In Classical Mechanics, both Goldstein and Taylor (authors of different books with the same title) talk about the centrifugal force term when solving the Euler-Lagrange equation for the two body problem, and I'm a little confused about what it exactly means - is it a real centrifugal force or a mathematical consequence of using polar coordinates for solving the Euler-Lagrange equation.

Their derivations of the Lagrangian $$L=\frac{1}{2}\mu(\dot{r}^{2}+r^{2}\dot{\theta}^{2})-U(r)$$ would lead to one motion of equation (theta) showing that angular momentum is constant and one radial equation of motion shown as $$\mu\ddot{r}=-\frac{dU}{dr}+\mu r\dot{\phi}^{2}=-\frac{dU}{dr}+F_{cf}.$$ They call $\mu r\dot{\phi}^{2}$ the fictitious force or the centrifugal force. I'm quite hazy on my memory of non-inertial frames, but I was under the assumption that fictitious forces only appear in non-inertial frames. The frame of reference in the two body problem was chosen such that the Center of Mass of the two bodies would be the origin so that would be an inertial frame, and I'm assuming that there are no non-inertial frames involved since neither author had talked about it in the previous chapters.

Would calling $\mu r\dot{\phi}^{2}$ an actual centrifugal force be incorrect then? Isn't it a term that describes the velocity perpendicular to the radius? From this two-body problem, it appears as though if I were to use polar coordinates when solving the Euler-Lagrange equations for any other problem, the centrifugal force term will always appear, so it would be a mathematical consequence of the choice of coordinate system rather than it being an actual fictitious force. Is that term being called a centrifugal force because it actually is a centrifugal force or is it because it has a mathematical form similar to it?


2 Answers 2


There are two equivalent descriptions$^1$ of the reduced two-body problem with a central potential $V(r)$:

  1. In an inertial frame with no fictitious forces: Here $\frac{1}{2}\mu r^{2}\dot{\theta}^{2}$ is the angular part of the kinetic energy.

  2. In a rotating frame following the reduced particle with fictitious forces and only 1D radial kinematics: Here $-\frac{1}{2}\mu r^{2}\dot{\theta}^{2}$ is the centrifugal potential (notice the minus sign!).

Each description leads to the same Lagrangian


$^1$ The reduced particle is confined to an orbit plane, so the problem is effectively just two-dimensional described by two coordinates $r$ and $\theta$. In both descriptions, the center-of-mass serves as the origin of the reference frame. The motion of the center-of-mass itself is trivial, as there is no external forces.

  • $\begingroup$ I see what happened. So there are two interpretations. The first is kinetic energy when deriving using polar coordinates. The second arises when changing to a 1-D problem. For this there are more than one way: in Goldstein one involves integration that is a hassle (Goldstein 3rd ed p.75) and the other is to consider it as a 1-D fictitious problem (p.76-77), which is easier. This fictitious problem changes the interpretation of $\frac{1}{2}\mu r^{2}\dot{\theta}^{2}$ from kinetic energy to potential energy contributed by potential energy. I guess this is mostly reiterating what you've said. $\endgroup$ Sep 22, 2012 at 1:18
  • $\begingroup$ @Qmechanic♦ Why did you say that the center of mass is the origin of the reference frame? If you use the reduced mass $\mu$, should't one of the two bodies be the origin of the system? $\endgroup$
    – Sørën
    Apr 9, 2016 at 20:05
  • 1
    $\begingroup$ $\uparrow$ @Sørën: No. (Note that the formulation of Kepler's first law on Wikipedia treats the ratio $m/M_{\odot}$ as zero, thereby ignoring the motion of the Sun, which is possibly confusing in above context.) $\endgroup$
    – Qmechanic
    Apr 9, 2016 at 20:46
  • $\begingroup$ @Qmechanic♦ So actually the center of mass is coincident with $M_{\odot}$ ? $\endgroup$
    – Sørën
    Apr 9, 2016 at 21:00
  • 2
    $\begingroup$ $\uparrow$ @Sørën: No, only to a very good approximation. $\endgroup$
    – Qmechanic
    Apr 9, 2016 at 21:25

I think you skipped a step...

Once you show that angular momentum $l$ is a constant of the motion you can eliminate $\dot{\theta}$ from the two-dimensional equation of motion and arrive at an equation in 1 independent variable only ($r$): $$ \mu \ddot{r} - \frac{l^2}{\mu r^3} = -\frac{dU}{dr}$$ If you think about it, this new 1-dimensional space is in fact a non-inertial frame (if it were inertial the mass would fall to $r=0$), so it's no surprise to see a new force term in addition to the original potential gradient.

Goldstein then interprets the new term as "the familiar centrifugal force" by re-writing it: $$ \frac{l^2}{\mu r^3} = \mu r \dot{\theta}^2 = \frac{\mu v_\theta^2}{r} $$

And it's true: that new force is indeed one of the non-inertial forces (the centrifugal) that arise if one analyzes the problem in a rotating frame with angular velocity $\omega=\dot{\theta}$.

Aside: The other non-inertial forces in a rotating frame arise from time-varying quantities. It can be shown that the entire non-inertial potential is a velocity-dependent one analogous to that of electromagnetism: $$U_{rot} = \mu \left\{ \left[ - \frac{\left( \boldsymbol{\omega \times r} \right)^2}{2} \right] - \boldsymbol{ \dot{r} \cdot \left( \boldsymbol{\omega \times r} \right) } \right\}$$

The first term gives the centrifugal force, the second the Coriolis and Euler forces (if $\boldsymbol{\omega}$ is non-constant).

  • $\begingroup$ What a coincidence! I finally replied and it was the same time as you did. I didn't think about the inertial frame and mass falling to $r = 0$. More food for thought. Thanks! $\endgroup$ Sep 22, 2012 at 1:29
  • $\begingroup$ @user1604449: yes, it looks like you got to the same place... $\endgroup$
    – Art Brown
    Sep 22, 2012 at 1:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.