Question

How would I convert kinetic energy for one particle in six dimensions (x, y, z, and their time derivatives) from Cartesian to spherical coordinates? Note: `dotq = (delq)/(delt)`, and this is from a 1962 Statistical Mechanics textbook by Norman Davidson.

In Cartesian, `K = K(x,y,z,(delx)/(delt), (dely)/(delt), (delz)/(delt))`, and in spherical, `K = K(r,theta,phi, (delr)/(delt), (deltheta)/(delt), (delphi)/(delt))`.

For convenience, we set `dotq = (delq)/(delt)`.

Not sure if anyone actually needs this, but this was quite crazy to do, and I'm willing to share how I did it so that if anyone ever needs to do this... they can? There is a lot of nested chain rule and product rule!

Answer 1

For spherical coordinates (where `phi` is on the `xy`-plane and `theta` is on the `yz`-plane),

`x = rsinthetacosphi`,
`y = rsinthetasinphi`,
`z = rcostheta`.

To express `K` for one particle in terms of `r,theta,phi,dotr,dottheta,dotphi`, we first begin with:

`K = 1/2m(dotx^2 + doty^2 + dotz^2)`

Let's focus on `x`, then `y`, then `z`, taking their time derivatives:

`dotx = (delx)/(delt) = r(dotthetacosthetacosphi - dotphisinthetasinphi) + dotrsinthetacosphi`
`doty = (dely)/(delt) = r(dotphisinthetacosphi + dotthetacosthetasinphi) + dotrsinthetasinphi`
`dotz = (delz)/(delt) = -rdotthetasintheta + dotrcostheta`

Note that `dotx^2 ne (del^2x)/(delt^2) = ddotx`, but actually, `dotx^2 = (dotx)^2`. In comparing two methods of squaring each term vs. integrating `mdotq` over six coordinates... yeah, I'm going with the first method.

`dotx^2 = [r(dotthetacosthetacosphi - dotphisinthetasinphi) + dotrsinthetacosphi]^2`

`= r^2(dotthetacosthetacosphi - dotphisinthetasinphi)^2 + dotr^2sin^2thetacos^2phi + 2rdotrsinthetacosphi(dotthetacosthetacosphi - dotphisinthetasinphi)`

`= r^2dottheta^2cos^2thetacos^2phi - 2r^2dotthetadotphisinthetacosthetasinphicosphi + r^2dotphi^2sin^2thetasin^2phi + dotr^2sin^2thetacos^2phi + 2rdotrdotthetasinthetacosthetacos^2phi - 2rdotrdotphisin^2thetasinphicosphi`

`= (r^2dottheta^2costheta + 2rdotrdotthetasintheta)costhetacos^2phi - (2r^2dotthetadotphisinthetacostheta + 2rdotrdotphisin^2theta) sinphicosphi + (r^2dotphi^2sin^2phi + dotr^2cos^2phi)sin^2theta`

Now, we repeat for `doty^2`...

`doty^2 = r^2(dotphisinthetacosphi + dotthetacosthetasinphi)^2 + dotr^2sin^2thetasin^2phi + 2rdotrsinthetasinphi(dotphisinthetacosphi + dotthetacosthetasinphi)`

`= r^2dotphi^2sin^2thetacos^2phi + r^2dottheta^2cos^2thetasin^2phi + 2r^2dotthetadotphisinthetacosthetasinphicosphi + dotr^2sin^2thetasin^2phi + 2rdotrdotphisin^2thetasinphicosphi + 2rdotrdotthetasin^2phisinthetacostheta`

`= (r^2dottheta^2costheta + 2rdotrdotthetasintheta)costhetasin^2phi + (2r^2dotthetadotphisinthetacostheta + 2rdotrdotphisin^2theta)sinphicosphi + (r^2dotphi^2cos^2phi + dotr^2sin^2phi)sin^2theta`

There is hope, as some `sin^2u + cos^2u` terms will show up once we add `dotx^2` and `doty^2`. Now, for `dotz^2`:

`dotz^2 = (-rdotthetasintheta + dotrcostheta)^2`

`= dotr^2cos^2theta - 2rdotrdotthetacosthetasintheta + r^2dottheta^2sin^2theta`

And now, adding these together gives us `(2K)/m`:

`(2K)/m = dotx^2 + doty^2 + dotz^2`

`= (r^2dottheta^2costheta + 2rdotrdotthetasintheta)costhetacos^2phi - cancel((2r^2dotthetadotphisinthetacostheta + 2rdotrdotphisin^2theta) sinphicosphi) + (r^2dotphi^2sin^2phi + dotr^2cos^2phi)sin^2theta + (r^2dottheta^2costheta + 2rdotrdotthetasintheta)costhetasin^2phi + cancel((2r^2dotthetadotphisinthetacostheta + 2rdotrdotphisin^2theta)sinphicosphi) + (r^2dotphi^2cos^2phi + dotr^2sin^2phi)sin^2theta + dotr^2cos^2theta - 2rdotrdotthetacosthetasintheta + r^2dottheta^2sin^2theta`

Regroup the `sin^2theta` terms:

`=> (r^2dottheta^2costheta + 2rdotrdotthetasintheta)costhetacos^2phi + (r^2dotphi^2sin^2phi + dotr^2cos^2phi + r^2dotphi^2cos^2phi + dotr^2sin^2phi)sin^2theta + (r^2dottheta^2costheta + 2rdotrdotthetasintheta)costhetasin^2phi + dotr^2cos^2theta - 2rdotrdotthetacosthetasintheta + r^2dottheta^2sin^2theta`

Getting some `sin^2u + cos^2u` action going on within the second term, and between the first and third terms! Thus, those reduce to give the factor in front of the `sin^2u` and `cos^2u`.

`=> r^2dottheta^2cos^2theta + 2rdotrdotthetasinthetacostheta+ (r^2dotphi^2 + dotr^2)sin^2theta + dotr^2cos^2theta - 2rdotrdotthetacosthetasintheta + r^2dottheta^2sin^2theta`

And some more happening with the first and last terms, and the third and fourth terms (upon multiplying the `sin^2theta` through the third term); also, the cross-terms cancel.

`=> r^2dottheta^2sin^2theta + r^2dottheta^2cos^2theta + cancel(2rdotrdotthetasinthetacostheta) + r^2dotphi^2sin^2theta + dotr^2sin^2theta + dotr^2cos^2theta - cancel(2rdotrdotthetasinthetacostheta)`

`= dotr^2 + r^2dottheta^2 + r^2dotphi^2sin^2theta`

Finally, multiply by `m/2` to get `K`:

`color(blue)(K = 1/2mdotr^2 + 1/2mr^2dottheta^2 + 1/2mr^2dotphi^2sin^2theta)`