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

 

[rwisz]

In order to prove the given equation for kinetic energy in parabolic coordinates, we can use the transformation equations for velocity in these coordinates. In parabolic coordinates, the velocity components are given by:

v_\alpha = \dot\alpha = \dot r + \dot x = \dot r + \dot\beta

v_\beta = \dot\beta = \dot r - \dot x = \dot r - \dot\alpha

Using these equations, we can rewrite the kinetic energy equation in terms of the velocity components in parabolic coordinates:

T = \frac{1}{2} m (v_\alpha^2 + v_\beta^2) = \frac{1}{2} m (\dot\alpha^2 + \dot\beta^2) = \frac{1}{2} m [(\dot r + \dot\beta)^2 + (\dot r - \dot\alpha)^2]

Expanding and simplifying, we get:

T = \frac{1}{2} m [2\dot r^2 + 2\dot\beta^2 + 2\dot r\dot\beta - 2\dot r\dot\alpha - 2\dot\alpha\dot\beta]

Using the definition of parabolic coordinates, we can substitute \dot r = \dot\alpha + \dot\beta and simplify further to get:

T = \frac{1}{2} m [2(\dot\alpha + \dot\beta)^2 + 2\dot\beta^2 + 2(\dot\alpha + \dot\beta)\dot\beta - 2(\dot\alpha + \dot\beta)\dot\alpha - 2\dot\alpha\dot\beta]

T = \frac{1}{2} m [2(\dot\alpha^2 + \dot\beta^2 + 2\dot\alpha\dot\beta) + 2\dot\beta^2 - 2\dot\alpha^2 - 2\dot\alpha\dot\beta]

T = \frac{1}{2} m [2(\dot\alpha^2 + \dot\beta^2) + 2\dot\beta^2 - 2\dot\alpha^2]

T = m(\dot\alpha^2 + \dot\beta^2)

Finally, substituting the definitions