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The Time-Spent Probability
Distribution for a Particle
Under Relativity

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)
v(x) = [(2/m)(E−V(x)]½

This can be rewritten as

v(x) = [(2/m)K½

where K is kinetic energy.

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

ψ(x) = 1/(Tv(x))½ = (m/2T)½/K¼

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 = cK½/(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/((m0½)K½/(1 + K/(m0c² ))½


The Time-Spent Probability
Density Function

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

P(z) = (m0½/T)(1 + K/(m0c² ))½/K½

Thus when kinetic energy K is small compared with (m0c² ) density is inversely proportional to K½ 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) = (m0¼/T½)(1 + K/(m0c² ))¼/K¼

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