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

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SUMMARY

The discussion focuses on expanding the expression \(\nabla f \cdot (p - p_0)\) in spherical polar coordinates. The gradient \(\nabla f\) is expressed as \(\frac{\partial f}{\partial r} e_r + \frac{1}{r \sin \theta} \frac{\partial f}{\partial \phi} e_{\phi} + \frac{1}{r} \frac{\partial f}{\partial \theta} e_{\theta}\). The specific case analyzed is on the unit sphere where \(r = r_0 = 1\), leading to the result \(\left[\nabla f\right]_0 \cdot (p - p_0) = \left[\frac{1}{\sin \theta} \frac{\partial f}{\partial \phi}\right]_0 \left[\sin \theta (\phi - \phi_0)\right] + \left[\frac{\partial f}{\partial \theta}\right]_0 (\theta - \theta_0)\). The necessity of the second \(\sin \theta\) factor is clarified as it accounts for the varying lengths of longitude and latitude on the sphere.

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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.
 
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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:
 

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