Let $S$ be a $3$-dimensional inertial frame. The basis vectors of $S$ are $x$, $y$ and $z$. The origin of $S$ is $O$.

Let $T$ be another frame of reference with origin $C(t)$, and basis vectors $u(t)$, $v(t)$ and $w(t)$. Notice that $T$ is moving relative to $S$.

My question is what are the necessary and sufficient conditions for $T$ to be inertial (meaning $F=ma$)? (assuming the bases are orthonormal).

In the simple case when $C(t)=O+b(t)$, $x=u$, $y=v$ and $z=w$, (so there is no rotation and the corresponding base vectors are parallel and point in the same direction) the condition is $db/dt=\mathrm{const}$., so $T$ moves uniformly. What is the solution in the general case?

  • $\begingroup$ The Poincare symmetry group. $\endgroup$
    – safesphere
    Nov 29 '17 at 15:15
  • $\begingroup$ T must be moving with constant (or zero) velocity relative to S. $\endgroup$ Nov 30 '17 at 1:54

Since you seem to be referring to newtonian mechanics, the group of symmetries is the Galilean group. A general transformation given by this group can be written as: $$\begin{cases} \textbf{r}' = \textbf{r}_0+R\textbf{r}+\textbf{v}t \\ t'=t+t_0\end{cases}$$ The Galilean group therefore has $10$ parameters: $3$ for shifts of the origin in the three spatial directions (given by the vector $\textbf{r}_0$), $3$ for rotations about one the three axis (given by the orthogonal matrix $R$), $3$ for boosts in any of the three directions (given by $\textbf{v}$) and finally $1$ for a shift of the origin of the time coordinate.

If instead you're referring to full group of isometries of Minkowski spacetime, i.e. special relativity, you should look for the Poincaré group.


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