Newton's Third Law of Motion

Consider a system of $N$ mutually interacting point objects. Let the $i$th object, whose mass is $m_i$, be located at vector displacement ${\bf r}_i$. Suppose that this object exerts a force ${\bf f}_{ji}$ on the $j$th object. Likewise, suppose that the $j$th object exerts a force ${\bf f}_{ij}$ on the $i$th object. Newton's third law of motion states that these two forces are equal and opposite, irrespective of their nature. In other words,

$${\bf f}_{ij} = - {\bf f}_{ji}.$$ (121)

One corollary of Newton's third law is that an object cannot exert a force on itself.

In an inertial frame, Newton's second law of motion applied to the $i$th object yields

$$m_i\,\frac{d^2 {\bf r}_i}{dt^2} = \sum_{j=1,N}^{j\neq i}\! {\bf f}_{ij}.$$ (122)

Note that the summation on the right-hand side of the above equation excludes the case $j=i$, since the $i$th object cannot exert a force on itself. Let us now take the above equation and sum it over all objects. We obtain

$$\sum_{i=1,N} \!m_i\,\frac{d^2 {\bf r}_i}{dt^2}=\sum_{i,j=1,N}^{j\neq i}\! {\bf f}_{ij}.$$ (123)

Consider the sum over forces on the right-hand side of the above equation. Each element of this sum--${\bf f}_{ij}$, say--can be paired with another element--${\bf f}_{ji}$, in this case--which is equal and opposite. In other words, the elements of the sum all cancel out in pairs. Thus, the net value of the sum is zero. It follows that the above equation can be written

$$M\,\frac{d^2 {\bf r}_{cm}}{dt^2}= {\bf0},$$ (124)

where $M = \sum_{i=1}^N m_i$ is the total mass. The quantity ${\bf r}_{cm}$ is the vector displacement of the center of mass of the system, which is an imaginary point whose coordinates are the mass weighted averages of the coordinates of the objects which constitute the system. Thus,

$${\bf r}_{cm} = \frac{\sum_{i=1}^N m_i\,{\bf r}_i}{\sum_{i=1}^N m_i}.$$ (125)

According to Eq. (124), the center of mass of the system moves in a uniform straight-line, in accordance with Newton's first law of motion, irrespective of the nature of the forces acting between the various components of the system.

Now, if the center of mass moves in a uniform straight-line, then the center of mass velocity,

$$\frac{d{\bf r}_{cm} }{dt}= \frac{\sum_{i=1}^N m_i\, d{\bf r}_i/dt}{\sum_{i=1}^N m_i},$$ (126)

is a constant of the motion. However, the momentum of the $i$th object takes the form ${\bf p}_i = m_i\,d{\bf r}_i/dt$. Hence, the total momentum of the system is written

$${\bf P} = \sum_{i=1}^N m_i\,\frac{d {\bf r}_i}{dt}.$$ (127)

A comparison of Eqs. (126) and (127) suggests that ${\bf P}$ is also a constant of the motion. In other words, the total momentum of the system is a conserved quantity, irrespective of the nature of the forces acting between the various components of the system. This result (which only holds if there is no net external force acting on the system) is a direct consequence of Newton's third law of motion.

Taking the vector product of Eq. (122) with the position vector ${\bf r}_i$, we obtain

$$m_i\,{\bf r}_i\times \frac{d^2{\bf r}_i }{dt^2}= \sum_{j=1,N}^{j\neq i} {\bf r}_i\times {\bf f}_{ij}.$$ (128)

However, it is easily seen that

$$m_i\,{\bf r}_i\times \frac{d^2{\bf r}_i }{dt^2}= \frac{d (m_i\,{\bf r}_i\times d{\bf r}_i/dt)}{dt} = \frac{d{\bf l}_i}{dt},$$ (129)

where

$${\bf l}_i = m_i\,{\bf r}_i\times \frac{d{\bf r}_i}{dt}$$ (130)

is the angular momentum of the $i$th particle about the origin of our coordinate system. The total angular momentum of the system (about the origin) takes the form

$${\bf L} = \sum_{i=1,N} {\bf l}_i$$ (131)

Hence, summing Eq. (128) over all particles, we obtain

$$\frac{d {\bf L}}{dt} = \sum_{i,j = 1,N}^{i\neq j} {\bf r}_i\times {\bf f}_{ij}.$$ (132)

Consider the sum on the right-hand side of the above equation. A general term, ${\bf r}_i\times {\bf f}_{ij}$, in this sum can always be paired with a matching term, ${\bf r}_j\times {\bf f}_{ji}$, in which the indices have been swapped. Making use of Eq. (121), the sum of a general matched pair can be written

$${\bf r}_i\times {\bf f}_{ij}+{\bf r}_j\times {\bf f}_{ji} = ({\bf r}_i-{\bf r}_j)\times {\bf f}_{ij}.$$ (133)

Suppose, now, that the forces acting between the various components of the system are central in nature, so that ${\bf f}_{ij}$ is parallel to ${\bf r}_i-{\bf r}_j$. In other words, the force exerted on object $j$ by object $i$ either points directly toward, or directly away from, object $i$, and vice versa. This is not a particularly onerous constraint, since most forces in nature are of this type (e.g., gravity). It follows that if the forces are central in nature then the vector product on the right-hand side of the above expression is zero. We conclude that

$${\bf r}_i\times {\bf f}_{ij}+{\bf r}_j\times {\bf f}_{ji} = {\bf0},$$ (134)

for all values of $i$ and $j$. Thus, the sum on the right-hand side of Eq. (132) is zero for any kind of central force. We are left with

$$\frac{d {\bf L}}{dt} = {\bf0}.$$ (135)

In other words, the total angular momentum of the system is a conserved quantity, provided that the different components of the system interact via central forces (and there is no net external torque acting on the system).