Difference between revisions of "Forest UCM LEq"

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for Example
 
for Example
  
:<math> z^2 = R^2-x^2 + y^2</math>
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:<math> z^2 = R^2-x^2 - y^2</math>
  
 +
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:<math>\vec r = x \hat i + y \hat j \pm sqrt{R^2-x^2 - y^2} \hat k</math>
  
  

Revision as of 12:54, 26 October 2014

Lagrange's Foramlism for Classical Mechanics

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 \dot x} \right ) \right ] =0 [/math]


or

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

here [math]x^{\prime} = \frac{d x}{d t} \equiv \dot x[/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 \dot x} \right ) = \frac{d}{dt} \left ( \frac{\partial }{\partial \dot x} \right ) \left ( T(\dot x) - U(x) \right ) [/math]
[math] = \frac{d}{dt} \left ( \frac{\partial }{\partial \dot x} \right ) T(\dot x) = \frac{d}{dt} m \dot x = \frac{d}{dt} p_x = \dot{p}_x = F_x [/math] Newton's second law in an Inertial reference frame


thus

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

Lagrange's Equations in generalized coordinates

Generalized coordinates [math](q_i)[/math] are a set of coordinates that uniquely specify the instantaneous state of a dynamical system.

The number of independent generalized coordinates [math](N_q)[/math] is given by subtracting the number of constraints [math](N_C)[/math] from the number of degrees of freedom [math]N_{DF}[/math].

[math]N_q = N_{DF} - N_C[/math]

example

Consider a particle constrained to move on a sphere of radius [math]R[/math].

[math]N_{DF} =3 [/math] There are 3 degrees of freedom for the 3-D problem
[math] N_C =1 [/math] The particle is constrained to the surface of the sphere
[math]N_q = 3-1 = 2 [/math] The motion of the particle may be described using two components

The constraint expressed in terms of cartesian coordinates is

[math] x^2 + y^2 + z^2 = R^2[/math]

The above constraint equation can be used to reduce the degrees of freedom by using the constraint to eliminate one of the above components

for Example

[math] z^2 = R^2-x^2 - y^2[/math]


[math]\vec r = x \hat i + y \hat j \pm sqrt{R^2-x^2 - y^2} \hat k[/math]


As shown above, Hamilton's principle leads to a re-expression of Newton's second law through the Euler-Lagrange Equation in a differential form known as Lagrange's equations.


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


The above is used to determine a stationary path followed by a particle obeying Newton's second law.

This path should be the same in all coordinate systems that are Inertial reference frames.



Forest_Ugrad_ClassicalMechanics