Radial Equation for Two-Body Central Forces

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SUMMARY

The discussion centers on deriving the radial equations for two-body central forces using the Lagrangian mechanics framework. Two methods yield different forms of the radial equation, specifically regarding the sign of the centrifugal term. Method 1, which substitutes the angular momentum into the Lagrangian, results in the equation μẋ = -l²/(μr³) - dU/dr. In contrast, Method 2 leads to μẋ = l²/(μr³) - dU/dr. The inconsistency arises from the treatment of constants during differentiation, emphasizing the importance of holding φ constant when deriving the radial equation.

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  • Understanding of Lagrangian mechanics
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I'm getting two different radial equations depending on when I plug in the angular momentum piece. Here's the Lagrangian:

L = \frac{1}{2} \mu (\dot{r}^2 + r^2 \dot{\phi}^2) - U(r)

The Euler-Lagrange equation for phi gives angular momentum (conserved), which can be solved for \dot{\phi}:

\dot{\phi} = \frac{l}{\mu r^2}

Now, let's find the radial equation (that is, the Euler-Lagrange equation for r):

Method 1: Substitute angular momentum piece into Lagrangian, then find the Euler-Lagrange equation for r.

L = \frac{1}{2} \mu \dot{r}^2 + \frac{l^2}{2 \mu r^2} - U(r)

\mu \ddot{r} = \frac{-l^2}{\mu r^3} - \frac{dU}{dr}

Method 2: Find the Euler-Lagrange equation for r, then substitute angular momentum piece into the radial equation.

\mu \ddot{r} = \mu r \dot{\phi}^2 - \frac{dU}{dr}

\mu \ddot{r} = \frac{l^2}{\mu r^3} - \frac{dU}{dr}

These two radial equations have opposite signs for the "centrifugal term." Which is correct, and why is the other wrong?
 
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Remember that when you solved for \dot{\phi}, you held r constant. Similarly, \phi should be held constant when you solve for the radial equation.
 

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