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Finding the Lagrangian for an elastic collision

  1. May 26, 2015 #1
    1. The problem statement, all variables and given/known data
    a. Suppose two particles with mass $m$ and coordinates $x_1$, $x_2$ collides elastically, find the lagrangian and prove that the linear momentum is preserved.
    b. Find new coordiantes (and lagrangian) s.t. the linear momentum is conjugate to the cyclical coordinate.

    2. Relevant equations

    3. The attempt at a solution
    For (a) I thought that $L=m(\dot{x_1}^2+\dot{x_2}^2)/2$ but I don't know how to show the preservating of momentum.
    For (b) I thought taking the new coordnates to be the center of mass and the distance of each particle from it but somehow I can't formulate it.

    Can anybody assist with proving the momentum preservation and its conjugation in (b) for cyclical coordinates?
  2. jcsd
  3. May 27, 2015 #2


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    The question uses the word "preserved" in an odd context. It seems to mean that total linear momentum is conserved.

    To include the collision you will need an interaction of some kind. I am hoping that the context of this question will provide you some guidance as to what is expected. An elastic collision means a conservative potential of some kind. This is "advanced physics homework" so it is probably not too much to expect you to know what that means with regard to a potential. So, what does it mean for a collision to be elastic, and what does a potential have to be such that it will produce an elastic collision?

    Once you have that you show that total linear momentum is conserved by showing it is an invariant. And you do that by showing its bracket with the Hamiltonian is zero.

    For part b) you need to remember what it means for the momentum to be conjugate. The question seems to be worded in slightly sloppy terms, possibly deliberately so. I think they are trying to get you to note that the conjugate variable and the momentum are not necessarily the same thing in different coordinate systems.
  4. May 29, 2015 #3
    May can use: ## \int L\,dt = 0 ##
  5. May 29, 2015 #4


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    Except that you forgot a bit. You forgot a ## \delta ## in front of that. Then you get ## \delta \int L\,dt = 0 ## which just gets you the Lagrange equation of motion. The integral you wrote is not usually zero.
  6. Jul 20, 2017 #5
    I would consider the two masses to be connected by a spring of infinite spring constant at the point of collision.
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