The Time Spent in
the Allowable Locations

Consider a particle moving in a periodic path. The time spent dt in an interval dx at x is equal to dx/|v(x)|, where v(x) is the velocity of the particle at x. The probability of finding the particle in the interval dx at x at a randomly chosen time is then proportional to dx/|v(x)|. Thus the probability density function for x is

P(x) = (1/|v(x)|)/T = 1/(T|v(x)|)

where T is the total time spent in a cycle; i.e.,

T = ∫ dx/|v(x)|.

The Probability Density
Function for Velocity

The time dt the system spends in a velocity interval dv is given by

dt = dv/|(dv/dt)| = dv/|a(v)|

where a(v) is acceleration (dv/dt) when the particle velocity is v.

Thus the probabilty of finding the particle having a velocity between b and c is

∫_b^cdv/[T|(dv/dt)|]

The time period for v is the same as the time period for x. See time period for the proof.

A Change in the
Variable off Integration

Now consider a change in the variable of integration from v to x, where v=v(x) in the expression for finding the velocity in the interval from b to c. Thus dv=(dv/dx)dx and hence the probability integral becomes

∫_B^C(dv/dx)dx/[T|(dv/dt)|]

where v(B)=b and v(C)=c.

But

(dv/dt) = (dv/dx)(dx/dt) = (dv/dx)v

So the probability integral reduces to

∫_B^C(dv/dx)dx/[T|(dv/dx)v|] = ∫_B^Cdx/(T|v|)

This says that the probability of finding velocity in the interval b to c is the same as finding x between location B where v=b and location C where v=c.

Thus the probability distribution of velocity as a function of v is the same as the probability distribution for location as a function of x. For an illustration of this result take the case of a harmonic oscillator.

Harmonic Oscillator

A harmonic oscillator is a mass subject to a restoring force of −kx, where x is the deviation from equilibrium and k is the stiffness coefficient. The energy of a harmonic oscillator is given by

E = ½mv² + ½kx²

where m is the mass of the particle.

Therefore

v(x) = (2/m)^½(E − ½kx²)^½
and hence
1/[T|v(x)|] = (m/2)^½/[T(E − ½kx²)^½]

The acceleration is easily computed as

a = (dv/dt) = (dv/dx)(dx/dt) = (dv/dx)v

The energy equation E = ½mv² + ½kx² can be differentiated with respect to x to give

0 = mv(dv/dx) + kx
and thus
(dv/dx)v = − (k/m)x = −kx/m

The restoring force F for the harmonic oscillator is −kx so the above equation is essentially

a = F/m

What is needed is acceleration as a function of v rather than of x. Note that the energy equation may be solved for x; i.e.,

x = (2/k)^½(E − ½mv²)^½

Thus

|a(v)| = (2k)^½(E − ½mv²)^½/m

Therefore the probability density function for velocity is

1/[T|a(v)|] = m/[T(2k)^½(E − ½mv²)^½]
= (m/k)^½(m/2)^½[T(E − ½mv²)^½]

The probability density function for x is

1/[T|v(x)|] = (m/2)^½/[T(E − ½kx²)^½]

This is essentially the same as 1/[Ta(v)|] with x and k substituted for v and m, respectively. Differences is constant cefficients do not matter be such differences get eliminated in the normalization process.

The precise derivation of 1/[Tv(x)] from 1/[Ta(v)] is as follows. First, in the expression for 1/[Ta(v)] (E − ½mv²)^½ is replaced by (k/2)^½x. This gives

1/[Ta(v)] = m/(kxT) = (m/k)(1/(Tx))

The constant factor of (m/k) is irrelevant in that it cancels out in the process of normalization of the probability density function.

Thus the probability density function for velocity as a function of velocity is the same as the probability density function for location as a function of location. .

(To be continued.)