What Is the Equation of Motion for a Particle with a Given Lagrangian?

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SUMMARY

The discussion focuses on deriving the equation of motion for a particle with a specified Lagrangian, given by L = (m²ẋ⁴)/12 + mẋ²V(x) - V²(x). The participants successfully apply the Euler-Lagrange equation, leading to the equation of motion expressed as dV/dx(mẋ² - 2V) - mẋẋ(mẋ² + 2V) = 0. Additionally, they emphasize the importance of correctly differentiating terms, as noted in the correction regarding the term m²ẋ².

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  • Knowledge of calculus, particularly differentiation
  • Basic concepts of classical mechanics, including energy conservation
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Homework Statement



A particle of mass m moves in one dimensional such that it has the Lagrangian

[itex]L=\frac{m^2\dot{x}^4}{12} + m\dot{x}^2V\left(x\right)- V^2\left(x\right)[/itex]

where V is some differentiable function of x only.
(i) Find the equation of motion for x(t)

(ii) Find an expression for the total energy and hence describe the physical nature of the system on the basis of this equation.


The Attempt at a Solution



[itex] \frac{\partial{L}}{\partial{x}}= m\dot{x}^2\frac{dV}{dx}- 2V\frac{dV}{dx}[/itex]

[itex]\frac{d}{dt}\left(\frac{\partial{L}}{\partial{\dot{x}}}\right) = m^2\dot{x}^2 + 2mV\ddot{x}[/itex]

[itex]m\dot{x}^2\frac{dV}{dx}- 2V\frac{dV}{dx} - m^2\dot{x}^2\ddot{x} - 2mV\ddot{x} = 0[/itex]

[itex]\frac{dV}{dx}\left(m\dot{x}^2-2V\right)- m\ddot{x}\left(m\dot{x}^2 +2V\right) =0[/itex]
 
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In your second line, m^2xdot^2 is wrong.
You took the d/dt wrongly.
 

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