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.

  • 1
    $\begingroup$ I think that, so long as $t(x) \ne 0, \forall x$ then, so long as $t$ is linear, the kernel is three dimensional (so the function you're chosen is the only bad one). I think this is the rank-nullity theorem, which is that, if $f$ is a linear function from $V\to W$, then $\dim(\ker g) + \dim(\mathop{\mathrm{im}} f) = \dim(V)$? If so, Arnold has missed out the statement that $t$ can't be zero everywhere, which would be something that would be obvious to physicists, perhaps less so to mathematicians. Note I'm a physicist who has done enough maths to be dangerous so this may all be wrong. $\endgroup$ – tfb Aug 6 '18 at 20:19
  • $\begingroup$ In my above comment, an $f$ changed to a $g$: it should read '[...] $\dim(\ker f) + \dim(\mathop{\mathrm{im}} f) = \dim(V)$ [...]'. Sorry. $\endgroup$ – tfb Aug 6 '18 at 20:33
  • $\begingroup$ I knew what you meant, and all you wrote is correct. I was more disturbed by the change of coordinates required to apply the 3 dimensional dot product, as is done in the text. $\endgroup$ – P. May Aug 6 '18 at 20:35
  • $\begingroup$ I believe it is as @DinosaurEgg wrote below, and Arnold meant to reserve a dimension for time in his proposed 4-space, in which case all of my problems are largely resolved. $\endgroup$ – P. May Aug 6 '18 at 20:36
  • $\begingroup$ And your prescription for that is correct: choose a basis for $\ker t$ and use the appropriate restriction of the original inner product in that subspace (not sure if that is the right term). $\endgroup$ – tfb Aug 6 '18 at 20:39

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.

  • $\begingroup$ You are trying to "define" a single simple concept by using a number of more complex concepts. This is hardly how definition works. To understand the Galilean spacetime, there is no need to study the subjects of affine space, surjective linear map, or symmetric bilinear form. There are simpler ways to define the Galilean spacetime mathematically without ambiguity or unnecessary complexity. $\endgroup$ – safesphere Aug 7 '18 at 9:01
  • $\begingroup$ @safesphere As a math person, this was just what I was looking for, and I perceive no ambiguity. If you have a definition you prefer, please leave an answer and I would love to read it. $\endgroup$ – P. May Aug 7 '18 at 13:07
  • $\begingroup$ @safesphere I agree that this definition is not the right way to understand what Gallilean spacetime is about. However, if you want a definition in the abstract algebra style (which is what P. May was looking for) then this is the way to go. $\endgroup$ – Phil Tosteson Aug 7 '18 at 14:35
  • $\begingroup$ @P.May It is great that this answer works for you. However, many people read this site, not only original posters, and based on this answer they may get a wrong impression that the Galilean spacetime is more complex than it actually is. Affine space is a generalization that is helpful in general cases. However, it is rather pointless first to go to a deep generalization and then all the way back by defining a very special case of this generalization. Sort of like undressing bare just to dress up again in exactly the same clothes. I didn't downvote or say the answer was wrong, just commented. $\endgroup$ – safesphere Aug 7 '18 at 15:59
  • $\begingroup$ @P.May BTW I did leave a link to an earlier answer for you to see if it is of any help, but someone deleted it. Not sure why moderators would do such things. It may or may not be helpful to you, but it may bring more clarity to others. So here it is again: physics.stackexchange.com/questions/368680/… and this with a bit more details:physics.stackexchange.com/questions/372843/… $\endgroup$ – safesphere Aug 7 '18 at 16:06

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.

  • $\begingroup$ All of that makes sense, if in your definition of space-time you "reserve" a dimension for time, as you did with your vector (t, x, y z), and I agree this is the intuitive way to do it. But Arnold does not do this in his definition, leaving the possibility that the "time axis" could be traversing the space diagonally, in which case there needs to be much more machinery in his time and distance metrics. $\endgroup$ – P. May Aug 6 '18 at 20:31
  • $\begingroup$ I'll continue through the book assuming he means what you have written. Thank you. $\endgroup$ – P. May Aug 6 '18 at 20:32
  • $\begingroup$ @P.May: don't do that. Getting tied into a particular set of coordinates / basis for things is a trap from which it takes people a long time to recover. That's why rnold is trying to avoid doing so. $\endgroup$ – tfb Aug 6 '18 at 20:36
  • $\begingroup$ @tfb Well if I'm not mean to do so, then the second portion of my initial question remains. A change of coordinates is required before the 3-dimensional dot product can be applied. $\endgroup$ – P. May Aug 6 '18 at 20:39
  • $\begingroup$ @P.May In answer to your first comment, I think a better way for Arnold to state that, without defining the specific mapping, would be: "The mapping t is DEFINED to have a kernel of dimension 3" . This would allow for "rotations" in 4-space, but the matrix would still always be of projective nature (rank=1). $\endgroup$ – DinosaurEgg Aug 6 '18 at 21:15

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