# Lagrangian for the Kepler Problem

1. Aug 26, 2013

### Philosophaie

In a two-body solution ( Kepler Problem) how do you find the Lagrangian, L, if the position vector is:

$$x = (v_x * t + x_0, v_y * t + y_0, v_z * t + z_0)$$

Action from Wikipedia is:

$$S = \int_{t_1}^{t_2} L * dt$$

Last edited: Aug 26, 2013
2. Aug 26, 2013

### Staff: Mentor

That is certainly not true for all t, as this would mean linear motion (no acceleration).

The Lagrangian can be found similar to all other Lagrangians for a particle moving in a potential, the potential is GMm/r where $r=|x|$.

3. Aug 26, 2013

### Astrum

If the central mass is large enough, the Lagrangian is

$$\mathscr{L} = \frac{1}{2}m(\dot{r}+ r \dot{\theta})^{2} - U(r)$$

Plugging it into the Euler-Lagrange equation:

$$\frac{\partial \mathscr{L}}{\partial q} = \frac{d}{dt}\frac{\partial \mathscr{L}}{\partial \dot{q}}$$

For the radial component, we have $m \ddot{r} = mr \dot{\theta}^{2}-U'(r)$. The $\theta$ component gives us $0$, because $\frac{\partial \mathscr{L}}{\partial \theta} = 0 = \frac{d}{dt}(mr^2 \dot{\theta})$, we can call $mr^2 \dot{\theta} = L$

At this point, you can pretty much solve the equations of motion with respect to time to get the position.

I hope that wasn't too convoluted of an answer.

Edit: Here's a link you might be interested in reading: MIT - Kepler Problem

Last edited: Aug 26, 2013
4. Aug 26, 2013

### Philosophaie

I found a solution for Classical Mechanics

http://www.wolframalpha.com/input/?i=kepler+problem

The Kinetic Energy came out to be $$\frac {1}{2} (\frac{dx}{dt})^2$$
The Potential Energy came out to be $$\frac {\lambda}{\frac {dx}{dt}}$$

Not sure what lambda is. Can someone explain?

5. Aug 26, 2013

### voko

None of that is correct. First, Kepler's problem is usually set in at least 2 dimensions. Second, the potential energy does not depend on velocity; it depends on displacement.

Last edited: Aug 26, 2013
6. Aug 26, 2013

### Astrum

I've got no idea what you're doing.

Look, the Lagrangian IS the kinetic minus the potential. $\mathscr{L} = KE - U$

In the Kepler Problem, where $M >> m$ $$\mathscr{L} = \frac{1}{2}m(\dot{r}+ r \dot{\theta})^{2} - \frac{GMm}{r}$$

As Voko says, this can be reduced to a one dimensional problem by playing noting that ---->

$$\frac{\partial \mathscr{L}}{\partial \theta} = 0 = \frac{d}{dt}(mr^2 \dot{\theta})$$ where we can set $mr^2 \dot{\theta} = L$ and play with the equations from here.

What EXACTLY are you trying to figure out? State your question CLEARLY, because I don't have a clue what you're asking for.

7. Aug 27, 2013

### vanhees71

The potential is spherically symmetric and thus angular momentum is conserved. The trajectory of the planet thus is in a plane perpendicular to the angular momentum and thus in a plane.

The kinetic energy is
$$T=\frac{m}{2} \dot{\vec{x}}^2.$$
Using polar coordinates in the plane perpendicular to angular momentum
$$\vec{x}=r \begin{pmatrix} \cos \varphi \\ \sin \varphi \end{pmatrix},$$
you get
$$T=\frac{m}{2} (\dot{r}^2+ r^2 \dot{\varphi}^2).$$
The Lagrangian thus is
$$L=T-V=\frac{m}{2} (\dot{r}^2+ r^2 \dot{\varphi}^2 )+ \frac{G mM}{r}.$$
Now you see that $\varphi$ is cyclic and thus
$$J=\frac{\partial L}{\partial \dot{\varphi}}=m r^2 \dot{\varphi}=\text{const}.$$
This is the conservation of angular momentum in this special frame of reference.

Rewriting the equation of motion for $r$ using this first integral as a function of $\varphi$ and doing the substitution $s=1/r$ you get an easy to solve differential equation, leading to ellipses, parabolas, or hyperbolas as trajectories.

The case $J=0$ has to be discussed separately.

8. Aug 27, 2013

### Philosophaie

Thanx

Last edited: Aug 27, 2013