Why the X-ray beam is polarized

$z,\ t$

Distance along the beam, and time.

$\hat{\mathbf{x}},\ \hat{\mathbf{y}}$

Unit vectors: horizontal across the beam, and vertical.

$E_x,\ E_y$

Amplitudes of the horizontal and vertical parts of the field.

$k=2\pi/\lambda$

Wavenumber. The field repeats every wavelength λ along the beam.

$\omega=2\pi\nu=ck$

Angular frequency. The field repeats ν times per second at a fixed point.

$\varphi$

How far the vertical part lags the horizontal part.

What it saysAny polarization is two perpendicular swings added together. With φ = 0 they stay in step and the tip moves along a line tilted by arctan(E_y/E_x). With φ = 90° and equal amplitudes the tip moves on a circle. For the APS beam E_y is nearly zero, so only the first term is left.

$h$

Planck's constant, 6.626×10⁻³⁴ J·s, or 4.136×10⁻¹⁵ eV·s.

$c$

Speed of light, 2.998×10⁸ m/s.

$hc$

Their product, 12.398 keV·Å. This is where the 12.398 comes from.

$\nu,\ \lambda$

Frequency and wavelength of the wave in equation 1.1.

What it saysPhoton energy and wavelength are the same information. Shorter wavelength means higher energy.

In numbershc = 4.136×10⁻¹⁵ eV·s × 2.998×10⁸ m/s = 1.2398×10⁻⁶ eV·m = 12.398 keV·Å. At the Se K edge, 12.66 keV, λ = 12.398 / 12.66 = 0.979 Å and ν = c/λ = 3.06×10¹⁸ Hz.

$I_h$

Intensity carried by the horizontal field component, proportional to E_x².

$I_v$

Intensity carried by the vertical component, proportional to E_y².

What it saysP = 1 means all the intensity is horizontal. P = 0 means the two share it equally, as in unpolarized light. P = −1 means all vertical.

In numbersP = 0.98 means I_h = (1 + P)/2 = 99% of the intensity is horizontal and 1% is vertical.

$q$

The charge. For an electron q = −e.

$\mathbf{a}$

Its acceleration, taken at the earlier moment when the light now arriving was emitted.

$\hat{\mathbf{n}}$

Unit vector from the charge to the observer: the line of sight.

$R$

Distance from charge to observer.

$\mathbf{a}_{\perp}$

The part of a perpendicular to the line of sight: $\mathbf{a}-(\mathbf{a}\cdot\hat{\mathbf{n}})\hat{\mathbf{n}}$.

$\varepsilon_0$

Permittivity of free space, the constant that sets the strength of electric forces.

What it saysTake the acceleration, throw away the piece that points at you, and what is left gives the direction of the field you receive. Its size is a·sin ψ, where ψ is the angle between your line of sight and a, and it falls off as 1/R. This form holds for a slow charge. Step 4 handles the fast one.

$dP/d\Omega$

Radiated power per unit solid angle in a given direction.

$\psi$

Angle between that direction and the acceleration.

$P_{\text{total}}$

The same pattern summed over all directions (the Larmor formula).

What it saysIntensity is field squared, so it goes as sin²ψ: greatest at right angles to the acceleration and zero along it. In three dimensions the pattern is a doughnut with its hole along a. Power also grows as a², so harder acceleration radiates much more.

$-e,\ m$

The electron's charge and rest mass.

$\mathbf{v}$

Its velocity, essentially c along the beam.

$\gamma$

The Lorentz factor, 11,742 at 6 GeV. It makes the electron γ times harder to push sideways.

$\ddot{x}$

Horizontal acceleration.

What it saysThe cross product of a forward velocity and a vertical field is horizontal. There is no vertical force at all, so the electron never leaves the horizontal plane. This one line is the origin of the beam's polarization.

$\theta'$

The angle of a light ray from the forward direction, measured in the electron's frame.

$\theta$

The angle of the same ray in the lab frame.

What it saysA ray that leaves sideways in the electron's frame (θ′ = 90°) comes out at only 1/γ in the lab. Half of everything the electron emits is squeezed into that forward cone.

In numbers1/γ = 85 µrad. That is 2.6 mm of beam radius at 30 m from the source.

$\theta$

Angle from the direction of travel.

$\phi$

Angle around the direction of travel. φ = 0 is the wiggle plane (horizontal), φ = 90° is vertical.

What it saysSet β = 0 and the left side reduces to the sin²ψ doughnut of equation 2.2. For large γ everything depends on the product γθ, which is why the right-hand plot in the figure stops changing. The first factor is the forward cone. The bracket carries the dipole's hole: it reaches zero at γθ = 1 in the horizontal plane, and it never reaches zero in the vertical plane.

$\theta_x,\ \theta_y$

Horizontal and vertical viewing angles from the axis, with $\theta^2=\theta_x^2+\theta_y^2$.

$E_x$

The wanted, horizontal field component.

$E_y$

The unwanted, vertical component.

What it saysThis is equation 2.1 after the relativistic squeeze. On the axis E_y = 0. It is also zero anywhere on the horizontal line (θ_y = 0) and on the vertical line (θ_x = 0), because the product θ_xθ_y vanishes there. It changes sign from one quadrant to the next, which gives the four leaves. Squaring and adding the two components gives back equation 4.2.

$\chi$

The angle between the local field direction and the horizontal.

What it saysClose to the axis the tilt is about 2γ²θ_xθ_y, so it grows with the square of the distance from the axis. Light very near the axis is very pure.

In numbersAt γθx = γθy = 0.3 (26 µrad each way at this ring): tan χ = −0.18, a tilt of 10°. The vertical intensity there is 0.18² = 3% of the horizontal.

What it saysThis is equation 1.3 with I_h and I_v added up over every direction the aperture lets through. The figure does this sum for the blue box. A small box collects only the pure center and P stays near 1. A large box takes in the leaves and P falls.

In numbersIn this model, a half-angle of 0.1/γ gives P = 99.99%, 0.5/γ gives 97%, and 1/γ gives 86%.

$\sigma_{x'},\ \sigma_{y'}$

The rms spread in the directions of the electrons, horizontally and vertically, in the undulator straight section.

$\theta_{\text{cen}}$

The single-electron central cone from equation 5.3 in the guide How NE-CAT works.

$\Sigma_{x'},\ \Sigma_{y'}$

The rough angular size of the real photon beam.

What it saysEach electron lays down the pattern of equation 5.1 around its own direction of travel. The beam is the sum of those patterns, so its angular size is the single-electron cone widened by the electrons' own spread. This is where the upgrade shows. The old ring's horizontal spread was as large as the cone itself. The new ring's spread is several times smaller than the cone in both directions, so the real beam is close to the single-electron picture.

In numbersFor 24-ID-C after the upgrade: electron spread 2.8 µrad horizontally and 1.3 µrad vertically, against a cone of about 10 µrad. Combined: about 10 µrad × 10 µrad.

$d$

Spacing of the reflecting atomic planes: 3.1356 Å for Si(111) and 1.9201 Å for Si(220).

$\theta_B$

The Bragg angle between the beam and those planes. The beam is turned through 2θ_B.

What it saysReflections from successive planes add in step only at one angle for each wavelength. Rotating the crystal selects the energy.

In numbersFor 24-ID-C after the upgrade: Si(111), d = 3.1356 Å. At 12.662 keV, sin θB = 12.398 / (2 × 3.1356 × 12.662) = 0.156, so θB = 8.98° and the beam turns through 18.0°.

$R_\sigma$

Integrated reflectivity for the field perpendicular to the reflection plane.

$R_\pi$

Integrated reflectivity for the field lying in the reflection plane.

What it saysThis is step 2 again. The crystal's electrons are driven along the incoming field and must radiate into the reflected direction. For σ, the reflected direction is always at 90° to their motion, where a dipole radiates best. For π, the reflected direction is only 90° − 2θ_B away from their motion, so the field is reduced by sin(90° − 2θ_B) = cos 2θ_B. At 2θ_B = 90° a π reflection would vanish completely. The value C per bounce applies to a thick perfect crystal.

In numbersAt that angle C = cos 18.0° = 0.951. Two bounces give C² = 0.905.

$P$

Polarization entering the mono. The horizontal share is (1 + P)/2 and the vertical share is (1 − P)/2.

$m$

The number of reflections: 2 for a channel cut or double-crystal mono, 1 for a single bounce.

$C^m$

The total π factor after m reflections.

vertical, horizontal

The plane in which the mono deflects the beam.

What it saysWhichever component is π is multiplied by C once per reflection, and then equation 1.3 is applied again. With vertical deflection the unwanted vertical light is π and is trimmed, so P rises slightly. With horizontal deflection the wanted horizontal light would be π and P would fall slightly.

In numbersP = 0.980 in gives 0.9819 out for 24-ID-C after the upgrade (2 reflections, vertical deflection). The other deflection plane would give 0.9779.

$\theta_g$

The grazing angle of the beam on the mirror, a few milliradians.

What it saysThe same geometric factor applies to a mirror, but the beam is turned through such a small angle that σ and π are treated almost identically.

In numbersAt θ_g = 3 mrad, C = 1 − 2 × (0.003)² = 0.99998.

What it saysPut ψ = 0 and 2θ = 90° into equation 7.3 in the guide X-ray fluorescence at NE-CAT and only the unpolarized half-term is left: that is the horizontal count. Put ψ = 90° and the polarized term adds in full: that is the vertical count. Their difference over their sum is P. A finite sensor cone leaks a little extra into the horizontal count (equation 8.2 in the guide X-ray fluorescence at NE-CAT), so this slightly underestimates P unless you use a small aperture.