1. Sep 14, 2014

### fuserofworlds

I'm currently going over some mechanics notes and am confused about the following situation:

In the book I'm looking at, it describes two particles absent of external forces, only exerting a force on each other. In deriving a potential energy equation for the two, it goes on to say that if the force is conservative and the second particle placed at the origin, one can say $$\vec{F}_{12} = -\vec{\nabla}_1 U(r_1) = -\frac {\partial U(r_1)}{\partial x_1} \hat{x} - \frac {\partial U(r_1)}{\partial y_1} \hat{y} - \frac {\partial U(r_1)}{\partial z_1} \hat{z}$$ where $x_1, y_1, z_1$ are the coordinates of particle 1, and $\vec{F}_{12}$ is the force on particle 1 due to particle 2.

My problem is, conceptually, I don't understand what $\partial U/\partial x_1$ is supposed to mean. What's the difference between $\partial U/\partial x_1$ and $\partial U/\partial x$? Is there any?

Last edited: Sep 14, 2014
2. Sep 14, 2014

### voko

$\vec r_1 = (x_1, y_1, z_1)$ is the displacement of particle 1. So $x_ 1$ and $\partial f \over \partial x_1$ are well defined, they have to with the $x$ coordinate of particle 1.

3. Sep 14, 2014

### Staff: Mentor

I think the 1 refers to particle one which has coordinates x1, y1, z1 vs the other particle at x2,y2,z2

4. Sep 14, 2014

### fuserofworlds

I understand that it refers to the coordinates of particle one, but I don't see why this is necessary. Isn't the derivative taken with respect to the axes? x, y, and z are variables and $\vec{r}_1$ is the position of particle one. If I take the derivative with respect to $x_1$, aren't I taking the derivative with respect to a number? Can I even do that?

5. Sep 14, 2014

### Staff: Mentor

If you had a system of a single particle like a projectile from a cannon. You would describe the system in terms of x,y,z understanding them to be variables that change in time.

Now you have two particles each with an x,y,z that change in time and so we number them with the understanding that 1 refers to particle 1 and 2 refers to particle 2. The system has six changing variables that are linked by the equations of motion for the system.

6. Sep 14, 2014

### voko

Is that how you were taught in multivariate calculus? Have you covered functions of more than three variables?

7. Sep 14, 2014

### fuserofworlds

I realize I phrased that wrong. What I mean is, in general your function for the potential will be a function of $\vec{r}$ in general, or U(x,y,z). Mathematically, how does one then take $(\partial U/\partial x_1, \partial U/\partial y_1, \partial U/\partial z_1)$, and how is that different from taking $(\partial U/\partial x, \partial U/\partial y, \partial U/\partial z)$? I just don't understand how the distinction plays out mathematically.

8. Sep 14, 2014

### Staff: Mentor

I guess you didn't read my post, oh well...

9. Sep 14, 2014

### voko

No, that is not true in general. What you say is correct in the case of one particle interacting with some potential field; the potential energy of a multi-particle system will generally be a function of the coordinates of all the particles, pretty much by definition: $$\vec F_i = -{\partial U \over \partial \vec r_i},$$ where $\vec r_i$ is the displacement of the i-the particle, and $\vec F_i$ is the total force acting on the i-th particle.

10. Sep 14, 2014

### fuserofworlds

I did, but am still a little confused, though I think I am closer to understanding. If there were a system of two particles each with their own coordinates, and you wanted to compute ∂U/∂x, you would just do it with respect to a given origin, correct?

What confuses me is that, if (as in the book's example) particle 2 is taken to be at the origin, then in my mind the potential should just depend on the position of particle 1. If you take the derivative with respect to $\vec{r}_1$, isn't that just like setting particle 1 at the origin? And doesn't that contradict the statement that particle 2 is at the origin?

Sorry about all the questions. I'm not sure why I'm having such a hard time understanding this.

11. Sep 14, 2014

### voko

That is true and that is a useful method of reducing the dimensional of the problem. It essentially transforms any two-particle system into a one-particle system in external field. With more particles, however, gains are not so spectacular.

No, why? Even assuming that $\vec r_2 = 0 = \mathrm{constant}$, it is still perfectly fine to differentiate $U$ with respect to $r_1$.