Understanding Galilean Structure

Question

I’m a student with a pure math background starting to work through Arnold’s “Mathematical Methods...” and I’m struggling right of the bat with Section 1.2 on Galilean Structure. (pg 4 - 6)

So we have this affine space $A^4$ accompanied by a space of displacements $\mathbb{R}^4$. Fine.

On page 5, Arnold defined Time as a linear mapping $t:\mathbb{R}^4 \to \mathbb{R}$, and says two events $a,b\in A^4$ are simultaneous if $t(b-a) = 0$. Fine.

Then Arnold says the set of events simultaneous with a given event is a three dimensional subspace $A^3$, to which I say "Not necessarily". The mapping $t(a)= 0$ for all $a\in A^4$ satisfies Arnold's definition of a time mapping, yet clearly has a four-dimensional kernel. Is a three-dimensional kernel a requirement for a Time mapping $t$? If so, Arnold is certainly not clear about that.

But let's say I accept that for now, meaning I believe we have some Time mapping $t$ with a three dimensional kernel. The text then says that we can define the distance between two simultaneous events $a,b\in A^3$ as $\rho(a,b)=\sqrt{\langle a-b, a-b \rangle}$ where $\langle, \rangle$ is the dot product in $\mathbb{R}^3$. But vector $a-b$ still has the same representation as it did in $\mathbb{R}^4$, (something like $(x_1, x_2, x_3, x_4)$, perhaps) so it does not make sense to directly apply the three dimensional dot product. I feel we would need to choose a basis for $\text{Ker}(t)$ and then we could use the coordinate representation of $b-a$.

I hope my gripes make sense. What I could really use a extremely rigorous definition of Galilean structure.

Answer 1

Here is a "mathematical" definition.

Let $V$ be a real vector space and $A$ be an affine space for $V$. A Galilean structure on $(A,V)$ is a surjective linear map $t: V \to \mathbb R$ and a symmetric bilinear form ${\rm ker} t \times {\rm ker } t \to \mathbb R$ which is positive definite.

However, just as every finite dimensional real vector space admits a basis (is isomorphic to $\mathbb R^n$), every finite dimensional Galilean space is is isomorphic to $\mathbb R^n$ with the standard projection $f(x_1, \dots, x_n) = x_1$ and the inner product given by the dot product.

Answer 2

Arnold here seems to be defining the so-called spacetime 4-vector $(t,x,y,z)$ in the way physicists use it to define events in special relativity. I would like to first give you the "physical" picture and then the mathematical abstraction/generalization of it will seem more accessible.

A physicist will say that two events happen at the same time (simultaneous) in their frame if their clock reading is the same for both events. Their time coordinate is the same in other words. Therefore if you define 4-vectors for the two events $E=(T,X,Y,Z), E'=(T',X',Y',Z')$ then they are simultaneous iff $T=T'$.

One linear mapping that realizes the condition $ t(E-E')=0$ iff $E,E'$ are simultaneous is the matrix:

$t= \begin{bmatrix} 1&0&0&0\\ 0&0&0&0\\ 0&0&0&0\\ 0&0&0&0\\ \end{bmatrix}$

You can easily check that the kernel of this matrix is spanned by three vectors and therefore its dimension is clearly 3. The vectors are 4-dimensional for sure, but the number of vectors in the basis describing the kernel is 3 and that's what matters. What this basically means is that defining a plane of constant time in spacetime defines a 3-D surface, which is the three-dimensional space we all live in and love.

For the second part $\rho(a,b)$ is just the Euclidean distance measure. Yes, the vectors can have 4 components but the dot product definition doesn't need to necessarily include the time component to be good enough for the purposes of Galilean invariant mechanics. For a definition of a 4-vector space where the dot product is not Euclidean and 3-D anymore, look at this reference.