Equation for finding the gradient in spherical coordinates

In summary, the conversation discusses the difficulty of using the equation for finding the gradient in spherical coordinates and the frustration of the speaker when trying to solve a problem using this equation. The problem being solved involves finding the gradient of a scalar field in spherical coordinates. The speaker expresses confusion over why the result of the calculation is negative and seeks clarification on this issue. The summary also mentions the speaker's realization that the exponent of r in the equation for the gradient is -1, which explains the negative result.
  • #1
KUphysstudent
40
1
<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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  • #2
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
 

1. What is the equation for finding the gradient in spherical coordinates?

The equation for finding the gradient in spherical coordinates is:
∇f = (∂f/∂r)er + (1/r)(∂f/∂θ)eθ + (1/rsinθ)(∂f/∂φ)eφ

2. How is the gradient represented in spherical coordinates?

In spherical coordinates, the gradient is represented by the vector ∇f, which is a combination of the partial derivatives with respect to the radial distance (r), polar angle (θ), and azimuthal angle (φ).

3. What is the purpose of finding the gradient in spherical coordinates?

The gradient in spherical coordinates is used to determine the direction and magnitude of the steepest increase of a function in 3-dimensional space. It is also useful in solving problems involving heat flow, fluid dynamics, and other physical phenomena.

4. How is the gradient calculated in spherical coordinates?

The gradient in spherical coordinates is calculated by taking the partial derivative of the function with respect to each coordinate (r, θ, φ) and multiplying it by the unit vector in that direction. The resulting vectors are then added together to form the gradient vector ∇f.

5. Can the gradient in spherical coordinates be used to find the directional derivatives?

Yes, the gradient in spherical coordinates can be used to find the directional derivative of a function in a specific direction. This is done by taking the dot product of the gradient vector and the unit vector representing the desired direction.

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