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Lagrangian in cartesian and polar

  • Thread starter jameshaley
  • Start date
  • #1

Homework Statement



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.


Homework Equations



d/dt(∂L/∂x')-∂L/∂x
L=T-V


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
 

Answers and Replies

  • #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
 
  • #3
Dick
Science Advisor
Homework Helper
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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
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.
 
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  • #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)
 
  • #5
Dick
Science Advisor
Homework Helper
26,258
618
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)
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:
  • #6
Yep it's just all clicked!

I understand now. Thank you
 

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