# Why are the transformations from the Galilean transformations affine?

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?

• The $vt$ terms cancel out in the difference
– LPZ
Commented Jul 6, 2023 at 21:33
• How so? I show that in the affine setting the distance terms do not cancel out. Commented Jul 6, 2023 at 23:54
• @Chordx I think you may be confusing yourself by letting $g_1$ act on a coordinate displacement $a-b$. A transformation acts on coordinates, and you should calculate the displacement afterwards.
– Amit
Commented Jul 7, 2023 at 1:20
• @Amit can you expand on that? I am interpreting the book as $g_i$ operating on the vector space $\mathbb{R} \times \mathbb{R}^3$ rather than the affine space $A^4$ which is part of the Galilean structure. The affine transformation sends elements of $A^4$ to $A^4$ but the corresponding well defined linear operator $f: \mathbb{R} \times \mathbb{R}^3 \to \mathbb{R} \times \mathbb{R}^3$ should be such that $f(x-y)=f(x)-f(y)?$ Commented Jul 7, 2023 at 1:36
• @Chordx affine is not the same as linear
– Amit
Commented Jul 7, 2023 at 1:42

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.

• @Chordx sure, in general the Galilean space is not, but $\Bbb{E}^1\times\Bbb{E}^3$ is one such example. I didn’t want to have to do this initially because I already gave a very general, detailed answer here, but one can say “suppose there exists an affine isomorphism $\Phi:A\to\Bbb{E}^1\times\Bbb{E}^3$ which preserves time intervals and distances of simultaneous events”. Then one can do everything I wrote. Anyway, this is not really important here, because the role of Galilean transformations is to identify the “isomorphisms” of $\Bbb{E}^4$. Commented Jul 7, 2023 at 12:24
• that’s why in the very example Arnold gives he talks about “Galilean coordinate space $\Bbb{R}\times\Bbb{R}^3$”. Anyway, these remarks don’t take away from the fact that as defined, $g_1,g_3$ are linear and $g_2$ is clearly affine. Also, they all preserve distances between simultaneous events, and your main mistake is that you didn’t take into account simultaneity which is why you got that extra $\mathbf{v}t$ term. Commented Jul 7, 2023 at 12:25
• @Chordx yes, and you see that in the case $a_1=b_1$, you get (modulo notation) essentially what I said in my second last paragraph. Commented Jul 7, 2023 at 13:02
• Forget the specifics of this problem. Do you understand the definition of an affine map between affine spaces and its relation to linear maps of the underlying vector spaces? It seems you simple need to review definitions and try out some simple examples. Commented Jul 7, 2023 at 16:12
• in general if $A$ is an affine space and $V$ the underlying vector space and $g:V\to V$ linear, then there are infinitly many affine maps $f:A\to A$ having $g$ as the linear part. Namely, for each $p\in A$, we have a map $f_p:A\to A$ defined as $f_p(x)=p+g(x-p)$. If you really want to check this works, then simply note that $f_p(x)-f_p(y)=[p+g(x-p)]-[p+g(y-p)]=(p-p)+[g(x-p)-g(y-p)]=g(x-y)$. Once again in words: literally pick a point in the affine space. Doesn’t matter which one. Add to that point your linear map evaluated at $x-p$. That’s it. Commented Jul 7, 2023 at 17:28

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.

• I understand what you are saying until the end of the bullet points. My issue comes with what I wrote in my original post to show why the differences are not the same, why is that wrong? Also, the distances between two affine points is a vector so the $g$ will increase this vector which causes the difference to increase. Commented Jul 7, 2023 at 12:18
• @Chordx I added the same thing but expressed in mathematical form. Is it clearer now? Commented Jul 7, 2023 at 17:44