Archive for August 25th, 2008

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?

 

 

 

 

Want to use Wind Power? You can by building it yourself

Monday, August 25th, 2008

For you environMENTAL cases you can build your own wind generator.  You can see all the information you need here: www.otherpower.com/otherpower_wind.html

Very interesting and fun designs.  What’s stopping you liberals?  The earth is warming!! How can you NOT build one of these? 

The nice thing about this set of likely Bush - Haters is that at least they appear to be walking the walk instead of just telling me what I ought to do.  And they appear to be scientifically truthful…. it’s not likely to be cheap or easy or cost effective if you already have grid power.

The engineer in me wishes he had time and space to build one.

Step away from the Microphone Madonna GrandMama

Monday, August 25th, 2008

Geesus when is this antique going to stop?  She’ll be mumified and still dancing.  And stop commenting on politics.   You commenting on politics is like George Bush gyrating and singing on stage in a weakly themed  leather dominatrix outfit with fishnets. Totally inappropriate. 

Madonna GrandMama.   She puffs out chalk dust when spanked.