1-D Lagrangian Mechanics

Consider a simple 1-D system consisting of two identical massive particles (with mass m m ). One is fixed at the origin, and the other is free to move and is located a distance + x +x from the origin.

If each particle only experiences the gravity force from the other particle, the differential equation describing the moving particle's position is:

m x ¨ = G m 2 x 2 \Large{m \ddot{x} = \frac{-G m^{2}}{x^{2}}}

We can derive this expression by formulating the behavior of the system in terms of Lagrangian mechanics . In that case, which part of the Euler equation does the m x ¨ \large{m \ddot{x}} term correspond to?

Notes: The dot notation denotes time-differentiation. The term L L represents the Lagrangian of the system. The term G G is the universal gravitational constant.

L x {\frac{\partial L}{\partial x}} 2 L t 2 {\frac{\partial^{2} L}{\partial t^{2}}} d d x L t {\frac{d}{dx}\frac{\partial L}{\partial t}} d d t L x ˙ {\frac{d}{dt}\frac{\partial L}{\partial \dot{x}}}

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1 solution

Steven Chase
Dec 13, 2016

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