# Vector Field Transformation to Spherical Coordinates

• Teclis
In summary, the conversation is about solving a problem from a textbook involving a vector field in spherical coordinates. The solution involves using matrix equations and transforming scalar functions. However, there is a discrepancy with one of the answers provided in the textbook, which may be due to a typo or an unintentional truncation.
Teclis
Homework Statement
Formulate the vector field

$$\mathbf{\overrightarrow{a}} = x_{3}\mathbf{\hat{e_{1}}} + 2x_{1}\mathbf{\hat{e_{2}}} + x_{2}\mathbf{\hat{e_{3}}}$$

in spherical coordinates.
Relevant Equations
$$\begin{bmatrix} \hat{e_{r}} \\ \hat{e_{\upsilon}} \\ \hat{e_{\phi}} \end{bmatrix} = \begin{bmatrix} \sin\upsilon\cos\phi &\sin\upsilon\sin\phi & \cos\upsilon \\ \cos\upsilon\cos\phi & \cos\upsilon\sin\phi & -\sin\upsilon \\ -\sin\phi & \cos\phi & 0 \end{bmatrix} \begin{bmatrix} \hat{e_{x}} \\ \hat{e_{y}} \\ \hat{e_{z}} \end{bmatrix}$$

and

$$\begin{bmatrix} \hat{e_{x}} \\ \hat{e_{y}} \\ \hat{e_{z}} \end{bmatrix} = \begin{bmatrix} \sin\upsilon\cos\phi &\cos\upsilon\cos\phi & -\sin\phi \\ \sin\upsilon\sin\phi & \cos\upsilon\sin\phi & \cos\phi \\ \cos\upsilon & -\sin\upsilon & 0 \end{bmatrix} \begin{bmatrix} \hat{e_{r}} \\ \hat{e_{\upsilon}} \\ \hat{e_{\phi}} \end{bmatrix}$$
I am trying to solve the following problem from my textbook:

Formulate the vector field

$$\mathbf{\overrightarrow{a}} = x_{3}\mathbf{\hat{e_{1}}} + 2x_{1}\mathbf{\hat{e_{2}}} + x_{2}\mathbf{\hat{e_{3}}}$$

in spherical coordinates.My solution is the following:

For the unit vectors I use the following matrix equations:
$$\begin{bmatrix} \hat{e_{r}} \\ \hat{e_{\upsilon}} \\ \hat{e_{\phi}} \end{bmatrix} = \begin{bmatrix} \sin\upsilon\cos\phi &\sin\upsilon\sin\phi & \cos\upsilon \\ \cos\upsilon\cos\phi & \cos\upsilon\sin\phi & -\sin\upsilon \\ -\sin\phi & \cos\phi & 0 \end{bmatrix} \begin{bmatrix} \hat{e_{x}} \\ \hat{e_{y}} \\ \hat{e_{z}} \end{bmatrix}$$

and

$$\begin{bmatrix} \hat{e_{x}} \\ \hat{e_{y}} \\ \hat{e_{z}} \end{bmatrix} = \begin{bmatrix} \sin\upsilon\cos\phi &\cos\upsilon\cos\phi & -\sin\phi \\ \sin\upsilon\sin\phi & \cos\upsilon\sin\phi & \cos\phi \\ \cos\upsilon & -\sin\upsilon & 0 \end{bmatrix} \begin{bmatrix} \hat{e_{r}} \\ \hat{e_{\upsilon}} \\ \hat{e_{\phi}} \end{bmatrix}$$

Transforming the scalar functions into spherical coordinates I have the following equation:

$$\mathbf{\overrightarrow{a}} = r\cos\upsilon\hspace{1mm}\mathbf{\hat{e_{1}}} + 2r\sin\upsilon\cos\phi \hspace{1mm} \mathbf{\hat{e_{2}}} + r\sin\upsilon\sin\phi \hspace{1mm}\mathbf{\hat{e_{3}}}$$

Substituting the values for the standard basis vectors and rearranging to the following form

$$\mathbf{\overrightarrow{a}} = a_{r}\mathbf{\hat{e_{r}}} + a_{\upsilon}\mathbf{\hat{e_{\upsilon}}} + a_{\phi}\mathbf{\hat{e_{\phi}}}$$

I arrive at

$$\mathbf{\overrightarrow{a}} = r \cos{\upsilon}(\sin{\upsilon}\cos{\phi} \hspace{1mm}\mathbf{\hat{e_{r}}}+ \cos{\upsilon} \cos{\phi} \hspace{1mm} \mathbf{\hat{e_{\upsilon}}}- \sin{\phi} \hspace{1mm} \mathbf{\hat{e_{\phi}}})+ \\ \hspace{25mm}2r \sin{\upsilon} \cos{\phi}( \sin{\upsilon} \sin{\phi} \hspace{1mm} \mathbf{\hat{e_{r}}}+ \cos{\upsilon} \sin{\phi} \hspace{1mm} \mathbf{\hat{e_{\upsilon}}}+ \cos{\phi} \hspace{1mm} \mathbf{\hat{e{\phi}}}) + \\ \hspace{1mm}r \sin{\upsilon} \sin{\phi} ( \cos{\upsilon} \hspace{1mm} \mathbf{\hat{e_{r}}}- \sin{\upsilon} \hspace{1mm} \mathbf{\hat{e_{\upsilon}}})$$

and solving for the scalar functions

$$a_{r} = r\cos\upsilon \sin\upsilon \cos \phi + 2r \sin^2{ \upsilon} \cos \phi \sin \phi + r \sin \upsilon \sin \phi \cos \upsilon$$

$$a_{\upsilon} = r \cos^2 \upsilon \cos \phi + 2r \sin \upsilon \cos \phi \cos \upsilon \sin \phi -r \sin^2 \upsilon \sin \phi$$

$$a_{\phi} = 2r \sin \upsilon \cos^2 \phi - r \cos \upsilon \sin \phi$$

All my answers match those given in the back of the textbook except for ##a_{r}## which is given as

$$a_{r} = \cos \phi + r \sin \upsilon \cos \upsilon \sin \phi$$

I have tried different trigonometric identities but am unable to rearrange my solution for ##a_{r}## to match the answer given in the text. Could someone please point out the error in my calculations or confirm for me that the answer provided by the textbook is incorrect?

Last edited:
I found only one "typo" that didn't propagate: It should say ## a=...r \sin{v} \, sin{\phi} \, e_3 ## in the middle of the page, with ## \sin{\phi} ##. I didn't check everything, but I think your result might be correct.

Teclis
I found only one "typo" that didn't propagate: It should say ## a=...r \sin{v} \, sin{\phi} \, e_3 ## in the middle of the page, with ## \sin{\phi} ##. I didn't check everything, but I think your result might be correct.
Thanks, you are correct it is a typo. I do in fact have ##\sin{\phi}## in my paper calculations and not ##\cos{\phi}##

Teclis said:
$$a_{r} = r\cos\upsilon \sin\upsilon \cos \phi + 2r \sin^2{ \upsilon} \cos \phi \sin \phi + r \sin \upsilon \sin \phi \cos \upsilon$$
All my answers match those given in the back of the textbook except for ##a_{r}## which is given as
$$a_{r} = \cos \phi + r \sin \upsilon \cos \upsilon \sin \phi$$
Your answer can be written (over two lines) as
$$a_{r} = r\cos\upsilon \sin\upsilon \cos \phi + 2r \sin^2{ \upsilon}\sin \phi$$ $$\cos \phi+ r \sin \upsilon \cos \upsilon \sin \phi$$
Notice something?

haruspex said:
Your answer can be written (over two lines) as
$$a_{r} = r\cos\upsilon \sin\upsilon \cos \phi + 2r \sin^2{ \upsilon}\sin \phi$$ $$\cos \phi+ r \sin \upsilon \cos \upsilon \sin \phi$$
Notice something?

Yes, so you think the answer in the Textbook has just been unintentionally truncated?

Teclis said:
Yes, so you think the answer in the Textbook has just been unintentionally truncated?
Looks like it. What's left is not even dimensionally correct.

## 1. What is a vector field transformation to spherical coordinates?

A vector field transformation to spherical coordinates is a mathematical process used to convert a vector field from Cartesian coordinates (x, y, z) to spherical coordinates (r, θ, φ). This allows for a more intuitive understanding of the vector field and can simplify calculations in certain situations.

## 2. Why is it useful to transform a vector field to spherical coordinates?

Transforming a vector field to spherical coordinates can be useful in situations where the vector field is more naturally described in terms of radial distance, polar angle, and azimuthal angle. It can also simplify calculations in certain problems involving spherical objects or systems.

## 3. How do you perform a vector field transformation to spherical coordinates?

To perform a vector field transformation to spherical coordinates, you first need to identify the coordinates of the vector field in Cartesian form (x, y, z). Then, you can use the following equations to convert to spherical coordinates:
r = √(x² + y² + z²)
θ = arccos(z/√(x² + y² + z²))
φ = arctan(y/x)

## 4. What are some common applications of vector field transformation to spherical coordinates?

Vector field transformation to spherical coordinates is commonly used in physics and engineering applications, such as in electromagnetism, fluid dynamics, and celestial mechanics. It is also used in computer graphics and image processing for 3D transformations.

## 5. Are there any limitations to using vector field transformation to spherical coordinates?

One limitation of using vector field transformation to spherical coordinates is that it can only be applied to vector fields that are symmetric around a central point. It is also important to note that the transformation may introduce errors or distortions in the vector field, depending on the complexity of the original field and the accuracy of the calculations.

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