1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    d/dt(∂L/∂x')-∂L/∂x
    L=T-V


    3. The attempt at a solution

    d/dt(∂L/∂x')-∂L/∂x=x''-x/sqrt(x^2+y^2)=0
    d/dt(∂L/∂y')-∂L/∂y=y''-y/sqrt(x^2+y^2)=0

    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

    -cost-cost/sqrt(sint^2+cost^2)=-2cost
    -sint-sint/sqrt(sint^2+cost^2)=-2sint

    for a this is where i think i get stuck.

    b) x=rcosθ,x'=cosθr'-rsinθθ'
    y=rsinθ,y'=sinθr'+rcosθθ'

    x'^2+y'^2=r'^2+r^2θ'^2
    sqrt(x^2+y^2)=r

    there for
    L(r,θ,r',')=(1/2)(r'^2+r^2θ'^2)-r

    and so

    d/dt(∂L/∂r')-∂L/∂r=r''-rθ'^2+1
    d/dt(∂L/∂θ')-∂L/∂θ=r'^2θ''

    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
    d/dt(∂L/∂x')-∂L/∂x=x''-x/sqrt(x^2+y^2)=0
    d/dt(∂L/∂y')-∂L/∂y=y''-y/sqrt(x^2+y^2)=0

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

    d/dt(∂L/∂x')-∂L/∂x=x''+x/sqrt(x^2+y^2)=0
    d/dt(∂L/∂y')-∂L/∂y=y''+y/sqrt(x^2+y^2)=0
     
  4. Oct 23, 2013 #3

    Dick

    User Avatar
    Science Advisor
    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.

    r(t)=1
    =sqrt(x^2+y^2)
    =sqrt((cost)^2+)sint)^2)

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

    Dick

    User Avatar
    Science Advisor
    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Lagrangian in cartesian and polar
  1. Polar to Cartesian (Replies: 1)

  2. Cartesian to polar (Replies: 4)

  3. Cartesian to polar (Replies: 2)

Loading...