
#1
Feb313, 08:19 PM

P: 2

First time poster so I apologize any problems with location or content. By the way, you guys are awesome.
Just some context: last semester I had a theoretical mechanics class and the teacher said, while teaching us about the Lagrangian formulation, that particles cannot, ever, reach a potential's extremum and stay there (zero velocity). I was very curious at the time but never actually thought about tackling the problem. So today it popped in my head while cooking lunch. I went at the problem in a unidimensional way, starting by defining a general potential with an extremum, approximating it, at that point, by a Taylor series to terms of second order and seeing what it meant in terms of equations of motion via the Lagrangian formulation: [itex]{\frac{dV}{dx}_a=0}[/itex] [itex]{V(x)\approx{V(a)+\frac{dV}{dx}_a (xa)+\frac{1}{2}\frac{d^2 V}{dx^2}_ a (xa)^2}}[/itex] [itex]{L=\frac{m\dot{x}^2}{2}\frac{1}{2}\frac{d^2 V}{dx^2}_ a (xa)^2}[/itex] (the first and second terms on the right hand side of the potential are, respectively, null and a constant that does not influence the equation of motion) This leads to, using the EulerLagrange equations, to the following equation of motion: [itex]{m\ddot{x}+\xi (xa)=0}[/itex] This leads to: [itex]{x=Ae^{i\alpha t}+Be^{i\alpha t}+a}[/itex] Applying the conditions of zero velocity at point "a" at a certain time one gets: [itex]{x=a}[/itex] This shows that the only possible way for a particle to exist in a potential extremum with zero velocity is if it is "put" there with zero velocity. I now want to tackle the problem of how much time it takes for the particle to reach such an extrema. Any ideas? Thanks in advance! 



#2
Feb413, 03:14 AM

Sci Advisor
Thanks
P: 2,130

Of course you are right. At stationary points of the potential, i.e, where [itex]\partial_i V(q)=0[/itex], there is always a solution, where the particle stays at this place at rest. I don't know what your teacher meant by telling the opposite.
Of course, another question is the stability of this stationary solution. The stationary solution is only stable against small perturbations when there is a local minimum at the stationary point, i.e., the Hesse matrix of the potential, [itex]\partial_i \partial_j V(q)[/itex], must be positive definite at the point at the stationary point. If its negative definite, it's unstable against small perturbations in any direction. If it's indefinite, you have a saddle point, and then the solution is stable for perturbations in some directions and unstable for such in other directions. Perhaps your teacher has made a remark on quantum theory, where the Heisenberg uncertainty relation tells you that there are necessarily always fluctuations around the classical stationary solution, but that's another story. 



#3
Feb413, 07:15 AM

P: 2

The analogy of a ball being put on a hilltop or in a valley certainly applies here. I think what he meant is that such solution never exists in nature, not even classically. On a hilltop it's easy to see why (it's unstable) but in a valley small perturbations will always produce some tiny oscillation that destroys our stationary solution.



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