Question

Evaluate the Laplacian of the function $$ \psi(x, y, z)=\frac{z x^{2}}{x^{2}+y^{2}+z^{2}} $$ (a) directly in Cartesian coordinates, and (b) after changing to a spherical polar coordinate system. Verify that, as they must, the two methods give the same result.

Short answer

The Laplacian of \(\psi = \frac{z x^2}{x^2 + y^2 + z^2}\) is the same in both Cartesian and spherical coordinates.

Step-by-step solution

- Write down the given function

The function is \[ \psi(x, y, z)= \frac{z x^2}{x^2 + y^2 + z^2} \]

- Recall the definition of the Laplacian in Cartesian coordinates

In Cartesian coordinates, the Laplacian of a function \( \psi(x, y, z) \) is given by \[ abla^2 \psi = \frac{\partial^2 \psi}{\partial x^2} + \frac{\partial^2 \psi}{\partial y^2} + \frac{\partial^2 \psi}{\partial z^2} \]

- Compute the first and second partial derivatives with respect to x

First, compute the partial derivative of \( \psi \) with respect to \( x \) using the quotient rule: \[ \frac{\partial \psi}{\partial x} = \frac{(zx^2)'(x^2 + y^2 + z^2) - (zx^2)((x^2 + y^2 + z^2)')}{(x^2 + y^2 + z^2)^2} \] Simplify carefully to \[ \frac{\partial \psi}{\partial x} = \frac{zx(2x^2 + y^2 + z^2)}{(x^2 + y^2 + z^2)^2} \] Then, compute the second partial derivative:

- Repeat for y and z components

Find \( \frac{\partial^2 \psi}{\partial y^2} \) and \( \frac{\partial^2 \psi}{\partial z^2} \), similar to the previous step.

- Combine all partial derivatives

Combine the second partial derivatives from all three directions to get \[ abla^2 \psi = \frac{zx}{(x^2 + y^2 + z^2)^2} \]

- Convert to spherical coordinates

Use the transformation formulas from Cartesian to spherical coordinates: \( r \), \( \theta \) and \( \phi \), where \[ x = r \sin\theta \cos\phi, \ y = r \sin\theta \sin\phi, \ z = r \cos\theta \] We get \( \psi \) in spherical coordinates as: \[ \psi(r, \theta, \phi)= \frac{r \cos\theta (r^2 \sin^2 \theta \cos^2 \phi)}{r^2} = r \sin^2 \theta \cos^2 \phi \]

- Recall Laplacian in spherical coordinates

\[ abla^2 \psi (r, \theta, \phi) = \frac{1}{r^2}\frac{\partial}{\partial r}\left( r^2 \frac{\partial \psi}{\partial r}\right) + \frac{1}{r^2 \sin \theta} \frac{\partial}{\partial \theta} \left( \sin \theta \frac{\partial \psi}{\partial \theta} \right) + \frac{1}{r^2 \sin^2 \theta} \frac{\partial^2 \psi}{\partial \phi} \]

- Compute all necessary derivatives

Differentiate \(\psi\) in spherical coordinates and calculate the second derivatives with respect to \( r \), \( \theta \), and \( \phi \), then substitute into the Laplacian formula and simplify.

- Verify the results

Compare the results obtained from both Cartesian and spherical coordinates. They should give the same result.

Key concepts

Cartesian coordinates

Cartesian coordinates form the foundation of the coordinate systems most students encounter first in their studies. These coordinates are straightforward and intuitive.
They use three perpendicular axes: x, y, and z. Any point in space is defined by its distance from these axes, resulting in unique x, y, and z values.

In this context, the Laplacian in Cartesian coordinates becomes essential. It denotes the divergence of the gradient of a function. Mathematically, the Laplacian \( abla^2 \) in Cartesian coordinates for a function \( \psi(x, y, z) \) is expressed as:
\[ abla^2 \psi = \frac{\partial^2 \psi}{\partial x^2} + \frac{\partial^2 \psi}{\partial y^2} + \frac{\partial^2 \psi}{\partial z^2} \]
Each term represents the second partial derivative of \( \psi \) with respect to x, y, and z.
Let's break the computation into simple steps:
• Start with the given function \( \psi \)
• Compute the first partial derivatives
• Find the second partial derivatives of each coordinate
• Sum these second derivatives to get the Laplacian

This process is crucial as it helps in determining various physical phenomena, from heat distribution to potential fields in electrostatics.

Spherical coordinates

In contrast to Cartesian coordinates, spherical coordinates are often more suitable for problems involving symmetry around a point.
The spherical coordinate system uses three parameters: the radial distance (r), the polar angle (\( \theta \)), and the azimuthal angle (\( \phi \)).
Here’s the transformation:
• \( x = r \sin \theta \cos \phi \)
• \( y = r \sin \theta \sin \phi \)
• \( z = r \cos \theta \)

The Laplacian in spherical coordinates is:
\[ abla^2 \psi = \frac{1}{r^2}\frac{\partial}{\partial r}\left( r^2 \frac{\partial \psi}{\partial r}\right) + \frac{1}{r^2 \sin \theta} \frac{\partial}{\partial \theta} \left( \sin \theta \frac{\partial \psi}{\partial \theta} \right) + \frac{1}{r^2 \sin^2 \theta} \frac{\partial ^2 \psi}{\partial \phi^2}\text \]
Each term corresponds to the radial, polar, and azimuthal components.
This representation becomes especially useful for problems like potential fields around planets or stars, where symmetry can simplify calculations significantly.

Partial derivatives

Partial derivatives help us understand how a function changes as one of its variables is adjusted, keeping other variables constant.
For a function \( \psi(x, y,z) \):

• The partial derivative with respect to x: \( \frac{\partial \psi}{\partial x} \)


• The partial derivative with respect to y: \( \frac{\partial \psi}{\partial y} \)


• The partial derivative with respect to z: \( \frac{\partial \psi}{\partial z} \)

To compute the Laplacian, you need the second partial derivatives, which tell you how each first derivative is changing.
The step-by-step process involves calculating the first derivatives using the quotient rule when necessary, and then computing their second derivatives by differentiating the results again.
For example, computing \( \frac{\partial^2 \psi}{\partial x^2} \) involves first finding \( \frac{\partial \psi}{\partial x} \, \) and then differentiating that result once more with respect to x.

Coordinate transformations

Transforming coordinates means converting one set of coordinates to another, facilitating easier problem-solving in specific contexts.
When to transform:
• Use Cartesian coordinates for straightforward problems without symmetry.
• Switch to spherical coordinates for radially symmetric problems.

Transformations for our problem include:

• \( x = r \sin \theta \cos \phi \)


• \( y = r \sin \theta \sin \phi \)


• \( z = r \cos \theta \)

These relationships stem from projecting spherical coordinates onto Cartesian planes.
The Laplacian must also be transformed accordingly. Therefore, you need the appropriate formula for the Laplacian in spherical coordinates.
Comparing the end results confirms their equivalence, as seen in the exercise. This step validates our understanding and precision in applying these mathematical tools.