Register to reply

Effective/Centrifugal Potential

by mishima
Tags: potential
Share this thread:
mishima
#1
May1-12, 09:37 PM
P: 303
Hi, I'm trying to better understand these terms as they relate to central force motion. Am I correct in that the centrifugal potential could be considered the "angular kinetic energy" expressed in terms of radius? Then since potential only depends on radius its considered part of the effective potential? Does the negative gradient of these 2 terms correspond to forces?

Is this easier to understand using Lagrangian mechanics? Thanks.
Phys.Org News Partner Physics news on Phys.org
Physicists discuss quantum pigeonhole principle
First in-situ images of void collapse in explosives
The first supercomputer simulations of 'spin?orbit' forces between neutrons and protons in an atomic nucleus
K^2
#2
May1-12, 11:58 PM
Sci Advisor
P: 2,470
No, that's not what effective potential means. You get effective potentials when you go into a accelerating reference frame. In case of central potentials, it's usually rotating frame, but it's more general than that. Central potential is most commonly used when talking about orbital motion. So lets look at such a case. Imagine a body orbiting in central potential along some elliptic path. In general, you have r(t) and θ(t). Now, lets go into a frame that rotates at the same rate as the orbiting body. In that frame, θ'(t)=0 and r'(t)=r(t). So now we only need one coordinate to describe the motion.

What forces act on the body for r(t) to vary? There is still the gravitational force, Fg=-GMm/r, but because we are now in a rotating frame, there are also the centrifugal and Coreolis forces. The later is irrelevant, since it's along θ. That gives us the total force acting on the body F=ωr-GMm/r. But ω can depend on r, so that needs to be looked after. Fortunately, we know that angular momentum L=mωr is a constant, so ω=L/(mr), which lets you rewrite formula for force. F=L/(mr)-GMm/r. This force does happen to be conservative, which means there is a potential that satisfies F=-dU/dr. Usually, you also take U→0 as r→∞. A potential that satisfies both is Ueff=L/(2mr)-GMm/r. That's your effective potential.

Basically, it's the potential that allows motion in an accelerated frame of reference as if it was an inertial frame of reference. Specifically, in this case, it's a potential in which body would move in one dimension, along r, the way a body is supposed to move when in orbit.

Edit: It arises a little more naturally in Lagrangian mechanics, but it's the same exact principle, and you'd use roughly the same kind of logic to derive it. Unfortunately, in my experience, both high school and university mechanics courses are really inadequate in covering accelerated reference frames. This is just an example of such inadequacy. You should have better foundation in accelerated frames before you dive into orbital mechanics using effective potentials.
mishima
#3
May2-12, 12:38 AM
P: 303
Thanks, you're right that my book didn't mention a rotating frame in its derivation. It used only the conservation of angular momentum and energy equation, algebraically eliminating the theta dot dependency.

It's much clearer considering θ'(t)=0 and the fictitious forces involved though, thank you.


Register to reply

Related Discussions
Why do we not call the effective potential(QFT) being density of effective action? Quantum Physics 1
Effective potential Classical Physics 2
Effective potential and Uehling potential Quantum Physics 0
Effective Potential Advanced Physics Homework 1
Effective potential Introductory Physics Homework 2