# S03. Galilean Relativity - SOLUTIONS

- Page ID
- 3921

### Exercise 3.1.1

**Answer**-
\[\begin{equation*}

\begin{aligned}

dx = dx' + vdt'

\end{aligned}

\end{equation*}\]dividing by \(dt\) gives

\[\begin{equation*}

\begin{aligned}

u= \frac{dx}{dt} = \frac{dx' + vdt'}{dt}

\end{aligned}

\end{equation*}\]and since \(dt = dt'\) this simplifies to

\[\begin{equation*}

\begin{aligned}

u = u' + v

\end{aligned}

\end{equation*}\]

### Exercise 3.2.1

**Answer**-
Starting from the law in the unprimed frame and using the coordinate transformation we get

\[\begin{equation*}

\begin{aligned}

-k((x' + vt') - (x_c' + vt')) = m\ddot{x}'

\end{aligned}

\end{equation*}\]which simplifies to

\[\begin{equation*}

\begin{aligned}

-k(x' - x_c') = m\ddot{x}'

\end{aligned}

\end{equation*}\]thus it has the same form and is therefore invariant under the Galilean transformation.

It might be worth pointing out that \( d/dt = d/dt' \) and since \(x'\) only differs by a term linear in \(t\) from \(x\), their second derivatives with respect to time are the same (hence the right-hand side in the first line of equation in this solution).

### Exercise 3.3.1

**Answer**-
Because Newton's laws are the same in both coordinate systems, one can not do an experiment to tell whether one is in the moving frame or the unmoving frame. All experiments, in either frame, will be consistent with Newton's laws, giving no indication of any motion with respect to some absolute rest frame.

Newton's laws -- and certainly the very specific one we examined in the above exercise -- are consistent with the principal of Galilean relativity. The principal of relativity is that all motion is relative. The Galilean boost leaves the form of the law invariant, guaranteeing that there is nothing special about either of the two frames; there is no objective meaning to a statement that one is moving and the other is not.