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Assuming all forces derive form a conservative source and that all forces observe the strong form of the third law, how do we arrive at the following equation?

\begin{equation} V=\sum _i V_i+\frac 12 \sum _i \sum _{j,j\neq i}V_{ij} \end{equation}

Okay so here are my thoughts.

Firstly we divide the work up into internal and external components: \begin{equation} \sum _i\int ^{r_2}_{r_1}\vec{F_i}^{(e)}\cdot d\vec s _i+\sum _{i,j}\int ^{r_2}_{r_1}\vec {F_{ij}}\cdot d\vec s _i \end{equation} The factor of a half comes in since we are summing over both $i$ and $j$, and (I think) we can assume that $V_{ij}=V_{ji}$ (why?).

let, \begin{equation} \vec F_i ^{(e)}=-\nabla_iV_i(\vec r_1,...,\vec r_n)\\ \vec F_{ij}=-\nabla_{ij}V_{ij}(|\vec r_i -\vec r_j|) \end{equation}

The $|\vec r_i -\vec r_j|$ is just the magnitude of the particles separation.

Now the work is given by: \begin{equation} W=\sum _i\int ^{r_2}_{r_1}-\frac{\partial }{\partial \vec r_i}V_i\cdot d\vec r_i+\frac 12 \sum _{i,j}\int^{r_2}_{r_1}-\frac{\partial }{\partial \vec r_{ij}}V_{ij}\cdot d\vec r_{ij} \end{equation} \begin{equation} W=-\sum _i\int^{V_2}_{V_1}dV_i-\frac 12 \sum _{i,j}\int ^{V_2}_{V_1}dV_{ij} \end{equation} So my question is how we got to the end step? This is all out of Goldstein in the 1st chapter. Unfortunately I can't follow the derivation at all ...

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Firstly we divide the work up into internal and external components: \begin{equation} \sum _i\int ^{r_2}_{r_1}\vec{F_i}^{(e)}\cdot d\vec s _i+\sum _{i,j}\int ^{r_2}_{r_1}\vec {F_{ij}}\cdot d\vec s _i \end{equation} The factor of a half comes in since we are summing over both $i$ and $j$, and (I think) we can assume that $V_{ij}=V_{ji}$ (why?).

They don't have to be equal. They just have to be equal to within an arbitrary constant. Adding an arbitrary constant to a potential doesn't make a bit of difference (at least in classical mechanics), so one might as well call that constant zero.

To see why they have to be equal (to within a constant), assume $V_{ij} = V_{ji} + \Delta V_{ij}$, where $\Delta V_{ij}$ is some non-constant function. Then $\vec F_{ij} = -\vec\nabla_{ij} V_{ij}$ $= -\vec\nabla_{ij} V_{ji} - \vec\nabla_{ij} \Delta V_{ij}$. Note that $\vec\nabla_{ij} V_{ji} = -\vec\nabla_{ji}V_{ji} = \vec F_{ji}$ Thus $\vec F_{ij} = - \vec F_{ji} - \vec\nabla_{ij} \Delta V_{ij}$. The only way this can satisfy Newton's third law is if $\vec\nabla_{ij} \Delta V_{ij} = 0$. This contradicts the assumption that $\Delta V_{ij}$ is some non-constant function.

let, \begin{equation} \vec F_i ^{(e)}=-\nabla_iV_i(\vec r_1,...,\vec r_n)\\ \vec F_{ij}=-\nabla_{ij}V_{ij}(|\vec r_i -\vec r_j|) \end{equation}

The $|\vec r_i -\vec r_j|$ is just the magnitude of the particles separation.

You have a misunderstanding here with respect to the external forces. The force $\vec F_i^{(e)}$ and hence the potential $V_i$ depends only on $\vec r_i$. There is no dependence on $\vec r_j$ where $j \ne i$. That first line should be $\vec F_i^{(e)} = -\nabla_i V_i(\vec r_1)$.

Now the work is given by: \begin{equation} W=\sum _i\int ^{r_2}_{r_1}-\frac{\partial }{\partial \vec r_i}V_i\cdot d\vec r_i+\frac 12 \sum _{i,j}\int^{r_2}_{r_1}-\frac{\partial }{\partial \vec r_{ij}}V_{ij}\cdot d\vec r_{ij} \end{equation} \begin{equation} W=-\sum _i\int^{V_2}_{V_1}dV_i-\frac 12 \sum _{i,j}\int ^{V_2}_{V_1}dV_{ij} \end{equation} So my question is how we got to the end step? This is all out of Goldstein in the 1st chapter. Unfortunately I can't follow the derivation at all ...

Goldstein uses $\int_1^2$, not $\int_{r_1}^{r_2}$ or $\int_{V_1}^{V_2}$ This may be part of your problem. Your own notation may be confusing you.

The second line follows from the first as a consequence of the fact that $dV = \vec{\nabla} V \cdot d\vec r$ for some potential $V$.

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  • $\begingroup$ Thank you for your answer! I understand now why the potentials are equal up to a constant! Is the force only a function of the interaction between one particle and the origin of the coordinates system? I guess that makes more sense! Could I instead write $\vec F ^{(e)}_{i}=-\nabla _iV_i(\vec r_i)$ instead? Your last point hits on my big misunderstanding !! I hope you can maybe expand slightly... What is Goldsteins $1$ and $2$ ... the big problem I'm having is converting the integrals from $r$ to $V$ ... this is the crux of my question. $\endgroup$ – user58536 Oct 23 '14 at 18:33
  • $\begingroup$ The external force on the ith particle is an interaction between that particle and stuff (other particles) that lies outside the system boundary. Keep in mind that Newtonian mechanics is ultimately a "particle" based system (where "particles" aren't quite well-defined) and interactions are particle-to-particle. Newton's third law doesn't work with a three-particle interaction (and there are some three particle interactions in QM). So, yes, you could write $vec F_i^(e) = -\nabla_i V_i(\vec r_i)$ instead. That's exactly what you first work integral says. $\endgroup$ – David Hammen Oct 23 '14 at 19:10
  • $\begingroup$ With regard to Goldstein's 1 and 2: Those represent initial and final configurations of all the particles. Since the external potential $V_i$ depends only on $r_i$, that first integral is better written as $\int_{\vec r_i;1}^{\vec r_i;2} -\nabla_i V_i(\vec r_i) \cdot d\vec r_i$ where $\vec r_{i,1}$ and $\vec r_{i,2}$ represents the initial and final positions of the ith particle. As an integral over potential, this becomes $\int_{V_{i;1}}^{V_{i;2}} d V_i$ where $V_{i;1}$ and $V_{i;2}$ are the initial and final values of $V_i$. The internal potentials are treated similarly. $\endgroup$ – David Hammen Oct 23 '14 at 19:17
  • $\begingroup$ Thank you very much for your time... I feel I'm very close to getting this... one last point .... (and although this may be simple it has just really stumped me... to change from $\int^{\vec {r}_i;2}_{\vec {r}_i;1}-\nabla _iV_i(\vec {r}_i)\cdot d\vec r_i$ to $\int ^{V_i;2}_{V_i;1}dV_i$ do you literally just cancel the $\frac {1}{\partial r_i}$ for the $d\vec r_i$? $\endgroup$ – user58536 Oct 23 '14 at 19:26
  • $\begingroup$ No, you do not literally just cancel the $1/\frac \partial r_i$ for the d\vec $f_i$. That's what I call "physics math" and it will get you in trouble. It appears to work in this case for two (or three) reasons: (1) $V_i$ is a function of $\vec r_i$ only, (2) $d V_i$ is an exact differential, and (3) the force that results is conservative and hence path independent. Note however that (3) is a consequence of (2). $\endgroup$ – David Hammen Oct 23 '14 at 19:58

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