I am having trouble with the following problem. I keep on getting a long unmanagable result - so any suggestion as to where I've gone wrong/how to do this would be a lifesaver! Please?

Consider a Vector Field in $\mathbb R^3 $ Given By F($\bar{x}$)=$\bar{\varepsilon} \times \bar{x}$
Where $\bar{\varepsilon}$ is a fixed non-zero vector and $\bar{x}$ is some variable vector

Compute this Vector Field in Spherical coordinates. I have assumed that they want me to express this field using spherical coordinates bases $\bar{e }_{p}$ $\bar{e }_{\phi}$ $\bar{e }_{\theta}$

$\bar{e }_{p}$ =$\cos \theta \sin \phi $ $\bar{i}$ + $\sin \theta \sin \phi $$\bar{j}$ + $\cos \phi$$\bar{k}$

$\bar{e }_{\phi}$ = $\cos \theta$ $\cos \phi $$\bar{i}$ + $\cos \phi $$ \sin \theta $$\bar{j}$ - $ \sin \phi $$\bar{k}$

$\bar{e }_{\theta}$ = $ \sin \theta $$\bar{i}$ +$\cos \theta $$\bar{j}$

Let $\bar{x}$ = x$\bar{i}$+y$\bar{j}$ + z$\bar{k}$

In Spherical Coordinates x=p $\cos \theta \sin \phi $
y=p $\sin \theta \sin \phi $
z=p $\cos \phi$

$\bar{x}$=p $\cos \theta \sin \phi $$\bar{i}$ += p $\sin \theta \sin \phi $$\bar{j}$ +p $\cos \phi$$\bar{k}$

Then $\bar{x}$= p$\bar{e }_{p}$

I then calculated the Fixed Vector in relation to the spherical coordinate base

$\bar{\varepsilon}$ = a$\bar{i}$+b$\bar{j}$ + c$\bar{k}$ (Vector in relation to Cartesian Base) where a,b,c, are constants

$\bar{\varepsilon}$ = a($\cos \theta \sin \phi $$\bar{e }_{p}$ + $\cos \phi \cos\theta $$\bar{e }_{\phi}$ - $\sin \theta $$\bar{e }_{\theta}$) + b($\sin \theta \sin \phi $$\bar{e }_{p}$ + $\cos \phi \sin \theta $$\bar{e }_{\phi}$ + $\sin \theta $$\bar{e }_{\theta}$) + c($\cos\phi $$\bar{e }_{p}$ - $\sin \phi $$\bar{e }_{\phi}$ (used inverse relation between bases)

$\bar{\varepsilon}$ = (a$\cos \theta \sin \phi $ + b$\sin \theta \sin \phi $+ c($\cos\phi $)$\bar{e }_{p}$ + (a$\cos \phi \cos\theta $+b$\cos \phi \sin \theta $ - c$\sin \phi $)$\bar{e }_{\phi}$ +(b$\sin \theta $-a$\sin \theta $)$\bar{e }_{\theta}$

I then took the cross-product of these two vectors.(I know how to do the cross-product so that's not an issue - however I am not sure that what I have done here is correct and I am extremely uncomfortable with the result. Where have I gone wrong? Any help, suggestion or comment would very very gratefully recieved

  • $\begingroup$ Can you express this in Cartesian coordinates? If so, then the Jacobian of the coordinate change will give you the coordinates for the vector field in Spherical coordinates. $\endgroup$ – DBFdalwayse Aug 22 '13 at 5:18
  • $\begingroup$ BTW; I id a very small edit, changing your $\Re^3$ into $ \mathbb R^3$; I hope that's O.K. $\endgroup$ – DBFdalwayse Aug 22 '13 at 5:19
  • 1
    $\begingroup$ @Mreaca You say $F(\bar{i})= \bar{\varepsilon} \times \bar{x}$. What do you mean by this? (the $\bar{i}$ seems strange to me here) $\endgroup$ – James S. Cook Aug 22 '13 at 5:25
  • $\begingroup$ oops! Notation error! thanks for pointing it out. $\endgroup$ – Mreaca Aug 22 '13 at 5:44
  • $\begingroup$ @Mreaca I think my answer is the same as your calculation, the main difference being that I just label the coefficient rather than keeping it expanded while taking the $\times$-product. This notational slight of hand saves me some thinking. Now, if we put those coefficients in explicitly and we also convert the spherical frame back into its cartesian equivalent and we multiply it all out... as a check we ought to get $(a \widehat{i}+b \widehat{j}+c \widehat{k}) \times x \widehat{i}+y \widehat{j}+z \widehat{k}$ back. But, I'd really just like to believe in what we've done and end the suffering $\endgroup$ – James S. Cook Aug 22 '13 at 6:03

I'll take a different approach here. Suppose $\vec{\varepsilon}$ is fixed vector, or constant vector field with respect to the cartesian frame. Then as we write $\vec{\varepsilon}$ in the spherical frame the coefficients manifest a point-dependence. In particular, there exist functions $\varepsilon_{\rho},\varepsilon_{\phi},\varepsilon_{\theta}$ such that $$\vec{\varepsilon} = \varepsilon_{\rho}\widehat{e}_{\rho}+\varepsilon_{\phi}\widehat{e}_{\phi}+\varepsilon_{\theta}\widehat{e}_{\theta} $$ we can calculate these coefficient functions by $\varepsilon_{\rho}=\vec{\varepsilon} \cdot \widehat{e}_{\rho}$, $\varepsilon_{\phi}=\vec{\varepsilon} \cdot \widehat{e}_{\phi}$ and $\varepsilon_{\theta} =\vec{\varepsilon} \cdot \widehat{e}_{\theta}$. This follows from the orthonormality of the spherical frame. (you can check, and I think you already realize $\widehat{e}_{\rho} \cdot \widehat{e}_{\rho}=1, \widehat{e}_{\rho} \cdot \widehat{e}_{\phi}=0$ etc...). However, you should also realize that $\{ \widehat{e}_{\rho}, \widehat{e}_{\phi}, \widehat{e}_{\theta} \}$ forms a right-handed-triple in the sense that the cross-products of $\widehat{e}_{\rho}, \widehat{e}_{\phi}, \widehat{e}_{\theta}$ share the same patterns as that of the standard cartesian frame: $$ \widehat{e}_{\rho} \times \widehat{e}_{\phi} = \widehat{e}_{\theta}, \ \ \widehat{e}_{\phi} \times \widehat{e}_{\theta} = \widehat{e}_{\rho}, \ \ \widehat{e}_{\theta} \times \widehat{e}_{\rho} = \widehat{e}_{\phi} $$ Now, as you point out, $\vec{x} = \rho \widehat{e}_{\rho}$ thus, $$ \vec{F} = \vec{\varepsilon} \times \vec{x} = (\varepsilon_{\rho}\widehat{e}_{\rho}+\varepsilon_{\phi}\widehat{e}_{\phi}+\varepsilon_{\theta}\widehat{e}_{\theta} ) \times \rho \widehat{e}_{\rho} = -\rho \varepsilon_{\phi}\widehat{e}_{\theta} + \rho \varepsilon_{\theta}\widehat{e}_{\phi}$$ now, we just need to calculate those dot-products and we're done.

That said, I'd rather use some notation like $\vec{A}$ instead of $\vec{ \varepsilon}$ since $e$ and $\varepsilon$ look so similar. In $\vec{A}$-notation, $$ \vec{F} = \vec{A} \times \vec{x} = (A_{\rho}\widehat{e}_{\rho}+A_{\phi}\widehat{e}_{\phi}+A_{\theta}\widehat{e}_{\theta} ) \times \rho \widehat{e}_{\rho} = -\rho A_{\phi}\widehat{e}_{\theta} + \rho A_{\theta}\widehat{e}_{\phi}$$

  • $\begingroup$ @Mreaca if I read your calculation correctly, we can already see that $\varepsilon_{\phi}=a\cos \phi \cos\theta +b\cos \phi \sin \theta - c\sin \phi $ and $\varepsilon_{\theta} = b\sin \theta -a\sin \theta $. ( I haven't checked to see these are correct) $\endgroup$ – James S. Cook Aug 22 '13 at 6:07
  • $\begingroup$ Thank you so so much! Your reply has really helped. Very informative (better than my [insert swear word of choice] textbook)Again Thank you so so much $\endgroup$ – Mreaca Aug 22 '13 at 6:22
  • $\begingroup$ @Mreaca I have also had this feeling towards certain texts at various times. The treatment of non-cartesian coordinates is really lacking in most of the texts for the basic calculus sequence. For me, it was the physics text of Griffith's Electrodynamics that first opened my eyes to the right way to think about vectors... in short, I feel your pain. $\endgroup$ – James S. Cook Aug 22 '13 at 6:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.