I am following an elementary physics course book, namely W.E. Gettys, F.J. Keller and M.J. Skove's Physics (in an Italian translation). In exercises where no non-conservative force acts on a rigid body, if I correctly understand them, it is assumed that the forces that the parts of the body exerce on each other do null total work.

Is the total work done by such forces null in the usual mathematical model of a rigid body? If it is, how can it be mathematically proved? I am talking about a mathematically ideal rigid, unelastic and undeformable body

What I understand is that, if the rigid body is a discrete system composed by points having mass $m_i$, with point $i$ exercing the force $\mathbf{F}_{ij}$ on point $m_j$, if we call $\mathbf{r}_j:[t_0,t_f]\to\mathbb{R}^3$ the curve along which point $j$ moves (according to the external inertial frame) in the temporal interval $[t_0,t_f]$, the total work done by the internal forces exerced by the points on each other is $$\sum_{i,j}\int_{t_0}^{t_f}\mathbf{F}_{ij}(\mathbf{r}_{j}(t))\cdot\mathbf{r}_{j}'(t)dt$$which, since Newton's third law states that $\mathbf{F}_{ij}=-\mathbf{F}_{ji}$, is -if I correctly understand- equal to $$\sum_{i<j}\int_{t_0}^{t_f}\mathbf{F}_{ij}(\mathbf{r}_{j}(t))\cdot(\mathbf{r}_{j}'(t) -\mathbf{r}_{i}'(t))dt$$but I cannot prove it to myself that such an integral is null...

I $\infty$-ly thank you for any answer!

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    $\begingroup$ If the body is rigid all your $\bf r$s are zero, which makes the integral a bit easier. $\endgroup$ Apr 30, 2015 at 9:23
  • $\begingroup$ @JohnRennie Thank you for the comment! By $\mathbf{r}_j$ I mean the parametrisation of the path taken by point $j$, which isn't null with respect to the external inertial frame... $\endgroup$ Apr 30, 2015 at 10:01
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    $\begingroup$ Related: physics.stackexchange.com/q/23097/2451 and links therein. $\endgroup$
    – Qmechanic
    Apr 30, 2015 at 11:43

1 Answer 1


We consider the integral: $$\sum_{i\lt j}\int_{t_0}^{t_f} \mathbf{F}_{ij}(\mathbf{r}_j(t))\cdot(\mathbf{r}'_j(t) - \mathbf{r}'_i(t))dt$$ For a rigid body, the distance between any two masses is always held constant, a fact that we can express as: $$\vert\mathbf{r}_i(t) - \mathbf{r}_j(t)\vert^2 = \Delta_{ij}$$ or $$(\mathbf{r}_i(t) - \mathbf{r}_j(t))\cdot(\mathbf{r}_i(t) - \mathbf{r}_j(t)) = \Delta_{ij}$$ where all the $\Delta_{ij}$ are constant for all time. Differentiating with respect to the time on both sides gives: $$2(\mathbf{r}_i(t) - \mathbf{r}_j(t))\cdot(\mathbf{r}'_i(t) - \mathbf{r}'_j(t)) = 0$$

Thus, we conclude that $(\mathbf{r}'_i(t) - \mathbf{r}'_j(t))$ is perpendicular to $(\mathbf{r}_i(t) - \mathbf{r}_j(t))$.

At this point, we will need an additional assumption to proceed - we require not just that $\mathbf{F}_{ij} = -\mathbf{F}_{ji}$, as per Newton's third law, but also that $$\mathbf{F}_{ij} = f(\mathbf{r}_i, \mathbf{r}_j)\ \left(\mathbf{r}_i(t) - \mathbf{r}_j(t)\right)$$ where $f$ is a scalar function i.e. the forces act on the radial line joining any two particles. With this assumption, the integrand vanishes: $$\mathbf{F}_{ij}(\mathbf{r}_j(t))\cdot(\mathbf{r}'_j(t) - \mathbf{r}'_i(t)) = 0$$ and we have that no internal work is done by the rigid body on itself.

Thus, it is internal forces along the line joining each pair of points in a rigid body that do no work; this is because any "net" motion between the pair of points in this direction (i.e. the direction of the force) would alter the distance, and is therefore forbidden.

Internal equal and opposite forces that do not act along the same line indeed do work: as a simple example, consider a system of two points at a constant distance $R$, with their (equal and opposite) internal forces acting perpendicular to the separation $\mathbf{R}$. The forces then produce a torque, and therefore alter the angular momentum of the system. If the magnitude of these forces were functions of $R$ alone, you would have a rigid body that spontaneously starts spinning with an ever increasing speed.

  • $\begingroup$ Wow: exactly what I needed to understand! I $\infty$-ly thank you! $\endgroup$ Apr 30, 2015 at 14:57

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