Time-Spent Probability Distribution.s

Let dt be the time a particle spends in an interval ds of its trajectory. Then the probability of finding it in that interval is dt/T where T is the total time the particle takes to execute its periodic trajectory. But dt=ds/|v| where v is the velocity of the particle.

The Classical Case

For a particle of mass m in a potential field V(x)

E = ½mv² + V(x)
so
v(x) = [(2/m)(E−V(x)]^½

This can be rewritten as

v(x) = [(2/m)K^½

where K is kinetic energy.

Therefore the wavefunction ψ(x) associated with the time-spent probability density function P_TS(x) is given by

ψ(x) = 1/(Tv(x))^½ = (m/2T)^½/K^¼

The Relativistic Case

The total energy of a particle is mc² where m is relativistic mass m₀/(1−β²)^½. Therefore kinetic energy K is mc²−m₀c².

In the derivation below the dependence of K, v and β on particle position is ignored to simplify the algebraic expressions.

K/(m₀c²) = 1/(1−β²)^½ − 1
(K + m₀c²)/(m₀c²) = 1/(1−β²)^½
(1−β²)^½ = (m₀c²/(K + m₀c²)
1−β² =[(m₀c²/(K + m₀c²]²
β = [1 − ((m₀c²/(K + m₀c²))²]^½
v/c = [(m₀c² + K) − m₀c²]/(m₀c² + K)
v/c = [K/((m₀c² + K)]^½
v = cK^½/(m₀c² + K)^½

Note that as K→∞ v→c as it must under Relativity.

There a further derivation of v based upon factoring m₀c² out of the denominator of the above fraction; i.e.,

v = (1/((m₀^½)K^½/(1 + K/(m₀c² ))^½

The Time-Spent Probability
Density Function

Since the probability density function P(z)=1/(Tv(z)).

P(z) = (m₀^½/T)(1 + K/(m₀c² ))^½/K^½

Thus when kinetic energy K is small compared with (m₀c² ) density is inversely proportional to K^½ just as in the classical case.

The Wave Functions

If the wave function ψ(z) is such that ψ(z)²=P(z) then

ψ(z) = (m₀^¼/T^½)(1 + K/(m₀c² ))^¼/K^¼