Read polarization before drawing a magnetic field

This chapter uses the sky axes and Doppler geometry defined earlier. It introduces Stokes parameters so that light from several regions can be added correctly. We assume optically thin synchrotron radiation from an isotropic power-law electron population in a uniform magnetic patch, with no Faraday rotation or circular polarization.

A polarization line has no arrowhead

Linear polarization picks out an axis in the sky plane. An electric-vector position angle, or EVPA, specifies that axis relative to a chosen sky direction. Rotating the axis by half a turn produces the same polarization. Angles therefore have a half-turn ambiguity.

Take $\mathbf e_1$ along the approaching jet's sky projection and $\mathbf e_2$ as in the geometry chapter. If $\hat{\mathbf E}$ is the unit electric vector, define its sky components as $E_1=\hat{\mathbf E}\cdot\mathbf e_1$ and $E_2=\hat{\mathbf E}\cdot\mathbf e_2$ . Our angle increases from the first axis toward the second. A paper using the opposite second axis will reverse the sign of its second component and of Stokes $U$ .

For one component with total intensity $I$ , fractional linear polarization $\Pi$ and angle $\chi$ , the linear Stokes parameters are

\begin{aligned} Q&=\Pi I\cos 2\chi,\\ U&=\Pi I\sin 2\chi. \end{aligned}

The doubled angle preserves the half-turn symmetry. Positive $Q$ favors an electric vector along the projected approaching axis; negative $Q$ favors the perpendicular direction. Positive or negative $U$ distinguishes the two diagonal orientations. Every Stokes intensity uses the same units as $I$ .

Add the light before measuring its angle

Independent emitting regions contribute additive intensities $I$ , $Q$ and $U$ . Their angles do not add. After summing the Stokes values, recover the net linear fraction and angle from

\begin{aligned} \Pi_{\rm net}&=\frac{\sqrt{Q^2+U^2}}{I},\\ \chi_{\rm net}&=\frac12\operatorname{atan2}(U,Q). \end{aligned}

The two-argument angle function keeps track of the signs of both arguments. Reduce the resulting angle modulo a half turn. If both linear Stokes parameters vanish, there is no preferred linear axis and the EVPA is undefined.

For example, combine two equally bright patches with equal fractional polarization and electric vectors at right angles. One has positive $Q$ , the other the same negative $Q$ , and both have zero $U$ . The total intensity doubles, but the linear polarization cancels. A low observed fraction can come from averaging ordered regions with different directions.

Two opposite normalized Stokes vectors cancel at positive total intensity. — Original Stokes-plane construction. Two equally bright, equally polarized components separated by a right angle have opposite doubled-angle Stokes vectors. Their intensities add and their polarization vectors cancel. The axes divide Stokes intensities by the linear polarized intensity of one patch, $\Pi_0 I_0$ , so the endpoints are not fractions of the total intensity. This illustrates spatial mixing; it is a separate example from the single-patch knots.

First find the photon in the plasma frame

The magnetic field that guides the electrons is most simply specified in their bulk rest frame. The observer's photon direction in that frame is $\mathbf n'$ , obtained by aberration. A local synchrotron electric vector points along $\mathbf n'\times\mathbf B'$ . Normalize it when this cross product is nonzero.

The magnetic field of the propagating light wave is a different object from the background magnetic field that made the electrons turn. The wave fields must be transformed together to the source frame. Transforming only a drawing of the background field and then adding a perpendicular line generally gives the wrong electric vector.

A compact expression for the correctly transformed direction uses the dimensionless velocity vector $\boldsymbol\beta_s=s\beta\hat{\mathbf z}$ :

\begin{aligned} \mathbf q'&=\mathbf B'+\mathbf n\times(\boldsymbol\beta_s\times\mathbf B')\\ &\quad-\frac{\Gamma}{1+\Gamma}(\mathbf B'\cdot\boldsymbol\beta_s)\boldsymbol\beta_s. \end{aligned}

Then normalize $\mathbf n\times\mathbf q'$ to obtain $\hat{\mathbf E}$ . The prime on the auxiliary vector records its dependence on the supplied rest-frame field; we use it inside a formula for the observed direction. When the speed is zero, the velocity terms vanish and the familiar perpendicular construction returns.

This expression follows from a Lorentz transformation of the electric and magnetic wave fields. It is equation C3 of Lyutikov, Pariev and Blandford's 2003 Appendix C, and equation 6 of the pinned 2004 jet paper. It does not require the emitter to be a complete cylindrical jet.

Use the same intensity for every Stokes value

A power-law electron energy distribution with index $p=2\alpha+1$ has uniform-field optically thin maximum linear fraction

\Pi_0=\frac{p+1}{p+7/3}.

Our fixed spectral index gives $p=2.5$ and $\Pi_0=21/29$ , about $72.4$ percent. With the physical ordered-field intensity from the previous chapter, the component form is

\begin{aligned} Q&=\Pi_0 I(E_1^2-E_2^2),\\ U&=2\Pi_0 I E_1E_2,\qquad V=0. \end{aligned}

Intensity, polarization fraction and electric direction now describe the same radiation. Because each knot contains one uniform patch, rotating its magnetic field changes its directional intensity and EVPA but leaves its nonzero-emission fractional polarization at the uniform-field value. This model has no spatial disorder to cancel polarized light within one knot.

At a zero-emission direction, all Stokes values are zero and the EVPA has no physical value. For several unresolved components, a zero linear sum can instead occur at nonzero total intensity. These are different reasons for an undefined angle.

What could a magnetic map tell us?

A sky pattern of EVPAs depends on field geometry, motion, viewing angle and the regions weighted most heavily by their emissivity. Resolving a jet across its width gives more information than measuring one integrated angle. Even then, the chosen reference frame matters: under an ideal-conductor boost along a jet, a transverse background magnetic component grows relative to an axial one.

A globally helical field adds an azimuthal direction that changes around the jet. It can produce cancellation when its whole cross-section is unresolved. The final paper guide follows that additional spatial integration. Here, “pitch” sets the direction in one local patch; it does not by itself measure a jet's twist or the spin of its central object.

Practice

Reverse the sign of the electric vector. What happens to $Q$ , $U$ and EVPA?
Add three equal-intensity patches with the same fraction and angles of zero, sixty and one hundred twenty degrees. What is their net linear polarization?
Rotate the ordered magnetic pitch of a single knot and see a changing EVPA but a constant fractional polarization. What assumption accounts for the constant value?

Explained answers

Both electric components reverse. Their squares and product remain the same, so both linear Stokes parameters and the polarization axis remain unchanged.

Their doubled angles are separated by one hundred twenty degrees and make a closed triangle in the Stokes plane. Both linear sums vanish, the total intensity is positive, and the net linear fraction is zero. The net EVPA is undefined.

A single uniform magnetic patch with the same power-law electrons always has the same intrinsic maximum fraction for a nonzero emitting direction. To lower its integrated fraction by geometric cancellation, one would need several differently oriented contributions within the region being averaged.

Sources

Lyutikov, Pariev and Blandford, arXiv:astro-ph/0305410v2 (7 November 2003), Appendix C, printed pages 15–17, especially C2–C5. We use its radiation-field transformation, without adopting the paper's gamma-ray-burst interpretation. Pinned derivation.
Lyutikov, Pariev and Gabuzda, arXiv:astro-ph/0406144v1 (6 June 2004), Section 2, printed pages 7–11, equations 2,6–8 and18; Figure 7 is on printed page 38. Jet polarization source.

Try the prediction

Rotate an ordered field

Predict whether one uniform emitting patch loses polarization when its magnetic pitch changes.
Pause at a fixed speed and viewing angle. Vary magnetic pitch and read the electric-vector angle and the Stokes values in the data area.
Explain why intensity and electric-vector direction can change while the uniform-patch polarization fraction stays constant.