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I'm trying to comprehend the proof that for a crystal with translational symmetry only 1,2,3,4 or 6 rotational axes exist.

The proof I'm trying to follow however uses a weird notation I haven't seen before.

Let $R$ be an element of a point group and $\vec{t}_n$ a translation. What does the expression $( R | \vec{t}_n)$ mean? It is not simply applying $\vec{t}_n$ then $R$ to the lattice, as that is simply written as $R \vec{t}_n$.

I hope anybody has seen this notation at some point and can tell me what it means.

Cheers

Edit: added an excerpt from the script I'm trying to follow using the notation.

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    $\begingroup$ It will be easier to decipher this if you provide a link to a sample paper or book that uses the notation. $\endgroup$ – Emilio Pisanty Jul 21 '15 at 13:23
  • $\begingroup$ "Our proof start from the fact that if (R | ⃗t) is a member of the space group and ⃗$tn$ is a primitive translation, then $R⃗t$ must also be a primitive trans- lation. This follows because if $(R | ⃗t)$ is a member of the group, so is $(R | ⃗t)(E | ⃗t)(R | ⃗t) = (E | R⃗t)$" $\endgroup$ – user17574 Jul 21 '15 at 13:48
  • $\begingroup$ That is pretty decidedly not what I meant by "link". Provide a reference and let your prospective answerers be the judges of which excerpts are useful and which ones are not. $\endgroup$ – Emilio Pisanty Jul 21 '15 at 15:09
  • $\begingroup$ Note also that the equation in your comment is almost certainly incorrect. $\endgroup$ – Emilio Pisanty Jul 21 '15 at 15:22
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Since you have not provided a direct reference, it's hard to be completely sure (and particularly to pin down the details), but there's really only one general idea that this can refer to.

In general, any arbitrary isometry $S$ of euclidean space has the form $$ \mathbf x\mapsto R\mathbf x+\mathbf t, $$ where $\mathbf t$ is an arbitrary vector and $R\in\mathrm{O}(3)$ is an orthogonal matrix. The isometry $S$ is then indexed by both these objects, so you can think of $S$ as the ordered pair $(R,\mathbf t)$. It is fairly common for authors to come up with slightly off-beat notations such as $(R|\mathbf t)$ or $\{R,\mathbf t\}$ (example) to emphasize that this is what's happening.

This sort of typographical emphasis is somewhat justified, because the group law in this notation is somewhat complicated: if $S=(R,\mathbf t)$ and $T=(P,\mathbf u)$, then $$ \mathbf x \stackrel{S}{\mapsto} R\mathbf x+\mathbf t \stackrel{T}{\mapsto} P(R\mathbf x+\mathbf t)+\mathbf u =PR\mathbf x+(P\mathbf t+\mathbf u), $$ which means that $T\circ S=(R,\mathbf t)\circ(P,\mathbf u)=(PR,P\mathbf t+\mathbf u)$.

Note that the details of the implementation can change, for example, if one translates before the rotation, i.e. $\mathbf x\mapsto R(\mathbf x+\mathbf t)$, but this does not affect the general mechanics. However, it does make it impossible to know exactly what each author means by the notation, which is why the precise notation convention is always specified at the start of each text.

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I think I can see where this is going, but I don't understand your notation so I will use different, but I hope that this will make sense....

The point here is that if translation is a symmetry operation (call it $t$) and rotation is a symmetry operation (call it $C_n$ - where $n$ is the number of $C_n$ operations required for a full turn - i.e. an $n-$fold axis) then because the symmertry operations are part of a mathematical group of operations $tC_n$ and $C_nt$ must be symmetry operations as well. So if you can translate along the x axis and there is a $C_4$ rotation you must be able to translate along the y axis in either direction and also in the minus x direction. - if $t$ is translate along the x-axis and $C_4$ is a anticlockwise/counterclockwise quarter turn then $tC_4$ is a quarter turn and then translate along the y axis (assuming that we take operations from right to left).

Now I think the point of the proof is that in maths you can tesselate in 2 dimensions only triangles, squares and hexagons - not pentagons, heptagons and higher... so any crystal cannot be repeated translationally on a $C_5$ ($5-$fold) axis, so if there is translational symmetry there cannot be a $C_5$ axis.

Hope this helps figure out the proof you are looking at.

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  • $\begingroup$ Thanks for the comment. However I did understand the proof that followed afterwards (note that what I wrote in the comments is only an excerpt, not the whole proof) and was really only confused what my professor was trying to say with that notation I have never seen. Thanks anyway :) $\endgroup$ – user17574 Jul 21 '15 at 14:40
  • $\begingroup$ @user17574 - ok I understand. Interesting though, I learnt something from your question - thanks for posting. $\endgroup$ – tom Jul 21 '15 at 21:22

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