# Yukawa potential energy function

• astenroo
In summary, the conversation discusses the process of deriving an expression for the radial force function from a given Yukawa-function. There are two possibilities mentioned: using normal or partial differentiation. It is noted that in this case, the normal derivative is the same as the partial derivative since the function only has one variable, r. It is also mentioned that the system for spherical coordinates may not be necessary in this derivation since the radial force is dependent only on r.
astenroo

## 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)

RoyalCat said:
$$\vec F = -\nabla U$$

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 dependant 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 dependant 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 $$\frac{\partial f}{\partial r}\hat r$$
It doesn't matter what the other two axes are.

I can confirm that you are on the right track. To derive the expression for the radial force function, you can indeed use differentiation with respect to r. This will give you the derivative of the potential energy function, which can then be equated to the negative of the radial force function according to the relationship F(r) = -dU(r)/dr. Partial differentiation is not necessary in this case since the function only depends on one variable, r. Keep up the good work!

## What is the Yukawa potential energy function?

The Yukawa potential energy function is a mathematical formula used to describe the interaction between two particles due to the exchange of a virtual particle. It is commonly used in quantum field theory and can also be used to explain the behavior of particles in nuclear physics.

## What is the formula for the Yukawa potential energy function?

The formula for the Yukawa potential energy function is U(r) = (hbar*c/4*pi)*(e^(-lambda*r)/r), where hbar is the reduced Planck's constant, c is the speed of light, and lambda is a constant related to the range of the interaction.

## What are the variables in the Yukawa potential energy function?

The variables in the Yukawa potential energy function are U, which represents the potential energy, r, which represents the distance between the two particles, hbar, which is the reduced Planck's constant, c, which is the speed of light, and lambda, which is a constant related to the range of the interaction.

## What is the significance of the Yukawa potential energy function?

The Yukawa potential energy function is significant because it helps to explain the strong nuclear force, one of the four fundamental forces of nature. It also provides a better understanding of the behavior of particles in quantum field theory and nuclear physics.

## How does the Yukawa potential energy function differ from other potential energy functions?

The Yukawa potential energy function differs from other potential energy functions in that it takes into account the exchange of a virtual particle between two particles. This allows for a better understanding of the behavior of particles at small distances, such as in nuclear interactions.

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