# dot-product spherical

I want to calculate dotproduct of $e_r *e_\phi$ they are unit vectors in spherical. where my spherical coordinates is $(r,\phi,\theta)$

My attempt is first to convert them to cartesian:

which gives: after som algebra $sin \phi cos\phi(1-1)= 0$.

I was thinking that I could do this in spherical coordinates directly without converting to cartesian?

Because $e_r$=(1,0,0) , $e_\phi$=(0,1,0)

Like $e_r *e_\phi$= (1,0,0) dot (0,1,0)=0 ?

• What is your definition of $e_r, e_{\phi}$? – Travis Willse Oct 11 '15 at 19:55
• Just so. There are two different dot products to consider here: the one in $xyz$-space and the one in $r\phi\theta$-space. If you’re asking what the dot product of the images of $e_r$ and $e_\phi$ under the $xyz$ dot product is, then you’ll need to proceed as you did. If you’re asking about the dot product in $r\phi\theta$-space, it’s whatever you define it to be (most likely $0$ so that the basis vectors in that space are orthogonal). – amd Oct 11 '15 at 20:03

The unit vectors $\hat r$, $\hat \theta$, and $\hat \phi$ are mutually orthogonal. To show explicitly that $\hat r$ and $\hat \phi$ are orthogonal, we take their inner product and observe that it is zero.

To that end we first write the spherical unit vectors in Cartesian coordinates as

$\hat r=\hat x\sin \theta \cos \phi+\hat y\sin \theta \sin \phi+\hat z \cos \theta$

and

$\hat \phi=-\hat x \sin \phi+\hat y\cos \phi$

Therefore, their inner product is given by

\begin{align} \hat r \cdot \hat \phi&=\sin \theta(-\sin \phi \cos \phi+\sin \phi \cos \phi)\\\\ &=0 \end{align}

as expected!

• I have to write it in cartesian to truly show that they are orthogonal? or can you simply say that r dot theta = 0, r dot phi=0 etc, if they got no number in front of them – AustralianTurtle Oct 11 '15 at 21:18
• By definition, the inner product or orthogonal vectors is zero. So, one need not expand these orthogonal vectors in a different (e.g., Cartesian) orthogonal basis set. However, if one does so, they will, of course, still be orthogonal. – Mark Viola Oct 11 '15 at 21:31