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Dynamics Question-Parabolic Coordinates

  1. Sep 28, 2009 #1
    1. The problem statement, all variables and given/known data

    I am told that parabolic coordinates in a plane are defined by [tex]\xi[/tex] = r + x and [tex]\eta[/tex] = r - x. after this i am then asked to show that this leads to a given expresion for the kinetic energy (if i knew x and y i could find this without a problem). From this I am then told to find the equations of motion, an expression which i find very vague. I am in fact familiar with the Lagrangian and Hamiltonian formalisms (loosely) but I do not know what my expression for potential energy will be in this system (if i am on the right track here). Anyway, the main problem I am having at the moment is that I am very unsure as to what "r" is. Any help would be greatly appreciated.
     
  2. jcsd
  3. Sep 28, 2009 #2
    Is this a two-dimensional problem (i.e. what system are you looking at)? If so, then my guess is that [tex]r=\sqrt{x^2+y^2}[/tex]. From this, you could invert the equations and then recast your Lagrangian (or Hamiltonian) in these two new coordinates (since you should know what L=T-V looks like in Cartesian coordinates already). Maybe an obvious expansion will present itself when you see it in these new variables.
     
  4. Sep 28, 2009 #3
    so yeah i think this is right. so here is what i did; [tex]\xi[/tex] = [tex]\sqrt{x^2+y^2}+x[/tex] and then i squared both sides to obtain [tex]\xi^{2} = 2x^{2} + y^{2} + 2x[/tex] [tex]\sqrt{x^2+y^2}[/tex] and [tex]\eta^{2} = 2x^{2} + y^{2} -2x[/tex] [tex]\sqrt{x^2+y^2}[/tex] so now that i have these equations i have been playing around with them trying to look for a way to separate out the x and the y terms so i can get x and y in terms of [tex]\xi[/tex] and [tex]\eta[/tex]. the various i have tried to do this is to add and subtract these equations to each other, each resulting in a varying degree of failure. not sure where to go from here, if i have even went about this correctly.
     
  5. Sep 29, 2009 #4
    oops im dumb. clearly to find x i take [tex]\xi[/tex] - [tex]\eta[/tex] to find that [tex]\xi[/tex] - [tex]\eta[/tex] [tex]\ = 2x[/tex] and as such [tex]\ x = (1/2)\xi[/tex] - [tex]\eta[/tex] so i then plug in x and find that [tex]\ y =\sqrt{\xi\eta}[/tex] at which point i then showed that kinetic energy [tex]\ T=(m/8)(\xi+\eta)(\dot{\xi}^2/\xi +\dot{\eta}^2/\eta)[/tex]. the final thing that i now have to do for this problem is to write down the equations of motion. now since this is for a particle am i assuming that the potential energy is varying solely as a gravitational potential? it is not obvious to me at all. or rather should i go through the action integral formalism for this coordinate system?
     
    Last edited: Sep 29, 2009
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