Conserved Quantity

A physical system in three spatial coordinates ( x , y , z ) (x,y,z) is described by the following Lagrangian :

L = x 2 y ˙ + 2 z 3 x ˙ y ˙ 2 + 3 x x ˙ z ˙ \large{L = x^2 \dot{y} + 2 z^3 \dot{x} \, \dot{y}^2 + 3 x \dot{x} \dot{z}}

Which quantity must be conserved?

Details and Assumptions:
- The dot convention denotes time differentiation

2 x y ˙ + 3 x ˙ z ˙ \large{2 x \dot{y} + 3 \dot{x} \dot{z}} 2 z 3 y ˙ 2 + 3 x z ˙ \large{2 z^3 \dot{y}^2 +3 x \dot{z}} 6 z 2 x ˙ y ˙ 2 \large{6 z^2 \dot{x} \dot{y}^2} x 2 + 4 z 3 x ˙ y ˙ \large{x^2 + 4 z^3 \dot{x} \dot{y} } 3 x x ˙ \large{3 x \dot{x}}

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1 solution

Steven Chase
Jul 29, 2017

Consider the Euler-Lagrange equation for the y y variable (similar equations exist for the other two as well):

d d t L y ˙ = L y \large{\frac{d}{dt} \frac{\partial{L}}{\partial \dot{y}} = \frac{\partial{L}}{\partial y}}

Inspecting the Lagrangian, we observe that the variable y y does not appear, but the other two do. Therefore, the Euler-Lagrange equation for y y reduces to:

d d t L y ˙ = 0 \large{\frac{d}{dt} \frac{\partial{L}}{\partial \dot{y}} = 0}

Obviously then, the time derivative of L d y ˙ \Large{\frac{\partial{L}}{d \dot{y}}} is zero, meaning that this quantity is conserved.

L d y ˙ = x 2 + 4 z 3 x ˙ y ˙ \large{\frac{\partial{L}}{d \dot{y}} = x^2 + 4z^3 \dot{x} \dot{y}}

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