Equation for finding the gradient in spherical coordinates

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SUMMARY

The discussion focuses on calculating the gradient of a scalar field β = A/r in spherical coordinates, where r = (x²+y²+z²)^(1/2) and A is a constant. The gradient is expressed as ∇β = ∂β/∂r ir + (1/r) ∂β/∂θ iθ + (1/(r sinθ)) ∂β/∂φ iφ. The user expresses confusion regarding the negative results obtained, particularly in relation to the exponent of r in the scalar field, which is -1, leading to negative derivatives. This highlights the importance of understanding the implications of negative derivatives in the context of gradient calculations.

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  • Understanding of spherical coordinates and their conversion from Cartesian coordinates.
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  • Knowledge of partial derivatives and their application in scalar fields.
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Students and professionals in mathematics, physics, and engineering who are working with vector calculus and need to understand gradient calculations in spherical coordinates.

KUphysstudent
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<Mentor note: moved from a technical forum and therefore without template>So I´m trying to understand how to use the equation for finding the gradient in spherical coordinates, just going from cartesian to spherical seemed crazy. Now I´m at a point where I want to try out what I have read and I immediately run into problems, which clearly tells me I have no idea what I´m doing.

Problem I was trying to solve:
Given a scalarfield β = A/r where r = (x^2+y^2+z^2)^1/2 and A is a konstant, calculate the gradient in spherical coordinates.

∇β = ∂β/∂r ir + 1/r ∂β/∂θ iθ + 1/rsinθ ∂β/∂φ iφ

When I thought the solution was pretty simply and then I go to the back of my book to check the result and I´m not even close.
How on Earth does the result become negative? it is also negative in Cartesian coordinates which don´t understand either.
Well that is basically my frustration, how does this become negative?
 
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KUphysstudent said:
Well that is basically my frustration, how does this become negative
If something becomes smaller, the derivative is negative !
From your ##\nabla\beta## in spherical coordinates, all that remains is the ##\partial \over \partial r## and the exponent of ##r## in ##\beta## is -1
 

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