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[Mentors' note: This thread was split off from https://www.physicsforums.com/threads/system-potential-energy-and-nonconservative-forces.1009237/]
For a general mechanical system, you usually split the specified forces into external specified forces ##\mathbf{F}_a## and internal specified forces ##\mathbf{G}_{ab}## (indices enumerate particles). If the external forces are conservative, ##\mathbf{F}_a = - \nabla \phi_a##, then\begin{align*}
\sum_a \int_1^2 \mathbf{F}_a \cdot \mathbf{v}_a dt = \sum_a (\phi_a(1) - \phi_a(2)) = \Phi(1) - \Phi(2)
\end{align*}and ##\Phi \equiv \displaystyle{\sum_a} \phi_a## is the external potential function. If the internal forces are conservative, that is, if ##\mathbf{G}_{ab} = g_{ab}(r_{ab}) \hat{\mathbf{r}}_{ab}## where ##\mathbf{r}_{ab} = \mathbf{r}_a - \mathbf{r}_b## and the function ##g_{ab}(r_{ab})## depends only on the inter-particle separation, then using ##\mathbf{G}_{ab} + \mathbf{G}_{ba} = \mathbf{0}## you have\begin{align*}
\sum_a \sum_{b \neq a} \int_1^2 \mathbf{G}_{ab} \cdot \mathbf{v}_a dt &=\sum_a \sum_{b>a} \int_{1}^2 (\mathbf{G}_{ab} \cdot \mathbf{v}_a + \mathbf{G}_{ba} \cdot \mathbf{v}_b)dt \\
&= \sum_a \sum_{b>a} \int_{1}^2 (\mathbf{G}_{ab} \cdot (\mathbf{v}_a - \cdot \mathbf{v}_b) )dt
\end{align*}Work on the term \begin{align*}
\mathbf{G}_{ab} \cdot (\mathbf{v}_a - \mathbf{v}_b) &= \dfrac{g_{ab}(r_{ab})}{r_{ab}} \mathbf{r}_{ab} \cdot \dfrac{d\mathbf{r}_{ab}}{dt} \\
&=\dfrac{1}{2} \dfrac{g_{ab}(r_{ab})}{r_{ab}} \dfrac{d}{dt}(r_{ab}^2) \\
&= g_{ab}(r_{ab}) \dfrac{dr_{ab}}{dt}
\end{align*}Then the integral becomes\begin{align*}
\sum_a \sum_{b>a} \int_1^2 g_{ab}(r_{ab}) \dfrac{dr_{ab}}{dt} dt &= \sum_a \int_1^2 g_{ab}(r_{ab}) dr_{ab} \\
&= \sum_a \sum_{b>a} [\psi_{ab}(r_{ab}(1)) - \psi_{ab}(r_{ab}(2)) ] \\
&= \Psi(1) - \Psi(2)
\end{align*}where the function ##\psi_{ab}## is such that ##g_{ab} = -\dfrac{d\psi_{ab}}{dr_{ab}}##, and where ##\Psi## is the internal potential function. The sum ##U = \Phi + \Psi## is then the overall potential function of the system, taking into account both the external and internal conservative specified forces.
If you have some non-conservative specified forces ##\mathbf{H}_a##, then\begin{align*}
\sum_a \int_1^2 (\mathbf{F}_a + \mathbf{H}_a) \cdot \mathbf{v}_a dt + \sum_a \sum_{b \neq a} \int_1^2 \mathbf{G}_{ab} \cdot \mathbf{v}_a dt&= \sum_a \int_1^2 m_a\ddot{\mathbf{r}}_a \cdot \dot{\mathbf{r}}_a dt \\
&= \sum_a \int_1^2 m_a \dfrac{1}{2} \dfrac{d}{dt} (v_a^2) dt \\
&= \sum_a \dfrac{1}{2}m_a(v_a(2)^2 - v_a(1)^2)
\end{align*}so that\begin{align*}-\Delta(\Phi + \Psi) + \sum_a \int_1^2 \mathbf{H}_a \cdot \mathbf{v}_a dt &= \Delta T \\
\implies \sum_a \int_1^2 \mathbf{H}_a \cdot \mathbf{v}_a dt &= \Delta (\Phi + \Psi + T) = \Delta E
\end{align*}Here ##\displaystyle{\sum_a \int_1^2 \mathbf{H}_a \cdot \mathbf{v}_a dt}## is the work done by any other non-conservative specified forces on the system.
This is not a particularly helpful way to think about things;fog37 said:
For a general mechanical system, you usually split the specified forces into external specified forces ##\mathbf{F}_a## and internal specified forces ##\mathbf{G}_{ab}## (indices enumerate particles). If the external forces are conservative, ##\mathbf{F}_a = - \nabla \phi_a##, then\begin{align*}
\sum_a \int_1^2 \mathbf{F}_a \cdot \mathbf{v}_a dt = \sum_a (\phi_a(1) - \phi_a(2)) = \Phi(1) - \Phi(2)
\end{align*}and ##\Phi \equiv \displaystyle{\sum_a} \phi_a## is the external potential function. If the internal forces are conservative, that is, if ##\mathbf{G}_{ab} = g_{ab}(r_{ab}) \hat{\mathbf{r}}_{ab}## where ##\mathbf{r}_{ab} = \mathbf{r}_a - \mathbf{r}_b## and the function ##g_{ab}(r_{ab})## depends only on the inter-particle separation, then using ##\mathbf{G}_{ab} + \mathbf{G}_{ba} = \mathbf{0}## you have\begin{align*}
\sum_a \sum_{b \neq a} \int_1^2 \mathbf{G}_{ab} \cdot \mathbf{v}_a dt &=\sum_a \sum_{b>a} \int_{1}^2 (\mathbf{G}_{ab} \cdot \mathbf{v}_a + \mathbf{G}_{ba} \cdot \mathbf{v}_b)dt \\
&= \sum_a \sum_{b>a} \int_{1}^2 (\mathbf{G}_{ab} \cdot (\mathbf{v}_a - \cdot \mathbf{v}_b) )dt
\end{align*}Work on the term \begin{align*}
\mathbf{G}_{ab} \cdot (\mathbf{v}_a - \mathbf{v}_b) &= \dfrac{g_{ab}(r_{ab})}{r_{ab}} \mathbf{r}_{ab} \cdot \dfrac{d\mathbf{r}_{ab}}{dt} \\
&=\dfrac{1}{2} \dfrac{g_{ab}(r_{ab})}{r_{ab}} \dfrac{d}{dt}(r_{ab}^2) \\
&= g_{ab}(r_{ab}) \dfrac{dr_{ab}}{dt}
\end{align*}Then the integral becomes\begin{align*}
\sum_a \sum_{b>a} \int_1^2 g_{ab}(r_{ab}) \dfrac{dr_{ab}}{dt} dt &= \sum_a \int_1^2 g_{ab}(r_{ab}) dr_{ab} \\
&= \sum_a \sum_{b>a} [\psi_{ab}(r_{ab}(1)) - \psi_{ab}(r_{ab}(2)) ] \\
&= \Psi(1) - \Psi(2)
\end{align*}where the function ##\psi_{ab}## is such that ##g_{ab} = -\dfrac{d\psi_{ab}}{dr_{ab}}##, and where ##\Psi## is the internal potential function. The sum ##U = \Phi + \Psi## is then the overall potential function of the system, taking into account both the external and internal conservative specified forces.
If you have some non-conservative specified forces ##\mathbf{H}_a##, then\begin{align*}
\sum_a \int_1^2 (\mathbf{F}_a + \mathbf{H}_a) \cdot \mathbf{v}_a dt + \sum_a \sum_{b \neq a} \int_1^2 \mathbf{G}_{ab} \cdot \mathbf{v}_a dt&= \sum_a \int_1^2 m_a\ddot{\mathbf{r}}_a \cdot \dot{\mathbf{r}}_a dt \\
&= \sum_a \int_1^2 m_a \dfrac{1}{2} \dfrac{d}{dt} (v_a^2) dt \\
&= \sum_a \dfrac{1}{2}m_a(v_a(2)^2 - v_a(1)^2)
\end{align*}so that\begin{align*}-\Delta(\Phi + \Psi) + \sum_a \int_1^2 \mathbf{H}_a \cdot \mathbf{v}_a dt &= \Delta T \\
\implies \sum_a \int_1^2 \mathbf{H}_a \cdot \mathbf{v}_a dt &= \Delta (\Phi + \Psi + T) = \Delta E
\end{align*}Here ##\displaystyle{\sum_a \int_1^2 \mathbf{H}_a \cdot \mathbf{v}_a dt}## is the work done by any other non-conservative specified forces on the system.
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