Three Dimensional Motion

Now we must treat the motion of a molecule in three dimensions. We know that the gas is isotropic (all directions are equivalent) so that the velocity probability distribution function must be the same in all directions. This means that the distribution functions, f(vx), f(vy), f(vz), must have the same functional form in all directions. Furthermore, motion in any one of the three x, y, z, directions must be independent of the motion in the other two directions.

Define the three dimensional velocity probability distribution function, F(vx,vy,vz), as follows:

F(vx,vy,vz)dvxdvydvz = the probability that a molecule has
an x-component of velocity between vx and vx + dvx AND
a y-component of velocity between vy and vy + dvy AND
a z-component of velocity between vz and vz + dvz.
Since the three directions are independent of each other this probability must be written as the product of the individual probabilities in the three directions. That is,
F(vx,vy,vz)dvxdvydvz = f(vx)dvxf(vy)dvyf(vz)dvz.                (1)
All the differentials cancel so that we conclude that
F(vx,vy,vz) = f(vx)f(vy)f(vz).                (2)
We are now going to use this three-dimensional velocity probability distribution function to determine the form of the one-dimensional distribution function, f.

Following the derivation of James Clerk Maxwell - a giant of science - we take the natural logarithm of Equation 2,

(3)
Now take the derivative of this with respect to vx,
(4)
We have already defined the molecular "speed," v as
(5)
The function F can be regarded either as a function of the three Cartesian velocity coordinates or of the speed, v, and the two angles which determine its direction. (The latter would be the equivalent of spherical polar coordinates except in velocity space rather than in coordinate space.)

With this understanding, let's rewrite the left-hand derivative as a derivative with respect to the speed, v,

.                (6)
(One might ask why Equation 6 does not include derivatives with respect to the angles which determine the direction of the velocity. Since we know that the gas is isotropic, none of its properties can depend on direction so that the angular derivatives of lnF must both be equal to zero. This implies that F must be a function of v alone and not of the angles which determine the direction of the velocity vector.)

Now, we know (or can figure out easily from Equation 5) that,

(7)
Placing this result in Equation 6 we get
(8)
and then inserting this back into Equation 4 we find that,
(9)
Rearranging to place everything involving v on the left side and everything involving vx on the right side we get,
(10)
We could just as easily have carried out this derivation with f(vy) or f(vz) which would give a similar result except with vx replaced by vy or vz. We conclude that,
(11)
In Equation 11 vx, vy and vz are totally independent of each other. That is, you can put in any values you wish for vx, vy and vz and the equation must remain valid. There is only one way that this can be true and that is if all the terms in Equation 11 are equal to the same constant. For reason's which we will see in a moment, let's call this constant −b. Then
(12)
or
(13a, b)
We can easily integrate Equation 13b to obtain (calling the constant of integration lna),
(14)
or, taking the exponential of both sides,
(15)
Equation 15 gives the functional form of f. We can see right away that it has the properties we expect. For example, it is an even function of vx. The function goes to zero at vx = ± ∞ , which means that its integral should be finite. (If we had made the original constant +b instead of −b, with b positive, then the function would go to infinity at vx = ± ∞ and would not be integrable.)

The function 15 contains two unknown constants, but we have enough information to determine what these constants are. First, the function must be normalized,

(16)
and, second,
(17)
Let's look at the normalization first,
(18)
This integral is a member of a class of integrals called "Gaussian integrals" and can be found in any standard table of integrals or on this web site,
(19)
The normalization integral becomes,
(20)
from which we see that
(21)
The function, f , becomes,
(22)
We determine the constant, b, from
(23)
which becomes, after inserting our form for f,
(24)
This integral is another one of the Gaussian integrals and can be found in standard integral tables or at the reference given above,
(25)
which gives,
(26)
or
(27)
We can now put together the complete form of f(vx),
(28)
This equation is one of the fundamental equations of kinetic molecular theory. It will provide the basis for all of our other probability distribution functions. Notice that the exponent is a kinetic energy (or a mechanical energy) divided by a thermal energy, kT. We will use this equation to calculate other quantities of interest.

Applications of f(vx)

Our first application of our new-found one-dimensional velocity probability distribution function will be to calculate

(29a, b)
This integral you can do without an integral table. Change the variable, let
(30)
then
(31a, b)
When we change variables in the integral we have to change the limits also. When vx = 0, u = 0 and when vx = ∞ , u = ∞ so the integral is still from 0 to ∞ . The integral becomes,
(32a, b, c)
The integral,
(33)
so
(34)
You will see why we made this last change on the right-hand-side in a little while.

Three-dimensional Speed Distribution

(Recall that velocity is a vector quantity and has three components, but speed is the magnitude of the velocity and contains no directional information. We will be very careful to preserve this distinction in all of our discussions.)

Let's go back to our original three-dimensional velocity distribution function. Recall that we defined the three dimensional probability distribution function, F(vx,vy,vz), as:

F(vx,vy,vz)dvxdvydvz = the probability that a molecule has an
x-component of velocity between vx and vx + dvx AND
a y-component of velocity between vy and vy + dvy AND
a z-component of velocity between vz and vz + dvz.
Recall also that we can write F in terms of the speed and the angles which give the direction,
.                (35)
θ is the "polar angle" of the velocity vector (the angle the vector makes with the z-axis − the north pole), and φ is the "azimuthal angle (which would correspond to longitude). (Remember that in calculus when you changed from Cartesian coordinates to spherical polar coordinates the differentials transformed in a special way,
.                (36)
The same is true when we transform from a Cartesian velocity space to a polar velocity space,
.)                (37)
So,
(38)
We already know that the function, F, can be written as a product,
F(vx,vy,vx) = f(vx)f(vy)f(vz).                (39)
So let's see what F looks like now that we know what the individual f's look like,
(40a, b, c, d)
We can just as well define the velocity probability in terms of speed, v, and the angles θ , and φ :
F(v,θ ,φ )v2sinθdθ dφ dv = the probability that a molecule has
a speed between v and v + dv AND
a polar angle of the velocity vector between θ and θ + AND
an azimuthal angle of the velocity vector between φ and φ + .
This probability is, then,
(41)
If we sum (integrate) this probability over all angles (θ goes from 0 to π , and φ goes from 0 to 2π ) we will have the probability that the speed lies between v and v + dv. Let's call this probability, F(v)dv. So
(42a, b, c)
We conclude that the "speed probability distribution function" is,
(43)
We can use this speed probability distribution function to calculate two new things. First, note that the range of v is 0 to ∞ (because v is the magnitude of the velocity), and then that,
(44)
We will leave it to the reader to show that this is true so that the function, F, is normalized.

Applications of the Speed Distribution Function

With our speed distribution function we can calculate the average speed as follows,

(45a, b)
From the integral tables, or by working it out, we find that
(46)
so our integral
(47)
Then
(48)
Notice that,
(49)
We now have two different measures of the average velocity, the rms velocity,
(50)
and the average speed,
(51)
These two averages are very close to each other (√ (8/π ) = 1.60 and √ 3 = 1.73) even though they describe slightly different aspects of molecular average velocities.

Most probable Speed

One other application of the molecular speed distribution function is to find the "most probable speed." If we look carefully at this distribution function,

(52)
we see that it is zero at v = 0 and zero at v = ∞ , and that it is positive between 0 and ∞ . Therefore it must go through a maximum between 0 and ∞ . The value of v at this maximum is the "most probable speed." We can find the most probable speed by setting the derivative of F equal to zero and solving for v,
(53)
Note that there is no need to keep the constants in because the position of the maximum does not depend on the values of the constants. So we can write,
(54)
The exponential cancels out and we solve
(55a, b, c)
which gives
(56)
We now have three measures of various aspects of molecular speeds,
(57)

(58)

and
(59)
all of which are close to each other in magnitude.  Each of theses quantities gives us an estimate of how fast molecules are moving in a gas at some temperature, T.  The average speed, Equation 58, is probably our best measure of how fast the molecules are moving.  This quantity is useful in other contexts, such as transport properties (diffusion, heat conductivity, etc.).

WRS

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