I'm putting here my notes about the formula for polarization correction for X-ray beam intensity. Let's start with

📖 Chapter 4 of Principles of Protein X-Ray Crystallography by Jan Drenth.

4.14.1. The Polarization Factor
The polarization factor P [...] originates from the fact that an electron does not scatter along its direction of vibration.
In other directions, electrons radiate with an intensity proportional to (sin α)² [...]

This section ends with such a formula:

For a polarized beam, the polarization factor is (Azaroff, 1955; Kahn et al., 1982)
$$P = \frac{1}{2} (1 + \cos^2 2\theta - \tau \cos 2 \phi \sin^2 2\theta). \qquad (1)$$
For $\phi$ see Figure 4.31d;
$$\tau = \frac{I_h - I_v}{I_h + I_v}$$

where $I_h$ and $I_v$ are the intensities of respectively the horizontal and vertical
component of the polarized beam.

For unpolarized beam $\tau=0$ and the last component vanishes.

Now let's check the two mentioned references.

📖 L. V. Azároff, Polarization Correction for Crystal-Monochromatized X-radiation, Acta Cryst. (1955) 8, 701-704.

Whenever crystal-monochromatized X-radiation is used, the beam striking the specimen crystal is partially polarized owing to reflection by the monochromator crystal. The polarization factor for the twice diffracted beam is different, therefore, from the relation, $\frac{1}{2}(1+\cos^2 2\theta)$, normally used.

So before this paper, the Lorentz-polarization correction didn't take into account the polarization of the incident beam. The formula that the author derived is longer than (1), containing also the angle of reflection on the monochromator planes (the polarization fraction is calculated, not specified).

📖 R. Kahn, R. Fourme, A. Gadet, J. Janin, C. Dumas and D. André, Macromolecular Crystallography with Synchrotron Radiation: Photographic Data Collection and Polarization Correction, J. Appl. Cryst. (1982) 15, 330-337.

This one is about synchrotron radiation. The beam is polarized before it hits the monochromator, so the Azaroff's formula can't be applied. The authors transformed that formula to (1). Except that (1) uses different symbols:
$\rho \rightarrow \phi \qquad \mathscr{I}' \rightarrow \tau$ _{that symbol with prime that Kahn et al used is \mathscr{I}}

All seemed to be settled, but then I came across this:

📖 W. Kabsch, Evaluation of Single-Crystal X-ray Diffraction Data from a Position-Sensitive Detector, J. Appl. Cryst. (1988) 21, 916-924

(b) Polarization factor
The intensity of scattering is proportional to $\sin^2 \varphi$, where $\varphi$ is the angle between the electric-field vector of the incident beam and the direction of the diffracted beam. Assuming that the electric-field vector is found with probability
$p$ in the polarization plane having normal $\textbf{n}$ and with probability 1ドル - p$ to be parallel to $\textbf{n}$, the expectation value for $\sin^2 \varphi$ is

$$\begin{flalign} \langle \sin^2 \varphi \rangle & = p \sin^2 \measuredangle (\textbf{S}, \textbf{S}_0 \times \textbf{n}) + (1 - p) \sin^2 \measuredangle (\textbf{S}, \textbf{n}) \qquad (2) &\\\ & = (1-2p) [1 - (\textbf{n} \cdot \textbf{S}/ |\textbf{S}|)^2] + p~ \{1 + [\textbf{S} \cdot \textbf{S}_0/(|\textbf{S}| |\textbf{S}_0|)]^2\} \qquad (3) \end{flalign}$$

The polarization plane normal $\textbf{n}$ and the degree of polarization $p$ depend upon the set up of the diffraction experiment and must be specified as input parameters. In the case of an unpolarized direct beam, $p = 0.5$ and $\textbf{n}$ may be any vector perpendicular to $\textbf{S}_0$. The integrated intensities must be divided by $\langle \sin^2 \varphi \rangle$.

$\textbf{S}_0, \textbf{S}$ are incident- and diffracted-beam wave vectors of the reflection

It's not immediately clear if (1) and (2) are equivalent. For me, it wasn't even clear if (2) and (3) are equal.

❓ Are the expressions (2) and (3) equal?

The second addend in (2) is:

$$(1 - p) \sin^2 \measuredangle (\textbf{S}, \textbf{n}) = (1 - p) [1 - \cos^2 \measuredangle (\textbf{S}, \textbf{n})] = (1-p) [1 - (\textbf{n} \cdot \textbf{S}/ |\textbf{S}|)^2]$$

The first addend is more problematic. But if we express the sine as

$$\sin \measuredangle (\textbf{S}, \textbf{S}_0 \times \textbf{n}) = \frac{|\textbf{S} \times (\textbf{S}_0 \times \textbf{n})|}{|\textbf{S}| |\textbf{S}_0|}$$

we can use the vector triple product formula:

$$|\textbf{S} \times (\textbf{S}_0 \times \textbf{n})| = |(\textbf{S} \cdot \textbf{n}) \textbf{S}_0 - (\textbf{S} \cdot \textbf{S}_0) \textbf{n}|$$

And since $\textbf{S}_0 \perp \textbf{n}$

$$\sin^2 \measuredangle (\textbf{S}, \textbf{S}_0 \times \textbf{n}) = (\textbf{n} \cdot \textbf{S}/ |\textbf{S}|)^2 + [\textbf{S} \cdot \textbf{S}_0/(|\textbf{S}| |\textbf{S}_0|)]^2$$

therefore

$$p\sin^2 \measuredangle (\textbf{S}, \textbf{S}_0 \times \textbf{n}) = -p~ [1 - (\textbf{n} \cdot \textbf{S}/ |\textbf{S}|)^2] + p~ \{1 + [\textbf{S} \cdot \textbf{S}_0/(|\textbf{S}| |\textbf{S}_0|)]^2\}$$

We can see that both addends together give (3). So yes, (2) and (3) are equal.

❓ Is (2) equivalent to (1)?

Both papers use different symbols. From the definition, $\measuredangle (\textbf{S}, \textbf{S}_0) = 2 \theta$.

But how $\measuredangle (\textbf{S}, \textbf{n})$ is related to $\phi$? Drawing the geometry with all the angles (not included here because it's not readable) I got, from three right-angled triangles, that

$$\cos \measuredangle (\textbf{S}, \textbf{n}) = \sin \phi \sin 2\theta$$

Probability $p$ is most likely

$$p = \frac{I_h}{I_h + I_v} \implies p = \frac{\tau + 1}{2}$$

Now we can substitute these three variables in the Kabsch's formula:

$$\begin{flalign} \langle \sin^2 \varphi \rangle & = (1-2p) [1 - (\textbf{n} \cdot \textbf{S}/ |\textbf{S}|)^2] + p~ \{1 + [\textbf{S} \cdot \textbf{S}_0/(|\textbf{S}| |\textbf{S}_0|)]^2\} &\\\ & = (1-2p)[1-\cos^2 \measuredangle (\textbf{S}, \textbf{n})] + p~ (1 + \cos^2 2\theta) &\\\ & = -\tau (1 - \sin^2 \phi \sin^2 2\theta) + \frac{\tau+1}{2}(1 + \cos^2 2\theta) &\\\ &= \frac{1}{2} [-2\tau (1 - \sin^2 \phi \sin^2 2\theta) + (\tau + 1)(1 + \cos^2 2\theta)] &\\\ &= \frac{1}{2} \{- \tau [2 - 2\sin^2 \phi \sin^2 2\theta - (2 - \sin^2 2\theta)] + 1 + \cos^2 2\theta \} &\\\ &= \frac{1}{2} [1 + \cos^2 2\theta - \tau (1 - 2\sin^2 \phi) \sin^2 2\theta] &\\\ &= \frac{1}{2} (1 + \cos^2 2\theta - \tau \cos 2\phi \sin^2 2\theta) \end{flalign}$$

and we end up with (1). They are equivalent.

---

In the end, I found a more recent paper from Wolfgang Kabsch that mentions polarization.

📖 W. Kabsch, Processing of X-ray snapshots from crystals in random orientations, Acta Cryst. (2014) D70, 2204

It has the same formula as in the 1988 paper, except that the angle that was previously given as $\angle (\textbf{S}, \textbf{n})$, in this paper is

$$\varphi_2 = \angle [(\textbf{S}_0 \times \textbf{n}) \times \textbf{S}_0, \textbf{S}].$$

The author prefers this form because $\textbf{S}_0$ may change as a result of refinements during processing.