Research Project

Top Quark Decay Analysis for More Accurate Mass Measurements.

2/15

Prove that quantum particle decays do not conserve mass (total mass of daughters is less than parent particle).  Build an expression for the vector momentum of each daughter particle in the rest frame of the parent particle in terms of the particles' masses.  --no numbers yet! this is theory on decays in general--

Important stuff:  

    Energy is the hypotenuse of a right triangle of legs rest mass and momentum.  Energy is greater than or equal to rest mass and momentum individually.

    When solving equations with radicals, arrange the equation so when you square both sides, you avoid annoying FOILs.

2/22

Build an expression for the energy of one of the daughter particles in the lab frame in terms of the boost, masses, and the angle between the lab momentum and the lab frame velocity.  Assume that the bottom is massless.  Plot the energies given constant boost and a flat distribution of cos(θ).

Important stuff: 

    The energy Lorentz transform takes into account the angle between momentum and frame velocity.  Even so, only momentum components parallel to the lab frame velocity have to be 'unboosted.'

    Plot means histograms, not lines...

    The plot produces a rectangle between minimum and maximum energies, where the rest mass of the bottom at the geometric mean of the rectangle.  The minimum is where cos(θ)  is -1, the maximum is when cos(θ) is +1.  The rectangle widens as the boost increases.  

    When the plot is of the ln of the energy, then rest mass falls at the arithmetic mean of the rectangle.

2/29

    Repeat the prior week's procedure, no longer assuming the bottom is massless.  Also, sketch a histogram for arbitrary distribution of boost for the massless bottom

Important stuff: 

    The mass moves the rectangle widens the rectangle, moreso at greater values of the boost.  In addition, this widening moves the maximum value more than the minimum value.

    The geometric mean/arithmetic mean of the ln plot is now always greater than the rest mass.

    On the histogram for distributed boost, the rectangle always contains the rest mass, so there is a gaussian-like shape that has a peak at the rest mass.

3/7

    Sketch a histogram of the energies given an arbitrary distribution of the boost for massive bottom.  The minimum and maximum boost depends on the boost of the parent particle. Prove that the maximum energy is always greater than the rest mass.  Prove that the minimum energy is always less than the rest mass.

Important stuff:   

    The rectangles vary in width, but the rest mass is always inside of it, so there is still a peak at the rest mass.  However, the shape remains low at low boosts, increases quickly to the peak, then tapers away, sort of like a temperature energy distribution graph. 

    While proving the maximum energy is always greater than the rest mass energy is an easy proof, the minimum can actually exceed the rest mass energy at sufficiently high boosts. However, this only occurs at boosts on the order of .9999976.  These boost only occur in particle collisions about four times more energetic than the LHC collider, so this case can be safely ignored.

3/21

    Prove that there is a peak at the rest mass in the energy plot.  Also prove that the geometric mean is greater than the rest mass energy.  Also begin to code a Pythia simulation of the process at Tevatron, LHC, and at maximum energy Pythia allows.

Important stuff:

    Plot is an integral of a function from some value γ (ln of ratio between bottom energy and rest mass energy) to infinity of another function of γ.  Therefore, energy only plays the role as a chooser of the lower index.  Flip and derivative, show that it is zero, derivative again, show second derivative is negative = peak.

    The geometric proof is manipulations of the equations until one gets a statement that can be concluded upon.

3/28

   The cross section for ttbar production from pp collisions is 150 picobarns, so Pythia has to be ordered to only observe them to get any data at all.  Vineet is looking into adding specific tags to increase the number of events that produce data beyond one if we are really, really, really lucky.

    The peak proof got a little bit confused, as I got my graphs mixed up (i.e.  the integrated function has actual meaning, and is not just arbitrary).  The geometric proof was simple and easy.

4/4

    We managed to get ttbar collisions in about 90% of our events, but they are not decaying.  We don't know why.  It almost looks like pythia is trying to decay them, but doesn't.  Normally pythia ignores stable particles once it has recorded their information, but these top quarks are recorded as having a single daughter particle: itself.  Multiple times.

    Peak proof now sketched out.  The proof that it is a critical point is fine, but the peak bit is sort of weak.

4/11

    The top quarks are decaying now.  Vineet did ...something.  (He was surprised as me when it suddenly cooperated).  Now we just need the computation time to get some good data. 100000 events took two days before the job was canceled by idle timeout. 1000 and 10000 work fine, so we have to look into trimming computation.

    The peak can be proven by logical contradiction.  If it is a trough or inflection point, there would have to be a critical point after the rest frame energy so that the distribution returns to the x axis.  There isn't (the we know the shape of the derivative function = boost distribution), so it has to be a peak.

    Downloaded TEX so I can make convert my mad scrawlings into readable format

4/18

    Nearly all of the number crunching is in the event generation, so we can't really trim the computation time at all.  We simply did 10000 events 10 times and added it together (why didn't we think of that before?).  The plots line up perfectly;  the peaks do not move, but higher beam energies means the spike is less abrupt and the tail constitutes more of the distribution of the energies.

    Addition to the proof is that if the boost distribution does not start at (0,0) i.e. there is a possibility that the produced particle is at rest in the lab frame, there is still a peak.  Instead of a smooth peak, it would be a cusp peak.  It does not apply to t decay, as the two daughter particles cannot exist at the same point, but it is an important general proof.

    A typo? mistake? in my TEX input made proving that there was a cusp.  Dropped and E on the bottom of the taken derivative, so it was still defined at E = Erest (the critical point we are proving).