Kinetic energy in parabolic coordinates

[QuArK21343]

Homework Statement

Prove that in parabolic coordinates $\alpha,\beta$ the kinetic energy is $T=m/8(\alpha+\beta)(\dot\alpha^2/\alpha+\dot\beta^2/\beta)$

Homework Equations

Parabolic coordinates are defined as follows: $\alpha=r+x, \beta=r-x$ with $r=\sqrt{x^2+y^2}$

The Attempt at a Solution

I don't know how to proceed in this situation: in simpler case (spherical or cylindrical coordinates) I write down the three components of velocity using geometrical intuition (e.g. $v_\rho=\dot \rho, v_\phi=\rho \dot\phi,v_z=\dot z$, because I see they are right...). What if I get only the definition of the new coordinates?

 

[Redbelly98]

Welcome to physics forums.

Homework Statement

Prove that in parabolic coordinates $\alpha,\beta$ the kinetic energy is $T=m/8(\alpha+\beta)(\dot\alpha^2/\alpha+\dot\beta^2/\beta)$

Homework Equations

Parabolic coordinates are defined as follows: $\alpha=r+x, \beta=r-x$ with $r=\sqrt{x^2+y^2}$

The Attempt at a Solution

I don't know how to proceed in this situation: in simpler case (spherical or cylindrical coordinates) I write down the three components of velocity using geometrical intuition (e.g. $v_\rho=\dot \rho, v_\phi=\rho \dot\phi,v_z=\dot z$, because I see they are right...). What if I get only the definition of the new coordinates?

I think the key here is to get x and y in terms of $\alpha, \beta$. And you hopefully already know how to express the kinetic energy in x, y coordinates, right?

 

[QuArK21343]

That's right! I check explicitly and it works. Now that I have the kinetic energy, it will be doable to write down the equations of motion. Thank you.

 

[Redbelly98]

You're welcome

FYI, it looks like this is a college junior or senior level course in classical mechanics. If so, future posts would go in the Advanced Physics area:

https://www.physicsforums.com/forumdisplay.php?f=154