Lagrangian for the Kepler Problem

[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

 

[mfb]

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)

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 $r=|x|$.

 

[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: http://ocw.mit.edu/courses/physics/8-01sc-physics-i-classical-mechanics-fall-2010/central-force-motion/central-force-motion-and-the-kepler-problem/MIT8_01SC_coursenotes28.pdf

 

[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?

 

[voko]

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?

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.

 

[Astrum]

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?

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.

 

[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.

 

[Philosophaie]

Thanx