Forest UCM LEq

Lagrange's Equations

Hamilton's principle

Hamilton's principle falls out of the calculus of variations in that seeking the shortest time interval is the focus of the variations.

Of all possible paths along which a dynamical system may move from on point to another within a specified time interval, the actual path followed is that which minimizes the time integral of the difference between the kinetic and potential energies.

Casting this in the language of the calculus of variations

[math]S \equiv \int_{t_1}^{t_2} \left ( T(\dot x) - U(x) \right ) dt = \int_{t_1}^{t_2} f(x,\dot x ; t) dt = \int_{t_1}^{t_2} \mathcal {L} ( x, \dot x;t) dt [/math]

if you want the above "action" integral to be stationary then according to the calculus of variations you want the Euler-Lagrange equation to be satisfied where

[math] \left [ \left ( \frac{\partial f}{\partial x} \right ) - \frac{d}{dt} \left ( \frac{\partial f}{\partial x^{\prime}} \right ) \right ] =0 [/math]

or

[math] \left [ \left ( \frac{\partial \mathcal {L} }{\partial x} \right ) - \frac{d}{dt} \left ( \frac{\partial \mathcal {L} }{\partial x^{\prime}} \right ) \right ] =0 [/math]

[math] \left ( \frac{\partial \mathcal {L} }{\partial x} \right ) = \frac{\partial }{\partial x} \left ( T(\dot x) - U(x) \right ) = -\frac{\partial U(x)}{\partial x} = F_x[/math] if I have conservative forces

[math] \frac{d}{dt} \left ( \frac{\partial \mathcal {L} }{\partial x^{\prime}} \right ) = \frac{d}{dt} \left ( \frac{\partial }{\partial x^{\prime}} \right ) \left ( T(\dot x) - U(x) \right ) [/math]

[math] = \frac{d}{dt} \left ( \frac{\partial }{\partial x^{\prime}} \right ) T(\dot x) = \frac{d}{dt} m \dot x = \frac{d}{dt} p_x = \dot{p}_x = F_x [/math] Newtons second law

thus

[math] \left [ \left ( \frac{\partial \mathcal {L} }{\partial x} \right ) - \frac{d}{dt} \left ( \frac{\partial \mathcal {L} }{\partial x^{\prime}} \right ) \right ] = F_x - F_x = 0 [/math]

Lagrange's Equations

Forest_Ugrad_ClassicalMechanics