Answer to Question #94089 in Mechanics | Relativity for Joy

Question

A particle of mass L=M²X^4/12 +MX²V<x>-V²<x>
Where V is some differentiable function of X. Find the equation of motion for X(t) and describe the physical nature of the system on the basis of this equation

Answer 1

The Lagrangian for the particle with mass $m$ is:

$$L=\frac{m^2\dot{x}^4}{12}+m\dot{x}^2 V(x)-V^2(x).$$

Differentiate by $x$:

$$\frac{dL}{dx}=m\dot{x}^2\frac{\partial V}{\partial x}-2V(x)\frac{\partial V}{\partial x}.\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space(1)$$

Differentiate the last equation by $\dot{x}$:

$$\frac{\partial L}{\partial\dot{x}}=\frac{1}{3}m^2\dot{x}^3+2m\dot{x}V(x).$$

Now differentiate by time:

$$\frac{d}{dt}\Big(\frac{\partial L}{\partial\dot{x}}\Big)=m^2\dot{x}^2\ddot{x}+2m\ddot{x}V(x)+2m\dot{x}^2\frac{\partial V}{\partial x}\space\space\space\space\space\space(2).$$

The Lagrange's equation:

$$\frac{d}{dt}\Big(\frac{\partial L}{\partial\dot{x}}\Big)-\frac{\partial L}{\partial{x}}=0.$$

Substitute (1) and (2) in the last equation:

$$m^2\dot{x}^2\ddot{x}+2m\ddot{x}V(x)+2m\dot{x}^2\frac{\partial V}{\partial x}-\\ -m\dot{x}^2\frac{\partial V}{\partial x}+2V(x)\frac{\partial V}{\partial x}=0,$$

$$\space\\ m^2\dot{x}^2\ddot{x}+2m\ddot{x}V(x)+m\dot{x}^2\frac{\partial V}{\partial x}+2V(x)\frac{\partial V}{\partial x}=0,\\ \space\\ (m\dot{x}^2+2V)\Big(m\ddot{x}+\frac{\partial V}{\partial x}\Big)=0.\space\space\space\space\space\space\space\space\space\space(3)$$

That is the equation of motion.

A force can be expressed through potential as

$$\frac{\partial V}{\partial x}=-F_x,\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space(4)$$

In equation (3) either of the pairs of parentheses equal to 0, so from the first parentheses divided by 2 we see it's the total energy equal to zero:

$$1/2\cdot m\dot{x}^2+V=0,$$

from the second parentheses using equation (4)

$$m\ddot{x}=-\frac{\partial V}{\partial x},\\ ma=F_x.$$

That's Newton's second law.