Yukawa potential energy function

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Homework Help Overview

The discussion revolves around deriving an expression for the radial force function from the Yukawa potential energy function. The original poster seeks confirmation on their approach to this derivation.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster considers two methods for deriving the radial force function: normal differentiation and partial differentiation. They express uncertainty about the necessity of partial differentiation given that the function depends solely on the variable r.
  • Some participants clarify that the normal derivative is equivalent to the partial derivative in this context, given that the function is a single-variable function. They also question whether spherical coordinates are relevant for this derivation.

Discussion Status

The discussion is ongoing, with participants exploring the implications of using spherical coordinates and confirming the appropriateness of the differentiation methods. There is no explicit consensus yet, but guidance has been provided regarding the treatment of the radial axis.

Contextual Notes

Participants note that the problem specifies only one axis, the radial axis, which may influence the approach to the gradient calculation.

astenroo
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Homework Statement


This is not yet an attempt at solving a problem. I just need confirmation on that I'm on the right track. So, I am supposed to derive an expression for the radial force function from the given Yukawa-function.


Homework Equations



U(r) = -(r/r0)U0 exp-(-r/r0)

The Attempt at a Solution



I'm thinking about two possibilities here. First: Derivation of said function with respect to r (normal differential). Second: Partial differentiation to solve for the gradient (although I think this is not necessary since r is a fixed value, and no coordinates are given in the context of the problem)
 
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RoyalCat said:
[tex]\vec F = -\nabla U[/tex]

http://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates

In this case the normal derivative is the same as the partial derivative.

Ah true, since this function only has one variable which is r. What I am a bit concerned about is that if the system for spherical coordinates should be used in this derivation... Or they aren't needed since the radial force is dependent only on r (in this given situation)?
 
astenroo said:
Ah true, since this function only has one variable which is r. What I am a bit concerned about is that if the system for spherical coordinates should be used in this derivation... Or they aren't needed since the radial force is dependent only on r (in this given situation)?

It doesn't matter, you have only one axis specified in the problem, and that's the radial axis. The way to take the gradient with respect to the radial axis is [tex]\frac{\partial f}{\partial r}\hat r[/tex]
It doesn't matter what the other two axes are.
 

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