Norman Albers Posted September 23, 2007 Posted September 23, 2007 I argue that acceleration is not what is being figured here. Consider that the rim velocity is [math]\omega r[/math] but that acceleration is [math]\omega^2 r[/math]. I may keep the first product constant but work with a larger and larger wheel to reduce the acceleration to an arbitrarily small value. You'd like to think that acceleration is equivalent to gravitation and so we are doing general relativity, but this is not the case. I would like to know the answer to characterizing a smaller and smaller wheel... Running some ballpark numbers on the Wheel of Doom, what sort of rim velocity, or [math]\gamma[/math] would it take to get near creating a black hole from the increasing relativistic mass at fixed (you hope) radius? Schwarzschild radius is [math]m=GM/c^2[/math] where 'm' has units of length and 'M' units of mass. 'G' is the grav. constant. We are, in this circumstance, realizing larger and larger [math]M=\gamma M_o[/math]. We ask how much relativistic mass is necessary to reach BH intensity at our chosen radius and original construction mass. The factor [math]c^2/G[/math] is about [math]10^{27}[/math], so I can write: [math]\gamma M_0 = 10^{27} R[/math]. If our radius is a kilometer and the rim mass is a million kilograms, then solve: [math]\gamma \times 10^6 = 10^{27} \times 10^3 [/math], or [math]\gamma = 10^{24}[/math], rather high. We don't have to go here to see strongly relativistic effects, like a [math]\gamma=10[/math].
alan2here Posted September 24, 2007 Posted September 24, 2007 so, how fast in rpm to see odd stuff hapening?
Norman Albers Posted September 24, 2007 Posted September 24, 2007 Fast enough so [math]\omega r [/math] approaches c.
alan2here Posted September 24, 2007 Posted September 24, 2007 Any stuff done like this with really big flywheels?
losfomot Posted September 24, 2007 Author Posted September 24, 2007 Any stuff done like this with really big flywheels? I don't think so... the problem in doing it experimentally is the centrifugal force generated... For a 10 meter diameter Flywheel... just to get the rim to a speed of 300km/sec (or 0.1% of c) would require the wheel to turn at a rate of 572,958 RPM (or 9549 RPS) This would generate an acceleration, at the rim, of about 18 million km/s/s... I don't think there will ever be a material that can withstand that kind of force. Just for the heck of it... For a 2km diameter flywheel... just to get the rim to a speed of 300 km/sec (or 0.1% of c) would require the wheel to turn at a rate of about 2865 RPM (or 48 RPS). This would generate an acceleration, at the rim, of about 90,000 km/sec/sec... there might be a material that can withstand that kind of acceleration, so let's take this one a bit further... Let's say our 2km diameter flywheel was constructed on top of a scale that showed a base weight (flywheel not moving) of 1,000,000 kg (probably an unrealistically small number for this size of flywheel, but...). If we got that wheel up to 2865 RPM (and the wheel managed to hold together against the tremendous centrifugal force) our scale should now read 1,000,001 kg. Yep, 1 kg is all we would gain from our 2km wheel whipping around almost 48 times every second. Length would also only contract by 0.0001%. And Time would also only dilate by .0001% I hope my #s are right, it would have been quite a mess if I had tried to show my work as I did it... I'm sure Norm or someone will correct me if I am wrong... The point is, it would be very hard/expensive to construct an apparatus to test this. EDIT- Actually my 1 kg figure is even too much, since the mass increases to .0001% the closer you get to the rim, the mass of the flywheel, as a whole, would increase much less than 0.0001%.
Norman Albers Posted September 24, 2007 Posted September 24, 2007 OK I'll bite on talking about distance of circumference. It will be measured as longer than [math]2\pi r[/math] by someone on a relativistic wheel. This seems to say to me that tidal stress has torn things apart, unless we built with elastic materials...Thwwppp.
alan2here Posted September 24, 2007 Posted September 24, 2007 That disappointingly tricky. Thanks for working it out.
Norman Albers Posted September 24, 2007 Posted September 24, 2007 Disappointing? Sounds like great material, to me. Give the wheel some sort of Maxwell's demon so it keeps pulling energy from space... Seriously, I am not sure of the implications and want other people's comments. Then again, check out neutron stars, or even the electron as I model it. You really have to come up with some attitude to create strong circulations.
Spyman Posted September 25, 2007 Posted September 25, 2007 The first part of this is true- since the radius is perpendicular to the direction of motion, the radius will not change. But you do NOT gety a contradiction" that R= r and R< r. What happens is that C= 2 pi r is no longer true- the geometry is no longer Euclidean. Well, don't you think that an NON-euclidean object sitting in euclidian space would be rather strange ? Absolutely not. If you mean measuring rods moving with the circumference of the disk, from the point of view of a person in the "lab frame", in which the disk is spinning, both circumference and measuring rods, moving at the same rate, will be reduced by same amount and so the same number of rods will cover the circumference. From the point of view of a person ON the circumference of the disk, the neither rods nor circumference will have changed. From the point of view of a person on the disk but closer to the center, both circumference and rods will have reduced by the same amount but less than observed from the laboratory frame- he will see a non-Euclidean geometry but closer to Euclidean geometry than the laboratory frame. Of course all that change in "geometry" is because rotation is an acceleration, producing "fictitious forces". I think you misinterpreted this point, the gamma factor works both ways, it's relative... If the static observer outside the disc, views a ruler riding on the rim as shorter then an moving observer riding on the rim of the disc views a static ruler outside the disc as shorter by the same gamma factor. Now, since the observer riding on the disc knows he is the one accelerating and thus accnowledges that his frame is warped relative the rest frame, he would conclude that his ruler has grown relative the ruler at rest. -------------------- EDIT: Removed the part of Earth as a flywheel, since the calculations I made where terribly wrong. (Hehe, really funny how one can turn the numbers upside-down when hurrying.) EDIT II: Hopefully correct calculations shows only a ~48 micrometers difference of the equatorial circumference. (From the perspective of a spacecraft comoving with Earth, but not orbiting.) -------------------- BIG flywheels in nature, like fast spinning Neutron stars... XTE J1739-285 makes 1122 laps every second, thats a whopping 67 320 rpm. (If it gets confirmed.) http://www.esa.int/esaSC/SEMPADBE8YE_index_0.html Due to it's extremly rapid rotation it's probably very distorted into an oblate spheroid shape. (And it could be close to the limit of it's break-up speed, despite the crushing gravity.) It's thought to be some 10 km across, which would make it a pretty large flywheel. If assuming that the equatorial radius is 5000 meters, it's equatorial circumferens is ~31 416 meters. (It's likely a much greater radius at the equator, but just to get a figure of the Lorentz contraction.) That would make a surface speed of 35 248 752 m/s relative a hovering stationery spacecraft. Gamma is ~1.006985 -> Spacecraft would measure circumference to ~31 198 meters, thats 218 meters shorter.
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