INFLATION
1.5 The polarization signature of the Inflationary epoch
1.5.3 Polarization decomposition
In order to separate the polarization signature of inflation, we first need machinery to describe a polarized field as a function of position on the sky. We follow the description presented in Cabella and Kamionkowski 2004. A photon propagating along the zˆ direction can be described by two
24 electric field vectors:
Ex=axcos(!t ⇠x) (1.65)
Ey=aycos(!t ⇠y). (1.66)
These electric field vectors can be represented by Stokes parameters as:
I=a2x+a2y (1.67)
Q=a2x a2y (1.68)
U = 2axaycos(⇠x ⇠y) (1.69)
V = 2axaysin(⇠x ⇠y). (1.70)
The last term, Stokes V, reflects the circular polarization state of the radiation. As the process of Compton scattering is incapable of producing circular polarization, we ignore Stokes V. Stokes Q quantifies the polarization in the ‘plus’ orientation (left-right and top-bottom) while Stokes U quantifies the polarization in the ‘cross’ orientation (rotated 45 degrees from Q). From Qand U, we wish to calculate coordinate-independent measures of the curl- and curl-free components of the polarization field.
For our discussion, it is sufficient to proceed in a flat-sky approximation, describing the field in a local Cartesian coordinate system. This will allow us to illustrate the salient features of polarization separation before introducing spherical harmonics. As in Seljak and Zaldarriaga 1997, we find two linear combinations of Q and U that, under rotation about the origin of our local Cartesian coordinate system, transfom as:
(Q+iU)0=e2i (Q+iU) (1.71)
(Q iU)0 =e2i (Q iU). (1.72)
These two complex combinations, Q+iU and Q iU, transform as spin-two quantities (in the two-dimensional plane) with spin-weights 2 and -2, respectively. We can construct coordinate- independent (i.e. spin-zero) quantities from these two linear combinations. Since we are dealing with spin-two quantities in two dimensions, this is possible using the spin lowering and raising operators. In the flat-sky case, these operators reduce toS±=@x±i@y (White et al. 1999). Acting on the two spin-two quantities above, we find:
S0, 2= (@x+i@y)2(Q iU) =Qxx+ 2iQxy iUxx+ 2Uxy Qyy+iU yy (1.73) S0,2= (@x i@y)2(Q+iU) =Qxx 2iQxy+iUxx+ 2Uxy Qyy iU yy. (1.74)
HereQij, Uij represent partial differentiation with respect toiandj. These two quantities are now coordinate independent. By rotating our coordinate system, we naturally rotateQinto U, but the above quantities are guaranteed to be invariant under those rotations. Let us test this construction by rotating by 90 degrees. As we know, the spin two quantities (Equations 1.71 and 1.72) transform simply as(Q+iU)0= (Q+iU)and(Q iU)0= (Q iU), respectively. Our spin zero quantities, however, will translate as:
S0,0 2= (@x0+i@y0)2[(Q iU)0] (1.75)
= (@x0+i@y0)2[ (Q iU)] (1.76)
= Qx0x0 2iQx0y0+iUx0x0 2Ux0y0+Qy0y0 iU y0y0 (1.77)
S0,20 = (@x0 i@y0)2[(Q+iU)0] (1.78)
= (@x0 i@y0)2[ (Q+iU)] (1.79)
= Qx0x0 + 2iQx0y0 iUx0x0 2Ux0y0+Qy0y0+iU y0y0, (1.80) recognizing that by rotation by 90 degrees, we translatex0 =y andy0= x. The partial derivatives thus become:
Qx0x0 =Qyy, Qx0y0 = Qyx, Qy0y0 =Qxx, (1.81) and similarly for U. Translating the derivatives in Equations 1.77 and 1.80 back into un-primed coordinates, we find that we recover Equations 1.73 and 1.74.
As the spin-zero quantities are rotationally invariant, so too will be any linear combination that we choose to form. Here we recall our original intent: We wish to construct parity-even and parity- odd (curl and curl-free) quantities from Q and U. We also recall that parity inversion is not the same as rotation: S0, 2 and S0,2 are rotationally invariant, but have ambiguous parity properties.
Here we guess at a linear combination that will remain rotationally invariant and will be either parity-even or parity-odd, which we will callE andB, respectively:
E= (S0,2+S0, 2)/2 = Qxx 2Uxy+Qyy (1.82) B=i(S0,2 S0, 2)/2 =Uxx 2Qxy Uyy. (1.83) We test our construction by considering parity inversion about the y axis, letting x! x. Under inversion about the y axis, U ! U. Acting on E and B with the parity operator ⇧, it is clear
26
−2 −1.5 −1 −0.5 0 0.5 1 1.5
−2
−1.5
−1
−0.5 0 0.5 1 1.5 2
(a) PureE
−2 −1.5 −1 −0.5 0 0.5 1 1.5
−2
−1.5
−1
−0.5 0 0.5 1 1.5 2
(b) PureB
Figure 1.8: Polarization fields composed of pureE-modes and pureB-modes.
from Equations 1.82 and 1.83 that:
⇧E=E (1.84)
⇧B= B. (1.85)
We arrived at the desired result: the two linear combinations of the derivatives of Q and U that we have formed obey the parity relations we require. We also note that the analog of Helmholtz decomposition holds here: any polarization field can be separated into its E and B constituents, and these components are orthogonal.
To conclude this discussion, we consider a concrete example:
P1(x, y) =↵(x2 y2) +i xy (1.86)
P2(x, y) =↵(xy) +i (x2 y2). (1.87) These two polarization fieldsP1(x, y)andP2(x, y)are plotted in Figures 1.8a and 1.8b, respectively.
From Figure 1.8b, we can immediately see thatP1(x, y)is parity even, whileP2(x, y)is parity odd.
Correspondingly, if we calculate the quantities in Equations 1.82 and 1.83 we find:
PE1(x, y) = 4↵ 2i ; PB1(x, y) = 0; (1.88) PE2(x, y) = 0; PB2(x, y) = 4i 2↵. (1.89) As expected, theB component of P1 is zero while theB component of P2 is non-zero.
The derivatives corresponding to the raising and lowering operators are more easily handled in
the Fourier domain. We define the flat-sky Fourier transform ofQandU as:
Q(~x) =
Z Q(~˜ u)e2⇡i~u·~xd2u (1.90)
U(~x) =
Z U˜(~u)e2⇡i~u·~xd2u. (1.91)
Here~xis the position in thex, y plane, and~uis the position in theu, v plane. We can then write ourE and B constructions in terms of Q˜ and U. Partial derivatives with respect to˜ xand y drop down factors ofuandvrespectively, resulting in:
E= 2⇡
Z
( ˜Q(~u)(v2 u2) 2 ˜U(~u)(uv))d2u (1.92) B = 2i⇡
Z
( ˜Q(~u)(uv) 2 ˜U(~u)(v2 u2))d2u. (1.93) Setting|~u|= (u2+v2)1/2 and✓= arctan(u/v), we find6:
E(~x) = 2⇡
Z
( ˜Q(~u) cos(2✓) + ˜U(~u) sin(2✓))d2u (1.94) B(~x) = 2i⇡
Z
( ˜Q(~u) sin(2✓) U˜(~u) cos(2✓))d2u. (1.95) Furthermore, we can identify the Fourier transform ofE andB:
E(~˜ u) = ( ˜Q(~u) cos(2✓) + ˜U(~u) sin(2✓)) (1.96) B(~˜ u) = ( ˜Q(~u) sin(2✓) U˜(~u) cos(2✓)). (1.97)
It is thus extremely convenient to calculate E and B in the Fourier domain. In the flat-sky limit, our job here is done. We first calculate Q˜ and U˜, take linear combinations as per Equations 1.96 and 1.97, perform the inverse transforms onE˜ andB, and we have arrived at the desired map-space˜ component separation.