# Ewald sphere

The Ewald sphere is the surface, in reciprocal-space, that all experimentally-observed scattering arises from. (Strictly, only the elastic scattering comes from the Ewald sphere; inelastic scattering is so-called 'off-shell'.) A peak observed on the detector indicates that a reciprocal-space peak is intersecting with the Ewald sphere.

Although the Ewald sphere may seem like an abstract concept, it is simply the geometric representation of the scattering equation (i.e. it is the surface that satisfies the scattering condition).

Example of a reciprocal-lattice (blue dots) being probed in an x-ray scattering experiment. The signal observed on the detector comes from the intersection of the Ewald sphere with the reciprocal-space peaks.
Simplified 2D representation of a scattering experiment. The reciprocal-space of a crystal (blue), contains an array of peaks. The only peaks that are observed on the detector are those that intersect the Ewald sphere.

## Mathematics

In TSAXS of an isotropic sample, we only probe the magnitude (not direction) of the momentum transfer:

$q = \frac{4 \pi}{\lambda} \sin(\theta)$

Where $2\theta$ is the full scattering angle. In GISAXS, we must take into account the vector components:

\begin{alignat}{2} q_x & = \frac{2 \pi}{\lambda} \cos(\alpha_f)\sin(2\theta_f) \\ q_y & = \frac{2 \pi}{\lambda} \left[ \cos(\alpha_f)\cos(2\theta_f) - \cos(\alpha_i) \right ]\\ q_z & = \frac{2 \pi}{\lambda} \left[ \sin(\alpha_f) + \sin(\alpha_i) \right ] \end{alignat}

Where $\alpha_i$ is the incident angle (the grazing angle of the beam with respect to the substrate plane), $\alpha_f$ is the out-of-plane scattering angle (with respect to the substrate plane), and $2\theta_f$ is the full in-plane scattering angle.

## Derivation

### Definitions

Consider reciprocal-space in the incident beam coordinate system: $(q_x,q_y,q_z)$. The incident beam is the vector $\mathbf{k_i} = \langle 0,-k,0 \rangle$, where:

$k = \frac{2 \pi}{\lambda}$

where $\lambda$ is, of course, the wavelength of the incident beam. An elastic scattering event has an outgoing momentum ($\mathbf{k_f}$) of the same magnitude as the incident radiation (i.e. $|\mathbf{k_i}| = |\mathbf{k_f}| = k$). Consider a momentum vector, and resultant momentum transfer, $\mathbf{q}$, of:

\begin{alignat}{2} \mathbf{k_f} & = \begin{bmatrix} 0 \\ -k \cos(2 \theta_s) \\ +k \sin(2 \theta_s) \rangle \\ \end{bmatrix} \\ \mathbf{q} & = \mathbf{k_f} - \mathbf{k_i} \\ & = \begin{bmatrix} 0 \\ -k \cos(2 \theta_s) \\ +k \sin(2 \theta_s) \\ \end{bmatrix} - \begin{bmatrix} 0 \\ -k \\ 0 \\ \end{bmatrix} \\ & = \begin{bmatrix} 0 \\ k( 1 -\cos(2 \theta_s) ) \\ +k \sin(2 \theta_s) \rangle \\ \end{bmatrix} \\ & = \begin{bmatrix} 0 \\ 2 k \sin^2(\theta_s) \\ 2 k \sin(\theta_s) \cos(\theta_s) \rangle \\ \end{bmatrix} \\ \end{alignat}

The magnitude of the momentum transfer is thus:

\begin{alignat}{2} q & = | \mathbf{q} | \\ & = \sqrt{ [2 k \sin^2(\theta_s))]^2 + [ 2 k \sin(\theta_s)\cos(\theta_s)]^2 } \\ & = \sqrt{ 4k^2 [ \sin^4(\theta_s) + \sin^2(\theta_s)\cos^2(\theta_s)] } \\ & = 2 k \sqrt{ \sin^4(\theta_s) + \sin^2(\theta_s)\cos^2(\theta_s) } \\ & = 2 k \sqrt{ \sin^2(\theta_s)} \\ & = 2 k \sin(\theta_s) \\ & = \frac{4 \pi}{\lambda} \sin(\theta_s) \\ \end{alignat}

where $2 \theta_s$ is the full scattering angle. The Ewald sphere is centered about the point $(0,-k,0)$ and thus has the equation:

$q_x^2 + (q_y-k)^2 + q_z^2 - k^2 = 0$

### TSAXS

In conventional SAXS, the signal of interest is isotropic: i.e. we only care about $|\mathbf{q}| = q$, and not the individual (directional) components $(q_x,q_y,q_z)$. In such a case we use the form of q derived above:

$q = \frac{4 \pi}{\lambda} \sin(\theta_s)$

In the more general case of probing an anisotropic material (e.g. CD-SAXS), one must take into account the full q-vector, and in particular the relative orientation of the incident beam and the sample: i.e. the relative orientation of the Ewald sphere and the reciprocal-space.

### GISAXS

Now let us assume that the incident beam strikes a thin film mounted to a substrate. The incident beam is in the grazing-incidence geometry (e.g. GISAXS, and we denote the angle between the incident beam and the film surface as $\theta_i$. The reciprocal-space of the sample is thus rotated by $\theta_i$ with respect to the beam reciprocal-space coordinates. We convert to the sample's reciprocal coordinate space; still denoted by $(Q_x,Q_y,Q_z)$. The equation of the Ewald sphere becomes (the center of the sphere is at $\langle 0,k \cos\theta_i, k \sin\theta_i \rangle$):

\begin{alignat}{2} & q_x^2 + \left(q_y-k \cos\theta_i \right)^2 + \left(q_z-k \sin\theta_i\right)^2 - k^2 = 0 \\ & q_y = +k\cos\theta_i + \sqrt{ k^2 - q_x^2 - \left(q_z-k\sin\theta_i\right)^2 } \end{alignat}

#### Reflectivity

Consider for a moment that the q-vector is confined to the $(q_y,q_z)$ plane:

\begin{alignat}{2} \mathbf{q} = \begin{bmatrix} 0 \\ - q \sin(\alpha_f - \alpha_i) \\ + q \cos(\alpha_f - \alpha_i) \\ \end{bmatrix} \end{alignat}

Obviously the specular condition is when $\alpha_f=\alpha_i$, in which case one obtains:

\begin{alignat}{2} \mathbf{q} = \begin{bmatrix} 0 \\ 0 \\ + q \\ \end{bmatrix} \end{alignat}

That is, reflectivity is inherently only probing the out-of-plane ($q_z$) component:

$q_z = \frac{4 \pi}{\lambda} \sin \alpha_i$

TBD

TBD