# Tag Info

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Jkej was correct in stating that both systems are unstable as stated. Any shift, no matter how small would cause either system to flip. In order for a system to be stable there has to be a restorative force which would tend to bring the system back to equilibrium. In this case, there is no restorative force. Imagine that both systems are tilted by 1 degree. ...

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Though I agree with the logic, I find I cannot reproduce this quantitatively. I get an electron number density of $1.2 \times 10^{41}$ m$^{-3}$ (is it just a unit thing?) for a Fermi energy of 30MeV. In a carbon white dwarf with 2 mass units per electron, the Fermi energy of the electrons reaches 30MeV at densities of $4\times 10^{14}$ kg/m$^{3}$ - i.e. at ...

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The Schroedinger model does much more than people here give it credit for. It doesn't just solve the problem of why the ground state doesn't radiate. It solves the problem of why and how the excited states radiate, and it does so by using nothing other than Maxwell's equations. In the Schroedinger model, the states that don't radiate are the states with a ...

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One of the problems with Bohr's theory to describe the hydrogen atom, was that the electron orbiting around the nucleus has an acceleration. Therefore it radiates and loses energy, until it would collapse with the nucleus. This is a common error in physicist's history of physics. The problem was not with Bohr's model, but (as Bohr thought) with ...

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the magnetic field lines are re aligning as the electron orbits the nucleus. however, even though the electron is accelerating as it orbits the nucleus, it simply does not have enough energy in its orbital to radiate until it reaches the next quantum amount of energy associated with the next orbital.

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The comments basically answer the question. Another answer is found on HyperPhysics by Rod Nave: While magnetic confinement seeks to extend the time that ions spend close to each other in order to facilitate fusion, the inertial confinement strategy seeks to fuse nuclei so fast that they don't have time to move apart. Directed onto a tiny ...

1

There would be a slight restorative force from the gravitational pull of the atmosphere itself. First, imagine an ocean world with a very small core. If the core was more dense than the ocean it will be "sink" to the center. Similarly, if your "core" was a shell, it would also "sink" to the center. Thus, if you had both a sphere and a shell they would ...

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I think you are missing the positive feedback effects of winds flowing from high to low pressure areas as the shell moves. At best you would have a parachute effect to slow things down.

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Maybe in a perfect world with no thermal or atmospheric disturbance (yes simulation) this could happen. Zeno holds true regarding infinite time, but for all practical purposes the bob would appear to get there in a reasonably short time. In the real world control systems engineers accomplish this feat all the time using feedback to overcome the small ...

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If you initially give to the bob a velocity $\sqrt{4rg}$, it will actually take an infinite time for the bob to reach the top! A little lesser velocity will cause the bob to stop earlier and come back toward the initial point, while a little greater one will take the bob over the top (the motion will continue, with increasing velocity, to the other side). ...

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Since it is a rigid rod, you are probably right. If the rigid rod is replaced with a string, then your teacher would be right, as the velocity at the top must be non zero in order for the string to remain tight and not collapse before reaching the top. In real life scenarios, however, it is nearly impossible to maintain an unstable equilibrium.

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The answer is due to the area-Mach number relation for hydrodynamic shocks. G.B. Whitham has a great book (check out Chapter 8) on all sorts of various waves and has a good discussion of this topic. The idea is that one can define the Mach number as a function of the cross-sectional area of a ray tube. The simple form is:  \frac{ 1 }{ A } \frac{ d A }{ ...

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Well, when we talk of stability of systems, at least for equilibrium systems, we require the free energy to be bounded below and be convex. As the free energy is obtained by a Legendre transformation (which preserves convexity), the energy functional is required to be convex. This essentially allows us to minimize energies to find ground states. Within the ...

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