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Peter Higgs Talking about the Higgs interaction and testing using the LHC to search for the Higgs Boson

Friday, September 12th, 2008

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Derivation of the Normal Gaussian distribution from physical principles

Monday, August 25th, 2008

In many physical systems the question arises what is the probability distribution that describes a system with a given expected energy E over the interval from -infinity to + infinity? Again you will use the maximum entropy principle to determine this.

The constraints are as follows:

$sum{kappa=1}{N}{P(x_i)}=1$ …. sum over all probabities must = 1
$sum{kappa=1}{N}{P(x_i){x_i}}=mu$ …. given an average value AKA "mean"
$sum{kappa=1}{N}{P(x_i){x_i}^2}=E$ ….. given an "energy" or standard deviation AKA "variance"

The langrangian is formed as follows:

$L=sum{kappa=1}{N}{{-P(x_i)}{log_2 P(x_i)}}+lambda_0(1-sum{kappa=1}{N}{P(x_i)}�)+lambda_1(mu-sum{kappa=1}{N}{{P(x_i)}{x_i}})+lambda_2(E-sum{kappa=1}{N}{{P(x_i)}{x_i}^2})$

${partial L} / {partial P_i}= {-log_2 P(x_i)}-1-lambda_0-lambda_1{x_i}-lambda_2{x_i}^2=0$ ….setting equal to zero to find the extrema point

Now the problem is to solve for the lambda coefficients. I use a trick. I assume the curve centered at the Y axis. The curve must have the same amount of entropy on the left side of the Y axis as on the right or entropy will not be maximized. Because the polynomial is even degree the probability curve must be "even"….that is symmetric about the Y axis.

Solve for the coefficients using identity of the integral of normal -infin to + infin

Using the identity and setting it to 1: $int{-infty}{+infty}{e^{-{(x-mu)^2/{2sigma^2}}}}=sigma{sqrt{2pi}} =1$ yields: $lambda_2={pi}/ln2$

The resultant distribution is: ${e^{-{pi}x^2}}$ which is the normal distribution.

Now for the other coefficients the job is made easier by observing the distribution can only retain this even about the mean form if the polynomial is in the form : (x-mu)^2 : this form can propagate along the X axis without distribution shape change.

Observations

base state of quantum harmonic oscillator is gaussian : It is maximum entropy.
gaussian wave packet can not mishapen because it is already max entropy. If it changes shape it decreases entropy which requires force.
gaussian is the base state of the wave packet? Is it possible to have forms of the higher energy states?

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Use of Maximum Entropy to explain the form of Energy States of an Electron in a Potential Well

Friday, May 23rd, 2008

The base state of an electron in an infinite potential well has the most "space" for the electron state. Thus it has the maximum entropy. Take that same state and imagine pinching the electrons existence to nil in the middle of the trough. Now you have state-2. The electron now exists in a smaller entropic state and guess what? It contains exploitable energy now. This is like a spring compressed. The electron can decompress and exert force / expend energy. For example in an interaction with another atom possibly a recoil could occur. In a crystal lattice an electron can transfer its energy to the atom next door and in effect yield conduction. All these are preliminary suppositions subject to more scrutiny. As mentioned before since the electron exists in this potential well in the form of free fall it can not have any acceleration. Thus its distribution must thoroughly avoid the edges of the well were it would indeed experience accelerations by bouncing and recoiling off of the walls.