Understanding Equilibrium Stability in Classical Mechanics: Virtual Work Lecture

AI Thread Summary
In classical mechanics, the stability of equilibrium can be assessed using the second derivative of potential energy, U. If the second derivative, ∂²U/∂x², is positive, the potential is at a minimum, indicating stable equilibrium. Conversely, if it is negative, the potential is at a maximum, signifying unstable equilibrium. Objects in a potential field tend to move towards lower potential, aligning with the direction of force. Understanding this relationship is crucial for analyzing equilibrium stability in mechanical systems.
gulsen
Messages
215
Reaction score
0
In classical mechanics - virtual work lecture, for determining equilibrium stability we were told that second derivate of potential can be used. How?

I've made a quick google search, but couldn't find anything remarkable.
 
Physics news on Phys.org
If \frac{\partial^2U}{\partial x^2} is postive then U is at a minimum (basic calculus - second derivative test). If instead it is negative then U is at a maximum. A point of equilibrium is stable if U is minimum and unstable if U is maximum.
 
Euclid said:
A point of equilibrium is stable if U is minimum and unstable if U is maximum.

I guess this because objects in a potential field tend move through where their potentials get lower?
 
Absolutely! After all, which way does the force point?
 
Thread 'Gauss' law seems to imply instantaneous electric field'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Back
Top