# 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:

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?

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.

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