- #1

Teclis

- 25

- 2

- 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?

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: