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In Arnold's Mathematical Methods of Classical Mechanics, he says on page 6 the following are Galilean transformations on the Galilean coordinate space $\mathbb{R} \times \mathbb{R}^3$ where $\mathbb{R}$ is the time axis and $\mathbb{R}^3$ has a fixed Euclidian structure:

(1) Uniform motion with velocity $\mathbf{v} \colon$ $$g_1(t,\mathbf{x})=(t,\mathbf{x}+\mathbf{v}t)$$ (2) Translation of the origin: $$g_2(t,\mathbf{x})=(t+s,\mathbf{x}+\mathbf{s}),$$ (3) Rotation of the coordinate axes: $$g_3(t,\mathbf{x})=(t,G\mathbf{x}),$$ where $G \in O(3).$

He says that Galilean transformations are affine transformations of $A^4$ which preserve intervals of time and distance between simultaneous events. Why are the above transformations affine?

For transformation (1), above we can see that if $a,b \in A^4,$ then before the transformation we have, denoting $d$ to be the distance between two simultaneous events using the given Euclidian structure, $d(a,b)=\|a-b\|.$ But, from transformation (1) we have that, letting $f$ be the corresponding affine transformation, $f(a)-f(b)=g_1(a-b)$ implies that $d(f(a),f(b))=\|f(a)-f(b)\|=\|(x-y)+\mathbf{v}t\|$ which implies that $f$ does not preserve distance. What am I missing here?

Qmechanic
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Arpit Mittal
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2 Answers2

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The maps $g_1,g_3$ are linear, while $g_2$ is affine and it is linear if and only if $(s,\mathbf{s})=(0,0)$. Since you talk only about $g_1$, I’ll explain in detail why.

The quick proof, is actually just an application of a few standard linear algebra theorems. The first component of $g_1$ is $t$, i.e $\pi_1\circ g_1=\pi_1$, and the latter is a linear map. Next, we have $\pi_2\circ g_1= \pi_2+\mathbf{v}\pi_1$, which is a sum of linear maps, hence linear, so we’re done. Anyway, if in doubt, just go back to definitions: \begin{align} g_1\left(c\cdot(t,\mathbf{x})\right)&=g_1\left(ct,c\mathbf{x}\right)=(ct,c\mathbf{x}+\mathbf{v}(ct))=c(t,\mathbf{x}+\mathbf{v}t)=c\cdot g_1(t,\mathbf{x}), \end{align} so this proves homogeneity of $g_1$. Next, we can prove additivity of $g_1$: \begin{align} g_1\left((t,\mathbf{x})+(s,\mathbf{y})\right)&=g_1\left(t+s,\mathbf{x}+\mathbf{y}\right)\\ &=(t+s,(\mathbf{x}+\mathbf{y})+\mathbf{v}(t+s))\\ &=(t,\mathbf{x}+\mathbf{v}t)+(s,\mathbf{y}+\mathbf{v}s)\\ &=g_1(t,\mathbf{x})+g_1(s,\mathbf{y}). \end{align} Thus, $g_1$ is linear. If you want to define things at the level of the affine spaces, then you can say that a Galilean boost with velocity $\mathbf{v}\in\Bbb{R}^3$ is a map $f_1:\Bbb{E}^1\times\Bbb{E}^3\to\Bbb{E}^1\times\Bbb{E}^3$ for which there exist $\mathbf{v}\in\Bbb{R}^3$ and $t_0\in\Bbb{E}^1$ such that for all $(t,p)\in\Bbb{E}^1\times\Bbb{E}^3$, we have $f_1\left(t,p\right)=(t,p+\mathbf{v}(t-t_0))$.

Now, take two events $a,b\in A=\Bbb{E}^1\times\Bbb{E}^3$ which are simultaneous. So, we can write them in the form $a=(t,p)$ and $b=(t,q)$ for some $t\in\Bbb{E}^1$ (the same in both tuples due to simultaneity) and $p,q\in\Bbb{E}^3$. So, the distance between $a$ and $b$ is $d(a,b)=\|p-q\|$. Next, we have $f_1(a)-f_1(b)=(0,p-q)$ meaning that $f_1(a)$ and $f_1(b)$ are simultaneous events (clear from the definitions, but I just wanted to emphasize this) and the distance between these events is $\|p-q\|$, thus proving that $f_1$ is a distance-preserving map between simultaneous events.

Honestly, at this stage, you don’t even need to go back to the affine spaces. If you want to work with the $g_i$’s, then just keep your domain and target as vector spaces. The map $g_1$ is linear, and it restricts to a map $\{t\}\times\Bbb{R}^3 \to \{t\}\times\Bbb{R}^3$ (the fiber of events at time $t$) where it is simply a translation by amount $\mathbf{v}t$, and translations obviously preserve distances.

peek-a-boo
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First, a map being affine does not mean that it preserves distances. In the case of Galilean transformations this is true, but it's not what affinity is about. A transformation is affine if it can be written as a linear transformation plus a translation. This is true for all of the three mentioned transformations:

  • $g_1$ is linear since $g_1(t+ t',x+x')=(t+t',(x+x')+v(t+t'))=(t,x+vt)+(t',x'+vt')=g_1(t,x)+g_1(t',x')$ and $g_1(\lambda t,\lambda x)=(\lambda t,\lambda x+v\lambda t)=\lambda (t,x+vt)=\lambda g_1(t,x)$.
  • $g_2$ is just a translation
  • $g_3$ is clearly linear

But also, these all do preserve distances. It's probably obvious for translations and rotations. And for boosts we have that the spatial coordinates are translated by a vector $vt$ directly proportional to $t$. But the two events we are looking at are simultaneous, so their $t$ is equal, meaning that the spatial coordinates are translated by the exact same vector, so their distances must remain the same. Or spelling it out mathematically: Let $(t,x),(t,x')$ be two simultaneous events. Their spatial distance is $\vert x-x'\vert$. Now apply $g_1$ to both to get two boosted events: $g_1(t,x)=(t,x+vt)$ and $g_1(t,x')=(t,x'+vt)$. Their spatial distance is $\vert x+vt-(x'+vt)\vert=\vert x-x'\vert$, which is exactly the same.

You don't transform the difference vector. You transform both points independently and then take the difference of the transformed points.

Vercassivelaunos
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