View Single Post
 P: 1 According the the kinetic theory of gases, molecules moving along the x direction are given by Σx= (1/2) mv^2, where m = mass and vx is the velocity in the x direction The distribution of particles over velocities is given by the Boltzmann law p(x)=e^[(-mv^2)/ekT] where velocities range from -∞ to ∞ Here, the probability distribution p(v), needs a constant, c, that will normalize the distribution so that c∫e^[(-mv^2)/2kT]dv=1 ( -∞ to ∞ ) The publisher used a trick converting this equation into a double integral from I=∫e^(-ax^2)dx ( -∞ to ∞ ) where a = m/2kT to I^2= ∫e^(-ax^2)dx ∫e^(-ay^2)dy -∞ to ∞ and combined the exponentials from the double integral to get I^2=∫ ∫e^-a(x+y^)2 dxdy -∞ to ∞ then converted to polar coordinates r and θ since r^2= x^2 + y^2 to simplify to I^2= ∫rdr ∫e^-ar^2dθ from 0 to 2π simplifying further to exchange dθ for dr I^2= ∫dθ ∫e^-ar^2dr from 0 to 2π and integrated the first integrand to ∫dθ from 0 to 2π = 2π reducing the double integral to I^2=2π∫e^-ar^2dr 0 to 2π and finally used u substition u=-ar^2 & du = -2ardr to leave us with I^2= (-π/a)∫e^udu from 0 to ∞ ==> I^2= (-π/a)e^u integrating from 0 to ∞ gives I^2=π/a and I = (π/a)^-1/2 where a= m/2kT gives us I= [(2πkT)/m]^1/2 So when used as a constant, the normalized probability distribution equation becomes p(v)dv= [(m)/2kT)^1/2] * ∫e^[(-mv^2)/2kT]dv My questions is How are you able to take an Integral with respect to x and make it into a double integral with respect to x and y?