Translational Invariance

Suppose that we transform our coordinate system such that the origin shifts from ${\bf r}={\bf0}$ to ${\bf r}= {\bf r}_{\rm shift}$, where ${\bf r}_{\rm shift}$ is independent of time. It follows that

$${\bf r}_i \rightarrow {\bf r}_i - {\bf r}_{\rm shift},$$ (1.91)

$${\bf v}_i \rightarrow {\bf v}_i,$$ (1.92)

$${\bf f}_{ij} \rightarrow {\bf f}_{ij},$$ (1.93)

$${\bf F}_i \rightarrow {\bf F}_i.$$ (1.94)

The latter two equations follow because forces are obviously not affected by the transformation. It is clear that the linear equation of motion, (1.89), is invariant under the transformation. On the other hand, the angular equation of motion, (1.90), becomes

$$m_i\,\frac{d({\bf r}_i\times {\bf v}_i)}{dt}-m_i\,{\bf r}_{\rm sh... ...times \sum_{j=1,N}^{j\neq i}{\bf f}_{ij} - {\bf r}_{\rm shift}\times {\bf F}_i.$$ (1.95)

However, the vector product of ${\bf r}_{\rm shift}$ with Equation (1.89) yields

$$m_i\,{\bf r}_{\rm shift}\times \frac{d{\bf v}_i}{dt} = {\bf r}_{\... ...times \sum_{j=1,N}^{j\neq i} {\bf f}_{ij} + {\bf r}_{\rm shift}\times{\bf F}_i,$$ (1.96)

The previous two equations can be combined to give

$$m_i\,\frac{d({\bf r}_i\times {\bf v}_i)}{dt} = \sum_{j=1,N}^{j\neq i} {\bf r}_i\times {\bf f}_{ij} + {\bf r}_i\times {\bf F}_i.$$ (1.97)

Thus, we conclude that the angular equation of motion, (1.90), is also invariant under the transformation. Of course, all of this makes sense because the choice of the origin of a Cartesian coordinate system is completely arbitrary, and has no bearing on the motions of bodies in the universe. One corollary of the previous analysis is that it does not matter about which point we choose to take moments of momenta and forces to generate angular momenta and torques, respectively, as long as we choose the same point in all cases.