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I have been trying to derive the expressions of partial derivatives of unit vectors with respect to each other in the spherical coordinate system. I was able to get all of them except $\frac{\partial \hat{\phi}}{\partial\theta}$

Note: I am using the mathematics convention of denoting azimuthal angle as $\phi$ and polar angle as $\theta$

Here's the figure I've made but I'm not able to find the angle of interest marked with a question mark in the figure.

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The drawing in my initial figure is wrong. The figure attached below is the correct one and it saliently provides the answer. enter image description here The angle of our interest is $\angle BAC$ named $\alpha$. To find this, $$ \tan{\phi}=\frac{AB}{OB}=\frac{AB}{r}$$ $$ AB=r\tan{\phi} $$ Now, $$ \tan{\alpha}=\frac{BC}{AB}$$ From the figure it is clear that $BC=r\sin{\phi}d\theta$ and we found that $AB=r\tan{\phi}$. Substituting, $$ \tan{\alpha}=\frac{r\sin{\phi}\;d\theta}{r\tan{\phi}}=\cos{\phi}\;d\theta$$

Hence the magnitude of change in the unit vector $ \hat{\theta}$ is $1\cdot\alpha=\tan{\alpha}=\cos{\phi}\;d\theta\;$ for small angles. From the figure it is clear that the direction is $\hat{\theta}$. Putting all of this together, $$ \frac{\partial\hat{\phi}}{\partial\theta}=\cos{\phi}\;\hat{\theta}$$

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  • $\begingroup$ In your question you say you need $\partial\hat\theta/\partial\phi$, but in this answer you find $\partial\hat\phi/\partial\theta$. $\endgroup$
    – rob
    Commented Aug 29, 2020 at 19:08
  • $\begingroup$ Thanks for pointing that out. My apologies, it was a mistake on my part getting muddled up in the physics vs. math conventions about spherical coordinates. I have reflected the change in the question. $\endgroup$ Commented Aug 31, 2020 at 10:19

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