I What is the relationship between force and potential in particle interactions?

AI Thread Summary
The discussion centers on the relationship between force and potential in particle interactions, specifically for two repelling particles. The force vectors F_12 and F_21 are derived from the interaction potential u(r), with F_12 calculated as the negative gradient of the potential, resulting in components of (-du/dx, -du/dy, -du/dz). F_21 is simply the negative of F_12, confirming Newton's third law. Clarification is provided that the potential is a scalar function, which can generate vector forces through its gradient. The interaction potential depends on the relative positions of the particles, ensuring consistency in force calculations.
Tim667
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Slight confusion about vectors from a potential
Suppose I have some interaction potential, u(r), between two repelling particles. We will name them particles 1 and 2.

I want to find the force vectors F_12 and F_21. Would I be correct in saying that the x-component of F_12 would be given by -du/dx, y-component -du/dy etc? And to find the components for the other force vector, would this simply be the negative of the first vector?

So F_12 would be given by (-du/dx, -du/dy, -du/dz) and F_21= (du/dx, du/dy, du/dz)?

Thank you
 
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Tim667 said:
Summary: Slight confusion about vectors from a potential

Suppose I have some potential, u(r), between two repelling particles. We will name them particles 1 and 2.

I want to find the force vectors F_12 and F_21. Would I be correct in saying that the x-component of F_12 would be given by -du/dx, y-component -du/dy etc? And to find the components for the other force vector, would this simply be the negative of the first vector?

So F_12 would be given by (-du/dx, -du/dy, -du/dz) and F_21= (du/dx, du/dy, du/dz)?

Thank you
You need to be careful. Normally with a potential you have a particle moving under the influence of an external force. In this case technically you have a different potential for each particle.
 
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PeroK said:
You need to be careful. Normally with a potential you have a particle moving under the influence of an external force. In this case technically you have a different potential for each particle.
I see, I should specify that this is an interaction potential between the two particles
 
Tim667 said:
I see, I should specify that this is an interaction potential between the two particles
Okay. Of course. As long as you define ##r## the right way round, what you have looks right. I must be getting tired.
 
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PeroK said:
Okay. Of course. As long as you define ##r## the right way round, what you have looks right. I must be getting tired.
That's okay. I think I was confused because potentials are usually scalar functions, and I wasn't sure you could get vectors from them
 
Tim667 said:
That's okay. I think I was confused because potentials are usually scalar functions, and I wasn't sure you could get vectors from them
Yes, the gradient takes a scalar function of position and generates a vector function of position.
 
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An interactian potential usually takes the form ##V(\vec{x}_1-\vec{x}_2)##. The force on particle 1 is
$$\vec{F}_{12}=-\vec{\nabla}_1 V(\vec{x}_1-\vec{x}_2)$$
and on particle 2
$$\vec{F}_{21}=-\vec{\nabla}_2 V(\vec{x}_1-\vec{x}_2)=-\vec{F}_{12}.$$
Newton's 3rd Law holds, because the potential only depends on the relative vector ##\vec{r}=\vec{x}_1-\vec{x}_2##.
 
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