Calculating Gradient of 1/|r-r'|: Tips & Results

In summary, the conversation discusses the calculation of the function gradient with respect to r using spherical coordinates. The speaker initially solves the problem using Cartesian coordinates and gets a correct result. However, when attempting to solve using spherical coordinates, their calculation leads to an incorrect result of $$\nabla \frac{1}{|r-r'|}=-\frac{1}{2}(r-r')^{-\frac{3}{2}}$$. They request help and ask for tips on how to solve the problem correctly in spherical coordinates.
  • #1
TheGreatDeadOne
22
0
Homework Statement
calculate the function gradient with respect to r
Relevant Equations
gradient in spherical coordinates
Doing R=|r-r'|, i get the expected result: [tex] \nabla \frac{1}{|r-r'|} = -\frac{1}{R^2}\hat r=-\frac{(r-r')}{|r-r'|^3}[/tex]

But doing it this way seems extremely wrong, as I seem to be disregarding the module. So I tried to do it by the chain rule, and I got:

[tex] \nabla \frac{1}{|r-r'|}=-\frac{1}{2}(r-r')^{-\frac{3}{2}}[/tex]

but that doing so looks much more correct using Cartesian coordinates. So, does anyone have any tips?
 
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  • #2
Let us see physical dimension of the quantity. It should be ##L^{-2}## but your calculation says ##L^{-3/2}##.
 
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  • #3
... not to mention that ##\nabla## is a vector operator. I.e. ##\nabla f## is a vector.
 
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  • #4
TheGreatDeadOne said:
Homework Statement:: calculate the function gradient with respect to r
Relevant Equations:: gradient in spherical coordinates

So I tried to do it by the chain rule, and I got:

[tex] \nabla \frac{1}{|r-r'|}=-\frac{1}{2}(r-r')^{-\frac{3}{2}}[/tex]

but that doing so looks much more correct using Cartesian coordinates. So, does anyone have any tips?
Something must have gone terribly wrong in your calculation cause the end result doesn't even make sense: r and r' are vectors and so is their difference r-r', But how do you raise a vector to the power of (-3/2)?

Please show us your calculation in cartesian coordinates to allow us to see where you go wrong.
 
  • #5
Delta2 said:
Something must have gone terribly wrong in your calculation cause the end result doesn't even make sense: r and r' are vectors and so is their difference r-r', But how do you raise a vector to the power of (-3/2)?

Please show us your calculation in cartesian coordinates to allow us to see where you go wrong.

Spherical coordinates. Using Cartesian coordinates I solve correctly, i am trying for spherical coordinates.
 
  • #6
TheGreatDeadOne said:
Spherical coordinates. Using Cartesian coordinates I solve correctly, i am trying for spherical coordinates.
Ok, well then please show us your calculation in spherical coordinates how you got $$\nabla \frac{1}{|r-r'|}=-\frac{1}{2}(r-r')^{-\frac{3}{2}}$$ with as much detail as possible.
 

1. How do you calculate the gradient of 1/|r-r'|?

To calculate the gradient of 1/|r-r'|, you will need to use the quotient rule of differentiation. This involves taking the derivative of the numerator and denominator separately, and then using the formula (f'g - fg')/g^2 to find the gradient.

2. What is the purpose of calculating the gradient of 1/|r-r'|?

The gradient of 1/|r-r'| is used in vector calculus to find the direction and magnitude of the steepest ascent or descent of a scalar field. This is useful in various fields such as physics, engineering, and economics.

3. Can you provide an example of calculating the gradient of 1/|r-r'|?

For example, if we have the function f(x,y) = 1/|x-y|, the gradient would be calculated as (1/|x-y|)' = ((1)'|x-y| - (1)|x-y|')/|x-y|^2 = (0 - 1/|x-y|^2)/|x-y|^2 = -1/|x-y|^4. Therefore, the gradient of f(x,y) is -1/|x-y|^4.

4. Are there any tips for calculating the gradient of 1/|r-r'|?

One tip is to simplify the expression as much as possible before differentiating. This can make the calculation easier and reduce the chances of making a mistake. Additionally, make sure to carefully apply the quotient rule and double check your work.

5. What are some potential results when calculating the gradient of 1/|r-r'|?

The result of calculating the gradient of 1/|r-r'| will always be a vector, as it represents both the direction and magnitude of change in the scalar field. Depending on the specific function, the gradient may be a constant vector or a vector-valued function.

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