- #1
squenshl
- 468
- 4
How do I find \nabla in Spherical Coordinates. Please help.
How do I Find \nabla in Spherical Coordinates?
- Context: Undergrad
- Thread starter squenshl
- Start date
Click For Summary
Discussion Overview
The discussion revolves around finding the gradient operator (\nabla) in spherical coordinates. Participants explore the theoretical and mathematical aspects of this topic, including definitions and derivations from first principles.
Discussion Character
- Technical explanation
- Mathematical reasoning
- Homework-related
Main Points Raised
- One participant asks for help in finding \nabla in spherical coordinates, indicating a need for clarification on the topic.
- Another participant suggests referring to a Wikipedia page for guidance, implying that the information is available but may not be straightforward.
- A participant expresses a desire to derive the components i, j, and k from the definition of \nabla, indicating a focus on foundational understanding.
- It is noted that \nabla can be expressed as a sum of partial derivatives with respect to x, y, and z multiplied by the unit vectors i, j, and k.
- One participant emphasizes the use of the chain rule in deriving the components, providing a partial derivation for \rho and suggesting similar approaches for other variables.
Areas of Agreement / Disagreement
There is no consensus on a single method for finding \nabla in spherical coordinates, and multiple approaches are discussed. Some participants provide references and partial solutions, while others seek more detailed explanations.
Contextual Notes
The discussion includes references to specific mathematical steps and assumptions, such as the definitions of spherical coordinates and the application of the chain rule, but does not resolve the overall method for finding \nabla.
Who May Find This Useful
Participants interested in vector calculus, particularly in the context of spherical coordinates, may find this discussion relevant.
- #2
tiny-tim
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Last edited by a moderator:
- #3
squenshl
- 468
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How do I go about doing it from scratch.
How do I find i, j, & k from the definition of \nabla
How do I find i, j, & k from the definition of \nabla
Last edited:
- #4
squenshl
- 468
- 4
\nabla = del/del(x) i + del/del(y) j + del/del(z) k
I found del/del(x), del/del)(y), del/del(z) but how do I find i, j, k. Help please.
I found del/del(x), del/del)(y), del/del(z) but how do I find i, j, k. Help please.
- #5
HallsofIvy
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Have you looked at the site tiny-tim gives? No one is going to go through the whole thing just for you! It's not terribly deep but very tedious!
Here's a start only:
Since \nabla u= \frac{\partial u}{\partial x}\vec{i}+ \frac{\partial u}{\partial y}\vec{j}+ \frac{\partial u}{\partial z}\vec{k}
so you need to use the chain rule
\frac{\partial u}{\partial x}= \frac{\partial u}{\partial \rho}\frac{\partial \rho}{\partial x}+ \frac{\partial u}{\partial \theta}\frac{\partial \theta}{\partial x}+ \frac{\partial u}{\partial \phi}\frac{\partial \phi}{\partial x}
Since \rho= (x^2+ y^2+ z^2)^{1/2},
\frac{\partial \rho}{\partial x}= (1/2)(x^2+ y^2+ z^2)^{-1/2}(2x}= \frac{\rho cos(\theta)sin(\phi)}{\rho}= cos(\theta)sin(\phi)
and similarly for the others.
Here's a start only:
Since \nabla u= \frac{\partial u}{\partial x}\vec{i}+ \frac{\partial u}{\partial y}\vec{j}+ \frac{\partial u}{\partial z}\vec{k}
so you need to use the chain rule
\frac{\partial u}{\partial x}= \frac{\partial u}{\partial \rho}\frac{\partial \rho}{\partial x}+ \frac{\partial u}{\partial \theta}\frac{\partial \theta}{\partial x}+ \frac{\partial u}{\partial \phi}\frac{\partial \phi}{\partial x}
Since \rho= (x^2+ y^2+ z^2)^{1/2},
\frac{\partial \rho}{\partial x}= (1/2)(x^2+ y^2+ z^2)^{-1/2}(2x}= \frac{\rho cos(\theta)sin(\phi)}{\rho}= cos(\theta)sin(\phi)
and similarly for the others.
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