10
$\begingroup$

Consider the coordinate transformation \begin{align*} x &= r\sin\theta\cos\phi \\ y &= r\sin\theta\sin\phi \\ z &= r\cos\theta \end{align*} from spherical coordinates $(r,\theta,\phi)$ to rectangular coordinates $(x,y,z)$. Here $r$ is the radius, $\theta$ is the inclination, and $\phi$ is the azimuth. Its Jacobian $$\frac{\partial(x,y,z)}{\partial(r,\theta,\phi)} = r^2\sin\theta$$ vanishes on the $z$-axis.


According to C. Lanczos in The Variational Principles of Mechanics :

[The Jacobian of a coordinate transformation may vanish] at certain singular points, which have to be excluded from consideration. For example, [for the coordinate transformation above] special care is required at the values $r = 0$ and $\theta = 0$, for which the Jacobian of the transformation vanishes.

Question :

Are points at which the Jacobian of a coordinate transformation vanishes "excluded from consideration" altogether or included in the analysis but handled with "special care"?


Perhaps a problem (from the same book) will clarify the question.

Characterize the position of a spherical pendulum of length $l$ by spherical coordinates $r$, $\theta$, $\phi$ and obtain : \begin{align*} T &= \frac{m}{2}l^2\Big(\dot{\theta}^{\,2} + \sin^2\!\theta \,\dot{\phi}^{\,2}\Big), \\ V &= mgl(1 - \cos\theta). \end{align*} Form the Lagrangian equations of motion.

The Lagrangian is $$L = T - V = \frac{m}{2}l^2\Big(\dot{\theta}^{\,2} + \sin^2\!\theta \,\dot{\phi}^{\,2}\Big) + mgl(\cos\theta - 1).$$ Since \begin{gather*} \partial_{\dot{\theta}}L = ml^2\dot{\theta}, \\ \partial_\theta L = ml^2\sin\theta\cos\theta\,\dot{\phi}^{\,2} - mgl\sin\theta, \\ \partial_{\dot{\phi}}L = ml^2\sin^2\!\theta\,\dot{\phi}, \\ \partial_\phi L = 0, \end{gather*} the Lagrangian equations of motion are \begin{gather*} \frac{d}{dt}\Big(ml^2\dot{\theta}\Big) - ml^2\sin\theta\cos\theta\,\dot{\phi}^{\,2} + mgl\sin\theta = 0 \\ \frac{d}{dt}\Big(ml^2\sin^2\!\theta\,\dot{\phi}\Big) = 0 \end{gather*}


In the solution to this problem, should it be stated that points on the $z$-axis are "excluded from consideration"? Or should they be included, but treated with "special care"?

In particular:

  1. Do the equations in this problem (for kinetic and potential energy; the Euler-Lagrange equations) hold on the $z$-axis?
  2. Do the partial derivatives $\partial_r$, $\partial_\theta$, $\partial_\phi$ (as defined in differential geometry) exist on the $z$-axis?
$\endgroup$
1
$\begingroup$

For the very first question: they are excluded. The change of coordinates needs to be a (al least) $C^1$ diffeomorphism.

https://en.wikipedia.org/wiki/Change_of_variables#Formal_introduction

For all $\theta$ and $\phi$, the coordinates $(0,\theta,\phi)$ represent the same point in $(x,y,z)$ (the origin).

The same happens with $(R,\theta, 0)$. Fixing $R$ and varying $\theta$ yields to the same point, $(0,0,R)$.

Consider the case where we want to compute an integral of a function over a sphere $S$: $$\int_Sf(x,y,z)\textrm{d}V.$$

Now let $S'$ be the sphere without the points where $r=0$, $\theta=0$, $\phi=0$ and $\phi=2\pi$ (so that the change of variables is a diffeomorphism). These sets differ by a set of zero volume, so

$$\int_Sf(x,y,z)\textrm{d}V=\int_{S'}f(x,y,z)\textrm{d}V$$

for any $f$ integrable on $S$. So when we use the Change of variables Theorem, we actually use it on the second integral.

$\endgroup$
  • $\begingroup$ Of course you need to be careful with that. Your $f(x,y,z)$ might be a delta function $\delta(x)\delta(y)\delta(z)$ $\endgroup$ – Andrei Aug 24 '17 at 2:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.