How te expand [tex] \nabla f \cdot (p-p_0) [/tex]in spherical polar coordinates

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
1 reply · 2K views
zheng
Messages
1
Reaction score
0
how to expand grad f * (p-p_0) in spherical polar coordinates

in spherical polar coordinates:
[tex]\nabla f[/tex] = [tex]\frac{\partial f}{\partial r} e_r[/tex]+ [tex]\frac{1}{r sin\theta}\frac{\partial f}{\partial \phi} e_{\phi}[/tex]+ [tex]\frac{1}{r}\frac{\partial f}{\partial \theta} e_{\theta}[/tex]

[tex]p=(r,\phi,\theta)[/tex] and [tex]p_0=(r_0,{\phi}_0,{\theta}_0)[/tex] is the position vectors.

in [tex]r=r_0=1[/tex] surface, what is [tex]\left[\nabla f\right]_0 \cdot (p-p_0)[/tex], where [tex]\left[\nabla f\right]_0[/tex] is the gradient of f in position [tex]p_0[/tex]

in one paper, the answer is [tex]\left[\nabla f\right]_0 \cdot (p-p_0)=\left[\frac{1}{sin\theta}\frac{\partial f}{\partial \phi} \right]_0 \left[sin\theta (\phi-{\phi}_0)\right]+\left[ \frac{\partial f}{\partial \theta} \right]_0 (\theta-{\theta}_0)[/tex]. I do not know why the second [tex]sin\theta[/tex] is needed.
 
Last edited:
Physics news on Phys.org
welcome to pf!

hi zheng! welcome to pf! :smile:

we're on the unit sphere, the difference in longitude is (θ - θ0) and the difference in latitude is (φ - φ0)

but you get more longitude than latitude for the same length, by a factor of sinθ :wink: