# Work and forces in systems of many particles

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1. Feb 17, 2016

### avorobey

I'm reading Goldstein's "Classical Mechanics", first chapter, and am confused about what's going on in equations of forces and work in systems of particles. For example, Goldstein calculates work done by all the forces, external and internal, in evolving the system from state $1$ to state $2$ (page 9):

$$W_{12} = \sum_{i}\int_{1}^{2}{\bf F}_i^{(e)}\cdot d{\bf s}_i + \sum_{i\ne j}\int_{1}^{2}{\bf F}_{ji}\cdot d{\bf s}_i$$

I want to understand this equation rigorously from the mathematical point of view. What is the Euclidean space in which the path integrals happen? The forces ${\bf F}_{i}^{(e)}$ and ${\bf F}_{ji}$ are functions of how many variables?

I thought, on my first reading, that everything happens in ${\bf R}^3$ and forces are vector fields ${\bf F}:{\bf R}^3\to {\bf R^3}$. But on second thought it seems that ${\bf F}_{ji}$ should depend on both ${\bf r}_j$ and ${\bf r}_i$. And also on the next page under the assumption that forces are conservative they're derived from potentials: ${\bf F}_{ji} = -\nabla_iV_{ij}$, ${\bf F}_i^{(e)} = -\nabla_iV_i$. Attempting to understand this mysterious to me $\nabla_i$ notation leads me to think that the $V$s are functions of all $N$ position vectors as $3N$ independent variables, and $\nabla_i$ picks the $3$ variables to vary for gradient.

But if the $F$s and the $V$s are functions of all $3N$ coordinate variables, what does $d{\bf s}_i$ mean exactly in the integral? And wouldn't it make e.g. $\int_{1}^{2}{\bf F}_i^{(e)}\cdot d{\bf s}_i$ a function of all the *remaining* $3N-3$ coordinates rather than a scalar?

- What the $F$s and the $V$s are functions of, in the most general setting;
- Following that, how are the work integrals to be understood geometrically;
- If some force is not conservative and depends, say, directly on time or on velocity of one of the particles, how does that change the meaning of the integral? What does that make $d{\bf s}_i$?

Thank you!

2. Feb 17, 2016

### drvrm

path of the system

by looking at your equations it appears that the N particle system is being described in a 3N dimensional space and try to figure out the geometry say by taking a two particle system i.e. N=2 initially - then the path of the system in the Six dimensional space between points 1 and 2 can be taken and a differential element of the path can be drawn as ds - now you can move ahead and see the formalism mathematically.

3. Feb 17, 2016

### jbriggs444

I read that as the total work done moving from state 1 to state 2 ($W_{12}$) is given by the sum over all particles in the system (indexed by i) of the work done on that particle by external forces ${\bf F}_i^{(e)}$ plus the sum over all particles in the system of the work done on that particle (again indexed by i) by each and every other particle in the system (indexed by j not equal to i) where the force on particle i by particle j is given by ${\bf F}_{ij}$

Nothing mysterious or six dimensional going on. Nothing involving fields, conservative or otherwise. Just three dimensional forces-at-a-distance in three space.

Edit: Technically, this does not have to be three space. The formulation does not depend on dimensionality. It would work equally well in any higher dimensional space. All you need to have available is a vector dot product for force times incremental displacement.

Last edited: Feb 17, 2016
4. Feb 17, 2016

### BvU

Second jBriggs: Goldstein is clear in $$W_{12} = \sum_i \int_1^2 {\bf F}_i \cdot d{\bf s}_i = \sum_{i}\int_{1}^{2}{\bf F}_i^{(e)}\cdot d{\bf s}_i + \sum_{i\ne j}\int_{1}^{2}{\bf F}_{ji}\cdot d{\bf s}_i\ :$$he simply splits up ${\bf F}_i$ in an external and an internal part.

5. Feb 17, 2016

### avorobey

But $F_{ji}$ depends (in general) on the position of particle j as well as i, right? If particle j changes position as the system moves from state $1$ to state $2$, then the integrand depends on both ${\bf r}_i$ and ${\bf r}_j$. But then, if the integral is simply over the path of particle i in three-dimensional space, as you're saying, then after evaluating the integral we're still left with a function of ${\bf r}_j$ rather than a scalar! That can't be right, I think?

Maybe I'm missing something completely trivial? Consider the usual definition of path integral from a purely mathematical point of view. We have a function ${\bf F}:{\mathbb R}^n\to{\mathbb R}^n$ and a curve $S$ in ${\mathbb R}^n$; then $\int_S{\bf F(r)}\cdot d{\bf r} = \int_a^b{\bf F}(r(t))\cdot r'(t)dt$. There is no place in this definition for ${\bf F}$ to depend on anything other than the $n$ coordinates of ${\bf r}$. If ${\bf F}$ depends on something else, for example, on coordinates of some other particle, then it's no longer a function ${\bf F}:{\mathbb R}^n\to{\mathbb R}^n$, but instead it's a function, say, ${\bf F}:{\mathbb R}^{2n}\to{\mathbb R}^n$, and the definition of what it means to calculate $\int_S{\bf F(r)}\cdot d{\bf r}$ needs to specify where ${\bf F}$ is getting the other half of its arguments.

And then later on, if it turns out that the force $F_{ji}$ is generated by a scalar potential $V_{ji}$, that scalar potential is a function ${V}:{\mathbb R}^{2n}\to{\mathbb R}$, because you need coordinates of two particles to specify what the potential is. So then it eludes me how I can say that in this case
$\int_1^2{\bf F(r)}\cdot d{\bf r} = -\int_1^2{\nabla_i V_{ji}}\cdot d{\bf r} = -V\Big|_1^2$. I seem to be invoking the gradient theorem to reach this conclusion, but I don't have a scalar function of $n$ variables as that theorem requires, I have a function of $2n$ coordinates. The notation "hides" the fact that $V$ depends on the coordinate of the other particle, too, but it still does - so why is the conclusion still true, doesn't this require some sort of argument?

6. Feb 17, 2016

### BvU

I think I see your point. While particle $i$ moves from 1 to 2, so does particle $j$. Herbie discusses it on p. 10 and 11 -- but restricts to conservative forces in order to introduce an internal potential energy.