# CMB polarization: the E and B modes

1. Aug 7, 2013

### fairy._.queen

Hi all!

In Ruth Durrer's book "The Cosmic Microwave Background" there is this picture (Fig. 5.2 page 193) about the E and B mode of the CMB polarization:

http://imageshack.us/photo/my-images/11/s7o6.png/
http://imageshack.us/photo/my-images/11/s7o6.png/

They represent the E and B pattern around the photon direction. I probably don't fully understand what the E and B mode represent, because the tangential E mode looks like circular polarization to me, whereas circular polarization is not expected in the CMB and the E and B mode definitely describe linear polarization only.
As for the B mode, I can't see how one can have the electric field oscillating in so many different directions.

Summing up: do those lines represent oscillating electric field? If so, how can the E mode be tangential without circular polarization? How can there be multiple directions of oscillations for the B mode?

2. Aug 7, 2013

### Allday

hi fairy._.queen,

I'm not an expert on CMB polarization but I think I can clear a few things up. When folks refer to the "E" mode and "B" mode of the CMB polarization field they aren't refering directly to electric and magnetic fields. They are simply decomposing the vector field that represents polarization into a component that has zero curl and a component that has zero divergence. This is always possible in 3-D and is known as the Helmholtz decomposition ( I think it must also always be possible in 2-D but not sure if its called the same thing ). In any case, the divergence free component is called "B" mode to remind you that it is divergence free like the magnetic field and the curl free component is called "E" mode to remind you that it is curl free like the E-field of a point charge.

3. Aug 8, 2013

### fairy._.queen

Hi, thank you very much for replying.
I knew that the E and the B mode are the divergence-free and curl-free part of the polarization, what I need to know is how those pattern can be reproduced around the direction of the photon using linear polarization only.
Thank you anyway! :-)

4. Aug 8, 2013

### Allday

For anything related to the CMB I usually check out something by Wayne Hu. This will probably be useful ...
http://astro.berkeley.edu/~mwhite/polar

I'm interested in your question so I'll scan through and see if I can find your answer

5. Aug 12, 2013

### lpetrich

You need to start with the Stokes parameters (Wikipedia). They are for describing incoherent electromagnetic waves, like most visible light or the CMB. You need to average over the squares of their electric vectors, and those vectors being vectors produces certain complications. For convenience, I'll do

Ei -> Ei*ei*w*t + E*i*e-i*w*t

where w is the angular frequency. This is to express shifted phases more conveniently.

Let the direction of propagation be z. The electric field will have nonzero components in the x and y directions. Here is the average over the outer product of the electric vector with its complex conjugate:

Jij = <EiE*j>

The Stokes parameters are often defined as
I = overall intensity
Q, U = linear polarization
V = circular polarization
I = <|Ex|2 + |Ey|2|>
Q = <|Ex|2 - |Ey|2|>
U = 2*Re(Ex*E*y)
V = -2*Im(Ex*E*y)
J becomes
Jxx = (I + Q)/2
Jxy = (U - i*V)/2
Jyx = (U + i*V)/2
Jyy = (I - Q)/2
or J = {{I+Q, U-i*V}, {U+i*V},{I-Q}}/2

J can be expressed in tensor form:
Jij = (I*δij - i*V*εij + Lij)/2
where L is symmetric and traceless: Lji = Lij and Tr(L) = Ʃk=1,2 Lkk = 0
L = {{Q, U}, {U, -Q}}

L thus expresses the linear polarization.

However, I and V multiply tensors without built-in direction information:
δ = {{1, 0}, {0, 1}}
ε = {{0, 1}, {-1, 0}}
So I and V are essentially scalar quantities, rather than components of a tensor, as Q and U are.

Once you understand Stokes parameters, you can proceed to the next step.

6. Aug 12, 2013

### fairy._.queen

Thank you very much for replying!
I understand the Stokes parameters, what is the next step?

7. Aug 12, 2013

### lpetrich

Let's now consider electromagnetic waves traveling in the z direction but varying over the x and y directions. How can we express that variation? We can do so using gradients of scalar functions.

The coherent case is easy:
Ex = ∂(W1)/∂x + ∂(W2)/∂y
Ey = ∂(W1)/∂y - ∂(W2)/∂x
or
Ei = ∇i(W1) + εijj(W2)
(summed over j, since it is a repeated index)

W1 term: zero curl, "electric"

The incoherent case is more difficult, and one has to use the Stokes parameters. Of the four Stokes parameters, I and V are already scalars, so they are easy. The difficult one is L, the linear polarization, since it's a 2-tensor. It can be expressed as

Lij = (∇ij(W1) - εikεjlkl(W1)) + (εikkj(W2) + εjkki(W2))
Q = (∂2(W1)/∂x2 - ∂2(W1)/∂y2) + (2∂2(W2)/∂x∂y)
U = (2∂2(W1)/∂x∂y) - (∂2(W2)/∂x2 - ∂2(W2)/∂y2)
The first term can also be expressed as (2∇ij(W1) - δij2(W1))

W1 term: zero curl, "electric"

How to picture the polarizations? Imagine a hump in the potential value. Here is where the polarizations point:

Vector, electric: up and down hill
Vector, magnetic: sideways around the hill

Tensor, electric: (up and down hill) + ... (sideways around the hill) -
Tensor, magnetic: (diagonal 1) + ... (diagonal 2) -

To go from the flat to the spherical case, you treat the surface of the sphere as a curved 2D object. For that, you can use the differential-geometry formalism that general relativity uses. But it works much the same.

Continuing with the spherical case, one can expand I, V, W1, and W2 in spherical harmonics, because that's convenient for expressing functions with some size scale. Much like a Fourier transform over flat space. One then calculates L (Q and U) from W1 and W2.

8. Aug 12, 2013

### lpetrich

CMB Polarization at cosmology.berkeley.edu

In the usual inflation models, there are two sorts of perturbations left over from inflation:
Scalar -- density fluctuations
Tensor -- gravitational waves (field is a 2-tensor)

Intensity - scalar
E modes - scalar, tensor
B modes - tensor

However, gravitational lensing can cause the E and B modes to mix, complicating the data analysis.

9. Oct 9, 2013

### fairy._.queen

Ok, I think I got what you mean. Now, back to my old question, could you please explain what the figure is showing?

Thanks a lot.

10. Oct 9, 2013

### lpetrich

What I described in post #7. The dot in the center is the peak of the hill of W1 or W2.

E (electric): (up and down hill) + ... (sideways around the hill) -
B (magnetic): (diagonal 1) + ... (diagonal 2) -

11. Oct 13, 2013

### fairy._.queen

I can't see the difference between a polarization pointing "sideways around the hill" and circular polarization... can you please help?
Sorry...

12. May 9, 2015

### warrenchu000

B-mode polarization is a group of linear polarizations arranged in a pinwheel pattern. It is not the same as circular polarization.