|San José State University|
& Tornado Alley
A Theorem Concerning|
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
where T is the total time spent in a cycle; i.e.,
The time dt the system spends in a velocity interval dv is given by
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
The time period for v is the same as the time period for x. See time period for the proof.
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
where v(B)=b and v(C)=c.
So the probability integral reduces to
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.
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
where m is the mass of the particle.
The acceleration is easily computed as
The energy equation E = ½mv² + ½kx² can be differentiated with respect to x to give
The restoring force F for the harmonic oscillator is −kx so the above equation is essentially
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.,
Therefore the probability density function for velocity is
The probability density function for x is
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
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.)
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