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On treating systems of particles, Goldstein starts with the consideration that whenever there are $k$ particles on a system, the $i$-th one obeys the relation

$$\dfrac{d}{dt}{\bf p}_i = {\bf F}_i^{(e)}+\sum_{j=1}^k {\bf F}_{ji}$$

Where ${\bf F}_{i}^{(e)}$ is the external force on the $i$-th particle and ${\bf F}_{ji}$ is the force that the $j$-th particle applies on the $i$-th one. Now if we compute the work on moving the system along a curve $\gamma$ we have:

$$W=\sum_{i=1}^k\int_\gamma {\bf F}_i = \sum_{i=1}^k \int_{\gamma}{\bf F}_i^{(e)}+\sum_{i,j=1}^k\int_{\gamma}{\bf F}_{ji}$$

Now he says the following (on p. 10)

In the special case that the external forces are derivable in terms of the gradient of a potential, the first term can be written as $$\sum_{i=1}^k \int_{\gamma} {\bf F}_i^{(e)} = -\sum_{i=1}^k \int_{\gamma} {\bf \nabla}_iV_i=-\sum_{i=1}^k V_i\bigg|_{a}^b$$ where the $i$ subscript on the del symbol indicates that the derivatives are with respect to the coordinates of $\mathbf{r}_i$.

Now I can't understand why this $i$ on the del symbol. Basically we are considering the configuration manifold to be $\mathbb{R}^3$ so the natural coordinate functions are $x,y,z: \mathbb{R}^3\to \mathbb{R}$ which are just projections onto the axis and with those coordinates, there's just one del operator, namely:

$${\bf \nabla} = \dfrac{\partial}{\partial x} e_1 + \dfrac{\partial}{\partial y}e_2+\dfrac{\partial}{\partial z}e_3$$

So what really means "differentiate with respect to the coordinates of $\mathbf{r}_i$? And why do we need it? Does it have any relationship with the idea of taking the configuration space for a system of $k$ particles as $\mathbb{R}^{3k}$?

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I think it's using spherical coordinates and the $r_i$ coordinates are centered on the i-th particle – Jim Aug 6 '14 at 19:23
but that's just a guess – Jim Aug 6 '14 at 19:24
@Jim, the vector $\mathbf{r}_i$ is really the position vector of the $i$-th particle. But as I understood his derivation was in cartesian coordinates. And that $\nabla_i V_i$ should be a vector, since it's equal to $-F_{i}^{(e)}$ which is a vector. – user1620696 Aug 6 '14 at 19:27
Also, last I checked, $\nabla\psi$ is a vector quantity. – Kyle Kanos Aug 6 '14 at 19:32
Yes @KyleKanos, I absorved it into the integral signs. And I know $\nabla \psi$ is a vector quantity, that's why I don't understand those subscripts. I thought that $\nabla_i = \partial/\partial x^i$ but it doesn't make sense at all, since $\nabla_i V_i$ would be a scalar while the force $F_{i}^{(e)}$ is a vector. – user1620696 Aug 6 '14 at 19:34

What Goldstein means by $\nabla_iV_i$ is $$ \nabla_iV_i=\left(\frac{\partial}{\partial x_{1,i}}\hat{x}_{1,i}+\frac{\partial}{\partial x_{2,i}}\hat{x}_{2,i}+\frac{\partial}{\partial x_{3,i}}\hat{x}_{3,i}\right)V_i $$ which is indeed a vector. Here, $\mathbf r_i=(x_{1,i},\,x_{2,i},\,x_{3,i})$ is the position of the $i$th particle (with respect to the origin), so $\nabla_i$ is the gradient of the potential from the particle.

Thinking of $\nabla_iV_i$ as some sort of "dot product" (to make it a scalar quantity) is erroneous because a dot product needs two vectors and $V_i$ is clearly a scalar quantity.

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This is typical to see in situations where the potential is a function of the coordinates of more than one particle: $$ V=V(\mathbf r_1,\ldots,\mathbf r_N)=V(x_1,y_1,z_1,\ldots,x_N,y_N,z_N). $$ The force produced by such a potential on the $i$th particle is the gradient of this function with respect to that particle's coordinates, while holding all the other particle coordinates constant. That's the meaning behind the notation $$\nabla_i = \dfrac{\partial}{\partial x_i} e_x + \dfrac{\partial}{\partial y_i}e_y+\dfrac{\partial}{\partial z_i}e_z.$$ It is still a vector-valued quantity, but it takes different values for the different particles.

If you want to systematize the use of subscripts, then the coordinate space of your system is described by coordinate functions $x_{ki}$, where $k=1,2,3$ describes the three spatial dimensions, and $i=1,\ldots,N$ labels the particles. In that notation, $$ \nabla_i=\dfrac{\partial}{\partial x_{ik}}e_k $$ with Einstein summation understood.

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Later on pages 10-11, Goldstein considers potentials of internal forces $V_{ij}$, and he writes the work of the internal forces as a sum over pairs of particles: $$-\int_1^2 {\bf \nabla}_iV_{ij}\cdot d{\bf s}_i+{\bf \nabla}_jV_{ij}\cdot d{\bf s}_j$$ I can parse this according to your answer, treating $V_{ij}$ as a function of 6 variables ${\bf r}_i,{\bf r}_j$. But then he designates ${\bf r}_{ij}={\bf r}_i-{\bf r}_j$ and "$\nabla_{ij}$ stands for the gradient w.r.t. ${\bf r}_{ij}$" (???), and $$\nabla_{i}V_{ij} = \nabla_{ij}V_{ij} = -\nabla_{j}V_{ij}$$ and I can't parse this at all! Help? – AnatolyVorobey Feb 15 at 19:34
Internal forces are always have potentials which are functions of the form $$V_{ij}=V_{ij}(\mathbf r_i,\mathbf r_j)=V_{ij}(\mathbf r_i-\mathbf r_j),$$ which should be enough to resolve your question. If not, ask it separately. – Emilio Pisanty Feb 15 at 20:29

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