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Distribution for a Particle
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.
For a particle of mass m in a potential field V(x)
This can be rewritten as
where K is kinetic energy.
Therefore the wavefunction ψ(x) associated with the time-spent probability density function PTS(x) is given by
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.
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.,
Since the probability density function P(z)=1/(Tv(z)).
Thus when kinetic energy K is small compared with (m0c² ) density is inversely proportional to K½ just as in the classical case.
If the wave function ψ(z) is such that ψ(z)²=P(z) then
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