<< Statistics`ContinuousDistributions`

This is the Maxwell distribution of speeds, from kinetic theory of gases (see eq. 21.4 in Atkins_8):

f(v)=4π(\(TraditionalForm\`\(\(\(M\/\(2  πRT\)\)\()\)\)\^1.5\) \(v\^2\) \(exp(\(-\(Mv\^2\/\(2  RT\)\)\))\)\)

This is the Maxwell-Boltzmann distribution of energies:

f(ϵ)=((2  π)/(((π(Subscript[k, B]) T))^1.5))(ϵ^0.5)exp(-(ϵ/((Subscript[k, B]) T)))

avogadroConstant = 6.022*10^23;  
kB = 
      1.38065*10^-23;   
molecularMass = 
      0.044/avogadroConstant;    
molWt = 
      0.044;    
temp1 = 
      298;   
temp2 = 1000;
temp3 = 2000;
distribution1 = 
        4*Pi*(molecularMass/(2  Pi*kB*temp1))^1.5*v^2*
          Exp[-molecularMass*
              v^2/(2  kB*temp1)];
distribution2 = 
        4*Pi*(molecularMass/(2  Pi*kB*temp2))^1.5*v^2*
          Exp[-molecularMass*
              v^2/(2  kB*temp2)];
distribution3 = 
        4*Pi*(molecularMass/(2  Pi*kB*temp3))^1.5*v^2*
          Exp[-molecularMass*
              v^2/(2  kB*temp3)];
Plot[{distribution1, distribution2, distribution3},  {v, 0, 2000}, 
        PlotStyle -> {{RGBColor[1, 0, 0]}, {RGBColor[0, 1, 
                0]}, {RGBColor[0, 0, 1]}}];
Integrate[distribution1, {v, 0, 10000}]
Integrate[distribution2, {v, 0, 10000}]
Integrate[distribution3, {v, 0, 10000}]
meanSpeed1 = 
      Integrate[v*distribution1, {v, 0, 10000}]
meanSpeed1a = (8*
            kB*temp1/Pi/molecularMass)^0.5
meanSpeed2 = 
      Integrate[v*distribution2, {v, 0, 10000}]
meanSpeed2a = (8*
            kB*temp2/Pi/molecularMass)^0.5
meanSpeed3 = 
      Integrate[v*distribution3, {v, 0, 10000}]
meanSpeed3a = (8*kB*temp3/Pi/molecularMass)^0.5

So x^2*Exp(-x^2) is a slightly skewed (non-symmetric) distribution function, similar to log-normal distribution (with scale parameter μ and shape parameter σ):
f(x)=(1/(xσ (Sqrt[ 2  π]))) 
        (e^(-((((lnx - μ))^2)/(2  (σ^2)))))

distribution4 = x^2*Exp[-x^2];
Plot[distribution4,  {x, 0, 3}];
Integrate[distribution4, {x, 0, 10}] // N

LNrml[μ_, σ_, x_] = 
        PDF[LogNormalDistribution[μ, σ], 
          x];
Plot[{LNrml[0, 0.3, x], LNrml[0, 0.5, x], LNrml[0, 1, x]}, {x, 0, 5}, 
        PlotStyle -> {{RGBColor[1, 0, 0]}, {RGBColor[0, 1, 
                0]}, {RGBColor[0, 0, 1]}}];

Mean[LogNormalDistribution[0, 0.3]]
Exp[0 + 1/2*0.3^2]  
Variance[LogNormalDistribution[0, 0.3]]
Exp[
        2*0 + 0.3^2]*(Exp[0.3^2] - 
          1)  
Skewness[LogNormalDistribution[0, 0.3]]

Mean[LogNormalDistribution[0, 0.5]]
Variance[LogNormalDistribution[0, 0.5]]
Skewness[LogNormalDistribution[0, 0.5]]
Mean[LogNormalDistribution[0, 1]] // N
Variance[LogNormalDistribution[0, 1]] // N
Skewness[LogNormalDistribution[0, 1]] // N

LNrml[μ_, σ_, x_] = 
        PDF[LogNormalDistribution[μ, σ], 
          x];
Plot[{LNrml[2, 1, x], LNrml[2.5, 1, x], LNrml[3, 1, x]}, {x, 0, 50}, 
        PlotStyle -> {{RGBColor[1, 0, 0]}, {RGBColor[0, 1, 
                0]}, {RGBColor[0, 0, 1]}}];
Mean[LogNormalDistribution[2, 1]] // N
Variance[LogNormalDistribution[2, 1]] // N
Skewness[LogNormalDistribution[2, 1]] // N
Mean[LogNormalDistribution[2.5, 1]]
Variance[LogNormalDistribution[2.5, 1]]
Skewness[LogNormalDistribution[2.5, 1]] // N
Mean[LogNormalDistribution[3, 1]] // N
Variance[LogNormalDistribution[3, 1]] // N
Skewness[LogNormalDistribution[3, 1]] // N

This is to be distinguished from the normal (often called Gaussian) distribution function, which is symmetric. Why a symmetric distribution of speeds in 1D ends up being skewed in 3D is an interesting 'statistical' issue (but beyond the scope of our interest)... See, for example, http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html.

f1(x)=(1/(Sqrt[2  π])) (e^(-((x^2)/2)))

f2(x)=(1/(σ (Sqrt[ 2  π]))) 
        (e^(-((((x - μ))^2)/(2  (σ^2)))))

f((Subscript[v, z]))=(Sqrt[(m/(2  π(Subscript[k, B]) T))]) 
        (e^(((-(Subscript[mv, z]^2))/(2  (Subscript[k, B]) T))))

gaussianDistr1 = 
        1/((2  Pi)^0.5)*Exp[-x^2/2];
Plot[gaussianDistr1, {x, -5, 5}];
Integrate[gaussianDistr1, {x, -10, 10}]

Nrml[μ_, σ_, x_] = 
        PDF[NormalDistribution[μ, σ], 
          x];
Plot[Nrml[0, 1, x], {x, -5, 5}];

distribution1a = (molecularMass/(2  Pi*kB*temp1))^0.5*
          Exp[-molecularMass*
              vZ^2/(2  kB*temp1)];
Plot[
        distribution1a, {vZ, -1000, 1000}];
Integrate[distribution1a, {vZ, -1000, 1000}]

For the intriguing evolution of nomenclature surrounding this distribution function (De Moivre, Laplace, Gauss, 'normal', 'bell', 'error'), see  Chapter 22 of Stigler's "Stastistics on the Table: The History of Statistical Concepts and Methods," 1999.

What is the origin of the name 'gaussian' in the Nobel-prize-winning Gaussian software (www.gaussian.com)?

alpha = 1;
gaussianS = (2  alpha/Pi)^0.75*
          Exp[-alpha*r^2];
gausianPy = (128  alpha^5/Pi^3)^0.25*y*
          Exp[-alpha*r^2];
gaussianDxy = (2048*alpha^7/Pi^3)^0.25*x*y*
          Exp[-alpha*r^2];
Plot[gaussianS, {r, 0, 2}];

This is the so-called "half-normal distribution":

f(x)=\(TraditionalForm\`\(\(2  θ\)\/π\) \(exp(\(\(-θ\^2\) x\^2\)\/π)\)\)

HalfNrml[theta_, x_] = 
        PDF[HalfNormalDistribution[theta], x];
Plot[HalfNrml[1, x], {x, 0, 5}];

Relationship to random walk and Einstein's diffusion (Brownian motion) equation (with λ=(2Dt)^0.5):
f(x,t)=(1/(Sqrt[4  πDt]))Exp[-((x^2)/(4  Dt))]
D=(RT/N) (1/(3  πμ(Subscript[d, p])))
λ(x)=\(TraditionalForm\`\(\@t\) \@\(\(RT\/N\) 1\/\(1.5  πμd\_p\)\)\)

meanFreePathSquare=Integrate[x*(1/(Sqrt[4  πDt]))*Exp[-x^2/4/D/t],{x, 0, Infinity}]

viscH2O = 1.35*10^-3;  
diamPart = 10^-6;   
temp = 
      17 + 273;
t = 60;   
gasC = 
      8.314;   
meanFreePath = 
      Sqrt[t*gasC*temp/(6.022*10^23)/3/Pi/
              viscH2O/(diamPart/2)]

lrr3@psu.edu (10/27/2006)