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!

1 Answer
Jan 14, 2017

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

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

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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)#