can’t just ignore that. Here’s another way of looking at it. Imagine you had a scrith sphere of the same radius. You agree there’s no gravitional pull inside the sphere, right? It all cancels. Now remove almost all the sphere, leaving a ribbon the width of the Ringworld. Virtually all of that mass removed is above you. Since there previously was no gravitational pull, there will be now. The mass you removed would have been pulling you up, so what’s left must be pulling you down. ---
**Author:** ![standingwave](https://avatars.discourse-cdn.com/v4/letter/s/9de0a6/32.png) [@standingwave](https://boards.straightdope.com/u/standingwave)\ **Post date:** [December 21, 2013, 1:10am UTC](https://boards.straightdope.com/t/under-what-conditions-is-nivens-ringworld-unstable/676755/50 "2013-12-21T01:10:51Z")
> [@ZenBeam](#): > > Well, you’re wrong here. What makes you think you can just ignore some directions? When you’re on the Ringworld, and look down, all you see is the Ringworld surface. When you look up, _most_ of what you see has no Ringworld. You can’t just ignore that. > > Here’s another way of looking at it. Imagine you had a scrith sphere of the same radius. You agree there’s no gravitional pull inside the sphere, right? It all cancels. Now remove almost all the sphere, leaving a ribbon the width of the Ringworld. Virtually all of that mass removed is above you. Since there previously was no gravitational pull, there will be now. The mass you removed would have been pulling you up, so what’s left must be pulling you down. But you’re also ignoring all the nearby mass you removed at the same time you removed everything else. ---
**Author:** ![Andy\_L](https://avatars.discourse-cdn.com/v4/letter/a/c67d28/32.png) [@Andy\_L](https://boards.straightdope.com/u/Andy_L)\ **Post date:** [December 21, 2013, 1:11am UTC](https://boards.straightdope.com/t/under-what-conditions-is-nivens-ringworld-unstable/676755/51 "2013-12-21T01:11:23Z")
> [@Trinopus](#): > > I’m almost certain this is wrong. You can simply ignore those directions, and worry only about the directions where there is mass. You can take it all the way down to a mathematical limit or infinitesimal. > > Inside finite cylinders, just like inside spheres, there is no “gravity.” It all cancels out. Inverse square law. Isaac Newton figured this all out. > > IANA Mathematician, but I was a maths major in college. Newton dealt with spheres - and Gauss with infinite cylinders. The real world of finite and not completely symmetric items is more complicated. I found this paper online [Radware Bot Manager Captcha](http://iopscience.iop.org/1742-6596/154/1/012038/pdf/1742-6596_154_1_012038.pdf) (titled “Gravitational disturbances in drag-free spacecraft”) which states “s mentioned above, the gravitational field inside the outer spherical shell geometry vanishes whatever the position of the inner proof mass inside the shell. If we consider an infinitely long hollow cylinder the gravitational field is also zero whatever the proof mass position inside the shell. **This is not the case when we consider a hollow cylinder of finite length**.” (bolding mine) ---
**Author:** ![eschereal](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/eschereal/32/18939_2.png) [@eschereal](https://boards.straightdope.com/u/eschereal)\ **Post date:** [December 21, 2013, 1:14am UTC](https://boards.straightdope.com/t/under-what-conditions-is-nivens-ringworld-unstable/676755/52 "2013-12-21T01:14:52Z")
Seems to me the sensible approach to keeping the ring stable would be huge fins at the edges canted toward the axis of spin.the fins would have to be large enough to sail against the solar wind, using its pressure to keep any part of the ring from being pushed inward and would have to act independently for varying wind pressure, because sending a signal to the opposite side would take over ten minutes. ---
**Author:** ![eschereal](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/eschereal/32/18939_2.png) [@eschereal](https://boards.straightdope.com/u/eschereal)\ **Post date:** [December 21, 2013, 1:16am UTC](https://boards.straightdope.com/t/under-what-conditions-is-nivens-ringworld-unstable/676755/53 "2013-12-21T01:16:08Z")
> [@BMalion](#): > > Wanna make a baby? Good luck with that. ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [December 21, 2013, 1:22am UTC](https://boards.straightdope.com/t/under-what-conditions-is-nivens-ringworld-unstable/676755/54 "2013-12-21T01:22:24Z")
> [@standingwave](#): > > But you’re also ignoring all the nearby mass you removed at the same time you removed everything else. I’m not ignoring anything. Think it through: 1. When you have a hollow uniform sphere, the gravitational pull inside the sphere is zero. 2. The sum of the gravitational pull of a Ringworld-sized strip of the sphere, **plus** the gravitational pull of the rest of the sphere is equal to the gravitational pull of the whole sphere. So those two must have equal and opposite gravitational fields inside the sphere. 3. The sphere-minus-Ringworld mass is _all_ above you. So it must pull you upwards. 4. Since the Ringworld gravitational field is equal and opposite to the rest of the sphere, it must pull you downwards. ---
**Author:** ![eburacum45](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/eburacum45/32/8690_2.png) [@eburacum45](https://boards.straightdope.com/u/eburacum45)\ **Post date:** [December 21, 2013, 4:36pm UTC](https://boards.straightdope.com/t/under-what-conditions-is-nivens-ringworld-unstable/676755/55 "2013-12-21T16:36:34Z")
The gravity of the nearby sections of the Ringworld act like an infinite plane - the edges are half a million kilometers away at the centre. The gravity of the far side is negligible with respect to anyone standing on that near-infinite plain. However the total mass of the Ringworld is quite low, comparable to the mass of Neptune stretched out into a ribbon 970,000,000 km long. So you can basically ignore it compared to the spin. ---
**Author:** ![Andy\_L](https://avatars.discourse-cdn.com/v4/letter/a/c67d28/32.png) [@Andy\_L](https://boards.straightdope.com/u/Andy_L)\ **Post date:** [December 21, 2013, 7:05pm UTC](https://boards.straightdope.com/t/under-what-conditions-is-nivens-ringworld-unstable/676755/56 "2013-12-21T19:05:31Z")
> [@ZenBeam](#): > > I’m not ignoring anything. Think it through: > > 1. When you have a hollow uniform sphere, the gravitational pull inside the sphere is zero. > 2. The sum of the gravitational pull of a Ringworld-sized strip of the sphere, **plus** the gravitational pull of the rest of the sphere is equal to the gravitational pull of the whole sphere. So those two must have equal and opposite gravitational fields inside the sphere. > 3. The sphere-minus-Ringworld mass is _all_ above you. So it must pull you upwards. > 4. Since the Ringworld gravitational field is equal and opposite to the rest of the sphere, it must pull you downwards. Here’s the kind of picture I was looking for [Gravitational Field](http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/GravField.htm) - I’m speaking of the illustration just below the heading “Field Inside a Spherical Shell”. This picture shows that in order for the cancellation to working inside a sphere, each spherical cap on the “near” side, there needs to be a larger spherical cap on the “far” side - and so for the Ringworld, that kind of balance can’t work. Another way to think about it. The Shell theorem applies only to shells of uniform density - and the Ringworld can be described as a shell of very variable density - it’s high density in spots, and 0 density in other spots… ---
**Author:** ![md2000](https://avatars.discourse-cdn.com/v4/letter/m/73ab20/32.png) [@md2000](https://boards.straightdope.com/u/md2000)\ **Post date:** [December 21, 2013, 8:01pm UTC](https://boards.straightdope.com/t/under-what-conditions-is-nivens-ringworld-unstable/676755/57 "2013-12-21T20:01:40Z")
No, I’ll correct myself and agree Inside an infinite hollow cylinder, like a hollow sphere, net gravity is zero anywhere. If you have a finite cylinder, the symmetry argument disappears. Logically there is Zero gravity, by symmetry, along the plane defined by the disc described by inside centerline of the ring. Off the center circumference line disc plane (but within the cylinder), the gravity would tend to pull you back to the middle? That center circumference line plane? As there is more mass on the side toward the centerline, not balanced by the less than half of cylinder on the side of you away from the centerline… ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [December 21, 2013, 9:06pm UTC](https://boards.straightdope.com/t/under-what-conditions-is-nivens-ringworld-unstable/676755/58 "2013-12-21T21:06:49Z")
For a finite cylinder, take its axis to be the Z-axis, and take it to be centered in the X-Y plane. If you are on the Z-axis, there won’t be any force in the X or Y direction, by symmetry. If you are on the X-Y plane, there won’t be force in the Z direction, by symmetry. The only place there will be zero total force by symmetry is at the origin. in general, on the X-Y plane you can have X and Y forces. And in general, on the Z-axis, you can hzve force in the Z-direction. [Previous page](https://boards.straightdope.com/t/under-what-conditions-is-nivens-ringworld-unstable/676755.md?page=2)