微分学雅可比行列式求解：球坐标变换结果验证疑问

Let's break down the calculation of $\frac{\delta(x, y, z)}{\delta(r, \theta, \phi)}$ step by step, and figure out where that extra $\cos\phi$ term might have slipped in.

First, remember that the Jacobian determinant is the determinant of the matrix of partial derivatives:
$$
\frac{\delta(x,y,z)}{\delta(r,\theta,\phi)} = \begin{vmatrix}
\frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} & \frac{\partial x}{\partial \phi} \
\frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} & \frac{\partial y}{\partial \phi} \
\frac{\partial z}{\partial r} & \frac{\partial z}{\partial \theta} & \frac{\partial z}{\partial \phi}
\end{vmatrix}
$$

Step 1: Calculate all partial derivatives #

From the given spherical-to-Cartesian conversions:

• For $x = r\sin\theta\cos\phi$:
  • $\frac{\partial x}{\partial r} = \sin\theta\cos\phi$
  • $\frac{\partial x}{\partial \theta} = r\cos\theta\cos\phi$
  • $\frac{\partial x}{\partial \phi} = -r\sin\theta\sin\phi$
• For $y = r\sin\theta\sin\phi$:
  • $\frac{\partial y}{\partial r} = \sin\theta\sin\phi$
  • $\frac{\partial y}{\partial \theta} = r\cos\theta\sin\phi$
  • $\frac{\partial y}{\partial \phi} = r\sin\theta\cos\phi$
• For $z = r\cos\theta$:
  • $\frac{\partial z}{\partial r} = \cos\theta$
  • $\frac{\partial z}{\partial \theta} = -r\sin\theta$
  • $\frac{\partial z}{\partial \phi} = 0$

Step 2: Compute the determinant (using the third row for simplicity) #

The third row has a 0, which reduces the number of calculations. Expanding along the third row:
$$
\cos\theta \cdot \begin{vmatrix} \frac{\partial x}{\partial \theta} & \frac{\partial x}{\partial \phi} \ \frac{\partial y}{\partial \theta} & \frac{\partial y}{\partial \phi} \end{vmatrix} - (-r\sin\theta) \cdot \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \phi} \ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \phi} \end{vmatrix} + 0 \cdot \text{(remaining term)}
$$

First minor determinant:

$$
\begin{vmatrix} r\cos\theta\cos\phi & -r\sin\theta\sin\phi \ r\cos\theta\sin\phi & r\sin\theta\cos\phi \end{vmatrix}
$$
Calculate the product of the diagonals and subtract:
$$
(r\cos\theta\cos\phi)(r\sin\theta\cos\phi) - (-r\sin\theta\sin\phi)(r\cos\theta\sin\phi)
$$
Factor out common terms and use $\cos^2\phi + \sin^2\phi = 1$:
$$
r^{2\cos\theta\sin\theta(\cos}2\phi + \sin^2\phi) = r^2\cos\theta\sin\theta
$$

Second minor determinant:

$$
\begin{vmatrix} \sin\theta\cos\phi & -r\sin\theta\sin\phi \ \sin\theta\sin\phi & r\sin\theta\cos\phi \end{vmatrix}
$$
Again, compute the diagonal products and subtract:
$$
(\sin\theta\cos\phi)(r\sin\theta\cos\phi) - (-r\sin\theta\sin\phi)(\sin\theta\sin\phi)
$$
Factor out common terms and use the Pythagorean identity:
$$
r\sin^2\theta(\cos2\phi + \sin^2\phi) = r\sin^2\theta
$$

Step 3: Combine the results #

Substitute the minor determinants back into the expansion:
$$
\cos\theta \cdot r^2\cos\theta\sin\theta + r\sin\theta \cdot r\sin^2\theta
$$
Simplify each term, then factor out $r^2\sin\theta$:
$$
r^2\cos2\theta\sin\theta + r^2\sin3\theta = r^{2\sin\theta(\cos}2\theta + \sin^2\theta) = r^2\sin\theta
$$

Where you might have gone wrong #

That extra $\cos\phi$ term suggests you probably missed applying the Pythagorean identity $\cos^2\phi + \sin^2\phi = 1$ at some step, or made a sign error when expanding the determinant. Double-check your minor determinant calculations—it's easy to mix up signs or forget to combine the trigonometric terms into 1!

内容的提问来源于stack exchange，提问作者Ravi Kumar