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

In summary, the conversation discusses expanding the gradient of a function in spherical polar coordinates and calculating its value at a specific position. The resulting equation includes a term with an extra sinθ factor to account for the difference in longitude and latitude on a unit sphere.
  • #1
zheng
1
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.
 
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  • #2
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:
 

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

1. What is the meaning of [tex] \nabla f \cdot (p-p_0) [/tex] in spherical polar coordinates?

This expression represents the directional derivative of a function f at a point p in spherical polar coordinates, where p0 is the reference point. It measures the rate of change of f in the direction of the vector (p-p0).

2. How is [tex] \nabla f \cdot (p-p_0) [/tex] calculated in spherical polar coordinates?

In spherical polar coordinates, the gradient operator [tex]\nabla[/tex] is expressed as [tex]\nabla = \hat{r} \frac{\partial}{\partial r} + \hat{\theta} \frac{1}{r} \frac{\partial}{\partial \theta} + \hat{\phi} \frac{1}{r \sin \theta} \frac{\partial}{\partial \phi}[/tex], where [tex]\hat{r}[/tex], [tex]\hat{\theta}[/tex], and [tex]\hat{\phi}[/tex] are unit vectors in the radial, polar, and azimuthal directions, respectively. Therefore, [tex] \nabla f \cdot (p-p_0) = \frac{\partial f}{\partial r} (p-p_0)_r + \frac{1}{r} \frac{\partial f}{\partial \theta} (p-p_0)_\theta + \frac{1}{r \sin \theta} \frac{\partial f}{\partial \phi} (p-p_0)_\phi[/tex].

3. What is the significance of expanding [tex] \nabla f \cdot (p-p_0) [/tex] in spherical harmonics?

Expanding [tex] \nabla f \cdot (p-p_0) [/tex] in spherical harmonics allows us to decompose the directional derivative into its spherical components, which can provide insight into the directional properties of the function f. It also allows us to simplify the expression and make it easier to work with in certain cases.

4. Can [tex] \nabla f \cdot (p-p_0) [/tex] be expanded in any other coordinate system?

Yes, the gradient operator [tex]\nabla[/tex] and the dot product can be expressed in different coordinate systems, such as cylindrical or Cartesian coordinates. However, the specific form of the expansion may vary depending on the coordinate system.

5. How is [tex] \nabla f \cdot (p-p_0) [/tex] used in practical applications?

Expanding [tex] \nabla f \cdot (p-p_0) [/tex] can be useful in various fields of science, such as physics, engineering, and mathematics. It can be used to solve differential equations, analyze the properties of physical systems, and optimize functions. It is also commonly used in vector calculus and multivariate calculus courses to understand the concept of directional derivatives and gradients in different coordinate systems.

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