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Sonor@12
Sonor@12 is offline
#1
Jan31-13, 02:41 PM
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?
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