# How is energy dissipated if force exists in action and reaction pairs?

If force exists in action-reaction pairs, then how is energy not always conserved? To illustrate, I would like to give the example of an inelastic collision in an isolated system. All forces acting on the objects colliding are internal, and the KE of the COM remains constant. Now, when the objects collide, while one object exerts a given force, the other object also exerts an equal and opposite reaction force. If we think of kinetic energy as work done by a force acting on that object, how is mechanical energy ever dissipated in such a case?

If a collision is not elastic, one or both objects are deformed during the collision. The force causing the deformation is greater than the force of the rebound. Work is put into the object which does come back out, and its temperature goes up.

This is because the total energy of a mechanical system is a sum of contributions, $$E=\sum_{i}\frac{1}{2}m_{i}\boldsymbol{v}_{i}^{2}+U_{\textrm{internal}}+U_{\textrm{external}}$$ Your system may be isolated, $$\sum_{i}U_{\textrm{external}}\left(\boldsymbol{x}_{i}\right)=0$$ But still, your internal potential energy is a sum, $$\sum_{\textrm{pairs}}U_{ij}\left(\boldsymbol{x}_{i}-\boldsymbol{x}_{j}\right)$$ ("Pairs" is an important word here.) So that, even though, as you correctly say, $$-\frac{\partial U_{ij}}{\partial\boldsymbol{x}_{i}}=\boldsymbol{F}_{ij}=-\boldsymbol{F}_{ji}=\frac{\partial U_{ij}}{\partial\boldsymbol{x}_{j}}$$ If you have microscopic re-arrangements of the particles making up your system, you have a change in the internal potential energy, because of these re-arrangements. Some molecules may get closer together, others may slip apart, etc. So there is a change in this complicated function which depends on re-arrangements in position between the pairs. This is because the relative positions in the internal configuration change, even if nothing else changes. If your system dissipates energy as radiation, etc., it is even clearer that that can happen, but even in the simplest naive case, it is bound to happen.

If force exists in action-reaction pairs, then how is energy not always conserved?

I assume by this that you actually mean, how is mechanical energy not always conserved. Locally energy is always conserved, but it can be transformed from mechanical energy to other forms.

I believe that your confusion comes from the idea that since internal forces always come in equal and opposite pairs that the mechanical work done by one should always be equal and opposite to the mechanical work done by the other. This neglects the fact that the rate of work is $$P=\vec F \cdot \vec v$$.

Even though the $$\vec F$$ has an equal and opposite counterpart, their respective $$\vec v$$ may differ. In particular, in continuum mechanics the stress power density is given by $$\mathbf T \cdot \mathbf D$$ where $$\mathbf T$$ is the Cauchy stress tensor and $$\mathbf D$$ is the rate of deformation tensor.

The rate of deformation is the key to understanding your question. The rate of deformation is essentially* the difference in velocity for neighboring parts of the material. The forces are equal and opposite, and if the velocity is the same then there is no mechanical work and no deformation. On the other hand, if the material is deforming then the velocities are not the same and so the mechanical work is not equal and mechanical energy is lost.

I would like to give the example of an inelastic collision in an isolated system.

So in an inelastic collision there is deformation of the colliding bodies. At some point the front stops while the rear continues moving forward. This difference in velocity is what drives the loss of mechanical energy despite all of the internal forces being equal and opposite.

• More specifically $$\mathbf D=(\mathbf L+\mathbf L^T)/2$$ where $$\mathbf L= \nabla \vec v$$

Besides the other nice answers, there is a mathematical way of understanding why conservation of mechanical energy doesn't follow from the fact the forces come in pairs.

Forces come in pairs means that: $$\mathbf F_{12} + \mathbf F_{21} = \frac{d\mathbf p_2}{dt} + \frac{d\mathbf p_1}{dt} = \frac{d(\mathbf p_2+\mathbf p_1)}{dt} = 0$$ So, for any collision: $$\mathbf p_1 + \mathbf p_2 = \mathbf p_1' + \mathbf p_2'$$.

But $$E_k = \frac{1}{2}mv^2 = \frac{p^2}{2m}$$. Supposing equal masses to simplify, what we can say from the momentum conservation that $$(\mathbf p_1 + \mathbf p_2)^2 = (\mathbf p_1' + \mathbf p_2')^2 \implies (p_1^2 + 2\mathbf {p_1.p_2}+ p_2^2) = (p_1'^2 + 2\mathbf {p_1'.p_2'} + p_2'^2)$$

For the kinetic energy be conserved the dot products must be equal: $$\mathbf {p_1.p_2} = \mathbf {p_1'.p_2'}$$, which is an additional requirement, not assured by momentum conservation.