Q. 52 - Calculus

Question

The formulas for converting from spherical coordinates to rectangular coordinates are $x=p\sin \varphi \cos \theta , y=p\sin \varphi \sin \theta , z=p\cos \varphi$. Prove that the Jacobian  $\frac{\partial (x, y, z)}{\partial (p, \theta , \varphi )}=-p^2 \sin \varphi$.

Step-by-Step Solution

Answer

It is proven that $\frac{\partial (x, y, z)}{\partial (p, \theta , \varphi )}=-p^2 \sin \varphi$.

Step 1: Given information

The transformation equations are,

$x=p\sin \varphi \cos \theta , y=p\sin \varphi \sin \theta , z=p\cos \varphi$

Step 2: Proof

The definition of Jacobian of a transformation using partial derivatives is given as

$$\begin{gathered} L.H.S=\frac{\partial (x, y, z)}{\partial (p, \theta , \varphi )} \\ \frac{\partial (x, y, z)}{\partial (p, \theta , \varphi )}=\left|\begin{array}{ccc}\frac{\partial x}{\partial p}& \frac{\partial y}{\partial p}& \frac{\partial z}{\partial p}\\ \frac{\partial x}{\partial \theta }& \frac{\partial y}{\partial \theta }& \frac{\partial z}{\partial \theta }\\ \frac{\partial x}{\partial \varphi }& \frac{\partial y}{\partial \varphi }& \frac{\partial z}{\partial \varphi }\end{array}\right| \\ \frac{\partial (x, y, z)}{\partial (p, \theta , \varphi )}=\left|\begin{array}{ccc}\sin \varphi \cos \theta & \sin \varphi \sin \theta & \cos \varphi \\ -p \sin \varphi \sin \theta & p\sin \varphi \cos \theta & 0\\ p\cos \varphi \cos \theta & p\cos \varphi \sin \theta & -p\sin \varphi \end{array}\right| \\ \frac{\partial (x, y, z)}{\partial (p, \theta , \varphi )}=-p^2{\sin }^3\varphi -p^2\sin \varphi {\cos }^2\varphi \\ \frac{\partial (x, y, z)}{\partial (p, \theta , \varphi )}=-p^2\sin \varphi ({\sin }^2\varphi +{\cos }^2\varphi ) \\ \frac{\partial (x, y, z)}{\partial (p, \theta , \varphi )}=-p^2\sin \varphi \left(1\right) \\ \frac{\partial (x, y, z)}{\partial (p, \theta , \varphi )}=-p^2\sin \varphi =R.H.S \end{gathered}$$