Prove forces derived from a velocity-dependent potential are not central

In summary, the conversation discusses the validity of a claim made in an attachment regarding the central force derived from a function V that is also a function of the difference of velocities of particles. The individual speaking believes that the claim is incorrect and provides a proof to support their belief. They use a mathematical definition for the gradient operator and show that the force derived from V satisfies the conditions in question. However, they also acknowledge and correct a mistake made in their initial reasoning.
  • #1
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I don't see why the last sentence in the attachment is true. It claims that if ##V_{ij}## or simply ##V## is also a function of the difference of velocities of particles ##i## and ##j##, then the force derived from ##V## is not central. In other words, if ##V=V(|r_i-r_j|, |v_i-v_j|)##, then (1.34) is not satisfied.

Let ##p=|r_i-r_j|## and ##q=|v_i-v_j|## and ##\partial_x=\frac{\partial}{\partial x}##.

##\nabla_iV(p, q)=\partial_{x_i}V(p, q)\vec{e_x}+\partial_{y_i}V(p, q)\vec{e_y}+\partial_{z_i}V(p, q)\vec{e_z}##. Definition of ##\nabla_i## is given by the first sentence of the attachment.

Since ##\partial_{x_i}V(p, q)=\partial_{x_i}V(p)##, the RHS above should be the same as the RHS of (1.34). Then, (1.34) will be satisfied.

##\partial_{x_i}V(p, q)=\partial_{x_i}V(p)## for the same reason as ##\partial_x(x+\dot{x}^2)=1=\partial_xx##.

Screen Shot 2016-05-15 at 8.57.48 pm.png


EDIT: I found the mistake. ##\partial_{x_i}V(p, q)\neq\partial_{x_i}V(p)## for the same reason as ##\partial_x(x+x\dot{x})\neq\partial_xx##.
 
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  • #2
Nonetheless, I am still able to prove that the force derived from ##V=V(p, q)## satisfies (1.34), contrary to what is claimed by the book.

Recall that ##p=|r_i-r_j|=\sqrt{(x_i-x_j)^2+(y_i-y_j)^2+(z_i-z_j)^2}## and ##q=|v_i-v_j|##.

Thus, ##\frac{\partial p}{\partial x_i}=\frac{1}{2p}2(x_i-x_j)=\frac{1}{p}(x_i-x_j)##.

Let ##V'(p, q)=\partial_{p}V(p, q)##.

##\nabla_iV(p, q)=\partial_{x_i}V(p, q)\vec{e_x}+\partial_{y_i}V(p, q)\vec{e_y}+\partial_{z_i}V(p, q)\vec{e_z}##

##=V'(p, q)\frac{\partial p}{\partial x_i}\vec{e_x}+V'(p, q)\frac{\partial p}{\partial y_i}\vec{e_y}+V'(p, q)\frac{\partial p}{\partial z_i}\vec{e_z}##

##=\frac{V'(p, q)}{p}(x_i-x_j)\vec{e_x}+\frac{V'(p, q)}{p}(y_i-y_j)\vec{e_y}+\frac{V'(p, q)}{p}(z_i-z_j)\vec{e_z}##

##=\frac{V'(p, q)}{p}(\vec{r_i}-\vec{r_j})##,

which is equivalent to the RHS of (1.34).

What's wrong?
 
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1. What is a velocity-dependent potential?

A velocity-dependent potential is a type of potential energy that depends on the velocity of an object. This means that the potential energy of the object changes as its velocity changes.

2. How are forces derived from a velocity-dependent potential different from central forces?

Forces derived from a velocity-dependent potential are different from central forces because they are not directed towards a central point. Instead, they are directed in a way that depends on the velocity of the object.

3. Why is it important to prove that forces derived from a velocity-dependent potential are not central?

It is important to prove this because it helps us understand the behavior of objects under the influence of different types of forces. It also allows us to make accurate predictions and calculations in physics and other scientific fields.

4. How can it be proven that forces derived from a velocity-dependent potential are not central?

This can be proven through mathematical analysis and calculations using the equations and principles of classical mechanics. By examining the direction and magnitude of the force, we can determine if it is derived from a velocity-dependent potential or a central force.

5. Can forces derived from a velocity-dependent potential ever act as central forces?

No, forces derived from a velocity-dependent potential cannot act as central forces. This is because their direction is always dependent on the velocity of the object, and they do not point towards a central point like central forces do.

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