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I have some confusion on potential energy of a two particle system. I'm using Section 4.9 from 'Classical Mechanics' by John R. Taylor as reference.

Assume two particles are at location $\vec{r_1}$ and $\vec{r_2}$. My first question is that the section claims that the force (and potential energy) only depends on $\vec{r_1} - \vec{r_2}$ because the two particle interaction should be translationally invariant. Why not depending on $\lvert \vec{r_1} - \vec{r_2}\rvert$? Surely if you rotate your view point, the force and potential energy should be the same?

The second question is I don't know how to interpret the potential energy $U(\vec{r_2} - \vec{r_1})$ as function of $\vec{r_2} - \vec{r_1}$. For single particle's potential energy, $U(\vec{r})$ is defined to be the negation of the work from a reference point $r_0$, that is $U(\vec{r}) = -\oint_{r_0}^r \vec{F} \cdot d\vec{r}$. How to interpret the two particle potential energy in the same way? Where is the reference point? How is the integral defined? Or do we just say that work integral definition is not available in a two particle system, so that the potential energy is defined to be the function such that $\nabla_{r_1}U(\vec{r_2} - \vec{r_1}) = F_{12}$?

Finally I would like some clarity on the $\nabla$ operator with subscript. In the book it is defined to be

$$ \nabla_{r_1} = \frac{\partial}{\partial x_1} \hat{\vec{x}} + \frac{\partial}{\partial y_1} \hat{\vec{y}} + \frac{\partial}{\partial z_1} \hat{\vec{z}} $$

That operator seems should apply to a scalar function $U(\vec{r_1})$ instead of $U(\vec{r_1} - \vec{r_2})$. Maybe $U(\vec{r_1} - \vec{r_2})$ is a composite function of $U(\vec{r_1}, \vec{r_2})$ and $d(\vec{r_1}, \vec{r_2}) = \vec{r_1} - \vec{r_2}$? Some more mathematical precision would be appreciated here.

Thanks in advance!

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  • $\begingroup$ Single particle's potential energy is undefined. $\endgroup$ Commented Jun 25, 2019 at 15:45

2 Answers 2

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Surely if you rotate your view point, the force and potential energy should be the same?

In magnitude, sure. But force is a vector. If all you plug in to the force is a scalar (e.g. $|\vec r_1 - \vec r_2|$), then how do you know in which direction it should point?

More concretely, you can imagine two masses - one at the origin, and the other at the point $(1,0,0)$. The particle at the origin will feel a gravitational force in the $+\hat x$ direction. Now perform a rotation, so the second particle is at the point $(-1,0,0)$. The distance $|\vec r_1 - \vec r_2|$ is precisely the same, but now the force will be in the $-\hat x$ direction, implying that the force cannot be a function of the distance alone.

The potential $U$, on the other hand, is different, at generally depends only on $|\vec r_1 - \vec r_2|$.

For single particle's potential energy, $U(\vec r)$ is defined to be the negation of the work from a reference point $\vec r_0$ $[\ldots]$ How to interpret the two particle potential energy in the same way?

The notation may be confusing you. In the two particle case, $U$ is a function which eats a vector and spits out a number. The interpretation is that the vector we plug in should denote the separation between the particles, and the number it spits out should be the amount of work (relative to some reference separation) that it took to arrange the particles in that way.

If it helps, define $\vec R \equiv \vec r_1 - \vec r_2$, and just let $U=U( R)$. $\lim_{R\rightarrow \infty}U( R)$ is the potential energy of the two particle system when the particles are infinitely far apart, which is conventionally taken to be zero. It takes work to bring them to some finite distance from one another; the negative of that work is the potential energy.

As an alternative interpretation, if you choose a coordinate system where $\vec r_2$ is fixed at the origin, then $U(R) = U(r_1)$ and you can treat it the same way you would treat a single particle potential - just understand that the potential is due to the interaction with the particle you've fixed at the origin.

Maybe $U(\vec r_1 - \vec r_2)$ is a composite function of $U(\vec r_1,\vec r_2)$ and $d(\vec r_1,\vec r_2)=\vec r_1-\vec r_2$?

Close. That doesn't make sense, though - a function $f=f(\vec r_1-\vec r_2)$ is a function with one argument, not two. If you'd like, it is a composition of $f(\vec R)$ and $\vec R=\vec r_1-\vec r_2$.

Explicitly, you would have something like

$$\frac{\partial }{\partial x_1} f(\vec R) = \frac{\partial f}{\partial \vec R} \cdot \frac{\partial \vec R}{\partial x_1}$$

$\frac{\partial f}{\partial \vec R}$ is just the gradient of $f$, while $\frac{\partial \vec R}{\partial x_1} = \frac{\partial}{\partial x_1}\big( (x_1-x_2)\hat x + (y_1-y_2)\hat y + (z_1-z_2)\hat z\big) = \hat x$.

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  • $\begingroup$ Thank you so much! You explained it really well! $\endgroup$
    – Rui Liu
    Commented Jun 25, 2019 at 15:26
  • $\begingroup$ @J. Murray Isn't single particle's potential energy meaningless?I think it is undefined! $\endgroup$ Commented Jun 25, 2019 at 15:44
  • $\begingroup$ @Unique I have no idea what you're talking about. It's perfectly well-defined in Newontian physics to consider the case of a single particle immersed in an external potential. You're making it sound like potential energy doesn't make sense unless you include more than one particle in your model, which is false. $\endgroup$
    – J. Murray
    Commented Jun 25, 2019 at 17:55
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A general definition for potential energy of a system of particles is given below:-

The change in potential energy of a system is defined as the negative of work done by the internal conservative forces of the system $$dW_{int,con}=-dU_{system}$$

Remember only change in potential energy is defined but absolute potential energy is undefined. Potential energy is always defined for a system of particles.

For example:-Potential energy of this box is 10 joules.This statement is a wrong statement.The correct statement would be:-

Potential energy of earth-box system is 10 joules.Reference point is surface of earth.

Hope this helps!

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  • $\begingroup$ I think J Murray explained more specifically to my questions, but still thanks for your answer! $\endgroup$
    – Rui Liu
    Commented Jun 25, 2019 at 15:28
  • $\begingroup$ @Rui Liu I already know that he gave the correct explanation but what i wanted to tell you that potential energy is always defined for a system which you misunderstood as you wrote single particle's potential energy.😀😀😀 $\endgroup$ Commented Jun 25, 2019 at 15:33
  • $\begingroup$ If the problem only involves one particle. For example there's some sort of conservative force field and only one particle is in it. Doesn't it make sense to talk about the potential energy of one particle? It might be physically not meaningful, but mathematically meaningful? $\endgroup$
    – Rui Liu
    Commented Jun 25, 2019 at 15:46

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