Understanding Euclidean Group E(n) Elements

[ChrisVer]

Well I am not sure if this thread belongs here or in mathematics/groups but since it also has to do with physics, I think SR would be the correct place.
An element of the Euclidean group E(n) can be written in the form (O,\vec{b}) which acts:
\vec{x} \rightharpoondown O\vec{x}+\vec{b}
With O \in O(n) and \vec{b} \in R^{n}
This would mean that the vector \vec{x} would be rotated by some angle and then translated by a vector.

Now I'm having a certain problem. Since it's a group, the multiplication of two of its elements should be an element itself.
this I write:
(O_{2},\vec{b_{2}})(O_{1},\vec{b_{1}})\vec{x}
giving:
(O_{2},\vec{b_{2}}) (O_{1}\vec{x}+\vec{b_{1}})

From here I followed two paths which I think the first gives a wrong answer, while I'm pretty sure the 2nd gives the correct one... However I don't understand what's their difference.

wrong path
(O_{2},\vec{b_{2}}) (O_{1}\vec{x}+\vec{b_{1}})=(O_{2},\vec{b_{2}}) O_{1}\vec{x}+(O_{2},\vec{b_{2}}) \vec{b_{1}}
which gives:
O_{2} O_{1}\vec{x}+\vec{b_{2}} +O_{2} \vec{b_{1}} + \vec{b_{2}}
see the two times of \vec{b_{2}} appearing

correct path
I write that:
(O_{2},\vec{b_{2}}) (O_{1}\vec{x}+\vec{b_{1}})=(O_{2},\vec{b_{2}}) \vec{x_{2}}
So I get:
O_{2}\vec{x_{2}}+\vec{b_{2}}
and reentering the definition of x_{2}=O_{1}\vec{x}+\vec{b_{1}} I get:
O_{2}O_{1}\vec{x}+O_{2}\vec{b_{1}}+\vec{b_{2}}

So here we have \vec{b_{2}} only once...
I was able to distinguish the correct from the wrong due to physical imaging (by double rotations and translations), however I don't understand (or more precisely see) what's the difference (and so the wrong in the 1st case) between the two approaches...

 

[Bill_K]

In method 1 you're trying to use the distributive law where it does not apply. You're saying T_b(x + y) = (x + b) + (y + b), whereas it should be just T_b(x + y) = x + y + b.

 

[BruceW]

In other words, the transformation is not linear. f(x+y) not equal to f(x) + f(y) in general.

 

[ChrisVer]

I think the problem of non-linearity is in the translations part... that's kinda funny, thanks...

 

[BruceW]

yep. it is pretty interesting. try looking up affine transformation, that should give more info on this topic.