# Euclidean group E(n) elements

1. Jan 31, 2014

### 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...

Last edited: Jan 31, 2014
2. Jan 31, 2014

### Bill_K

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

3. Jan 31, 2014

### BruceW

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

4. Feb 1, 2014

### ChrisVer

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

5. Feb 1, 2014

### BruceW

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