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Homework Help: Lagrangian in cartesian and polar

  1. Oct 23, 2013 #1
    1. The problem statement, all variables and given/known data

    Consider the following Lagrangian in Cartesian coordinates:
    L(x, y, x', y') = 12 (x^ 2 + y^2) -sqrt(x^2 + y^2)

    (a) Write the Lagrange equations of motion, and show that x = cos(t);
    y =sin(t) is a solution.

    (b) Changing from Cartesian to polar coordinates, x = r cos ; y = r sin ,
    show that the Lagrangian becomes

    L(r;θ ; r';') = 1/2 (r'^2 + r^2θ^2)- r;

    and hence fi nd the Lagrange equations in polar coordinates.

    (c) Show that the solution given in Cartesian coordinates in (a) is still a
    solution when expressed in polar coordinates.

    2. Relevant equations


    3. The attempt at a solution


    got both as i think there is 2 degrees of freedom

    put in x=cost, y=sint
    x'=-sint, x''=-cost
    y'=cost y''=-sint

    put these it above i get


    for a this is where i think i get stuck.

    b) x=rcosθ,x'=cosθr'-rsinθθ'


    there for

    and so


    and thats where i get stuck. This is a practise exam question, and i want to make sure im going the right way about it. Any help would be much appropriated
  2. jcsd
  3. Oct 23, 2013 #2

    worked out where I went wrong on this, missed the minus'

  4. Oct 23, 2013 #3


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    Homework Helper

    Yes. And you should probably fix your initial Lagrangian up, it's got some confusing typos. You mean L=(1/2)*(x'^2+y'^2)-sqrt(x^2+y^2). Any other questions? Your Cartesian solution is r(t)=1 and θ(t)=t in polars. It should be pretty easy to show that satisfies your polar equations of motion.
  5. Oct 23, 2013 #4
    Thank you for the reply

    ''Your Cartesian solution is r(t)=1 and θ(t)=t in polars.''

    I'm struggling a bit to understand why this is.


    And for θ(t)=t
    I thought arctan(y/x)
  6. Oct 23, 2013 #5


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    Homework Helper

    I'm not sure what you are thinking. If you want x=cos(t) and y=sin(t) and x=r(t)cos(θ(t)) and y=r(t)sin(θ(t)) then surely putting r(t)=1 and θ(t)=t will do the job just by inspection. If you want to follow through with your calculation, arctan(y/x)=arctan(sin(t)/cos(t))=arctan(tan(t))=t.
    Last edited: Oct 23, 2013
  7. Oct 23, 2013 #6
    Yep it's just all clicked!

    I understand now. Thank you
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