- #1
squenshl
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How do I find \nabla in Spherical Coordinates. Please help.
How do I Find \nabla in Spherical Coordinates?
- Thread starter squenshl
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SUMMARY
The discussion focuses on calculating the gradient operator \nabla in spherical coordinates. The key formula presented is \nabla u = \frac{\partial u}{\partial x}\vec{i} + \frac{\partial u}{\partial y}\vec{j} + \frac{\partial u}{\partial z}\vec{k}, with an emphasis on using the chain rule for partial derivatives. The transformation from Cartesian to spherical coordinates involves calculating derivatives of the radius ρ, polar angle θ, and azimuthal angle φ. Specific derivatives such as \frac{\partial \rho}{\partial x} = \frac{\rho \cos(\theta) \sin(\phi)}{\rho} are derived, illustrating the tedious nature of the calculations.
PREREQUISITES- Understanding of vector calculus and the gradient operator.
- Familiarity with spherical coordinate transformations.
- Knowledge of partial derivatives and the chain rule.
- Basic proficiency in mathematical notation and symbols.
- Study the derivation of the gradient in spherical coordinates using resources like Wikipedia's spherical coordinate system page.
- Practice calculating partial derivatives in spherical coordinates.
- Explore applications of the gradient operator in physics, particularly in electromagnetism and fluid dynamics.
- Learn about the divergence and curl operators in spherical coordinates for a comprehensive understanding of vector calculus.
Students and professionals in mathematics, physics, and engineering who are working with vector calculus and need to apply the gradient operator in spherical coordinates.
- #2
tiny-tim
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- #3
squenshl
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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
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\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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