Points where the Jacobian of a coordinate transformation vanishes

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, are the formulas for the partial derivatives valid on the $z$-axis? If not, is this because the existence of of the partial derivatives $\partial_r$, $\partial_\theta$, $\partial_\phi$ (as defined in differential geometry) is not guaranteed on the $z$-axis?

3 Answers

In Cartesian coordinates all points in $R^2$ belong to the domain. However, when one converts to cylindrical or spherical coordinates, some points are excluded from the domain. For cylindrical coordinates, all the points on the $z$-axis are excluded and for the spherical coordinates the origin is excluded.

This leads to some interesting consequences. Consider for instance the Fourier transform of a Dirac delta function $$\iint_{-\infty}^{\infty} \exp[i 2\pi (x a+y b)]\ {\rm d}a {\rm d}b = \delta(x)\delta(y) .$$ Now convert this expression to cylindrical coordinates. The righthand side will become an integral over Bessel functions and it simply gives zero, because the origin is excluded.

Another interesting consequences is the Schwarzschild metric. Einstein derived general relativity with the aid of an assumption of continuity. However, the Schwarzschild metric contains a singularity. Isn't that a contradiction? No, because the Schwarzschild metric is formulated in spherical coordinates, which means that the location of the singularity at the origin is excluded from the domain.

• This isn't really what I was asking. I've made the question more specific. – Randy Randerson Aug 22 '17 at 8:07
• The delta "function" is a badly-chosen example because it isn't really a function and it is not obvious to me that "change of coordinates" makes any formal sense for a (delta) distribution. – ACuriousMind Aug 22 '17 at 11:46
• @ACuriousMind: Still illustrates the principle. – flippiefanus Aug 23 '17 at 4:23
• @RandyRanderson: can't give you much more sorry. – flippiefanus Aug 23 '17 at 4:23

The equations of motion don't apply on the z-axis. In particular, if you have a path that goes from $\theta=\pi, \phi=0$ to $\theta=\pi, \phi=\pi$ (i.e., crossing over the negative z-axis going in the negative x-direction), $\dot\phi$ is undefined. Formally, solutions only exist in the region where the Jacobian is nonzero. So, how does one handle solutions that cross the z-axis? You solve for the trajectory before the particle hits the z-axis, solve for the trajectory after the particle hits the z-axis, and then you match boundary conditions at the z-axis. You insist that, while $\dot\phi$ is undefined for a moment of time, the two-sided limits match. This is what is meant by treating the point "with special care."

It's important in the sense that all points on the $z$-axis do not have a well defined $\varphi$ value, so you can't use spherical coordinates. But you can always represent them in Cartesian coordinates.

• This isn't really what I was asking. I've made the question more specific. – Randy Randerson Aug 22 '17 at 8:07
• @RandyRanderson Well, for the specific example it really won't matter if you take "special care" or not. The reason is that if the pendulum crosses $z$-axis, its motion will stay in a 2D plane and there's no need to use spherical coordinates to describe it. – Salmonella mayonnaise Aug 22 '17 at 16:52
• Thats not the point. – Randy Randerson Aug 22 '17 at 18:40