Vector Field Transformation to Spherical Coordinates

[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?

 

[Charles Link]

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 $\cos{\phi}$

 

[haruspex]

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

 

[Teclis]

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?

 

[haruspex]

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.