Dynamics Question-Parabolic Coordinates

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Parabolic coordinates are defined as ξ = r + x and η = r - x, with r potentially representing the radial distance in a two-dimensional system. The discussion revolves around deriving the kinetic energy expression and finding equations of motion using these coordinates. The user successfully manipulates the equations to express x and y in terms of ξ and η, ultimately deriving the kinetic energy formula. There is uncertainty regarding the form of potential energy, with a suggestion that it may be gravitational, and a consideration of using the action integral formalism. Clarification on these concepts is sought to proceed further in the problem.
QuantumLuck
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Homework Statement



I am told that parabolic coordinates in a plane are defined by \xi = r + x and \eta = 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.
 
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Is this a two-dimensional problem (i.e. what system are you looking at)? If so, then my guess is that r=\sqrt{x^2+y^2}. 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.
 
cipher42 said:
Is this a two-dimensional problem (i.e. what system are you looking at)? If so, then my guess is that \sqrt{x^2+y^2}. 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.

so yeah i think this is right. so here is what i did; \xi = \sqrt{x^2+y^2}+x and then i squared both sides to obtain \xi^{2} = 2x^{2} + y^{2} + 2x \sqrt{x^2+y^2} and \eta^{2} = 2x^{2} + y^{2} -2x \sqrt{x^2+y^2} 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 \xi and \eta. 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.
 
QuantumLuck said:
so yeah i think this is right. so here is what i did; \xi = \sqrt{x^2+y^2}+x and then i squared both sides to obtain \xi^{2} = 2x^{2} + y^{2} + 2x \sqrt{x^2+y^2} and \eta^{2} = 2x^{2} + y^{2} -2x \sqrt{x^2+y^2} 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 \xi and \eta. 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.

oops I am dumb. clearly to find x i take \xi - \eta to find that \xi - \eta \ = 2x and as such \ x = (1/2)\xi - \eta so i then plug in x and find that \ y =\sqrt{\xi\eta} at which point i then showed that kinetic energy \ T=(m/8)(\xi+\eta)(\dot{\xi}^2/\xi +\dot{\eta}^2/\eta). 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?
 
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