Consider a transformation to a relatively uniformly moving frame of reference, where each position vector \(\boldsymbol{r}_{i}\) is replaced by \(\boldsymbol{r}_{i}^{\prime}=\boldsymbol{r}_{i}-\boldsymbol{v} t\). (Here \(\boldsymbol{v}\) is a constant, the relative velocity of the two frames.) How does a relative position vector \(\boldsymbol{r}_{i j}\) transform? How do momenta and forces transform? Show explicitly that if equations (1.1) to (1.4) hold in the original frame, then they also hold in the new one.

Using the given transformation for the position vectors, \(\boldsymbol{r}_i'=\boldsymbol{r}_i-\boldsymbol{v}t\), we can find the transformation of the relative position vector \(\boldsymbol{r}_{ij}\). For this, let's first express \(\boldsymbol{r}_{ij}\) in terms of the original position vectors. The relative position vector is defined as the difference between position vectors \(\boldsymbol {r}_{i}\) and \(\boldsymbol{r}_{j}\): \(\boldsymbol{r}_{ij} = \boldsymbol{r}_i - \boldsymbol{r}_j\) Now, we can express \(\boldsymbol{r}_{ij}\) in terms of the primed frame using the given transformation: \(\boldsymbol{r}_{ij}' = \boldsymbol{r}_i' - \boldsymbol{r}_j' = (\boldsymbol{r}_i - \boldsymbol{v}t) - (\boldsymbol{r}_j - \boldsymbol{v}t)\) After simplifying the equation, we get: \(\boldsymbol{r}_{ij}' = \boldsymbol{r}_i - \boldsymbol{r}_j = \boldsymbol{r}_{ij}\) So, the relative position vector \(\boldsymbol{r}_{ij}\) remains the same in the new frame.

The momentum of any particle can be expressed as the product of its mass and velocity: \(\boldsymbol{p}_i = m\boldsymbol{v}_i\) Now let's find the velocity of the particle in the new frame. Since \(\boldsymbol{r}_i' = \boldsymbol{r}_i - \boldsymbol{v}t\), taking a time derivative gives the velocity in the new frame: \(\boldsymbol{v}_i' = \frac{d}{dt}(\boldsymbol{r}_i - \boldsymbol{v}t)= \frac{d\boldsymbol{r}_i}{dt} - \boldsymbol{v}\) Hence, we have: \(\boldsymbol{v}_i' = \boldsymbol{v}_i - \boldsymbol{v}\) Now, we can find the momentum in the new frame: \(\boldsymbol{p}_i' = m\boldsymbol{v}_i'\) Substituting the expression for \(\boldsymbol{v}_i'\), we find: \(\boldsymbol{p}_i' = m(\boldsymbol{v}_i - \boldsymbol{v})\)

The force acting on a particle can be expressed as the time derivative of its momentum: \(\boldsymbol{F}_i = \frac{d\boldsymbol{p}_i}{dt}\) Now let's find the force acting on the particle in the new frame. We already have the momentum in the new frame: \(\boldsymbol{p}_i' = m(\boldsymbol{v}_i - \boldsymbol{v})\) Taking the time derivative, we get: \(\frac{d\boldsymbol{p}_i'}{dt} = m\left(\frac{d\boldsymbol{v}_i}{dt} - \frac{d\boldsymbol{v}}{dt}\right)\) Since \(\boldsymbol{v}\) is constant, its time derivative is zero: \(\frac{d\boldsymbol{v}}{dt} = 0\) So, we can write the force in the new frame as: \(\boldsymbol{F}_i' = m\frac{d\boldsymbol{v}_i}{dt} = \boldsymbol{F}_i\) The force, therefore, remains the same in the new frame.

We have now shown that the relative position vector \(\boldsymbol{r}_{ij}\), momentum \(\boldsymbol{p}_i'\), and force \(\boldsymbol{F}_i'\) transform the way we described in Steps 1, 2, and 3. Without loss of generality, if the original equations (1.1) to (1.4) hold in the original frame, they'll also hold in the new frame because, as we've shown, the relative position vector, momentum, and force transform coherently within the new frame.