A Physics Question

[Borodog]

Show that the Ringworld is unstable.

[thylacine]

[ QUOTE ]
Show that the Ringworld is unstable.

[/ QUOTE ]

The what?

[Borodog]

[ QUOTE ]
[ QUOTE ]
Show that the Ringworld is unstable.

[/ QUOTE ]

The what?

[/ QUOTE ]

Google if you must.

[Borodog]

Can someone with mathematica or maple or something similar integrate this puppy for me?

http://i27.photobucket.com/albums/c153/Borodog/eq.gif

[jason1990]

-(4Sqrt[2]((R - x)EllipticE (http://mathworld.wolfram.com/CompleteEllipticIntegraloftheSecondKind.html)[-4Rx/(R - x)^2] - (R + x)EllipticK (http://mathworld.wolfram.com/CompleteEllipticIntegraloftheFirstKind.html)[-4Rx/(R - x)^2]))/((R + x)Sqrt[-2 + R/x + x/R])

[ill rich]

you have to be kidding me.

[Borodog]

Hmm. That's uglier than I'd hoped. Care to approximate it for small values of x/R?

[jason1990]

If a = x/R, then the above expression can be written as

f(a)*4Sqrt[2a]/(1 - a^2),

where

f(a) = (-1 + a)EllipticE[-4a/(-1 + a)^2] + (1 + a)EllipticK[-4a/(-1 + a)^2].

Now,

f'(a) = 2aEllipticK[-4a/(-1 + a)^2]/(-1 + a), and
f''(a) = EllipticK[-4a/(-1 + a)^2]/(-1 + a) - EllipticE[-4a/(-1 + a)^2]/(1 + a).

This gives f(0) = f'(0) = 0 and f''(0) = -Pi. So for small a, f(a) ~= -(Pi/2)a^2, so the integral is approximately

4Sqrt[2]a^(1/2)f(a)
~= -2*Pi*Sqrt[2]a^(5/2)
~= -8.8858 a^(5/2).

[Borodog]

Thanks. Will post an update later.

[MegaloMialo]

[ QUOTE ]
If a = x/R, then the above expression can be written as

f(a)*4Sqrt[2a]/(1 - a^2),

where

f(a) = (-1 + a)EllipticE[-4a/(-1 + a)^2] + (1 + a)EllipticK[-4a/(-1 + a)^2].

Now,

f'(a) = 2aEllipticK[-4a/(-1 + a)^2]/(-1 + a), and
f''(a) = EllipticK[-4a/(-1 + a)^2]/(-1 + a) - EllipticE[-4a/(-1 + a)^2]/(1 + a).

This gives f(0) = f'(0) = 0 and f''(0) = -Pi. So for small a, f(a) ~= -(Pi/2)a^2, so the integral is approximately

4Sqrt[2]a^(1/2)f(a)
~= -2*Pi*Sqrt[2]a^(5/2)
~= -8.8858 a^(5/2).

[/ QUOTE ]

Correct.

[Borodog]

Ok, here's the update I promised.

Unlike the previous questions I've posed, I haven't worked this one out yet (haven't really had the time). However, I did show that the instability is neither due to perturbation of the ring in the plane of rotation, nor perpendicular to it, as both of these produce beautiful Hooke's Laws (thanks jason for the integral and approximation).

Which means that the only thing left is that it must be unstable due to the rotation, which is surprising, as rotation usually acts to stabilize things, not destabilize them.

I will calculate the torque later and see what pops up.

[danny]

Borodog,

I think you're wrong. From the form of the integral you asked about, it appears that you were calculating the restoring force for a small displacement of the ring in the plane of the ring. As we saw, that got somewhat messy.

If instead we calculate the potential energy of the system for a displacement in the plane, we get a simpler integral, with the result:

V = -(4/(x+R))*EllipticK(4xR/(x+R)^2)

Since EllipticK has a minimum at zero, we see that the potential has a maximum at x = 0, and the ringworld is unstable to displacement in the plane of the ring.

[thylacine]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Show that the Ringworld is unstable.

[/ QUOTE ]

The what?

[/ QUOTE ]

Google if you must.

[/ QUOTE ]

Could you post a link please?

[Borodog]

[ QUOTE ]
Borodog,

I think you're wrong. From the form of the integral you asked about, it appears that you were calculating the restoring force for a small displacement of the ring in the plane of the ring. As we saw, that got somewhat messy.

If instead we calculate the potential energy of the system for a displacement in the plane, we get a simpler integral, with the result:

V = -(4/(x+R))*EllipticK(4xR/(x+R)^2)

Since EllipticK has a minimum at zero, we see that the potential has a maximum at x = 0, and the ringworld is unstable to displacement in the plane of the ring.

[/ QUOTE ]

Hmm.

I may have dropped a negative. Bravo on the approach; the potential route is clearly the simpler one.

Let me see if I can't find a negative somewhere.

Edit:

Found it. I displaced the star by a distance x, but then calculated the restoring force on the ring. Duh. Newton's 3rd law: F_1,2 = - F_2,1

You win, and well done.

[Borodog]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Show that the Ringworld is unstable.

[/ QUOTE ]

The what?

[/ QUOTE ]

Google if you must.

[/ QUOTE ]

Could you post a link please?

[/ QUOTE ]

http://en.wikipedia.org/wiki/Ringworld

[Justin A]

So in English, what's the conclusion to all this math you've just done?

[Phil153]

[ QUOTE ]
[ QUOTE ]
Borodog,

I think you're wrong. From the form of the integral you asked about, it appears that you were calculating the restoring force for a small displacement of the ring in the plane of the ring. As we saw, that got somewhat messy.

If instead we calculate the potential energy of the system for a displacement in the plane, we get a simpler integral, with the result:

V = -(4/(x+R))*EllipticK(4xR/(x+R)^2)

Since EllipticK has a minimum at zero, we see that the potential has a maximum at x = 0, and the ringworld is unstable to displacement in the plane of the ring.

[/ QUOTE ]

Hmm.

I may have dropped a negative. Bravo on the approach; the potential route is clearly the simpler one.

Let me see if I can't find a negative somewhere.

Edit:

Found it. I displaced the star by a distance x, but then calculated the restoring force on the ring. Duh. Newton's 3rd law: F_1,2 = - F_2,1

You win, and well done.


[/ QUOTE ]
Thank Christ that's over. All this intellectual maths debating was making me tired. Congrats to Danny for finishing first, and to Borodog for taking it on the chin so gracefully.

[Borodog]

[ QUOTE ]
So in English, what's the conclusion to all this math you've just done?

[/ QUOTE ]

I calculated (with jason's gracious help) the restoring force on the ring for small displacements of the central star, and showed to be of the form of Hooke's Law:

F = -kx

This is just a spring force, and according to it the ring should just wobble back and forth with a period of about a year and a half.

The problem is that I dropped a negative; the displacement x is of the star, the force I caluclated is that applied to the ring. Hence you actually get an equation of the form:

F = +kx

Which is manifestly NOT a restoring force, and for small displacement to the side, the ring actually accelerates off center rather than returning to it.

Danny not only pointed out my sign error, he pointed out that calculating the gravitational potential energy of the system was simpler, and showed that it has a maximum at zero; meaning that the ring is unstable to small displacements.