.6]{fig3_3a} \label{fig:force}}
\subfigure[]{\includegraphics[scale=0.6]{fig3_3b} \label{fig:force_decomp}}
\caption{(a) shows the force on each particle in a system consisting of 3 particles which interact through a 3-body potential given in \eref{eqn:summation_split}. Since the potential is derived from an energy decomposition, we have $\bm{f}_1^{\rm{int}} + \bm{f}_2^{\rm{int}} +\bm{f}_3^{\rm{int}}=\bm{0}$. (b) shows the force decomposition of each $\bm{f}_{\alpha}^{\rm{int}}$ such that $\bm{f}_{\alpha\beta}=-\bm{f}_{\beta\alpha}$, but not necessarily parallel to the line joining particles $\alpha$ and $\beta$.}
\label{fig:force_3body}
\end{figure}

From \eref{eqn:phi_ij_antisym} it is clear that $\bar{\bm{f}}_{\alpha\beta}$ is anti-symmetric with respect to its arguments. Therefore, the integrand on the right-hand side of \eref{eqn:stress_force_differential_3body_***} satisfies all the necessary conditions for the application of Lemma \ref{lem1} given in Appendix \ref{ch:noll}. Conditions (1) and (2) in Appendix \ref{ch:noll} are satisfied through the regularity conditions on W. Therefore, using Lemma \ref{lem1}, we have
\begin{align}
\label{eqn:stress_general}
\bar{\bm{\sigma}}_{\rm{v}}(\bm{x},t) &= 
\frac{1}{2} \sum_{\substack{\alpha,\beta \\ \alpha \ne \beta}} \int_{\real{3}} \int_{s=0}^{1} \notag \\
& \left \langle -( \nabla_{\bm{x}_\beta} \phi_\alpha - \nabla_{\bm{x}_\alpha} \phi_\beta ) W \mid \bm{x}_\alpha = \bm{x} + s\bm{z}, \bm{x}_\beta = \bm{x} - (1-s) \bm{z} \right \rangle \, ds\otimes \bm{z} \, d\bm{z}.
\end{align}
The stress $\bar{\bm{\sigma}}_{\rm{v}}$ is non-symmetric in general because $\bar{\bm{f}}_{\alpha\beta}$, defined in \eref{eqn:phi_ij_antisym}, need not be parallel to the line joining particles $\alpha$ and $\beta$ as shown in \fref{fig:force_3body}. We therefore have two expressions for the stress for the same three-body potential. The symmetric expression in \eref{eqn:stress_force_general} and the non-symmetric expression in \eref{eqn:stress_general}. We show next that the non-central-force decomposition that led to the non-symmetric stress tensor is not physically meaningful since it violates the strong law of action and reaction.


\subsubsection*{Weak and strong laws of action and reaction\footnote{This derivation is due to Roger~Fosdick \cite{fosdick}.}}
The following derivation hinges on the fact that in a material system the balance laws of linear and angular momentum must be satisfied for any part of the body.

Consider a system of $N$ particles with masses $m_\alpha$ $(\alpha=1,\dots,N)$. The total force on particle $\alpha$ is 
\begin{equation}
\bm{f}_{\alpha} = \bm{f}^{\rm{ext}}_{\alpha} + \sum_{\substack{\beta \\ \beta \ne \alpha}} \bm{f}_{\alpha\beta},
\end{equation}
where $\bm{f}^{\rm{ext}}_{\alpha}$ is the external force on particle $\alpha$, and, as above, $\bm{f}_{\alpha\beta}$ is the contribution to the force on particle $\alpha$ due to the presence of particle $\beta$. No assumptions are made regarding the terms $\bm{f}_{\alpha\beta}$ or the interatomic potential from which they are derived.

A ``part'' $\wp_t$ of the system consists of $K \le N$ particles. We suppose $\bm{x}_0$ is a fixed point in space. Let $\bm{F}^{\rm{ext}}(\wp_t)$ denote the total force on the part $\wp_t$ external to the part. Let $\bm{M}^{\rm{ext}}(\wp_t; \bm{x}_0)$ denote the total external moment on $\wp_t$ about $\bm{x}_0$. Let $\bm{L}(\wp_t)$ be the linear momentum of the part $\wp_t$ and $\bm{H}(\wp_t; \bm{x}_0)$ be the angular momentum of $\wp_t$ about $\bm{x}_0$.

We adopt the following balance laws, valid for all parts of the system:\footnote{The view that the balance of linear momentum and the balance of angular momentum are fundamental laws of mechanics lies at the basis of continuum mechanics. See, for example, Truesdell's article ``Whence the Law of Moment and Momentum?'' in \cite{truesdell_essays}.}
\begin{align}
\bm{F}^{\rm{ext}}(\wp_t) &= \frac{d\bm{L}}{dt}(\wp_t), \label{eqn:force_balance}\\
\bm{M}^{\rm{ext}}(\wp_t;\bm{x}_0) &= \frac{d\bm{H}}{dt}(\wp_t;\bm{x}_0).\label{eqn:mom_balance}
\end{align}
We now show that by applying these balance laws to particular parts of the system, that the strong law of action and reaction can be established. As a first observation, let $\wp_t$ consist of the single particle $\alpha$. The external force and linear momentum for $\wp_t = \{\alpha\}$ is 
\begin{align}
\bm{F}^{\rm{ext}}(\{\alpha\}) &= \bm{f}^{\rm{ext}}_{\alpha}(t) + \sum_{\substack{\gamma \\ \gamma \ne \alpha}}\bm{f}_{\alpha\gamma}(t), \\
\bm{L}(\{\alpha\}) &= m_{\alpha}\dot{\bm{x}}_{\alpha}(t) \qquad \text{(no sum)}.
\end{align}
The balance of linear momentum in \eref{eqn:force_balance} requires
\begin{equation}
\label{eqn:newton_law_1}
\bm{f}^{\rm{ext}}_{\alpha} + \sum_{\substack{\gamma \\ \gamma \ne \alpha}} \bm{f}_{\alpha\gamma} = m_{\alpha} \ddot{\bm{x}}_{\alpha}.
\end{equation}
The external moment and angular momentum of $\wp_t$ is 
\begin{equation}
\bm{M}^{\rm{ext}}(\{\alpha\};\bm{x}_0) = (\bm{x}_{\alpha}(t) - \bm{x}_0) \times (\bm{f}^{\rm{ext}}_{\alpha} + \sum_{\substack{\gamma \\ \gamma \ne \alpha}} \bm{f}_{\alpha\gamma}) = m_{\alpha}(\bm{x}_{\alpha}(t) - \bm{x}_0) \times \ddot{\bm{x}}_{\alpha}(t),
\end{equation}
where we have used \eref{eqn:newton_law_1}, and 
\begin{equation}
\bm{H}(\{\alpha\};\bm{x}_0) = (\bm{x}_\alpha-\bm{x}_0) \times m_{\alpha} \dot{\bm{x}}_{\alpha}(t).
\end{equation}
The balance of angular momentum in \eref{eqn:mom_balance} is satisfied identically, since
\begin{align*}
m_{\alpha}(\bm{x}_{\alpha}(t) - \bm{x}_0) \times \ddot{\bm{x}}_\alpha(t) &= \frac{d}{dt} \left [ (\bm{x}_{\alpha}(t) - \bm{x}_0) \times m_{\alpha} \dot{\bm{x}}_{\alpha}(t) \right ] \\ 
&= \dot{\bm{x}}_{\alpha}(t) \times m_{\alpha} \dot{\bm{x}}_{\alpha}(t) + (\bm{x}_{\alpha}(t) - \bm{x}_0) \times m_{\alpha} \ddot{\bm{x}}_{\alpha}(t) \\
&= m_{\alpha} (\bm{x}_{\alpha}(t) - \bm{x}_0) \times \ddot{\bm{x}}_{\alpha}(t).
\end{align*}

\smallskip
As a second observation, let $\wp_t$ consist of the union of the two particles $\alpha$ and $\beta$. The external force and linear momentum are
\begin{align}
\bm{F}^{\rm{ext}}(\{\alpha,\beta\