(adsbygoogle = window.adsbygoogle || []).push({}); Consider systems described by Lagrangians

[tex] L_{1} = \frac{1}{2} m\dot{x}^2 [/tex]

[tex] L_{2} = \frac{1}{2} m\dot{x}^2 - a \dot{x} x [/tex]

when [tex] \dot{x} = \frac{dx}{dt} [/tex] and a is a constnat

a) Derive the momenta conjugate to x for each system and write down the coressponding hamiltonians. Are the hamiltonians constants of motion and why?

Momenta conjugate to x (not familiar with the term conjugate)... does this mean

[tex] p_{x} = \frac{\partial L}{\partial x} [/tex]

but isnt the momentum

[tex] p_{\dot{x}} = \frac{\partial L}{\partial \dot{x}} [/tex]

b) Obtain the Lagrange equations of motion for each system, work out their solutions and explain what htey represent.

simply using the Euler Lagrange equation

for L1

let x = x(t)

[tex] \frac{\partial L}{\partial x} - \frac{d}{dt} \frac{\partial L}{\partial \dot{x}} = 0 [/tex]

but dL/dx = 0 so the second term is zero

[tex] \frac{d}{dt} \frac{\partial L}{\partial \dot{x}} = 0 [/tex]

dL/dx dot is mx dot so

[tex] \frac{d}{dt} m \dot{x} = 0 [/tex]

intengrate both sides wrt t

[tex] m x(t) = C_{1}t + C_{2} [/tex]

so far so good?

for L2

[tex] \frac{\partial L}{\partial x} = -a\dot{x} [/tex]

[tex] \frac{\partial L}{\partial \dot{x}} = m \dot{x} - ax [/tex]

[tex] \frac{d}{dt} \frac{\partial L}{\partial \dot{x}} = m \ddot{x} - a \dot{x} [/tex]

Into Euler Lagrange

[tex] -a\dot{x} - (m\ddot{x} - a\dot{x}) = 0 [/tex]

[tex] m\ddot{x} = 0 [/tex]

smae solution as L1 above

[tex] m x(t) = C_{1}t + C_{2} [/tex]

i suspect i made some mistake in L2. In any case they are the same trajectories. They represent straight lines?

c) Show taht

[tex] L_{2} = L_{1} + \frac{d}{dt} F(x,\dot{x}) [/tex]

and find the function F. What does this imply above the actions

[tex] A_{i} = \int_{t_{1}}^{t_{2}} L_{l} dt [/tex]

and how does this relate to your results of b)?

well thats easy

[tex] \frac{d}{dt} F(x,\dot{x}) = -\int ax \dot{x} dt = -\frac{a x^2}{2} [/tex]

the action for L2 is

[tex] A_{i} = \int_{t_{1}}^{t_{2}} L_{2} dt = \int_{t_{1}}^{t_{2}} \frac{1}{2} m \dot{x}^2 + \int_{t_{1}}^{t_{2}} \frac{-ax^2}{2} dt [/tex]

not quite sure how to proceed from here... WOuld i use the solutions from b to solve the action for each lagrangian?

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# Homework Help: Variational Mechanics

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