Equation for finding the gradient in spherical coordinates

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The discussion revolves around calculating the gradient of a scalar field β = A/r in spherical coordinates, highlighting the confusion experienced when transitioning from Cartesian to spherical systems. The gradient formula presented is ∇β = ∂β/∂r ir + 1/r ∂β/∂θ iθ + 1/rsinθ ∂β/∂φ iφ. The user expresses frustration upon finding discrepancies between their calculated results and those provided in their textbook, particularly regarding the negativity of the gradient. It is noted that the negative value arises because the derivative of a decreasing function is negative, as seen in the exponent of r in the scalar field. The conversation emphasizes the importance of understanding the mathematical principles behind the gradient 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 basicly my frustration, how does this become negative?
 
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KUphysstudent said:
Well that is basicly 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
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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