tag:blogger.com,1999:blog-83086395789779790352024-03-05T06:49:33.908-08:00Neutrino's Adventures in CosmologyNeutrinohttp://www.blogger.com/profile/11148204219908257621noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-8308639578977979035.post-32064849478156829242012-03-04T22:36:00.000-08:002012-03-04T22:36:26.434-08:00E-B Leakage Part 2: Computational StoryNeutrinohttp://www.blogger.com/profile/11148204219908257621noreply@blogger.com0tag:blogger.com,1999:blog-8308639578977979035.post-71279833254831881552012-03-02T23:42:00.046-08:002012-03-05T10:46:03.723-08:00E-B Leakage Part 1: Maths Story<b><u>General Maths</u></b><br />
A function is <span style="color: red;">harmonic </span>if it obeys the Laplace equation.<br />
<div><br />
</div><div>To know every value of the function of a line segment you need to know:</div><div><ul><li>that the function satisfies the Laplace equation</li>
<li>the values of the function on the boundary</li>
</ul><div>To know every value of the function of a circle you need to know:</div></div><div><ul><li>that the function satisfies the Laplace equation</li>
</ul><div>For a circle there is no boundary and so the function must be a constant. This is also true for a sphere.<br />
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<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjY0Dg0WfTjZ2Hq6L4EZFJhxgEVR1dBVfGq63gVJiXSVmaV06TWxKHlfTDmjOHiUhFLH1IUq9H9y1dTZV2T9zd5a-w2Nhuy-fTLNYox9bPhyphenhyphenRikbBLhMOvurzJgC2j1M5C3uxdlORstnN6z/s1600/IMG_0420.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="240" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjY0Dg0WfTjZ2Hq6L4EZFJhxgEVR1dBVfGq63gVJiXSVmaV06TWxKHlfTDmjOHiUhFLH1IUq9H9y1dTZV2T9zd5a-w2Nhuy-fTLNYox9bPhyphenhyphenRikbBLhMOvurzJgC2j1M5C3uxdlORstnN6z/s320/IMG_0420.jpg" width="320" /></a></div><br />
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<b><u>CMB Case</u></b><br />
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 <span style="color: red;">projection </span>of Q, U into P_{E}, P_{B})<br />
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<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg3IQmBcD9GYzSGfq83x4Qqld6FlTZt1yktH3amlbPPIEoaIjbv6RtRCJJYZrsoXM5EFqNCDESMyZZLG_jc7CB3gVaVCj78KLO3aNuPgnBlkInIQ_6wToeIjWlrYQ3XTHwmPdnv3Cdgnw4j/s1600/IMG_0422.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="129" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg3IQmBcD9GYzSGfq83x4Qqld6FlTZt1yktH3amlbPPIEoaIjbv6RtRCJJYZrsoXM5EFqNCDESMyZZLG_jc7CB3gVaVCj78KLO3aNuPgnBlkInIQ_6wToeIjWlrYQ3XTHwmPdnv3Cdgnw4j/s320/IMG_0422.JPG" width="320" /></a></div><br />
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How can we transform the scalar fields P_{E} and P_{B} such that the physical quantity P_{ab} doesn't change?<br />
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eg for P_{E}<br />
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Adding any constant to P_{E} will preserve the physical quantity P_{ab}. Therefore the transformation is gauge invariant. <br />
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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:<br />
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhjNzbmxh73Ew7JsErfNUieeZxXvAQcoKkLZvU31kz3VEIdZAD1f_DgJKieN93zcJ3TfcpHr9OCru9TqfWBNXtF_tUv52p0D0KMuPL0F5U7CFoeA-oIelM9d2TqZ4jMcO93AXd-F_KVqwPD/s1600/IMG_0424.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="250" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhjNzbmxh73Ew7JsErfNUieeZxXvAQcoKkLZvU31kz3VEIdZAD1f_DgJKieN93zcJ3TfcpHr9OCru9TqfWBNXtF_tUv52p0D0KMuPL0F5U7CFoeA-oIelM9d2TqZ4jMcO93AXd-F_KVqwPD/s320/IMG_0424.jpg" width="320" /></a></div><div class="separator" style="clear: both; text-align: center;"><br />
</div>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.<br />
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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. <br />
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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. <br />
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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.<br />
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(Thanks Liam Fitzpatrick, Jared Kaplan)</div></div>Neutrinohttp://www.blogger.com/profile/11148204219908257621noreply@blogger.com0tag:blogger.com,1999:blog-8308639578977979035.post-66277716346003759452012-03-02T23:26:00.005-08:002012-03-05T10:44:10.842-08:00Fourier TransformsF=1/T and X=1/K (K is wave number)<br />
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<b><u>Fourier series motivation</u></b><br />
A Fourier transform is the continus limit of a Fourier series.<br />
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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. <br />
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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}.<br />
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There are 2 ways to describe the function f(x):<br />
<br />
<ul><li>Tell me the value of f(x) for every x</li>
<li>Tell me the Fourier coefficients a_{1}, a_{2}, a_{3}....</li>
</ul><br />
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<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEieCbql44hR3nrLDJeF1RdZ_I8Z3LlCilXEsGFLFinqqwz610pljG13ZBjNHWAv0ceJ_GKoamztvlTKtTUyJ2-Gq1ZT-R4wf9GB7nutB_nb-q8K7jpfOQvyjgp4kbuFfJzWgm-5lDmY4F25/s1600/IMG_0425.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="270" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEieCbql44hR3nrLDJeF1RdZ_I8Z3LlCilXEsGFLFinqqwz610pljG13ZBjNHWAv0ceJ_GKoamztvlTKtTUyJ2-Gq1ZT-R4wf9GB7nutB_nb-q8K7jpfOQvyjgp4kbuFfJzWgm-5lDmY4F25/s320/IMG_0425.jpg" width="320" /></a></div><br />
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<b><u>Waves in a box</u></b><br />
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.<br />
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjHO-69n7lbid1UOSsiNDFzZl0MzFenLdW7w_WaCewKI5eCmTNLEhhxV0oSD42R2CxyjRdo8RQHAYZXLimdyh5OvwdQjHxr19QZGEuh-DWTsvWyRCPuT9LCmAHfOhNffSk-H3TYpM9AmhBt/s1600/IMG_0426.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="159" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjHO-69n7lbid1UOSsiNDFzZl0MzFenLdW7w_WaCewKI5eCmTNLEhhxV0oSD42R2CxyjRdo8RQHAYZXLimdyh5OvwdQjHxr19QZGEuh-DWTsvWyRCPuT9LCmAHfOhNffSk-H3TYpM9AmhBt/s320/IMG_0426.JPG" width="320" /></a></div><div class="separator" style="clear: both; text-align: center;"><span style="color: #38761d;"><br />
</span></div><div class="separator" style="clear: both; text-align: left;"> 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.</div><div class="separator" style="clear: both; text-align: center;"><br />
</div><div class="separator" style="clear: both; text-align: center;"><br />
</div><div style="text-align: left;">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.</div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiJ8LkqkfP7OVIRldEBWU8BRGHtQzxWTktbDdDsXGQNhHBUXjzcViTuQo_eHiNpoSUnpARKzQuDM3W6q9FI2IXNzoOZjm9d7NFknWMMuZKQY2MqKuNNDw2aWN2t1r6Tld-7C8yYILwg7Z6h/s1600/IMG_0406.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="112" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiJ8LkqkfP7OVIRldEBWU8BRGHtQzxWTktbDdDsXGQNhHBUXjzcViTuQo_eHiNpoSUnpARKzQuDM3W6q9FI2IXNzoOZjm9d7NFknWMMuZKQY2MqKuNNDw2aWN2t1r6Tld-7C8yYILwg7Z6h/s320/IMG_0406.JPG" width="320" /></a></div><div class="separator" style="clear: both; text-align: center;"><span style="color: #38761d;">(Text reads: </span><span style="color: red;">Lowest frequency mode you can resolve<br />
Highest frequency mode you can resolve<br />
Can't resolve this very high frequency mode</span><span style="color: #38761d;">)</span></div><div style="text-align: left;"><br />
</div>What happens to the K space box if you cut out a piece of the X space box?<br />
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjdseR2fA7xa9e5_6rip0Szyvd0E4kjKgtDNqptcNba_Z3QeeRj0AHCtc8Gg_1dP8TGGGidL2n0PsxFei6oWEDmDsW-qyPaRUdrMPPaKi6eydPyqjX9RfnN1wVtTw2Ye65Scl11QFvXq9Xn/s1600/IMG_0427.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="238" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjdseR2fA7xa9e5_6rip0Szyvd0E4kjKgtDNqptcNba_Z3QeeRj0AHCtc8Gg_1dP8TGGGidL2n0PsxFei6oWEDmDsW-qyPaRUdrMPPaKi6eydPyqjX9RfnN1wVtTw2Ye65Scl11QFvXq9Xn/s320/IMG_0427.JPG" width="320" /></a></div><div class="separator" style="clear: both; text-align: center;"><span style="color: blue;"><br />
</span></div><div class="separator" style="clear: both; text-align: left;">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.</div><div class="separator" style="clear: both; text-align: center;"><span style="color: #38761d;"><br />
</span></div>(Thanks Jeremy Mardon, Adam Brown)Neutrinohttp://www.blogger.com/profile/11148204219908257621noreply@blogger.com0