To derive the heat conduction equation in spherical coordinates$(r, \theta, \phi)$ from Cartesian coordinates, we transform the Laplacian operator$\nabla^2 T = \frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} + \frac{\partial^2 T}{\partial z^2}$.
For a system with constant thermal conductivity $k$, the resulting spherical equation is:
$$\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial T}{\partial r}\right) + \frac{1}{r^2 \sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta \frac{\partial T}{\partial \theta}\right) + \frac{1}{r^2 \sin^2\theta}\frac{\partial^2 T}{\partial \phi^2} + \frac{\dot{q}}{k} = \frac{1}{\alpha}\frac{\partial T}{\partial t}$$
1. Define Coordinate Relations
Establish the geometric links using the standard physics convention (where $\theta$ is the polar/colatitude angle and $\phi$ is the azimuthal angle):
- $x = r \sin\theta \cos\phi$
- $y = r \sin\theta \sin\phi$
- $z = r \cos\theta$
The inverse relationships are:
$$r = \sqrt{x^2 + y^2 + z^2}$$
$$\theta = \cos^{-1}\left(\frac{z}{r}\right) = \tan^{-1}\left(\frac{\sqrt{x^2+y^2}}{z}\right)$$
$$\phi = \tan^{-1}\left(\frac{y}{x}\right)$$
2. Calculate Partial Derivatives
Differentiate $r$, $\theta$, and $\phi$ with respect to $x$, $y$, and $z$ using standard calculus rules:
For $r$:
$$\frac{\partial r}{\partial x} = \frac{x}{r} = \sin\theta \cos\phi, \quad \frac{\partial r}{\partial y} = \frac{y}{r} = \sin\theta \sin\phi, \quad \frac{\partial r}{\partial z} = \frac{z}{r} = \cos\theta$$
For $\theta$:
Using $\cos\theta = z/r$, implicit differentiation yields:
$$\frac{\partial \theta}{\partial x} = \frac{\cos\theta \cos\phi}{r}, \quad \frac{\partial \theta}{\partial y} = \frac{\cos\theta \sin\phi}{r}, \quad \frac{\partial \theta}{\partial z} = -\frac{\sin\theta}{r}$$
For $\phi$:
Because $\phi$ is independent of $z$:
$$\frac{\partial \phi}{\partial x} = -\frac{\sin\phi}{r \sin\theta}, \quad \frac{\partial \phi}{\partial y} = \frac{\cos\phi}{r \sin\theta}, \quad \frac{\partial \phi}{\partial z} = 0$$
3. Expand via Multi-Variable Chain Rule
The spatial transformation operator for any Cartesian direction follows the pattern:
$$\frac{\partial}{\partial x} = \frac{\partial r}{\partial x}\frac{\partial}{\partial r} + \frac{\partial \theta}{\partial x}\frac{\partial}{\partial \theta} + \frac{\partial \phi}{\partial x}\frac{\partial}{\partial \phi}$$
Substitute the partial derivatives into the chain rule for each axis:
$$\frac{\partial}{\partial x} = \sin\theta \cos\phi \frac{\partial}{\partial r} + \frac{\cos\theta \cos\phi}{r}\frac{\partial}{\partial \theta} – \frac{\sin\phi}{r \sin\theta}\frac{\partial}{\partial \phi}$$
$$\frac{\partial}{\partial y} = \sin\theta \sin\phi \frac{\partial}{\partial r} + \frac{\cos\theta \sin\phi}{r}\frac{\partial}{\partial \theta} + \frac{\cos\phi}{r \sin\theta}\frac{\partial}{\partial \phi}$$
$$\frac{\partial}{\partial z} = \cos\theta \frac{\partial}{\partial r} – \frac{\sin\theta}{r}\frac{\partial}{\partial \theta}$$
4. Compute Second Derivatives and Combine
To find $\nabla^2 T$, apply each operator to itself (e.g., $\frac{\partial^2 T}{\partial z^2} = \frac{\partial}{\partial z}\left(\frac{\partial T}{\partial z}\right)$) and sum the expressions:
$$\nabla^2 T = \frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} + \frac{\partial^2 T}{\partial z^2}$$
The Radial Component ($r$):
Summing the pure second-order radial derivatives yields:
$$(\sin^2\theta \cos^2\phi + \sin^2\theta \sin^2\phi + \cos^2\theta)\frac{\partial^2 T}{\partial r^2} = \frac{\partial^2 T}{\partial r^2}$$
Accounting for cross-derivatives that act on the $1/r$ terms adds a first-order radial term:
$$\frac{2}{r}\frac{\partial T}{\partial r}$$
Combined, these form the identity: $\frac{\partial^2 T}{\partial r^2} + \frac{2}{r}\frac{\partial T}{\partial r} = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial T}{\partial r}\right)$.
The Polar Component ($\theta$):
Summing the pure second-order $\theta$ derivatives yields:
$$\frac{1}{r^2}\frac{\partial^2 T}{\partial \theta^2}$$
Accounting for cross-derivatives acting on the trigonometric terms adds a first-order polar term:
$$\frac{\cot\theta}{r^2}\frac{\partial T}{\partial \theta}$$
Combined, these form the identity: $\frac{1}{r^2}\left(\frac{\partial^2 T}{\partial \theta^2} + \cot\theta \frac{\partial T}{\partial \theta}\right) = \frac{1}{r^2 \sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta \frac{\partial T}{\partial \theta}\right)$.
The Azimuthal Component ($\phi$):
Summing the pure second-order $\phi$ derivatives and simplifying the trigonometric identities gives:
$$\frac{1}{r^2 \sin^2\theta}\frac{\partial^2 T}{\partial \phi^2}$$
All remaining mixed cross-terms (like $\frac{\partial^2 T}{\partial r \partial \theta}$) cancel out perfectly due to opposing signs in the expansions.
✅ Derived Spherical Equation
Substituting the transformed Laplacian back into the general energy conservation equation yields the final spherical coordinate system profile:
$$\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial T}{\partial r}\right) + \frac{1}{r^2 \sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta \frac{\partial T}{\partial \theta}\right) + \frac{1}{r^2 \sin^2\theta}\frac{\partial^2 T}{\partial \phi^2} + \frac{\dot{q}}{k} = \frac{1}{\alpha}\frac{\partial T}{\partial t}$$