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Euler-Lagrange Equations for Two Body Problem

  1. Dec 13, 2014 #1
    1. The problem statement, all variables and given/known data
    I'm trying to do a little review of Lagrangian Mechanics through studying the two-body problem for a radial force. I have the Lagrangian of the system [tex] L=\frac{1}{2}m_1\dot{\vec{r_1}}^{2}+\frac{1}{2}m_2\dot{\vec{r_2}}^{2}-V(|{\vec{r_1}-\vec{r_2}}|) [/tex]. Now I'm trying to find the Euler-Lagrange Equations for [itex] r_1 [/itex] and [itex] r_2 [/itex] but I'm confused about taking the derivative of the potential portion with respect to either [itex] r_1 [/itex] or [itex] r_2 [/itex]. Please call me stupid and then tell me why I'm being stupid here.

    2. Relevant equations
    [tex] L=\frac{1}{2}m_1\dot{\vec{r_1}}^{2}+\frac{1}{2}m_2\dot{\vec{r_2}}^{2}-V(|{\vec{r_1}-\vec{r_2}}|) [/tex]

    [tex] \frac{dL}{dq}=\frac{d}{dt}\frac{dL}{d\dot{q}} [/tex]




    3. The attempt at a solution
    [tex] \frac{\partial L}{\partial r_1}=-\frac{\partial V(|{\vec{r_1}-\vec{r_2}}|)}{\partial r_1}=...? [/tex]
     
  2. jcsd
  3. Dec 13, 2014 #2

    ShayanJ

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    Gold Member

    You're a stupid because you're doing it in terms of [itex] \vec r_1 [/itex] and [itex] \vec r_2 [/itex]. Change coordinates to [itex] \vec R=\frac{m_1\vec r_1+m_2\vec r_2}{m_1+m_2} [/itex] and [itex] \vec r=\vec r_1-\vec r_2 [/itex](centre of mass coordinates) and you'll see that there is only one variable here!
     
  4. Dec 13, 2014 #3
    In central force problems, it is useful to instead use the coordinates:
    [tex]
    \begin{align}
    \vec{r} &= \vec{r}_2-\vec{r}_1\\
    \vec{R} &= \frac{m_1\vec{r}_1 + m_2\vec{r}_2}{M}
    \end{align}
    [/tex]
    The reason is that with these coordinates the Lagrangian reduces to something that looks like a single body with a "reduced mass" [itex]\mu[/itex]:
    [tex]
    L=\frac{1}{2}\underbrace{\frac{m_1m_2}{m_1+m_2}}_{\mu}\dot{\vec{r}}^2 + \underbrace{\frac{1}{2}M\dot{\vec{R}}^2}_{0} - U(\vec{r}) \ ,
    [/tex]
     
  5. Dec 13, 2014 #4
    I'm not switching to the easier coordinates just yet, although I do know they simplify things greatly. What would the Euler-Lagrange Equations be if I remained with the non-reduced coordinates for right now?
     
  6. Dec 13, 2014 #5

    ShayanJ

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    Gold Member

    In contrast to what you may think, its not only a bit harder, its much harder. Because [itex] |\vec r_1-\vec r_2| [/itex] depends on all six components of the position vectors and so you should consider all the components in both the kinetic and potential terms. Just expand the position vectors w.r.t. their Cartesian components. So you'll have 6 EL equations.
     
  7. Dec 14, 2014 #6
    This is not entirely correct. ##\vec r_1## was a the position of a body. What is ##r_1##? Usually, that denotes the magnitude of the corresponding vector. You could use this as one of the generalized coordinates, but then you would need two other for the body, and what would those be?

    I interpret your messages as if you really want to use Cartesian coordinates, in which case it is not ##r_1## that you should be using. It is ##r_{11}, \ r_{12}, \ r_{13} ##. Then you need to compute, for example, $$ \partial V(|\vec r_1 - \vec r_2|) \over \partial r_{11} . $$ The question is how?

    Observe that ##V(s)## is a scalar function of one scalar argument, so it has some derivative ##V'(s)##. Then you could use the chain rule and obtain $$ {\partial V(|\vec r_1 - \vec r_2|) \over \partial r_{11} } = V'(|\vec r_1 - \vec r_2|) {\partial |\vec r_1 - \vec r_2| \over \partial r_{11} }. $$

    Can you continue from here?
     
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