﻿ A Theorem Concerning Time-Spent Probability Distributions
San José State University

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Thayer Watkins
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A Theorem Concerning
Time-Spent Probability
Distributions

## 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.,

## 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

#### ∫bcdv/[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

#### ∫BC(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

#### ∫BC(dv/dx)dx/[T|(dv/dx)v|] = ∫BCdx/(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)½and 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.,

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.)