Visualizing the Gas Cloud
Imagine a gas cloud spreading outwards from the origin in all directions. The problem states that the number density of the gas molecules, n(r), is not uniform; it decreases as we move further away from the origin according to the relation n(r)=n0e−αr4.
To find the total number of molecules in this entire space, we cannot simply multiply density by total volume. Instead, we must use the power of calculus. We start by considering a very thin spherical shell of radius r and an infinitesimally small thickness dr. Because the shell is so thin, the density of the gas within this specific shell is practically constant.
Setting up the Integral
The volume of this thin spherical shell, dV, is simply its surface area multiplied by its thickness. Thus, dV=4πr2dr.
The number of molecules residing just within this shell, let's call it
dN, is the product of the local density and the shell's volume:
dN=n(r)⋅dV
Substituting our given expressions into this relation, we get:
dN=(n0e−αr4)⋅(4πr2dr)
To find the total number of molecules
N in the entire universe of this gas cloud, we must sum up all these tiny
dN contributions from the origin (
r=0) all the way to infinity (
r=∞). This gives us our master integral:
N=∫0∞4πn0e−αr4r2dr
The Art of Substitution
At first glance, integrating
e−αr4r2 looks intimidating. However, a clever algebraic substitution will break it down beautifully. Let's take the troublesome exponent and set it to a new variable
t:
αr4=t
From this, we can isolate
r:
r=t1/4α−1/4
Now, we need to find the differential
dr in terms of
dt. Differentiating our substitution equation yields:
4αr3dr=dt
We need to replace the
r2dr term in our integral. Let's manipulate the differential equation to isolate
r2dr:
r2dr=4αrdt
Substitute the expression for
r back into the denominator:
r2dr=4α(t1/4α−1/4)dt=4α3/4t1/4dt
The Final Proportionality
Now, we substitute everything back into our master integral. Notice that as r→0, t→0, and as r→∞, t→∞. So, our limits of integration remain unchanged.
N=∫0∞4πn0e−t4α3/4t1/4dt
Let's pull all the constants out of the integral. The
4 in the numerator and denominator cancel out perfectly:
N=πn0α−3/4∫0∞e−tt−1/4dt
The Gamma Function Connection
Look closely at the remaining integral: ∫0∞e−tt−1/4dt. This is a definite integral with constant limits. It does not depend on r, t, α, or n0. It evaluates to a pure, constant number (specifically, it is related to the Gamma function, Γ(3/4)).
Since the integral is just a constant multiplier, we can confidently conclude the proportionality of the total number of molecules:
N∝n0α−3/4
This elegant result shows how the total particle count scales with the central density n0 and the spatial decay parameter α.