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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: ${\bf F}_{ji} = -\nabla_iV_{ij}$, ${\bf F}_i^{(e)} = -\nabla_iV_i$. The search for the meaning of this mysterious to me $\nabla_i$ notation led me to this physics.SE answer, which claims 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?

I'm so confused! Please help me understand what's going on mathematically here:

  • 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!

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  • $\begingroup$ The indexes identify the particles; for the external forces there is one index, that of the affected particle; internal forces require two indices, identifying both particles; each particle is located by a position vector, r, which is an ordinary geometrical position. $\endgroup$ Feb 17, 2016 at 1:02
  • $\begingroup$ Continuing, the potentials are functions of position. The index on the gradient operator means that the gradient is with respect to the corresponding particle position. Note that the potentials, V, are defined everywhere, and by thecsums are evaluated for each particle. Similar for external forces; internal forces are defined for each possible particle pairing. The integration variable, s, is indexed by the corresponding particle, and is the arclength or integrated path length parameter. $\endgroup$ Feb 17, 2016 at 1:17
  • $\begingroup$ In my edition of Goldstein it says "where the subscript $i$ on the del operator indicates that the derivatives are with respect to the components of $\vec r_i$.” So I think that $\nabla_i V_{ji}$ means find the gradient of the potential due to particle $j$ at the position of particle $i$, $\vec r_i$. The index on the del operator is to emphasis as to where the gradient should be found. $\endgroup$
    – Farcher
    Feb 19, 2016 at 13:18

2 Answers 2

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Arnold's book Mathematical Methods of Classical Mechanics gives a detailed treatment of mechanics in a mathematically rigorous way. Sections 3C, 3D and 10 are most relevant for your question. I will present here the main points for a system of $n$ particles in Euclidean 3-dimensional space.

First, let us define

  • $t$ is a real variable in a prescribed interval $I$, for time
  • $\mathbf x_i : I \rightarrow \mathbb R^3$ ($i=1,\dots,n$). Given a instant of time $t \in I$, $\mathbf x_i(t)$ is the position in Euclidean space at time $t$.
  • $m_i$ are the masses.
  • $\mathbf F_i:\mathbb R^{3n}\times\mathbb R^{3n}\times I \rightarrow\mathbb R^3$ is the force on the $i$-th particle. The first $\mathbb R^{3n}$ argument denotes the positions of all particles, the second their velocities $\dot{\mathbf x}_i$ and the third the time.

Newton's equation for the motion of the system of particles is $m_i \ddot{\mathbf x}_i=\mathbf F_i(\mathbf x_1,\dots,\mathbf x_n,\dot{\mathbf x_1},\dots,\dot{\mathbf x_n},t)$.

Now, the force may be written as the sum of the external force and individual forces due to each other particle: $\mathbf F_i=\sum_{j \ne i}\mathbf F_{ji}+\mathbf F^{(e)}_i$, where $\textbf F_{ji}$ is the force on $m_i$ due to $m_j$ and its arguments are the same of $\textbf F$, and the other term gives the force acting on $m_i$. The kinetic energy theorem states:

$T(t) - T(t_0) = W_{t_0 \rightarrow t} = \sum_{i=1}^n \int_{t_0}^t \textbf F_i \cdot \dot{\textbf x}_i\,\mathrm d t$

By replacing $\mathbf F_i$, one arrives to the formula in the question, if we make the convention $d \mathbf x_i \rightarrow \dot{\mathbf x}_i dt$, or instead interpret the integrals of the form $\int \mathbf F_i\cdot d\mathbf x_i$ in the Riemann-Stieltjes sense, as $\int \mathbf F_i(t)\cdot d\mathbf x_i(t)$.

And the forces are conservative if and only if there exists potentials (scalar fields) $U_{ji}, U_i : \mathbb R^{3n} \rightarrow \mathbb R$ s.t. $\mathbf F_{ji} = -\nabla_i U_{ji},\;\mathbf F^{(e)}_i = -\nabla_i U_i$, where $\nabla_i$ is an operator denoting gradient with respect to $x_i$. This means $\nabla_i f$, where $f=f(\mathbf x_1,\dots,\mathbf x_n)$, is the vector field $\nabla_i f : \mathbb R^{3n} \rightarrow R^3$ defined by $\nabla_i f=\left( \frac{\partial f}{\partial x_i}, \frac{\partial f}{\partial y_i}, \frac{\partial f}{\partial z_i} \right)$. Here, $x_i,y_i,z_i$ are components of $\mathbf x_i$.

The work integrals are then understood geometrically as the integral of the dot product of $\mathbf F_i(t)$ and $d\mathbf x_i(t)$, as each particle moves in Euclidean space through time, summed for all particles.

And, finally, addressing your interpretation:

But if the Fs and the Vs are functions of all 3N coordinate variables, what does dsi 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?

Yes, it would. And this would not be a sensible work integral, since all particles move at the same time. This would only make sense if all particles stood still as the ith particle moved. This is why its much preferable to treat coordinates as function of time.

Even if we took all the remaining particles positions as functions of $\mathbf x$, we would still run into trouble, for the ith particle may return to the same place while others moved. So, the best way is to take all coordinates as explicit functions of time.

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  • $\begingroup$ I appreciate your detailed answer! One question: in this formalism, is it essential that the ${\bf F}_i^{(e)}$ and the ${\bf F}_{ji}$ depend on $6n+1$ arguments rather than on $6+1$ and $6*2+1$ arguments respectively (and the same can be asked about $U_{ji}, U_i$)? That is, is it physically relevant/realizable e.g. for $F_{ji}$ to depend on positions/velocities of particles other than $i,j$, or is it mere notational convenience to make the writing-up of integrals and sums and gradients neater? $\endgroup$ Feb 21, 2016 at 7:23
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The integrals are all in $\mathbb R^3$. The forces $\mathbf F_i^{(e)}$ and $\mathbf F_{ji}$ are just 3-vectors. (Otherwise, what would $\mathbf F_i^{(e)}\cdot d\mathbf s_i$ mean? The dot product between a vector field and an infinitesimal vector?)

$\mathbf F_i^{(e)}$ is perhaps the easiest to deal with, so I'll address that first. $\mathbf F^{(e)}$ is the external force field. It could be a function of many different things. Asking that has sent you down the rabbit hole of confusion. It doesn't matter. It's a vector field. $\mathbf F_i^{(e)}$ short for $\mathbf F^{(e)}(\vec x_i)$, the value of that vector field at the point $x_i$.

We want to integrate the value of that vector field at points along the path taken by particle $i$ as the system evolves from state 1 to state 2, $\int_{1_i}^{2_i} \mathbf F^{(e)}(\mathbf s_i)\cdot d\mathbf s_i$, or for short, $\int_1^2 \mathbf F_i^{(e)}\cdot d\mathbf s_i$. (Here I'm being even more pedantic that Goldstein. States 1 and 2 are members of $\mathbb R^{3N}$. The ith particle however follows a path from point $1_i$ to $2_i$, both of which are members of $\mathbb R^3$.)

With regard to the mysterious $\nabla_i V_i$, that again is a bit of shorthand. Each $V_i$ is a scalar field, a differentiable mapping from $\mathbb R^3$ to $\mathbb R$. The gradient of this scalar field, $\nabla V_i$, is a vector field. The value of that vector field at the position of the ith particle is $\nabla V_i(\mathbf s_i)$, or $\nabla_i V_i$ for short.

The notations $V_{ji}$ and $\nabla_i V_{ji}$ are similar. $V_{ji}$ is the potential field due to particle j as sensed by particle i. The gradient of this potential is a vector field, so once again we need to know where to evaluate it, and that would once again be the position of the ith particle, or $\nabla_i V_{ji}$ for short.

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  • $\begingroup$ Thank you! Although this is reassuring, I still remain confused. If each $V_{ij}$ is "just" a scalar field, how does that take into account the movement of particle $j$? E.g. suppose $V_{ij}$ is the gravitational potential due to $j$'s mass, and suppose that at some time $t\ne t_0$ the particle $i$ comes back to position $1_i$, due to various forces acting on it (i.e. its path self-intersects). Then if the formalism as you describe, $V_{ji}$ must have the same value at $t_0$ and $t$ because it's just a scalar field; but "in reality" $j$ will be elsewhere by that time and $V$ will change. $\endgroup$ Feb 17, 2016 at 8:06
  • $\begingroup$ @Avva I do not think that there is a conflict. Particle $j$ produces a field which it carries with it. At a particular time particle $i$ finds itself in the field due to particle $j$ and at that time the potential at position $\vec r_i$ is $V_{ji}$. Later in time particle $j$ has moved together with its field and particle $i$ has gone back to where it started from $\vec r_i$. The potential at position $\vec r_i$ is still labelled $V_{ji}$ but can now be a different value from before unless the body is rigid. $\endgroup$
    – Farcher
    Feb 19, 2016 at 13:08

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