Sunday, March 4, 2012
Friday, March 2, 2012
E-B Leakage Part 1: Maths Story
General Maths
A function is harmonic if it obeys the Laplace equation.
A function is harmonic if it obeys the Laplace equation.
To know every value of the function of a line segment you need to know:
- that the function satisfies the Laplace equation
- the values of the function on the boundary
To know every value of the function of a circle you need to know:
- that the function satisfies the Laplace equation
For a circle there is no boundary and so the function must be a constant. This is also true for a sphere.
CMB Case
On the CMB, physical observables are Q and U polarisation. The rank 2 polarisation tensor is defined in terms of Q and U. We can write this in terms of 2 scalar fields P_{E} and P_{B}. (This is a projection of Q, U into P_{E}, P_{B})
How can we transform the scalar fields P_{E} and P_{B} such that the physical quantity P_{ab} doesn't change?
eg for P_{E}
Adding any constant to P_{E} will preserve the physical quantity P_{ab}. Therefore the transformation is gauge invariant.
Q(E,B) and U(E,B) have both E and B components. We know that physically in the universe we expect the E mode to be much bigger than the B mode, so let's say Q(E,B) and U(E,B) are made up of ~99% E mode and ~1% B mode. The projection defined above gives us a method of writing down E and B in terms of Q and U. Eg for the flat sky we write B(l) as the linear combination:
Since the E mode dominates in Q(E,B) and U(E,B), we would expect most linear combinations of Q(E,B) and U(E,B) to be mostly E mode. The special way we have written our B mode (according to how it was defined in the projection), is such that the signal from the dominating E mode gets cancelled out and leaves just the B mode component.
We defined the E and B mode on a sphere, therefore with no information about a boundary. If we cut a piece of the full map out, the new cut map has a boundary, and our projection is not gauge invariant on the cut map.
Specifically, at the boundary of the cut map the projection is not valid and the special linear combination of Q(E,B) and U(E,B) which defines the B mode no longer projects out the dominating E mode.
This effect produces noise on the boundary of the cut map which is the "leakage" of the E mode into the B mode for non periodic boundary conditions.
(Thanks Liam Fitzpatrick, Jared Kaplan)
CMB Case
On the CMB, physical observables are Q and U polarisation. The rank 2 polarisation tensor is defined in terms of Q and U. We can write this in terms of 2 scalar fields P_{E} and P_{B}. (This is a projection of Q, U into P_{E}, P_{B})
How can we transform the scalar fields P_{E} and P_{B} such that the physical quantity P_{ab} doesn't change?
eg for P_{E}
Adding any constant to P_{E} will preserve the physical quantity P_{ab}. Therefore the transformation is gauge invariant.
Q(E,B) and U(E,B) have both E and B components. We know that physically in the universe we expect the E mode to be much bigger than the B mode, so let's say Q(E,B) and U(E,B) are made up of ~99% E mode and ~1% B mode. The projection defined above gives us a method of writing down E and B in terms of Q and U. Eg for the flat sky we write B(l) as the linear combination:
We defined the E and B mode on a sphere, therefore with no information about a boundary. If we cut a piece of the full map out, the new cut map has a boundary, and our projection is not gauge invariant on the cut map.
Specifically, at the boundary of the cut map the projection is not valid and the special linear combination of Q(E,B) and U(E,B) which defines the B mode no longer projects out the dominating E mode.
This effect produces noise on the boundary of the cut map which is the "leakage" of the E mode into the B mode for non periodic boundary conditions.
(Thanks Liam Fitzpatrick, Jared Kaplan)
Fourier Transforms
F=1/T and X=1/K (K is wave number)
Fourier series motivation
A Fourier transform is the continus limit of a Fourier series.
Think of a function f(x) which satisfies some boundary conditions f(0)=0 and f(pi)=0. I have drawn some example functions in blue. The function sin(nx) satisfies the boundary conditions for n=integer. Any sum of sin(nx) will also satisfy the boundary conditions.
The inverse is true, that any continuous function satisfying the boundary conditions can be written as a sum of sin(nx) with coefficients a_{n}.
There are 2 ways to describe the function f(x):
Waves in a box
One particular wavelength or distance X, corresponds to a single value in K space. Unless we have an infinite decomposition, our K spectrum is discretised.
What happens to the K space box if you cut out a piece of the X space box?
(Thanks Jeremy Mardon, Adam Brown)
Fourier series motivation
A Fourier transform is the continus limit of a Fourier series.
Think of a function f(x) which satisfies some boundary conditions f(0)=0 and f(pi)=0. I have drawn some example functions in blue. The function sin(nx) satisfies the boundary conditions for n=integer. Any sum of sin(nx) will also satisfy the boundary conditions.
The inverse is true, that any continuous function satisfying the boundary conditions can be written as a sum of sin(nx) with coefficients a_{n}.
There are 2 ways to describe the function f(x):
- Tell me the value of f(x) for every x
- Tell me the Fourier coefficients a_{1}, a_{2}, a_{3}....
Waves in a box
One particular wavelength or distance X, corresponds to a single value in K space. Unless we have an infinite decomposition, our K spectrum is discretised.
To fill in K space values for a continuous spectrum you have to count waves that go into the box a non-integer number of times. eg what wavelength x fits in the box 1.1 times, 1.2, 1.3, 1.4 etc and all the waves in between.
The size of the box in X space limits the largest wavelength mode you can see. The size of the pixels in X space limits the smallest wavelength mode you can see, which corresponds to the highest frequency mode you can resolve in K space.
(Text reads: Lowest frequency mode you can resolve
Highest frequency mode you can resolve
Can't resolve this very high frequency mode)
Highest frequency mode you can resolve
Can't resolve this very high frequency mode)
Size of box in X space not equal to size of box in K space, but you do get same number of pieces of information. Highest mode in K space is inverse to smallest spacing in X. Longest wavelength mode in X is inverse to smallest spacing in K. Cutting the box decreases resolution in K space and information about the longest wavelength modes is lost.
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