# Derivation of Laue equations

The Wikipedia article for the Laue equations presents them as follows:

Let $$\mathbf{a}\, ,\mathbf{b}\, ,\mathbf{c}$$ be the primitive vectors of the crystal lattice $$L$$, whose atoms are located at the points $$\mathbf x = p\,\mathbf a+q\,\mathbf b+r\,\mathbf c$$ that are integer linear combinations of the primitive vectors.

Let $$\mathbf{k}_{\mathrm{in}}$$ be the wavevector of the incoming (incident) beam, and let $$\mathbf k_{\mathrm{out}}$$ be the wavevector of the outgoing (diffracted) beam. Then the vector $$\mathbf k_{\mathrm{out}} - \mathbf k_{\mathrm{in}} = \mathbf{\Delta k}$$ is called the "scattering vector" (also called transferred wavevector) and measures the change between the two wavevectors.

The three conditions that the scattering vector $$\mathbf{\Delta k}$$ must satisfy, called the "Laue equations", are the following: the numbers $$h, k, l$$ determined by the equations

$$\mathbf{a}\cdot\mathbf{\Delta k}=2\pi h$$ $$\mathbf{b}\cdot\mathbf{\Delta k}=2\pi k$$ $$\mathbf{c}\cdot\mathbf{\Delta k}=2\pi l$$

must be integer numbers. Each choice of the integers $$(h,k,l)$$, called Miller indices, determines a scattering vector $$\mathbf{\Delta k}$$. Hence there are infinitely many scattering vectors that satisfy the Laue equations. They form a lattice $$L^*$$, called the reciprocal lattice of the crystal lattice. This condition allows a single incident beam to be diffracted in infinitely many directions. However, the beams that correspond to high Miller indices are very weak and can't be observed. These equations are enough to find a basis of the reciprocal lattice, from which the crystal lattice can be determined. This is the principle of x-ray crystallography.

The same article provides the mathematical derivation as follows:

The incident and diffracted beams are planar wave excitations

$${\displaystyle f_{\mathrm {in} }(t,\mathbf {x} )=A_{\mathrm {in} }\cos(\omega \,t-\mathbf {k} _{\mathrm {in} }\cdot \mathbf {x} )}$$

$${\displaystyle f_{\mathrm {out} }(t,\mathbf {x} )=A_{\mathrm {out} }\cos(\omega \,t-\mathbf {k} _{\mathrm {out} }\cdot \mathbf {x} ).}$$

of a field that for simplicity we take as scalar, even though the main case of interest is the electromagnetic field, which is vectorial.

The two waves propagate through space independently, except at the points of the lattice, where they resonate with the oscillators, so their phase must coincide. Hence for each point

$${\displaystyle \cos(\omega \,t-\mathbf {k} _{\mathrm {in} }\cdot \mathbf {x} )=\cos(\omega \,t-\mathbf {k} _{\mathrm {out} }\cdot \mathbf {x} ),}$$

or equivalently, we must have

$${\displaystyle \omega \,t-\mathbf {k} _{\mathrm {in} }\cdot \mathbf {x} =\omega \,t-\mathbf {k} _{\mathrm {out} }\cdot \mathbf {x} +2\pi n,}$$

for some integer $$n$$, that depends on the point $$\mathbf{x}$$. Simplifying we get

$${\displaystyle \mathbf {\Delta k} \cdot \mathbf {x} =(\mathbf {k} _{\mathrm {out} }-\mathbf {k} _{\mathrm {in} })\cdot \mathbf {x} =2\pi n.}$$

Now, it is enough to check that this condition is satisfied at the primitive vectors $${\displaystyle \mathbf {a} ,\mathbf {b} ,\mathbf {c} }$$ (which is exactly what the Laue equations say), because then for the other points $${\displaystyle \mathbf {x} =p\,\mathbf {a} +q\,\mathbf {b} +r\,\mathbf {c} }$$ we have

$${\displaystyle \mathbf {\Delta k} \cdot \mathbf {x} =\mathbf {\Delta k} \cdot (p\,\mathbf {a} +q\,\mathbf {b} +r\,\mathbf {c} )=p\,2\pi h+q\,2\pi k+r\,2\pi l=2\pi (hp+kq+lr)=2\pi n,}$$

where $$n$$ is the integer $${\displaystyle hp+kq+lr}$$.

This ensures that if the Laue equations are satisfied, then the incoming and outgoing wave have the same phase at all points of the crystal lattice, so the oscillation of the atoms, that follows the incoming wave, can at the same time generate the outgoing wave.

There are two points that I am unclear on:

1. The two waves propagate through space independently, except at the points of the lattice, where they resonate with the oscillators, so their phase must coincide.

What is meant by "except at the points of the lattice, where they resonate with the oscillators"? What oscillators is it referring to? And how is this exceptional case represented mathematically (that is, how is it represented differently to the typical situation)?

1. $${\displaystyle \omega \,t-\mathbf {k} _{\mathrm {in} }\cdot \mathbf {x} =\omega \,t-\mathbf {k} _{\mathrm {out} }\cdot \mathbf {x} +2\pi n,}$$

Why is $$2\pi n$$ being added here? I realise that $$\varphi = \omega \,t-\mathbf{k}_{\mathrm {out}} \cdot \mathbf{x}$$ is the phase of the wave, but I'm not totally clear on why $$2\pi$$ was added.

I would appreciate it if people would please take the time to clarify these two points.

## EDIT

I think I have found the answer to my second question. In a section on the Laue equations, my textbook says the following:

Constructive interference will occur in a direction such that contributions from each lattice point differ in phase by $$2\pi$$.

So would I be correct in thinking that the addition of $$2\pi n$$ is due to constructive interference? And I'm assuming that this is also true for integer multiples of $$2\pi$$ (so $$2\pi n$$), right?

• Your answer to second question is correct. The oscillators are the atoms and their corresponding electrons! To understand how the wave interacts with your lattice you can take a look at the Lorentz model and in case you have a metal: Lorentz-Drude model. In these, they model the interaction of EM wave with an atom by a damped harmonic oscillator driven by the EM wave. Commented Jan 21, 2020 at 9:45
• @Simon Thanks for the clarification. Are these the oscillators being referenced hyperphysics.phy-astr.gsu.edu/hbase/oscdr.html ? Commented Jan 21, 2020 at 9:50
• Yes but you might want to read this reference: ocw.mit.edu/courses/electrical-engineering-and-computer-science/… Commented Jan 21, 2020 at 12:00
• @Simon Thanks for that. Commented Jan 21, 2020 at 12:02

This is more a wording problem than a physics problem. In order to solve this, consider a simple harmonic oscillator, i.e. a particle of mass $$m$$ on a weightless spring with spring constant $$D$$. This system has an eigenfrequency $$f_0 = \frac{1}{2\pi}\sqrt{\frac{D}{m}}$$, which is also called resonance frequency. If we excite a harmonic oscillator, it always oscillates with the frequency of the excitation. Thus, if we apply an external force $$F(t) = F_0 \, \cos(\omega t)$$ the harmonic oscillator oscillates with the frequency $$f_{\textrm{force}} = \frac{\omega_{\textrm{force}}}{2\pi} = \frac{\omega}{2\pi}$$. The amplitude of the oscillation is substantially influence by the difference $$\Delta f = f_{\textrm{force}} - f_0$$. If this difference is zero, the amplitude of the oscillation is maximal. So, if somebody says "if the excitation frequency is in resonance with the oscillator", it means that the excitation frequency is equal to the resonance frequency of the oscillator. Therefore, the amplitude of the oscillator is maximal.

Now, let's consider your problem. You have two waves which propagate through space. At each point in space these two waves add together. Thus, the total amplitude is given by \begin{align} A_{tot}(t, x) &= A_1(t, x) + A_2(t, x) \\ &= A_1 \cdot \cos(k x - \omega t+ \varphi_{1}(x)) + A_2 \cdot \cos(k x - \omega t + \varphi_{2}(x)) \end{align} I write it in 1D, but this is true generally -- I just don't won't to write all the vectors. If we consider a certain point in space and in time, the expression $$k x - \omega t$$ is the same for both waves. Thus, this oscillating term is not of interest and we drop it -- mathematically, you can use trigonometric identities to get a common pre-factor. Thus, if we split the amplitude and only consider the common amplitude $$A = \min\{A_1, A_2\}$$ of each term, we obtain $$A_{common} = A \cdot \cos(\varphi_{1}(x)) + A \cdot \cos(\varphi_{2}(x))$$ Now, without loss of generality, we can choose $$\varphi_1(x) = 0$$ and $$\varphi_2(x) = \Delta \varphi(x) = \varphi_{2}(x) - \varphi_{1}(x)$$. This simplifies the relation to $$A_{common} = A \big[ 1 + \cos(\Delta \varphi) \big]$$ Hence, the phase difference $$\Delta \varphi$$ determines the maximal amplitude of the oscillation. Some people say, the two waves are in resonance if their relative phase difference vanishes and thus the amplitude of oscillation is maximal.

The last missing piece for your answer is, that we often imagine that space is made up from invisible oscillators.

The two waves propagate through space independently, except at the points of the lattice, where they resonate with the oscillators, so their phase must coincide.

It means, that two independent waves (as described above) satisfy $$\Delta \varphi = n \cdot 2\pi$$ at the positions in space, where they interfere constructively. However, instead of using this formulation, the author imagines invisible oscillators in space. Now, at the positions of constructive interferences, the invisible oscillator gets excited by the two independent waves. When these two waves are in phase (which means that their relative phase difference is $$\Delta \varphi = n \cdot 2\pi$$) they interfer constructively. Thus, the resulting amplitude of the harmonic oscillator is at it maximum.

• One essential point of the derivation given on wiki is the claim that the phase match between the incident and scattering waves must occur for every single grid. When such a condition is met, the intensity is at its maximum. It is, for some reason, not that apparent to me (the resonance is a good one but stays to be an analogy). One can actually set out to calculate (as a good exercise) the general expression for the intensity and identify its maxima to satisfy Laue's condition. Also, see physics.stackexchange.com/questions/526415/… Commented Oct 3, 2021 at 23:34

In the Laue equations, the path-length difference (PLD) can be expressed as: $$\text{PLD} = A_2M-A_1N = T\cdot(k'-k)=T\cdot \Delta k$$ See the schematic shown above. Please note that $$k'$$ and $$k$$ are both unit vectors representing the incident and scattered X-rays.

This derivation is quite straightforward when we follow the algebra of vectors. With the angles outlined in the schematic, how can we prove the Laue equation with trigonometric-function calculations? To help us get started, we define the magnitude of vector $$T$$ as $$r$$, so that $$A_2M-A_1N =r\cos \alpha - r\cos(\alpha + 2 \theta)$$