# Deriving mechanical energy balance law from assumed pairwise potential

I'm currently working through a book on continuum mechanics that derives mechanical balance laws by considering the particles that compose the continuum. One of the balance laws, pertaining to the rate of change of the sum of internal and kinetic energy for a configuration $$\Omega_t$$, is given as follows: $$\begin{gather} \frac{\text{d}}{\text{d}t}(U[\Omega_t] + K[\Omega_t]) = \sum_{i \in I} \mathbf{\dot{x}}\cdot\left[ \mathbf{f}_i^\text{env} + \sum_{j\notin I} \mathbf{f}_{ij}^{\text{int}} \right]. \end{gather}$$ Here, $$I$$ refers to the index of the particles that make up the configuration, whilst $$U[\Omega_{ij}] = \sum_{i,j \in I, i < j} U_{ij}, \hspace{10mm}\text{Internal Energy}\\ K[\Omega_{ij}] = \sum_{i \in I} \frac{1}{2}m\lvert\mathbf{\dot{x}}\lvert^2, \hspace{6mm}\text{Kinetic Energy}\\$$

The internal forces are given as $$\mathbf{f}_{ij}^\text{int} = -\nabla^{\mathbf{x_i}}U_{ij}$$, wheras the body forces $$\mathbf{f}_i^\text{env}$$ aren't specified.

One of the exercises directs the reader to obtain the first equation from Newton's second law, provided as $$m_i \mathbf{\ddot{x}}_i = \mathbf{f}_i^\text{env} + \sum_{j=1,j\neq i} \mathbf{f}^\text{int}_{ij}.$$ However, it also directs the reader to use the fact that the potential $$U_{ij}$$ is pairwise. Taking the time derivative of the kinetic energy is straightforward, but I'm having trouble seeing how the internal energy comes in. Any advice would be greatly appreciated.

From the first equation, $$U + K = \int(RHS)dt$$

Taking the derivatives of both sides with respect to $$\dot x$$, U vanishes in the LHS because only the kinetic energy in a function of $$\dot x$$.

Taking now the derivatives with respect to $$t$$, we get the final expression.

• That's a lovely solution, thank you for providing it. However, the problem requests the derivation of the first equation from the last equation and explicitly refers to the pairwise potential as a prerequisite for the derivation: my apologies for not being clear. Do you think there's any way of building the LHS of the first equation using the expressions that follow?
– user259499
Commented Nov 29, 2020 at 0:45

obtain the first equation from Newton's second law

$$m_i \mathbf{\ddot{x}}_i = \mathbf{f}_i^\text{env} + \sum_{j=1,j\neq i} \mathbf{f}^\text{int}_{ij}.$$

dot product with $$~\mathbf{\dot{x}}_i$$

$$\Rightarrow$$

$$m_i \mathbf{\dot{x}}_i\,\cdot\,\mathbf{\ddot{x}}_i = \mathbf{\dot{x}}_i\,\cdot\,\left[\mathbf{f}_i^\text{env} + \sum_{j=1,j\neq i} \mathbf{f}^\text{int}_{ij}\right].\tag 1$$

with:

$$m_i \mathbf{\dot{x}}_i\,\cdot\,\mathbf{\ddot{x}}_i= \frac{m_i}{2}\,\frac{d}{dt}\left(\mathbf{\dot{x}}_i\,\cdot\,\mathbf{\dot{x}}_i\,\right)$$

eq. (1):

$$\frac 12 m_i\,\frac{d}{dt}\left(\mathbf{\dot{x}}_i\,\cdot\,\mathbf{\dot{x}}_i\,\right)=\frac {d}{dt}\mathbf{dx}_i\,\cdot\,\left[\mathbf{f}_i^\text{env} + \sum_{j=1,j\neq i} \mathbf{f}^\text{int}_{ij}\right].\tag 2$$

multiply eq. (2) with $$dt$$ and integrate , you obtain:

$$\text{Kinetic energy}=-\text{Potential energy}$$