I Why do phase trajectories point upwards and downwards in a quadratic potential?

AI Thread Summary
In a quadratic potential, phase trajectories can point both upwards and downwards due to the nature of the potential's stability. The left side of the phase trajectory indicates downward motion, reflecting stable conditions, while the right side shows upward motion, which corresponds to unstable conditions where particles accelerate away from the potential minimum. As a test particle moves in this unstable region, its velocity increases, leading to a trajectory that points upwards despite the overall downward acceleration towards the potential. This behavior illustrates the complex dynamics of phase space in relation to potential energy landscapes. Understanding these trajectories is crucial for analyzing motion in phase space effectively.
Lambda96
Messages
233
Reaction score
77
Hi,

I am currently preparing for my exam and have just watched a video about motion in phase space.



From minute 4 a quadratic potential is introduced and then from minute 6 minute the phase trajectory.

Here are the pictures

quadratic potential
Bildschirmfoto 2023-02-02 um 12.41.54.png


phase trajectory
Bildschirmfoto 2023-02-02 um 12.42.28.png


Regarding phase trajectory on the left side, I understand that these are pointing downwards, but I don't understand why the phase trajectory on the right side is pointing upwards, I would say that these must also be pointing downwards, as the potential is unstable and no matter where I place a test particle, it will always accelerate downwards towards the potential.
 
Physics news on Phys.org
Take a point on the curve with v0<0, x0>a. At the next instant in time, 0>v1>v0 and a<x1<x0 so the curve is ”pointed“ upwards.
 
Hello! Let's say I have a cavity resonant at 10 GHz with a Q factor of 1000. Given the Lorentzian shape of the cavity, I can also drive the cavity at, say 100 MHz. Of course the response will be very very weak, but non-zero given that the Loretzian shape never really reaches zero. I am trying to understand how are the magnetic and electric field distributions of the field at 100 MHz relative to the ones at 10 GHz? In particular, if inside the cavity I have some structure, such as 2 plates...
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...
Back
Top