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1

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 }{ ...


0

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 ...


0

The sign of your interaction term is wrong. The energy is not bounded from below. You can not minimise the energy to find the most stable state. The energy is minimised by $|A| \to \infty$. If you change the sign of the non-linear term in your equation you get a similar linearised problem although with the replacement $a_o^2 \to -a_o^2$. Then your ...


-1

Actually with the lever sticking up above the balance scale shown; balance axis point; it adds gravitational weight to whichever side is initially positioned downward, and also since this side is down, it should weigh more being closer to the ground... and should tend to keep this side down... However is doesn't... If you put a weight, on some point of a ...


5

Fun question. The muon density inside a white dwarf is negligible, because Fermi suppression does not really apply. Fermi suppression is the technical name of the effect you were describing: the decrease in the rate of a process due to the fact that there are no free states to accommodate one of the decaying particles (an electron, in this case). The ...


1

The answer by Nathaniel is clear and correct, yet let me add: it is obvious by inspection that this is motion in a potential, which can be derived from a Hamiltonian $$ \frac{\dot r^2+ r^2\dot\theta^2}{2} + V(r,\theta)\;. $$ All you need to do now is to establish whether, around the equilibrium points, $V$ has a minimum or a maximum/saddle: in the first case ...


3

I take the core of the question to be Is it possible to do linear stability analysis on 2nd order differential equations by finding eigen values of Jacobian matrix? The answer is yes, but first you have to convert your second-order equations into first-order ones. This is actually pretty easy to do: every time you see a second derivative, e.g. ...


0

You need to linearize your equations around any stationary point. Then you can indeed treat it like by finding eigen values.


0

I suggest you to read http://arxiv.org/pdf/1408.0287.pdf for a nice discussion on the RG dependence of the effective potentials and the RG invariant statements that one can make. About your point, I think you are simply misunderstanding what people mean: they are calculating at what point in $\Phi$, not in $\mu$, the potential cross zero.



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