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.(adsbygoogle = window.adsbygoogle || []).push({});

An element of the Euclidean group [itex]E(n)[/itex] can be written in the form [itex](O,\vec{b})[/itex] which acts:

[itex] \vec{x} \rightharpoondown O\vec{x}+\vec{b}[/itex]

With [itex]O \in O(n)[/itex] and [itex] \vec{b} \in R^{n}[/itex]

This would mean that the vector [itex]\vec{x}[/itex] 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:

[itex](O_{2},\vec{b_{2}})(O_{1},\vec{b_{1}})\vec{x}[/itex]

giving:

[itex](O_{2},\vec{b_{2}}) (O_{1}\vec{x}+\vec{b_{1}})[/itex]

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

[itex](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}}[/itex]

which gives:

[itex]O_{2} O_{1}\vec{x}+\vec{b_{2}} +O_{2} \vec{b_{1}} + \vec{b_{2}}[/itex]

see the two times of [itex]\vec{b_{2}}[/itex] appearing

correct path

I write that:

[itex](O_{2},\vec{b_{2}}) (O_{1}\vec{x}+\vec{b_{1}})=(O_{2},\vec{b_{2}}) \vec{x_{2}}[/itex]

So I get:

[itex]O_{2}\vec{x_{2}}+\vec{b_{2}}[/itex]

and reentering the definition of [itex]x_{2}=O_{1}\vec{x}+\vec{b_{1}}[/itex] I get:

[itex]O_{2}O_{1}\vec{x}+O_{2}\vec{b_{1}}+\vec{b_{2}}[/itex]

So here we have [itex]\vec{b_{2}}[/itex] 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...

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# Euclidean group E(n) elements

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