Is the Euclidean metric the only one invariant under Galilean Transformations?

Question

Is $$ds^2=dx^2+dy^2+dz^2$$ the only metric that is invariant under Galilean transformations? And if yes how do you prove it?

Answer 1

I am not expert in this area but I think following should be a good enough proof.

Let us consider the general Galilean transformation: $$x'=Ax+a$$ where $A$ is $3\times 3$ matrix and $a$ is a vector. Then, we can implement this transformation as $$y'=By$$ where $B$ is a $4\times 4$ matrix and $y$ is a vector in 4 dimensions such that $$B\equiv\begin{pmatrix}A&a\\0&1\end{pmatrix}\quad,\quad y\equiv\begin{pmatrix}x\\1\end{pmatrix}$$

Now we are looking for invariants under this transformation. Assume that $I$ is such an invariant of the form $$I=y^TKy$$ for a $4\times4$ matrix $K$. Then, the invariance requires $$I'=I\rightarrow y^TB^TKBy=y^TKy\rightarrow B^TKB=K$$

Let us take $K$ to be of the form $$K\equiv\begin{pmatrix}p&q\\r&s\end{pmatrix}$$ for $3\times 3$ matrix $p$. Then, the invariance requires $$\begin{pmatrix}A^T&0\\a^T&1\end{pmatrix} \begin{pmatrix}p&q\\r&s\end{pmatrix} \begin{pmatrix}A&a\\0&1\end{pmatrix} =\begin{pmatrix}p&q\\r&s\end{pmatrix} $$

From this equation, we get the followings $$A^TpA=p\\ A^Tpa+A^Tq=q\\ a^TpA+rA=r\\a^Tpa+a^Tq+ra+s=s$$

Now, since $a$ and $A$ are independent, choosing $a=0$ in second and third equations forces $q=r=0$. Hence we are left with $$A^TpA=p\\a^Tpa=0$$

Since $A$ generates $SO(3)$ transformation, first equation implies that $p=1$. But the second equation then can only be satisfied if $a=0$. That means, for the generic Galilean transformation, there is no invariant quantitiy unless $p=1$ and $a=0$. Since $a$ should be independent, we somehow need to eliminate it. The straightforward method would be to use $dx$ instead of $x$, hence the only invariant is $$I=dx^Tdx$$ which is $dx^2+dy^2+dz^2$ in component form.