How to find the x-component of a spherical vector?

Question

I am given the point $P(r=0.89, \theta=30^\omicron, \phi=45^\omicron)$ and $\vec E=1/r^2(cos(\phi)\hat a_r +sin(\theta)\hat a_\phi)$.

Find the x-component of $\vec E$ at $P$.

I found the vector in spherical components ($0.8927\hat a_r + 0.6312\hat a_\phi$), but I am not sure how to turn this in to Cartesian components.

Answer 1

So, I figured it out. I was converting $\hat a_\phi$ wrong. From $\hat a_r$, $\hat a_x=sin(\theta)cos(\phi)$ and from $\hat a_\phi$, $\hat a_x=-sin(\phi)$. I was missing the negative sign on the $sin$.

So now you have $\vec E_x=1/r^2(sin(\theta)cos^2(\phi) - sin(\theta)sin(\phi))$.

Answer 2

I am not sure how $\hat{a}_r$ and $\hat{a}_{\phi}$ are defined. But suspect that there are two possibilities.

• Either they are the same as $\hat{r}$ and $\hat{\phi}$, which expressed in Cartesian coordinates look like, $$\hat{r}=\hat{x}\sin\theta\cos\phi+\hat{y}\sin\theta\sin\phi+\hat{z}\cos\theta$$ $$\hat{\phi}=-\hat{x}\sin\phi+\hat{y}\cos\phi$$ $$\hat{\theta}=\hat{x}\cos\theta\cos\phi+\hat{y}\cos\theta\sin\phi-\hat{z}\sin\theta$$
• Or they mean the second time derivative from $\hat{r}$ and $\hat{\phi}$, $$\ddot{\hat{r}}=-\hat{r}\left(\dot{\theta}^2+\dot{\phi}^2\sin\theta\right)+\hat{\theta}\left(\ddot{\theta}+\dot{\phi}^2\sin\theta\cos\theta\right)+\hat{\phi}\left(\ddot{\phi}\sin\theta+2\dot{\theta}\dot{\phi}\cos\theta\right)$$ $$\ddot{\hat{\phi}}=-\hat{r}\left(\ddot{\phi}\sin\theta+2\dot{\theta}\dot{\phi}\cos\theta\right)+\hat{\theta}\left(\ddot{\phi}\cos\theta-2\dot{\theta}\dot{\phi}\sin\theta\right)+\hat{\phi}\ddot{\phi}\left(\cos^2\theta-\sin^2\theta\right)$$

However I suspect they mean the first rather than the last.