- #1
Proofrific
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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:
[tex]L = \frac{1}{2} \mu (\dot{r}^2 + r^2 \dot{\phi}^2) - U(r)[/tex]
The Euler-Lagrange equation for phi gives angular momentum (conserved), which can be solved for [tex]\dot{\phi}[/tex]:
[tex] \dot{\phi} = \frac{l}{\mu r^2} [/tex]
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.
[tex]L = \frac{1}{2} \mu \dot{r}^2 + \frac{l^2}{2 \mu r^2} - U(r)[/tex]
[tex] \mu \ddot{r} = \frac{-l^2}{\mu r^3} - \frac{dU}{dr} [/tex]
Method 2: Find the Euler-Lagrange equation for r, then substitute angular momentum piece into the radial equation.
[tex] \mu \ddot{r} = \mu r \dot{\phi}^2 - \frac{dU}{dr} [/tex]
[tex] \mu \ddot{r} = \frac{l^2}{\mu r^3} - \frac{dU}{dr} [/tex]
These two radial equations have opposite signs for the "centrifugal term." Which is correct, and why is the other wrong?
[tex]L = \frac{1}{2} \mu (\dot{r}^2 + r^2 \dot{\phi}^2) - U(r)[/tex]
The Euler-Lagrange equation for phi gives angular momentum (conserved), which can be solved for [tex]\dot{\phi}[/tex]:
[tex] \dot{\phi} = \frac{l}{\mu r^2} [/tex]
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.
[tex]L = \frac{1}{2} \mu \dot{r}^2 + \frac{l^2}{2 \mu r^2} - U(r)[/tex]
[tex] \mu \ddot{r} = \frac{-l^2}{\mu r^3} - \frac{dU}{dr} [/tex]
Method 2: Find the Euler-Lagrange equation for r, then substitute angular momentum piece into the radial equation.
[tex] \mu \ddot{r} = \mu r \dot{\phi}^2 - \frac{dU}{dr} [/tex]
[tex] \mu \ddot{r} = \frac{l^2}{\mu r^3} - \frac{dU}{dr} [/tex]
These two radial equations have opposite signs for the "centrifugal term." Which is correct, and why is the other wrong?