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Q&A How could a bad random number generator explain baryon asymmetry?

This one's fun and easy. No biased coin needed. Assumptions: Matter attracts itself and if it hits its opposite it becomes energy. If you have energy $E=2mc^2$ pass by a particle you get another...

posted 9y ago by Black‭  ·  edited 4y ago by HDE 226868‭

Answer
#1: Post edited by user avatar HDE 226868‭ · 2020-06-03T01:33:15Z (over 4 years ago)
Fixed formatting (minus signs).
  • <p>This one's fun and easy. No biased coin needed.</p>
  • <p>Assumptions:</p>
  • <ul>
  • <li>Matter attracts itself and if it hits its opposite it becomes energy.</li>
  • <li>If you have e=2mc<sup>2</sup> pass by a particle you get another pair of normal and anti particles.</li>
  • </ul>
  • <p>If normal and anti particles are not in a perfect checkerboard pattern there's a cluster (This point doesn't matter if you have a finite universe since corners will have an imbalanced count regardless). If we have a cluster of normal matter then as long as it attracts anti-matter at a slow enough rate it will stay normal. Checkerboard patterns will tend to turn into energy and voids as explained below.
  • We can have a perfectly even starting distribution of matter and either a finite universe or starting jitters in gravity (those can be coin flips if you want).</p>
  • <p>Say we have a cluster like so:</p>
  • <p><code>++++−−</code>
  • <br/>
  • <code>++++−−</code>
  • <br/>
  • <code>++++−−</code>
  • <br/>
  • <code>++++−−</code>
  • <br/>
  • <code>++−−−−</code>
  • <br/>
  • <code>−−−−−−</code></p>
  • <p>Then break the gravity (it doesn't matter which way you break it), the anti particles are too thin in concentration to annihilate the cluster quick enough. If we have a particle take out another on the perimeter of the cluster then you can make a void. The energy will spread out and pop particles in and out of existence as energy waves overlap. But those energy waves have less energy the larger the circle they are and they need 2 particle's worth. Assuming you're <a href="https://en.wikipedia.org/wiki/Von_Neumann_neighborhood" rel="nofollow">4-Connected</a>, if the annihilation of two particles is right next to each other you have enough energy to remake the pairs involved (and you flip you coins if you want) otherwise at the next unit away you have less energy on a given point of the wave. To make it quick and intuitive just redistribute all the energy across the grid on any annihilation since that's likely the ultimate outcome of a chain of reactions. Since all the particles are giving 1/36 their energy to each space and you need a whole 2 particle's worth of energy to spawn a pair then you won't get any extra particles even after the last pair touch. So pure energy is the ultimate fate if gravity bound. But the universe is expanding and you have an Observable Universe. The expansion will pull gravity out of the equation and allow clusters to survive. If you have a large enough perimeter for your area for a given expansion you can survive until gravity doesn't come into play. From there the effects of having an Observable Universe will keep you in the dark about your matching anti cluster. If you up this to 3D it becomes easier since you now have more "internals" per "surface". The more dimensions your universe has the more likely you are to have clusters that can survive.</p>
  • <p>Prerequisites for this model:</p>
  • <ul>
  • <li>Observable Universe (max light speed or whatever requirements to have this, if you want the character to discover the truth you don't have to enforce this strictly.)</li>
  • <li>Expansion Constant large enough to overcome gravitational bonds between clusters of matter the size you want to use (you have pressure from annihilation helping you)</li>
  • <li>No "Perfectly Uniform Checkerboard" pattern and/or a finite universe.</li>
  • <li>An initial gravitational disturbance (at least 1)</li>
  • </ul>
  • <p><strong>P.S.</strong>
  • The finite universe argument is tougher but I'm fairly certain you'll have a corner survive even with a checkerboard pattern, as long as you don't have the initial disturbance dead center.</p>
  • <p>This one's fun and easy. No biased coin needed.</p>
  • <p>Assumptions:</p>
  • <ul>
  • <li>Matter attracts itself and if it hits its opposite it becomes energy.</li>
  • <li>If you have energy $E=2mc^2$ pass by a particle you get another pair of normal and anti particles.</li>
  • </ul>
  • <p>If normal and anti particles are not in a perfect checkerboard pattern there's a cluster (This point doesn't matter if you have a finite universe since corners will have an imbalanced count regardless). If we have a cluster of normal matter then as long as it attracts anti-matter at a slow enough rate it will stay normal. Checkerboard patterns will tend to turn into energy and voids as explained below.
  • We can have a perfectly even starting distribution of matter and either a finite universe or starting jitters in gravity (those can be coin flips if you want).</p>
  • <p>Say we have a cluster like so:</p>
  • <p><code>++++--</code>
  • <br/>
  • <code>++++--</code>
  • <br/>
  • <code>++++--</code>
  • <br/>
  • <code>++++--</code>
  • <br/>
  • <code>++----</code>
  • <br/>
  • <code>------</code></p>
  • <p>Then break the gravity (it doesn't matter which way you break it), the anti particles are too thin in concentration to annihilate the cluster quick enough. If we have a particle take out another on the perimeter of the cluster then you can make a void. The energy will spread out and pop particles in and out of existence as energy waves overlap. But those energy waves have less energy the larger the circle they are and they need 2 particle's worth. Assuming you're <a href="https://en.wikipedia.org/wiki/Von_Neumann_neighborhood" rel="nofollow">4-Connected</a>, if the annihilation of two particles is right next to each other you have enough energy to remake the pairs involved (and you flip you coins if you want) otherwise at the next unit away you have less energy on a given point of the wave. To make it quick and intuitive just redistribute all the energy across the grid on any annihilation since that's likely the ultimate outcome of a chain of reactions. Since all the particles are giving 1/36 their energy to each space and you need a whole 2 particle's worth of energy to spawn a pair then you won't get any extra particles even after the last pair touch. So pure energy is the ultimate fate if gravity bound. But the universe is expanding and you have an Observable Universe. The expansion will pull gravity out of the equation and allow clusters to survive. If you have a large enough perimeter for your area for a given expansion you can survive until gravity doesn't come into play. From there the effects of having an Observable Universe will keep you in the dark about your matching anti cluster. If you up this to 3D it becomes easier since you now have more "internals" per "surface". The more dimensions your universe has the more likely you are to have clusters that can survive.</p>
  • <p>Prerequisites for this model:</p>
  • <ul>
  • <li>Observable Universe (max light speed or whatever requirements to have this, if you want the character to discover the truth you don't have to enforce this strictly.)</li>
  • <li>Expansion Constant large enough to overcome gravitational bonds between clusters of matter the size you want to use (you have pressure from annihilation helping you)</li>
  • <li>No "Perfectly Uniform Checkerboard" pattern and/or a finite universe.</li>
  • <li>An initial gravitational disturbance (at least 1)</li>
  • </ul>
  • <p><strong>P.S.</strong>
  • The finite universe argument is tougher but I'm fairly certain you'll have a corner survive even with a checkerboard pattern, as long as you don't have the initial disturbance dead center.</p>