The principle which I have been taught is taking $d\hat{\phi}$,$d\hat{\theta}$ and $d\hat{r}$, and the cross product of any two for example $d\hat{\phi}$ and $d\hat{\theta}$ should give me vector which is in direction with the $d\hat{r}$ vector. That is the cross product between the initial cross product and $d\hat{r}$ is $0$. I tried it but I am not getting as $0$. Following is my attempt.
Spherical Coordinates:
$x=r\sin{\theta} \cos{\phi}$
$y=r\sin{\theta} \sin{\phi}$
$z=r\cos{\theta} $
$d\vec{l}_\phi=r\sin{\theta} d\phi\ \hat{\phi}$
$d\vec{l}_\theta=rd\theta\ \hat{\theta}$
$d\vec{r}=dr\ \hat{r}$
To calculate $d\vec{\phi}=|{d\phi}|\hat{\phi}$ and $d\vec{\theta}=|{d\theta}|\hat{\theta}$, making use of the spherical coordinates in terms of cartesian, I get $$d\vec{\phi}=\frac{-ydx}{x^2+y^2}\hat{i}+\frac{xdy}{x^2+y^2}\hat{j}$$ $$d\vec{\theta}=\frac{xzdx}{\sqrt{x^2+y^2}(x^2+y^2+z^2)}\hat{i}+\frac{yzdy}{\sqrt{x^2+y^2}(x^2+y^2+z^2)}\hat{j}-\frac{\sqrt{x^2+y^2}dz}{(x^2+y^2+z^2)}\hat{k}$$
then I put them in the $d\vec{l}_\phi$ and $d\vec{l}_\theta$ which I am getting as
$$\frac{xdydz}{\sqrt{x^2+y^2+z^2}}\hat{i}-\frac{ydxdz}{\sqrt{x^2+y^2+z^2}}\hat{j}-\frac{zdxdy}{\sqrt{x^2+y^2}\sqrt{x^2+y^2+z^2}}\hat{k}$$ further I cross product this with $\vec{r}$ I am supposed to get each term as $0$ but I don't.

  • 1
    $\begingroup$ Why not compute pairwise dot products instead? You method only tells you whether or not $d\phi$ and $d\theta$ are both orthogonal to $dr$. It says nothing about the orthogonality of the first two vectors, so you still have to check that some other way. $\endgroup$
    – amd
    May 1, 2020 at 16:30
  • $\begingroup$ yes that is a way, dot products didnt come to my mind. Still, how are the dot products $0$?how do the ${dx}^2$ and ${dy}^2$ terms vanish? $\endgroup$
    – 1500kook12
    May 1, 2020 at 16:39
  • $\begingroup$ Crossposted to physics.stackexchange.com/q/550643/2451 $\endgroup$
    – Qmechanic
    May 9, 2020 at 16:13
  • $\begingroup$ sorry for that, i guess i was too impatient. i wont repeat it. $\endgroup$
    – 1500kook12
    May 9, 2020 at 16:52

1 Answer 1


I'm here from your question in the Physics StackExchange. (Please keep questions such as these on this site.) So, let's get started. I'll use the $(r,\theta,\varphi)$ convention. We know that $$ \begin{array}{l} \hat{\mathbf{r}} =\sin( \varphi )\cos( \theta )\hat{\mathbf{i}} +\sin( \varphi )\sin( \theta )\hat{\mathbf{j}} +\cos( \varphi )\hat{\mathbf{k}}\\ \hat{\boldsymbol{\theta}} =-\sin( \theta )\hat{\mathbf{i}} +\cos( \theta )\hat{\mathbf{j}}\\ \hat{\boldsymbol{\varphi }} =\cos( \varphi )\cos( \theta )\hat{\mathbf{i}} +\cos( \varphi )\sin( \theta )\hat{\mathbf{j}} -\sin( \varphi )\hat{\mathbf{k}} \end{array}$$ You can take the dot products from here.

  • $\begingroup$ why am i not able to prove in the way i mentioned? or is it like it is getting proved but its too complex to prove it explicitly? also i apologise for the crossposting. $\endgroup$
    – 1500kook12
    May 9, 2020 at 16:56
  • 1
    $\begingroup$ It might be doable that way, but it's horribly complex. Doing it this way is much easier. $\endgroup$
    – K.defaoite
    May 9, 2020 at 20:56
  • $\begingroup$ Besides, take the dot product, not cross product. $\endgroup$
    – K.defaoite
    May 9, 2020 at 20:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .