# algebra precalculus – Confused about how we scale graph axis’ to make the axis’ dimensionless.

I am trying to understand the solution to part $$mathrm{(iii)}$$. But, for the question I’m asking to make sense I need to include the solutions to parts $$mathrm{(i)}$$ and $$mathrm{(ii)}$$ also:

Consider a triangular lattice where the sides of the triangles have length $$d$$. The figure gives a choice of unit cells (dashed lines).

$$mathrm{(i)}$$ Use the sides of the unit cells as the primitive lattice vectors, $$boldsymbol{a}_1$$ and $$boldsymbol{a}_2$$. Write down these vectors in Cartesian coordinates.

$$mathrm{(ii)}$$ Write down a pair of reciprocal space vectors $$b_{1,2}$$ satisfying the condition that
$$a_icdot b_j = 2pidelta_{ij}$$ .
(If you want to use the explicit formula in three dimensions given in the lectures,
then you should pick as $$boldsymbol{a}_3$$ the unit vector in the direction out of the page.)

$$mathrm{(iii)}$$ The reciprocal lattice vectors G are defined by $$G = h_1b_1 + h_2b_2$$ where $$h_{1,2}$$ are integers and $$boldsymbol{b}_1$$ and $$boldsymbol{b}_2$$. Sketch the lattice that is formed by the reciprocal lattice vectors $$boldsymbol{G}$$ of the triangular lattice.

Solutions:

$$mathrm{(i)}$$ The primitive lattice vectors are $$boldsymbol{a}_1 = (d, 0)$$ and $$boldsymbol{a}_2 = left(dfrac{d}{2},dfrac{sqrt{3}d}{2}right)$$.

$$mathrm{(ii)}$$ A choice of the primitive lattice vectors (bold arrows in
diagram) for the reciprocal lattice is $$boldsymbol{b}_1=left(dfrac{2pi}{d},-dfrac{2pi}{sqrt{3}d}right)$$ and $$boldsymbol{b}_2=left(0,dfrac{4pi}{sqrt{3}d}right)$$. Other choices are possible, such as $$−boldsymbol{b}_1$$ and $$−boldsymbol{b}_2$$.

$$mathrm{(iii)}$$ $$boldsymbol{G} = h_1boldsymbol{b}_1 + h_2boldsymbol{b}_2$$ with integers $$h_{1,2}$$.
The diagram shows all the $$boldsymbol{G}$$ vectors plotted as points in $$boldsymbol{k}$$-space.
All the $$boldsymbol{G}$$ vectors form a periodic array in reciprocal space. This ‘reciprocal lattice’ for a triangular lattice in real space is itself a triangular lattice in $$boldsymbol{k}$$-space.

When I asked my lecturer about this scaling on the $$x$$ and $$y$$ axis he just said (something like) that it is to “avoid having factors of $$dfrac{2pi}{d}$$ on each increment of the $$x$$ and $$y$$ axis”. This makes sense since having a dimensionless $$x$$-axis looks clearer than this:

and similarly for the $$y$$ axis.

So I will first factor out $$dfrac{2pi}{d}$$ then the reciprocal lattice vectors are $$boldsymbol{b}_1=left(dfrac{2pi}{d},-dfrac{2pi}{sqrt{3}d}right)=dfrac{2pi}{d}left(1,-dfrac{1}{sqrt{3}}right)$$ and $$boldsymbol{b}_2=left(0,dfrac{4pi}{sqrt{3}d}right)=dfrac{2pi}{d}left(0,dfrac{2}{sqrt{3}}right)$$ I thought the graph axis’ should look like this:

and similarly for the $$y$$-axis.

The reason I think the graph axis should read $$dfrac{2pi k_x}{d}$$ and not $$dfrac{k_x d}{2pi}$$ (in the solution) is simply because I have factored out the $$dfrac{2pi}{d}$$ above so that what is plotted does not depend on $$dfrac{2pi}{d}$$. Math is not my strong point and I just cannot figure out why the axis reads $$dfrac{k_x d}{2pi}$$ instead of $$dfrac{2pi k_x}{d}$$ (which is what it looks like it should be). Can anyone please explain what is going on here?