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Lagrangian of three bodies, using an exponential potential

  1. Dec 11, 2012 #1
    What I know:
    L = T - V
    V = e^(x1-x2) + e^(x2-x3) for n = 3 and a mass = 1

    What I believe:
    T = .5Ʃ (x'_i)^2 from 1 to n

    So lets say you have three bodys that can just be considered pints masses of 1 and on the same line:

    x1 x2 x3
    They have an exponential potential. What is L?
  2. jcsd
  3. Dec 11, 2012 #2
    Don't you think you should also consider interaction between 1 and 3.By the way,this is most suited in homework section.
  4. Dec 11, 2012 #3


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    Not necessarily. He said they are points on a line. So the interaction might be pair-wise. Think masses connected by springs, but with a different potential.

    jbay, if you take them to be unit mass, yes, that's correct expression for kinetic energy. So if your potential energy is correct, then you have your Lagrangian.
  5. Dec 11, 2012 #4
    what about something like force field in which there will be an interaction between 1 and 3,but it is not clear from context given.
  6. Dec 11, 2012 #5
    So I do not need to account for the interaction of 1 and 3 in V. Also great now my only problem is what is x'? Can I integrate V to get it since the e"s are basically acceleration?
  7. Dec 11, 2012 #6


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    We don't know if you need to account for interaction between 1 and 3, because you didn't state the problem clearly enough to deduce that. What kind of interaction are you looking at? Is it pair-wise only between neighbors? Is it a global interaction? You need to define that.

    You find your x' after you solve equations of motion. You have the Lagrangian, so you know that your equations of motion will be these.

    [tex]\frac{d}{dt}\frac{\partial L}{\partial \dot{x}_i} - \frac{\partial L}{\partial x_i} = 0[/tex]

    Write it out. It will give you a system of 3 partial differential equations. If they separate out nicely, you might be able to solve them to find x(t).
  8. Dec 13, 2012 #7
    But how would you do the d/dt portion if there is no t.
  9. Dec 13, 2012 #8


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    It's a total derivative, not a partial one.

    [tex]\frac{d}{dt}f(x, \dot{x}) = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial \dot{x}}\frac{d\dot{x}}{dt} = \frac{\partial f}{\partial x}\dot{x} + \frac{\partial f}{\partial \dot{x}}\ddot{x}[/tex]
  10. Dec 13, 2012 #9
    since lagrangian does not contaib explicit time dependence,it means energy is conserved.
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